Physics for Engineers and Scientists, Volume III, 3E

W. W. Norton & Company, Inc. • www.NortonEbooks.com

PHYSICS

Hans C. Ohanian, John T. Markert

THIRD EDITION

FOR ENGINEERS AND SCIENTISTS

Volume Three

Physics forEngineers andScientists Third Edition

W • W • NORTON & COMPANY B NEW YORK • LONDON

Volume 3 (Chapters 36–41)RELATIV ITY, QUANTA, AND

PARTICLES

H A N S C . O H A N I A N , U N I V E R S I T Y O F V E R M O N T

J O H N T . M A R K E R T , U N I V E R S I T Y O F T E X A S A T A U S T I N

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To Susan Ohanian, writer, who gently tried to teach me some of her craft.—H.C.O.

To Frank D. Markert, a printer by trade; to Christiana Park, for her thirst for new knowledge; and to

Erin, Ryan, Sean, and Gwen, for their wonder and clarity.—J.T.M.

Copyright © 2007 by W.W. Norton & Company, Inc.

All rights reserved

Printed in the United States of America

Third Edition

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Brief Contents

PREFACE xiii

OWNER’S MANUAL xxv

PRELUDE: THE WORLD OF PHYSICS xxxv

PART I MOTION, FORCE, AND ENERGY 1

1. SPACE, TIME, AND MASS 2

2. MOTION ALONG A STRAIGHT LINE 28

3. VECTORS 69

4. MOTION IN TWO AND THREE DIMENSIONS 94

5. NEWTON’S LAWS OF MOTION 130

6. FURTHER APPLICATIONS OF NEWTON’S LAWS 173

7. WORK AND ENERGY 204

8. CONSERVATION OF ENERGY 235

9. GRAVITATION 271

10. SYSTEMS OF PARTICLES 305

11. COLLISIONS 338

12. ROTATION OF A RIGID BODY 365

13. DYNAMICS OF A RIGID BODY 394

14. STATICS AND ELASTICITY 429

PART II OSCILLATIONS, WAVES, AND FLUIDS 466

15. OSCILLATIONS 468

16. WAVES 507

17. SOUND 536

18. FLUID MECHANICS 565

PART III TEMPERATURE, HEAT, ANDTHERMODYNAMICS 600

19. THE IDEAL GAS 602

20. HEAT 628

21. THERMODYNAMICS 661

iii

PART IV ELECTRICITY AND MAGNETISM 692

22. ELECTRIC FORCE AND ELECTRIC CHARGE 694

23. THE ELECTRIC FIELD 721

24. GAUSS’ LAW 756

25. ELECTROSTATIC POTENTIAL AND ENERGY 789

26. CAPACITORS AND DIELECTRICS 828

27. CURRENTS AND OHM’S LAW 858

28. DIRECT CURRENT CIRCUITS 887

29. MAGNETIC FORCE AND FIELD 926

30. CHARGES AND CURRENTS IN MAGNETIC FIELDS 964

31. ELECTROMAGNETIC INDUCTION 993

32. ALTERNATING CURRENT CIRCUITS 1030

PART V WAVES AND OPTICS 1068

33. ELECTROMAGNETIC WAVES 1070

34. REFLECTION, REFRACTION, AND OPTICS 1111

35. INTERFERENCE AND DIFFRACTION 1168

PART VI RELATIVITY, QUANTA, AND PARTICLES 1214

36. THE THEORY OF SPECIAL RELATIVITY 1216

37. QUANTA OF LIGHT 1254

38. SPECTRAL LINES, BOHR’S THEORY, AND QUANTUMMECHANICS 1286

39. QUANTUM STRUCTURE OF ATOMS, MOLECULES,AND SOLIDS 1320

40. NUCLEI 1354

41. ELEMENTARY PARTICLES AND COSMOLOGY 1396

APPENDICES A-1

Chapters 1–21 appear in Volume 1; Chapters 22–36 appear in Volume 2; Chapters 36–41 appear in Volume 3.

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Table of Contents

PREFACE xiii

OWNER’S MANUAL xxv

PRELUDE: THE WORLD OF PHYSICS xxxv

PART I MOTION, FORCE, AND ENERGY 1

1. SPACE, TIME, AND MASS 2

1.1 Coordinates and Reference Frames 3

1.2 The Unit of Length 5

1.3 The Unit of Time 9

1.4 The Unit of Mass 11

1.5 Derived Units 13

1.6 Significant Figures; Consistency of Units and

Conversion of Units 14

SUMMARY / QUESTIONS / PROBLEMS / REVIEWPROBLEMS / ANSWERS TO CHECKUPS 20

2. MOTION ALONG A STRAIGHT LINE 28

2.1 Average Speed 29

2.2 Average Velocity for Motion along a

Straight Line 32

2.3 Instantaneous Velocity 35

2.4 Acceleration 39

2.5 Motion with Constant Acceleration 42

2.6 The Acceleration of Free Fall 49

2.7* Integration of the Equations of Motion 54


3. VECTORS 69

3.1 The Displacement Vector and Other Vectors 70

3.2 Vector Addition and Subtraction 72

3.3 The Position Vector; Components of a

Vector 76

3.4 Vector Multiplication 81


4. MOTION IN TWO AND THREE DIMENSIONS 94

4.1 Components of Velocity and Acceleration 95

4.2 The Velocity and Acceleration Vectors 98

4.3 Motion with Constant Acceleration 102

4.4 The Motion of Projectiles 104

4.5 Uniform Circular Motion 112

4.6 The Relativity of Motion and the Addition

of Velocities 115


5. NEWTON’S LAWS OF MOTION 130

5.1 Newton’s First Law 131

5.2 Newton’s Second Law 133

5.3 The Combination of Forces 138

5.4 Weight; Contact Force and Normal Force 141

5.5 Newton’s Third Law 144

5.6 Motion with a Constant Force 151


6. FURTHER APPLICATIONS OF NEWTON’S LAWS 173

6.1 Friction 174

6.2 Restoring Force of a Spring; Hooke’s Law 182

6.3 Force for Uniform Circular Motion 184

6.4* The Four Fundamental Forces 191


v

Chapters 1–21 appear in Volume 1; Chapters 22–36 appear in Volume 2; and Chapters 36–41 appear in Volume 3.

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7. WORK AND ENERGY 204

7.1 Work 205

7.2 Work for a Variable Force 211

7.3 Kinetic Energy 214

7.4 Gravitational Potential Energy 218


8. CONSERVATION OF ENERGY 235

8.1 Potential Energy of a Conservative Force 236

8.2 The Curve of Potential Energy 244

8.3 Other Forms of Energy 248

8.4* Mass and Energy 251

8.5 Power 253


9. GRAVITATION 271

9.1 Newton’s Law of Universal Gravitation 272

9.2 The Measurement of G 277

9.3 Circular Orbits 278

9.4 Elliptical Orbits; Kepler’s Laws 282

9.5 Energy in Orbital Motion 288


10. SYSTEMS OF PARTICLES 305

10.1 Momentum 306

10.2 Center of Mass 313

10.3 The Motion of the Center of Mass 323

10.4 Energy of a System of Particles 327


11. COLLISIONS 338

11.1 Impulsive Forces 339

11.2 Elastic Collisions in One Dimension 344

11.3 Inelastic Collisions in One Dimension 348

11.4* Collisions in Two and Three Dimensions 351


12. ROTATION OF A RIGID BODY 365

12.1 Motion of a Rigid Body 366

12.2 Rotation about a Fixed Axis 367

12.3 Motion with Constant Angular

Acceleration 374

12.4* Motion with Time-Dependent Angular

Acceleration 376

12.5 Kinetic Energy of Rotation; Moment of

Inertia 378


13. DYNAMICS OF A RIGID BODY 394

13.1 Work, Energy, and Power in Rotational Motion;

Torque 395

13.2 The Equation of Rotational Motion 399

13.3 Angular Momentum and its Conservation 406

13.4* Torque and Angular Momentum as

Vectors 410


14. STATICS AND ELASTICITY 429

14.1 Statics of Rigid Bodies 430

14.2 Examples of Static Equilibrium 433

14.3 Levers and Pulleys 441

14.4 Elasticity of Materials 445


PART II OSCILLATIONS, WAVES,AND FLUIDS 466

15. OSCILLATIONS 468

15.1 Simple Harmonic Motion 469

15.2 The Simple Harmonic Oscillator 476

15.3 Kinetic Energy and Potential Energy 480

15.4 The Simple Pendulum 484

15.5* Damped Oscillations and Forced

Oscillations 488


16. WAVES 507

16.1 Transverse and Longitudinal Wave Motion 508

16.2 Periodic Waves 509

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16.3 The Superposition of Waves 516

16.4 Standing Waves 520


17. SOUND 536

17.1 Sound Waves in Air 538

17.2 Intensity of Sound 540

17.3 The Speed of Sound; Standing Waves 543

17.4 The Doppler Effect 574

17.5* Diffraction 553


18. FLUID MECHANICS 565

18.1 Density and Flow Velocity 567

18.2 Incompressible Steady Flow; Streamlines 569

18.3 Pressure 573

18.4 Pressure in a Static Fluid 575

18.5 Archimedes’ Principle 580

18.6 Fluid Dynamics; Bernoulli’s Equation 582


PART III TEMPERATURE, HEAT,AND THERMODYNAMICS 600

19. THE IDEAL GAS 602

19.1 The Ideal-Gas Law 603

19.2 The Temperature Scale 609

19.3 Kinetic Pressure 613

19.4 The Internal Energy of an Ideal Gas 616


20. HEAT 628

20.1 Heat as a Form of Energy Transfer 629

20.2 Thermal Expansion of Solids and Liquids 633

20.3 Thermal Conduction 638

20.4 Changes of State 642

20.5 The Specific Heat of a Gas 644

20.6* Adiabatic Expansion of a Gas 647


21. THERMODYNAMICS 661

21.1 The First Law of Thermodynamics 663

21.2 Heat Engines; The Carnot Engine 665

21.3 The Second Law of Thermodynamics 675

21.4 Entropy 677


PART IV ELECTRICITY AND MAGNETISM 692

22. ELECTRIC FORCE AND ELECTRIC CHARGE 694

22.1 The Electrostatic Force 695

22.2 Coulomb’s Law 698

22.3 The Superposition of Electrical Forces 703

22.4 Charge Quantization and Charge

Conservation 706

22.5 Conductors and Insulators; Charging by

Friction or by Induction 708


23. THE ELECTRIC FIELD 721

23.1 The Electric Field of Point Charges 722

23.2 The Electric Field of Continuous Charge

Distributions 729

23.3 Lines of Electric Field 736

23.4 Motion in a Uniform Electric Field 740

23.5 Electric Dipole in an Electric Field 742


24. GAUSS’ LAW 756

24.1 Electric Flux 757

24.2 Gauss’ Law 762

24.3 Applications of Gauss’ Law 763

24.4 Superposition of Electric Fields 772

24.5 Conductors and Electric Fields 774


25. ELECTROSTATIC POTENTIAL AND ENERGY 789

25.1 The Electrostatic Potential 790

25.2 Calculation of the Potential from the Field 798

CONTENTS vii

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30. CHARGES AND CURRENTS IN MAGNETIC FIELDS 964

30.1 Circular Motion in a Uniform Magnetic

Field 965

30.2 Force on a Wire 969

30.3 Torque on a Loop 972

30.4 Magnetism in Materials 976

30.5* The Hall Effect 980


31. ELECTROMAGNETIC INDUCTION 993

31.1 Motional EMF 994

31.2 Faraday’s Law 997

31.3 Some Examples; Lenz’ Law 1001

31.4 Inductance 1008

31.5 Magnetic Energy 1013

31.6* The RL Circuit 1015


32. ALTERNATING CURRENT CIRCUITS* 1030

32.1 Resistor Circuit 1013

32.2 Capacitor Circuit 1035

32.3 Inductor Circuit 1038

32.4* Freely Oscillating LC and RLC Circuits 1041

32.5* Series Circuits with Alternating EMF 1046

32.6 The Transformer 1053


PART V WAVES AND OPTICS 1068

33. ELECTROMAGNETIC WAVES 1070

33.1 Induction of Magnetic Fields; Maxwell’s

Equations 1071

33.2* The Electromagnetic Wave Pulse 1075

33.3 Plane Waves; Polarization 1079

33.4 The Generation of Electromagnetic Waves 1088

33.5 Energy of a Wave 1092

33.6* The Wave Equation 1096


viii CONTENTS

25.3 Potential in Conductors 803

25.4 Calculation of the Field from the

Potential 806

25.5 Energy of Systems of Charges 811


26. CAPACITORS AND DIELECTRICS 828

26.1 Capacitance 829

26.2 Capacitors in Combination 834

26.3 Dielectrics 838

26.4 Energy in Capacitors 844


27. CURRENTS AND OHM’S LAW 858

27.1 Electric Current 859

27.2 Resistance and Ohm’s Law 863

27.3 Resistivity of Materials 868

27.4 Resistances in Combination 872


28. DIRECT CURRENT CIRCUITS 887

28.1 Electromotive Force 888

28.2 Sources of Electromotive Force 890

28.3 Single-Loop Circuits 893

28.4 Multi-Loop Circuits 897

28.5 Energy in Circuits; Joule Heat 901

28.6* Electrical Measurements 903

28.7* The RC Circuit 907

28.8* The Hazards of Electric

Currents 913


29. MAGNETIC FORCE AND FIELD 926

29.1 The Magnetic Force 928

29.2 The Magnetic Field 931

29.3 Ampére’s Law 938

29.4 Solenoids and Magnets 943

29.5 The Biot-Savart Law 948


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34. REFLECTION, REFRACTION, AND OPTICS 1111

34.1 Huygens’ Construction 1113

34.2 Reflection 1114

34.3 Refraction 1117

34.4 Spherical Mirrors 1128

34.5 Thin Lenses 1135

34.6* Optical Instruments 1144


35. INTERFERENCE AND DIFFRACTION 1168

35.1 Thin Films 1169

35.2* The Michelson Interferometer 1174

35.3 Interference from Two Slits 1177

35.4 Interference from Multiple Slits 1183

35.5 Diffraction by a Single Slit 1190

35.6 Diffraction by a Circular Aperture; Rayleigh’s

Criterion 1196


PART VI RELATIVITY, QUANTA,AND PARTICLES 1214

36. THE THEORY OF SPECIAL RELATIVITY 1216

36.1 The Speed of Light; the Ether 1218

36.2 Einstein’s Principle of Relativity 1220

36.3 Time Dilation 1224

36.4 Length Contraction 1230

36.5 The Lorentz Transformations and the

Combination of Velocities 1232

36.6 Relativistic Momentum and Energy 1239

36.7* Mass and Energy 1242


37. QUANTA OF LIGHT 1254

37.1 Blackbody Radiation 1255

37.2 Energy Quanta 1258

37.3 Photons and the Photoelectric Effect 1264

37.4 The Compton Effect 1269

37.5 X Rays 1273

37.6 Wave vs. Particle 1276


CONTENTS ix

38. SPECTRAL LINES, BOHR’S THEORY, AND QUANTUMMECHANICS 1286

38.1 Spectral Lines 1287

38.2 Spectral Series of Hydrogen 1291

38.3 The Nuclear Atom 1293

38.4 Bohr’s Theory 1295

38.5 Quantum Mechanics; The Schrödinger

Equation 1302


39. QUANTUM STRUCTURE OF ATOMS, MOLECULES,AND SOLIDS 1320

39.1 Principal, Orbital, and Magnetic Quantum

Numbers; Spin 1321

39.2 The Exclusion Principle and the Structure of

Atoms 1328

39.3* Energy Levels in Molecules 1333

39.4 Energy Bands in Solids 1336

39.5 Semiconductor Devices 1340


40. NUCLEI 1354

40.1 Isotopes 1355

40.2 The Strong Force and the Nuclear Binding

Energy 1359

40.3 Radioactivity 1365

40.4 The Law of Radioactive Decay 1372

40.5 Fission 1377

40.6* Nuclear Bombs and Nuclear Reactors 1379

40.7 Fusion 1384


41. ELEMENTARY PARTICLES AND COSMOLOGY 1396

41.1 The Tools of High-Energy Physics 1397

41.2 The Multitude of Particles 1403

41.3 Interactions and Conservation Laws 1405

41.4 Fields and Quanta 1409

41.5 Quarks 1412

41.6 Cosmology 1416


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APPENDIX 6: THE INTERNATIONAL SYSTEM OF UNITS(SI) A-21

A6.1 Base Units A-21

A6.2 Derived Units A-23

A6.3 Prefixes A-23

APPENDIX 7: BEST VALUES OF FUNDAMENTALCONSTANTS A-23

APPENDIX 8: CONVERSION FACTORS A-26

APPENDIX 9: THE PERIODIC TABLE AND CHEMICALELEMENTS A-31

APPENDIX 10: FORMULA SHEETS A-33

Chapters 1–21 A-33

Chapters 22–41 A-34

APPENDIX 11: ANSWERS TO ODD-NUMBEREDPROBLEMS AND REVIEW PROBLEMS A-35

PHOTO CREDITS A-39

INDEX A-41

x CONTENTS

APPENDICES

APPENDIX 1: GREEK ALPHABET A-1

APPENDIX 2: MATHEMATICS REVIEW A-1

A2.1 Symbols A-1

A2.2 Powers and Roots A-1

A2.3 Arithmetic in Scientific Notation A-2

A2.4 Algebra A-3

A2.5 Equations with Two Unknowns A-5

A2.6 The Quadratic Formula A-5

A2.7 Logarithms and the Exponential Function A-5

APPENDIX 3: GEOMETRY AND TRIGONOMETRYREVIEW A-7

A3.1 Perimeters, Areas, and Volumes A-7

A3.2 Angles A-7

A3.3 The Trigonometric Functions A-8

A3.4 The Trigonometric Identities A-9

A3.5 The Laws of Cosines and Sines A-10

APPENDIX 4: CALCULUS REVIEW A-10

A4.1 Derivatives A-10

A4.2 Important Rules for Differentiation A-11

A4.3 Integrals A-12

A4.4 Important Rules for Integration A-15

A4.5 The Taylor Series A-18

A4.6 Some Approximations A-18

APPENDIX 5: PROPAGATING UNCERTAINTIES A-19

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Preface

Our aim in Physics for Engineers and Scientists, Third Edition, is to present a modern

view of classical mechanics and electromagnetism, including some optics and quantum

physics. We also want to offer students a glimpse of the practical applications of physics

in science, engineering, and everyday life.

The book and its learning package emerged from a collaborative effort that began

more than six years ago. We adapted the core of Ohanian’s earlier Physics (Second

Edition, 1989) and combined it with relevant findings from recent physics education

research on how students learn most effectively. The result is a text that presents a

clear, uncluttered explication of the core concepts in physics, well suited to the needs

of undergraduate engineering and science students.

Organizat ion of TopicsThe 41 chapters of the book cover the essential topics of introductory physics: mechan-

ics of particles, rigid bodies, and fluids; oscillations, wave motion, heat and thermo-

dynamics; electricity and magnetism; optics; special relativity; and atomic and subatomic

physics.

Our arrangement and treatment of topics are fairly traditional with a few delib-

erate distinctions. We introduce the principle of superposition of forces early in

Chapter 5 on Newton’s laws of motion, and we give the students considerable expo-

sure to the vector superposition of gravitational forces in Chapter 9. This leaves the

students well prepared for the later application of vector superposition of electric and

magnetic forces generated by charge or current distributions. We place gravitation in

Chapter 9 immediately after the chapters on work and energy, because we regard

gravitation as a direct application of these concepts (instructors who prefer to post-

pone gravitation can, of course, do so). We introduce forces on stationary electric

charges in a detailed, complete exposition in Chapter 22, before proceeding to the

less obvious concept of the electric field in Chapter 23. We start the study of magnetism

in Chapter 29 with the force on a moving charged particle near a current, instead of

the more common practice of starting with a postulate about the magnetic field in

the abstract. With our approach, the observed magnetic forces on moving charges

lead naturally to the magnetic field, and this progression from magnetic force to

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magnetic field will remind students of the closely parallel progression from electric force

to electric field. For efficiency and brevity, we sometimes combine in one chapter

closely related topics that other authors elect to spread over more than one chapter.

Thus, we cover induction and inductance together in Chapter 31 and interference

and diffraction together in Chapter 35.

Conc ise Wri t ing wi th Sharp Focus on Core ConceptsOur goal is concise exposition with a sharp focus on core concepts. Brevity is desir-

able because long chapters with a large number of topics and excessive verbiage are

confusing and tedious for the student. In our writing, we obey the admonitions of

Strunk and White’s Elements of Style: use the active voice; make statements in posi-

tive form; use definite, specific, concrete language; omit needless words.

We strove for simplicity in organizing the content. Each chapter covers a small set

of core topics—rarely more than five or six—and we usually place each core topic in

a section of its own. This divides the content into manageable segments and gives the

chapter a clear and clean outline. Transitional sentences at the beginning or end of

sections spell out the logical connections between each section and the next. Within

each section, we strove for a seamless narrative leading from the discussions of concepts

to their applications in Example problems. We sought to avoid the patchy, cobbled

structure of many texts in which the discussions appear to serve as filler between one

equation and the next.

Emphas is on the Atomic S t r uc ture o f Mat terThroughout the book, we encourage students to keep in mind the atomic structure of

matter and to think of the material world as a multitude of restless electrons, protons,

and neutrons. For instance, in the mechanics chapters, we emphasize that all macro-

scopic bodies are systems of particles and that the equations of motion for macro-

scopic bodies emerge from the equations of motion of the individual particles. We

emphasize that macroscopic forces are the result of a superposition of the forces among

the particles of the system, and we consider atoms and their bonds in the qualitative

discussions of elasticity, thermal expansion, and changes of state. By exposing students

to the atomic structure of matter in the first semester, we help them to grasp the nature

of the charged particles that play a central role in the treatment of electricity and mag-

netism in the second semester. Thus, in the electricity chapters, we introduce the con-

cepts of positive and negative charge by referring to protons and electrons, not by

referring to the antiquated procedure of rubbing glass rods with silk rags.

We try to make sure that students are always aware of the limitations of the nine-

teenth-century fiction that matter and electric charge are continua. Blind reliance on

this old fiction has often been justified by the claim that, although engineering students

need physics as a problem-solving tool, the atomic structure of matter is of little con-

cern to them.This supposition may be adequate for a superficial treatment of mechan-

ical engineering. Yet much of modern engineering—from materials science to

electronics—hinges on understanding the atomic structure of matter. For this pur-

pose, engineers need a physicist’s view of physics.

Real -Wor ld Examples Begin Each ChapterEach chapter opens with a “Concepts in Context” photograph illustrating a practical

application of physics. The caption for this photo explores various core concepts in a

concrete real-world context. The questions included in the caption are linked to sev-

eral solved Examples or discussions later in the chapter. Such revisiting of the

xii PREFACE

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chapter-opening application provides layers of learning, as new concepts are carefully

built upon a foundation firmly planted in the real world. The emphasis on real-world

data is also evident throughout other Examples and in the end-of-chapter problems.

By exposing students to realistic data, we give them confidence to apply physics in

their later science or engineering courses.

Conceptual Discussions Precede and Motivate the MathOnly after a careful exposition of the conceptual foundations in a qualitative physical

context does each section proceed to the mathematical treatment. Thus, we ensure

that the mathematical formulas and their consequences and variations are rooted in a

firm conceptual foundation. We were very careful to provide clear, thorough, and accu-

rate explanations and derivations of all mathematical statements, to ensure that students

acquire a good intuition about why particular equations are applied. Immediately after

such derivations, we provide solved Examples to establish a firm connection between

theory and concrete practical applications.

Examples En l iven the Text We devote significant portions of each chapter to carefully selected Examples of solved

problems—about 390 altogether or 9 on average per chapter.These Examples are con-

crete illustrations of the preceding conceptual discussions. They build cumulatively

upon each other, from simple to more complicated as the chapter progresses.To enliven

the text, we employ realistic data in the Examples, such as students would actually

encounter outside the classroom. The solved Examples are designed to cover most

variations of possible problems, with solutions that include both general approaches and

specific details on how to extract the important information for the given problem.

For instance, when such keywords as initially or at rest occur in a solved Example, we

are careful to point out their importance in the problem-solving process. Comments

appended to some Examples draw attention to limitations in the solution or to wider

implications.

Checkup Ques t ions Implement Ac t ive Learn ingWe conclude each section of a chapter with a series of brief Checkup questions. These

permit students to test their mastery of core concepts, and they can be of great help in

clearing up common misconceptions. Checkup questions include variations and “flip

sides” of simple concepts that often occur to students but are rarely addressed. We give

detailed answers to each Checkup question at the back of the chapter. The entire book

contains roughly 5 Checkup questions per section—comprising a total of about 800

Checkup questions.

The final Checkup question of each section is always in multiple-choice forrmat—

specifically designed for interactive teaching. At the University of Texas, instructors

use such multiple-choice questions as classroom concept quizzes for welcome breaks

in conventional lecturing. When more than one answer is popular, the instructor and

class immediately know that more discussion or more examples are needed. Such occa-

sions lend themselves well to peer instruction, in which the students explain to one

another their reasoning before responding. This pedagogy implements an active, par-

ticipatory alternative to the traditional lecture format. In addition, several supplements

to the textbook, including the Student Activity Workbook, Online Concept Tutorials,

Smartwork online homework, and PhysiQuizzes also implement active learning and

a mastery-based approach.

PREFACE xiii

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Problem-So lv ing Techn iques Many chapters have inserts in the form of boxes devoted to Problem-Solving Techniques.

These 39 skill boxes summarize the main steps or approaches for the solution of

common classes of problems. Often deployed after several seemingly disparate Examples,

the Problem-Solving Techniques boxes underscore the unity and generality of the

techniques used in the Examples. The boxes list the steps or approaches to be taken,

providing a handy reference and review.

Math He lp We have placed a Math Help box wherever students encounter a mathematical con-

cept or technique that may be difficult or unfamiliar. These 6 skill boxes briefly review

and summarize such topics as trigonometry, derivatives, integrals, and ellipses. Students

can find more detailed help in Appendix 2 on basic algebra, 3 on trigonometry and

geometry, 4 on calculus, and 5 on propagation of uncertainties.

Phys ics in Prac t i ce Many chapters have a short essay on Physics in Practice that illustrates an application

of physics in engineering and everyday life. These 27 essay boxes discuss practical

topics, such as ultracentrifuges, communication and weather satellites, magnetic lev-

itation, etc. Each of these essays provides a wealth of interesting detail and offers a

practical supplement to some of the chapter topics. They have been designed to be

engaging, yet sufficiently qualitative to provide some respite from the more analytical

discussions, Examples, and Questions.

F igures and Ba l loon Capt ionsOver 1,500 figures illustrate the text. We made every effort to assemble a visual nar-

rative as clear as the verbal narrative. Each figure in a sequence carefully builds upon

the visual information in the figure that precedes it. Many figures in the text contain

a caption in “balloon” that points to important features within the figure. The bal-

loon caption is a concise and informative supplement to the conventional figure cap-

tion. The balloons make immediately obvious some details that would require a long,

wordy explanation in the conventional caption. Often the balloon captions are arranged

so that some cause-effect or other sequential thought process becomes immediately

evident. All drawn figures are available to instructors in digital form for use in the

course.

End–of–Chapter Summar y Each chapter narrative closes with several support elements, starting with a brief

Summary. The Summary contains the essential physical laws, quantities, definitions,

and key equations introduced in the chapter. A page reference, key equation number,

and often a thumbnail figure accompany these laws, definitions, and equations. The

Summary does not include repetition of the detailed explanations of the chapter. The

Summary is followed by Questions for Discussion, Problems, Review Problems, and

Answers to Checkups.

Ques t ions for D iscuss ionAfter the chapter’s Summary, we include a large selection of qualitative Questions for

Discussion — about 700 in the entire book or roughly 17 per chapter. We intend these

qualitative end-of-chapter Questions to stimulate student thinking. Some of these

questions are deliberately formulated so as to have no unique answer, which is intended

to promote class discussion.

xiv PREFACE

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ProblemsAfter the chapter’s qualitative Questions, we include computational Problems grouped

by chapter section — about 3000 in the entire book, or roughly 73 per chapter. Each

problem’s level of difficulty is indicated by no asterisk, one asterisk (*), or two asterisks

(**). Most no-asterisk Problems are easy and straightforward, only requiring students

to “plug in” the correct values to compute answers or to retrace the steps of an Example.

One-asterisk Problems are of medium difficulty. They contain a few complications

requiring the combination of several concepts or the manipulation of several formu-

las. Two-asterisk Problems are difficult and challenging. They demand considerable

thought and perhaps some insight, and occasionally require appreciable mathematical

skill. When an Online Concept Tutorial (see below) is available for help in mastering

the concepts in a given section, a dagger footnote (†) tells students where to find the

tutorial.

We tried to make the Problems interesting for students by drawing on realistic

examples from technology, science, sports, and everyday life. Many of the Problems

are based on data extracted from engineering handbooks, car repair manuals, Jane’s

Book of Aircraft, The Guinness Book of World Records, newspaper reports, research and

industrial instrumentation manuals, etc. Many other Problems deal with atoms and

subatomic particles. These Problems are intended to reinforce the atomistic view of

the material world. In some cases, experts will perhaps consider the use of classical

physics somewhat objectionable in a problem that really ought to be handled by quan-

tum mechanics. But we believe that the advantages of familiarization with atomic

quantities and magnitudes outweigh the disadvantages of an occasional naive use of clas-

sical mechanics.

Among the Problems are a smaller number of somewhat contrived, artificial

Problems that make no pretense of realism (for example, “A block slides on an inclined

plane tied by string...”). Such unrealistic Problems are sometimes the best way to bring

an important concept into sharp focus. Some Problems are formulated as guided prob-

lems, with a series of questions that take the student through an important problem-

solving procedure, step by step.

Rev iew Prob lems After the Problems section of each chapter, we offer an extra selection of Review

Problems — about 600 in the entire book or roughly 15 per chapter. We wrote these

Review Problems specifically to help students prepare for examinations. Hence, Review

Problems often test comprehension by requiring students to apply concepts from more

than one section of the chapter and occasionally from prior, related chapters. Answers

to all odd-numbered Problems and Review Problems are given in Appendix 11.

Uni t s and S ign i f i cant F iguresWe use the SI system of units exclusively throughout the text. In the abbreviations for

the units we follow the recommendations of the International Committee for Weights

and Measures (CIPM), although we retain some traditional units, such as revolution

and calorie that have been discontinued by the CIPM. In addition, for the sake of

clarity we spell out the name of the unit in full whenever the abbreviation is likely to

lead to ambiguity and confusion (for instance, in the case of V for volt, which is easily

confused with V for potential; or in the case of C for coulomb, which might be con-

fused with C for capacitance). We try to use realistic numbers of significant figures,

with most Examples and Problems using two or three. In cases where it is natural to

employ some data with two significant figures and some with three, we have been

careful to propagate the appropriate number of significant figures to the result.

PREFACE xv

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For reference purposes, we give the definitions of the British units. Currently only

the United States, Bangladesh, and Liberia still adhere to these units. In the United

States, automobile manufacturers have already switched to metric units for design and

construction.The U. S. Army has also switched to metric units, so soldiers give distances

in meters and kilometers (in army slang, the kilometer is called a “klick,” a usage that

is commendable itself for its brevity). British units are not used in examples or in prob-

lems, with the exception of a handful of problems in the early chapters. In the defini-

tions of the British units, the pound (lb) is taken to be the unit of mass, and the pound

force (lbf ) is taken to be the unit of force. This is in accord with the practice approved

by the American National Standards Institute (ANSI), the Institute of Electrical and

Electronics Engineers (IEEE), and the U. S. Department of Defense.

Opt ional Sec t ions and ChaptersWe recognize course content varies from institution to institution. Some sections and

some chapters can be regarded as optional and can be omitted without loss of conti-

nuity. These optional sections are marked by asterisks in the Table of Contents.

Mathemat i ca l Prerequis i tesIn order to accommodate students who are taking an introductory calculus course con-

currently, derivatives are used slowly at first (Chapter 2), and routinely later on. Likewise,

the use of integrals is postponed as long as possible (Chapter 7), and they come into

heavy use only in the second volume (after Chapter 21). For students who need a

review of calculus, Appendix 4 contains a concise primer on derivatives and integrals.

AcknowledgmentsWe have had the benefit of a talented author team for our support resources. In addi-

tion to their primary role in the assembly of the learning package, they all have also made

substantial contributions to the accuracy and clarity of the text.

Stiliana Antonova, Barnard College

Charles Chiu, University of Texas-Austin

William J. Ellis, University of California-Davis

Mirela Fetea, University of Richmond

Rebecca Grossman, Columbia University

David Harrison, University of Toronto

Prabha Ramakrishnan, North Carolina State University

Hang Deng-Luzader, Frostburg State University

Stephen Luzader, Frostburg State University

Kevin Martus, William Paterson University

David Marx, Illinois State University

Jason Stevens, Deerfield Academy

Brian Woodahl, Indiana University–Purdue University-Indianapolis

Raymond Zich, Illinois State University

And at Sapling Systems and Science Technologies in Austin, Texas, for content,

James Caras, Ph.D.; Jon Harmon, B.S.; Kevin Nelson, Ph.D.; John A. Underwood,

Ph.D.; and Jason Vestuto, M.S. and for animation and programming, Jeff Sims and

Nathan Wheeler.

Our manuscript was subjected to many rounds of peer review. The reviewers

were instrumental in identifying myriad improvements, for which we are grateful:

xvi PREFACE

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Yildirim Aktas University of North Carolina–Charlotte

Patricia E. Allen Appalachian State University

Steven M. Anlage University of Maryland

B. Antanaitis Lafayette College

Laszlo Baksay Florida Institute of Technology

Marco Battaglia University of California-Berkeley

Lowell Boone University of Evansville

Marc Borowczak Walsh University

Amit Chakrabarti Kansas State University

D. Cornelison Northern Arizona University

Corbin Covault Case Western Reserve University

Kaushik De University of Texas at Arlington

William E. Dieterle California University of Pennsylvania

James Dunne Mississippi State University

R. Eagleton California Polytechnic University-Pomona

Gregory Earle University of Texas-Dallas

William Ellis University of California-Davis

Mark Eriksson University of Wisconsin-Madison

Morten Eskildsen University of Notre Dame

Bernard Feldman University of Missouri–St. Louis

Mirela Fetea University of Richmond

J. D. Garcia University of Arizona

U. Garg University of Notre Dame

Michael Gurvitch State University of New York at Stony Brook

David Harrison University of Toronto

John Hernandez University of North Carolina–Chapel Hill

L. Hodges Iowa State University

Jean-Pierre Jouas United Nations International School

Kevin Kimberlin Bradley University

Sebastian Kuhn Old Dominion University

Tiffany Landry Folsom Lake College

Dean Lee North Carolina State University

Frank Lee George Washington University

Stephen Luzader Frostburg State University

Kevin Martus William Paterson University

M. Matkovich Oakton Community College

David McIntyre Oregon State University

Rahul Mehta University of Central Arkansas

Kenneth Mendelson Marquette University

Laszlo Mihaly State University of New York at Stony Brook

Richard Mistrick Pennsylvania State University

Rabindra Mohapatra University of Maryland

Philip P. J. Morrison University of Texas at Austin

Greg Mowry University of Saint Thomas

David Murdock Tennessee Technological University

Anthony J. Nicastro West Chester University

Scott Nutter Northern Kentucky University

Robert Oerter George Mason University

Ray H. O’Neal, Jr. Florida A & M University

Frederick Oho, Winona State University

Paul Parris University of Missouri–Rolla

Ashok Puri University of New Orleans

Michael Richmond Rochester Institute of Technology

John Rollino Rutgers University–Newark

David Schaefer Towson State University

Joseph Serene Georgetown University

H. Shenton University of Delaware

Jason Stevens Deerfield Academy

PREFACE xvii

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Jay Strieb Villanova University

John Swez Indiana State University

Devki N. Talwar Indiana University of Pennsylvania

Chin-Che Tin Auburn University

Tim Usher California State University-San Bernardino

Andrew Wallace Angelo State University

Barrett Wells University of Connecticut

Edward A.P. Whittaker Stevens Institute of Technology

David Wick Clarkson University

Don Wieber Contra Costa College

J. William Gary University of California-Riverside

Suzanne Willis Northern Illinois University

Thomas Wilson Marshall University

William. J. F. Wilson University of Calgary

Brian Woodahl Indiana University–Purdue University-Indianapolis

Hai-Sheng Wu Mankato State University

We thank John Belcher, Michael Danziger, and Mark Bessette of the Massachusetts

Institute of Technology for creating the cover image. It illustrates the magnetic field

generated by two currents in two copper rings. This is one frame of a continuous ani-

mation; at the instant shown, the current in the upper ring is opposite to that in the

lower ring and is of smaller magnitude. The magnetic field structure shown in this

picture was calculated using a modified intregration technique.This image was created

as part of the Technology Enabled Active Learning (TEAL) program in introduc-

tory physics at MIT, which teaches physics interactively, combining desktop experiments

with visualizations of those experiments to “make the unseen seen.”

We thank the several editors that supervised this project: first Stephen Mosberg,

then Richard Mixter, John Byram, and finally Leo Wiegman, who had the largest

share in the development of the text, and also gave us the benefit of his incisive line-

by-line editing of the proofs, catching many slips and suggesting many improvements.

We also thank the editorial staff at W. W. Norton & Co., including Chris Granville,

April Lange, Roy Tedoff, Rubina Yeh, Rob Bellinger, Kelly Mitchell, Neil Hoos, Lisa

Rand, and Sarah Mann, as well as the publishing professionals whom Norton engaged,

such as Paul Lacy, Richard K. Mickey, Susan McLaughlin, and John B. Woolsey for

their enthusiasm and their patience in dealing with the interminable revisions and

corrections of the text and its support package. In addition, JTM is grateful to Robert

W. Christy of Dartmouth University for various pointers on textbook writing.

xviii PREFACE

HANS C. OHANIAN JOHN T. MARKERT

Burlington, Vermont Austin, Texas

[email protected] [email protected]

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Publ i ca t ion FormatsPhysics for Engineers and Scientists comprises six parts. The text is published in two

hardcover versions and several paperback versions.

Hardcover Vers ionsThird Extended Edition, Parts I–VI, 1450 pages, ISBN 0-393-92631-1

(Chapters 1–41 including Relativity, Quanta and Particles)

Third Edition, Parts I–V, 1282 pages, ISBN 0-393-97422-7

(Chapters 1–36, including Special Relativity)

Paperback Vers ions Volume 1, (Chapters 1–21) 778 pages, ISBN 0-393-93003-3

Part I Motion, Force, and Energy (Chapters 1–14)

Part II Oscillations, Waves, and Fluids (Chapters 15–18)

Part III Temperature, Heat, and Thermodynamics (Chapter 19–21)

Volume 2, (Chapters 22–36) 568 pages, ISBN 0-393-93004-1

Part IV Electricity and Magnetism (Chapters 22–32)

Part V Waves and Optics (Chapters 33–35 and Chapter 36 on Special Relativity)

Volume 3, (Chapters 36–41) 250 pages, ISBN 0-393-92969-8

Part VI Relativity, Quanta, and Particles

In addition, to explore customized versions, please contact your Norton representa-

tive.

Two Nor ton ebook Opt ionsPhysics for Engineers and Scientists is available in a Norton ebook format that retains

the content of the print book. The ebook offers a variety of tools for study and review,

including sticky notes, highlighters, zoomable images, links to Online Concept Tutorials,

and a search function. Purchased together, the SmartWork with integrated ebook

bundle makes it easy for students to check text references when completing online

homework assignments.

The ebook may also be purchased as a standalone item. The downloadable PDF

version is available for purchase from Powells.com.

Package Opt ionsEach version of the text purchased from Norton—with or without SmartWork—will

come with free access to our website at Norton’s StudySpace that includes the valuable

Online Concept Tutorials. Each version of the text may be purchased as a stand-alone

book or as a package that includes—each for a fee—Norton’s new SmartWork online

homework system or the Student Activity Workbook by David Harrison and William

Ellis. Hence, several optinal packages are available to instructors:

• Textbook–StudySpace–Online Concept Tutorials + Student Activity Workbook

• Textbook–StudySpace–Online Concept Tutorials + SmartWork/ebook

• Textbook–StudySpace–Online Concept Tutorials + SmartWork/ebook + Student

Activity Workbook

PREFACE xix

nortonebooks.com

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The Suppor t ProgramTo enhance individual learning and also peer instruction, a carefully integrated support

program accompanies the text. Each element of the support program has two goals.

First, each support resource mirrors the text’s emphasis on sharply focused core con-

cepts. Second, treatment of a core concept in a support resource offers a perspective that

is different from but compatible with that of the text. If a student needs help beyond

the text, he or she would more likely benefit from a fresh presentation on the same

concept rather than from one that simply repeats the text presentation.

Hence, the text and its support package offers three or more different approaches

to the core concepts. For example, Newton’s First and Second Laws are rendered with

interactive animations in the Online Concept Tutorial “Forces,” with pencil-and- paper

exercises in Chapter 5 of the Student Activity Workbook crafted by David Harrison

and William Ellis, and with concept test inquiries in PhysiQuiz questions written by

Charles Chiu and edited by Jason Stevens.

Both printed and digital resources are offered within the support program.

Outstanding web-based resources for both instructors and students include tutorials

and a homework system.

Smar tWork Onl ine Homework Sys tem SmartWork—Norton’s online homework management system—provides ready-made

automatically graded assignments, including guided problems, simple feedback ques-

tions, and animated tutorials—all specifically designed to extend the text’s emphasis

on core concepts and problem-solving skills.

Developed in collaboration with Sapling Systems, SmartWork features an intu-

itive, easy-to-use interface that offers instructors flexible tools to manage assignments,

while making it easy for students to compose mathematical expressions, draw vectors

and graphs, and receive helpful and immediate feedback. Two different types of ques-

tions expand upon the exposition of concepts in the text:

Simple Feedback Problems present students with problems that anticipate common

misconceptions and offer prompts at just the right moment to help them discover the

correct solution.

Guided Tutorial Problems addresses more challenging topics. If a student answers

a problem incorrectly, SmartWork guides the student through a series of discrete tuto-

rial steps that lead to a general solution. Each step is a simple feedback question that

the student answers, with hints if necessary. After completing all of the tutorial steps,

the student returns to the original problem ready to apply this newly-obtained knowl-

edge.

SmartWork problems use algorithmic variables so two students are unlikely to see

exactly the same problem. Instructors can use the problem sets provided, or can customize

these ready-made questions and assignments, or use SmartWork to create their own.

SmartWork is available bundled with the Norton ebook of Physics for Engineers

and Scientists. Where appropriate, SmartWork prompts students to review relevant

sections in the textbook. Links to the ebook make it easy for students to consult the

text while working through problems online.

Onl ine Concept Tu tor ia l sDeveloped in collaboration with Science Technologies specifically for this course,

these 45 tutorials feature interactive animations that reinforce conceptual under-

standing and develop students’ quantitative skills. In-text icons alert students to

xx PREFACE

www.wwnorton.com/physics


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the availability of a tutorial. All Online Concept Tutorials are available on the free

StudySpace web site and are integrated into SmartWork. Tutorials can also be

accessed from a CD-ROM that requires no installation, browser tune-ups, or

plug-ins.

StudySpace Webs i teSTUDYSPACE www.wwnorton.com/physics. This free and open website is the portal

for both public and premium content. Free content at StudySpace includes the

Online Concept Tutorials and a Study Plan for each chapter in Physics for Engineers

and Scientists. Premium content at StudySpace includes links to the online ebook

and to SmartWork.

WebAss ignSelected end-of-chapter problems from Physics for Engineers and Scientists are avail-

able in WebAssign, with additional problems available to adopting instructors by

request to WebAssign.

Addi t iona l Ins t r uc tor Resources

TEST BANK by Mirela Fetea, University of Richmond; Kevin Martus, William Paterson

University; and Brian Woodahl, Indiana University-Purdue University-Indianapolis.

The Test Bank offers approximately 2000 multiple-choice questions, available in

ExamView, WebCT, BlackBoard, rich-text, and printed format.

INSTRUCTOR SOLUTIONS MANUAL by Stephen Luzader and Hang-Deng Luzader,

both of Frostburg State University, and David Marx of Illinois State University.The

Instructor Solution Manual offers solutions to all end-of-chapter Problems and

Review Problems, checked for accuracy and clarity.

PHYSIQUIZ “CLICKER” QUESTIONS by Charles Chiu, University of Texas at Austin,

with Jason Stevens, Deerfield Academy.The PhysiQuiz multiple-choice questions

are designed for use with classroom response, or “clicker”, systems. The 300

PhysiQuiz questions are available as PowerPoint slides, in printed format, and as

transparency masters.

NORTON MEDIA LIBRARY INSTRUCTOR CD-ROM The Media Library for instrutors includes

selected figures, tables, and equations from the text in JPEG and PowerPoint for-

mats, PhysiQuiz “clicker” questions, and PowerPoint-ready offline versions of the

Online Concept Tutorials.

INSTRUCTOR RESOURCE MANUAL offers a guide to the support package with descrip-

tions of the Online Concept Tutorials, information about the SmartWork home-

work problems available for each chapter, printed PhysiQuiz “clicker” questions,

and instructor notes for the workshop activities in the Student Activity

Workbook.

TRANSPARENCY ACETATES Approximately 200 printed color acetates of key figures from

the text.

BLACKBOARD AND WEBCT COURSE CARTRIDGES Course Cartridges for BlackBoard

and WebCT include access to the Online Concept Tutorials, a Study Plan for each

chapter, multiple-choice tests, plus links to the premium, password-protected con-

tents of the Norton ebook and SmartWork.

PREFACE xxi


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Addi t iona l S tudent Resources

STUDENT ACTIVITY WORKBOOK by David Harrison, University of Toronto, and William

Ellis, University of California Davis. The Student Activity Workbook is an important

part of the learning package. For each chapter of Physics for Engineers and Scientists, the

Workbook’s Activities break down a physical condition into constituent parts. The

Activities are pencil and paper exercises well suited to either individual or small group

collaboration.The Activities include both conceptual and quantitative exercises. Some

Activities are guided problems that pose a question and present a solution scheme via

follow up questions. The Workbook is available in two paperback volumes: Volume

1 comprises Chapters 1–21 and Volume 2 comprises Chapters 22–41.

STUDENT SOLUTIONS MANUAL by Stephen Luzader and Hang-Deng Luzader, both

of Frostburg State University, and David Marx of Illinois State University. The

Student Solutions Manual contains detailed solutions to approximately 25% of

the problems in the book, chosen from the odd-numbered problems whose answers

appear in the back of the book.The Manual is available in two paperback volumes:

Volume 1 comprises Chapters 1–21 and Volume 2 comprises Chapters 22–41.

ONLINE CONCEPT TUTORIALS CD-ROM The 45 Online Concept Tutorials (see above)

can also be accessed from an optional CD-ROM that requires no installation,

browser tune-up, or plug-in.

xxii PREFACE

About the Authors Hans C. Ohanian received his B.S. from the University of California, Berkeley, and

his Ph.D from Princeton University, where he worked with John A. Wheeler. He has

taught at Rensselaer Polytechnic Institute, Union College, and the University of

Vermont. He is the author of several textbooks spanning all undergraduate levels:

Physics, Principles of Physics, Relativity: A Modern Introduction, Modern Physics, Principles

of Quantum Mechanics, Classical Electrodynamics, and, with Remo Ruffini, Gravitation

and Spacetime. He is also the author of dozens of articles dealing with gravitation, rel-

ativity, and quantum theory, including many articles on fundamental physics published

in the American Journal of Physics, where he served as associate editor for some years.

He lives in Vermont. [email protected]

John T. Markert received his B.A. in physics and mathematics from Bowdoin College

(1979), and his M.S. (1984) and Ph.D. (1987) in physics from Cornell University,

where he was recipient of the Clark Award for Excellence in Teaching. After postdoc-

toral research at the University of California, San Diego, he joined the faculty at the

University of Texas at Austin in 1990, where he has received the College of Natural

Sciences Teaching Excellence Award and is currently Professor of Physics and Department

Chair. His introductory physics teaching methods emphasize context-based approaches,

interactive techniques, and peer instruction. He is author or coauthor of over 120 jour-

nal articles, including experimental condensed-matter physics research in supercon-

ductivity, magnetism, and nanoscience. He lives in Austin, Texas, with his spouse and

four children. [email protected]

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About the Book xxiii

Owner’s Manual for Physics for Engineers and ScientistsThese pages give a brief tour of the features of Physics for Engineers and Scientists and

its study resources. Some resources are found within the book. Others are located in

accompaning paperback publications or at the StudySpace web portal. Features on the

text pages shown here come chiefly from the discussion of friction in Chapter 6, but

are common in other chapters.

The learning resources listed below help students study by offering alternative

explanations of the core concepts found in the text.These student resources are briefly

described at the end of this owner’s manual:

• Online Concept Tutorials • Student Activity Workbook

• SmartWork Online Homework • Student Solutions Manual

• StudySpace

Each chapter of the textbook starts

with a real-world example of a

core concept. Chapter 6 opens

with the concept of friction and

uses automobile tires as an

example of friction that is revisited

in several different conditions. The

opening photograph, it’s caption

and the caption’s closing questions

all discuss this example.

In this chapter, the rubber tires of an

automobile are revisited to explore

concepts in friction on pages 176, 179,

180, and 186, as indicated.

Most chapters have six or

fewer sections. Most

sections are four or five

pages in length and cover

one major topic.

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xxiv OWNER’S MANUAL

The icon indicates an Online Concept Tutorial isavailable for a key concept. Each such icon includesthe identification number of the tutorial—8, in thiscase. These tutorials offer a visual guide and self-quiz for the concept at hand. Find all the Tutorialsat www.wwnorton.com/physics.

In mathematical expressions, such as ma=F, the boldtype indicates a vector and italic indicates variablesthat are not vectors.

Text in italic type indicates major definitions of lawsor statements of general principles.

Text in bold type highlights the first use of a keyterm and is generally accompanied by an explanation.

Key concepts orimportant variants ofthese concepts have akey-term label in themargin.

Short biographical sketches

appear in the margins of this

text. Each offer a brief glimpse

into the life of some major

contributor to our knowledge

about the physical world—in this

case, Italian artist and engineer

Leonardo da Vinci.

Highlighted equations are keyequations that express centralphysics concepts mathematically.

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About the Book xxv

O

y

x

N

P

w

(b)

fk

Py

Px

Py

Px

A push at an angle hasboth horizontal andvertical components.

(a)

P

30

Throughout the text, figures often build on eachother with a new layer of information.• Balloon comments often point out components

of special note in the figure.

The Concept in Context icon here indicatesthe chapter-opening example —automobiletires—is being revisited. In this Example, weexplore the slowing down of a skidding auto-mobile with a specific coefficient of kineticfriction for a rubber tire.

Solutions in Examples may cover bothgeneral approaches and specific detailson how to extract the information fromthe problem statement.

Comments occasionally close anExample to point out the particularlimitations and broader implications ofa Solution.

Examples are a critical part of each chapter.• Examples provide concrete illustrations of the conceptsbeing discussed.• As the chapter unfolds, Examples progress fromsimple to more complex.

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xxvi OWNER’S MANUAL

Problem-Solving Techniques boxesappear in relevant places throughoutthe book and offers tips on how toapproach problems of a particularkind—in this case, problems involvingthe use of friction or centripetal force.

Answers to Checkups appear at the veryback of each chapter, after the ReviewProblems.

A Checkup appears at the end of eachsection within a chapter.• Each Checkup is a self-quiz to test

the reader’s mastery of the concepts inthe preceding section.• Each Checkup has an answer (see

below).

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About the Book xxvii

Math Help boxes offer specific mathematical guidance at the initiallocation in the text where that tech-nique is most relevant.• In this case in Chapter 9, ellipses

are important in studying orbits.• Additional math help is available in

Appendices 2, 3, 4, and 5 at the backof the textbook.

Throughout the text, Physics inPractice boxes offer specific details ona real-world application of theconcept under discussion—in thiscase, forces at work in automobilecollisions in Chapter 11.

The text frequently offers tables of typical values ofphysical quantities.• Such tables usually are labeled “Some ...,” as in thiscase, from Chapter 5.• These tables give some impression of the magnitudes

encountered in the real world.

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xxviii OWNER’S MANUAL

A Summary lists the subjects and page refer-ences for any special content in thischapter—such as Math Help, Problem-SolvingTechniques, or Physics in Practice boxes.• Next the Summary lists the chapter’s core

concepts in the order they are treated. Theconcept appears on the left in bold.• The mathematical expression for the concept

appears in the middle column with an equationnumber on the far right.

Each chapter closes with a Summary followedby Questions for Discussion, Problems, ReviewProblems, and Answers to Checkups.

About 15 or more Questions forDiscussion follow the Summary in eachchapter.• These questions require thought, but not

calculation; e.g. “Why are wet streetsslippery?”• Some of these questions are intended as

brain teasers that have no unique answer,but will lead to provocative discussions.

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About the Book xxix

The dagger footnote (†) that accompanies aProblem heading—in this case, “6.1Friction”—indicates the availability of anOnline Concept Tutorial on this specific topicand states its web address.

About 70 Problems and 15 Review Problemsfollow each chapter’s Questions for Discussion.• The Problem’s statement contains data andconditions upon which a solution will hinge.• Problems are grouped by chapter section andproceed from simple to more complex withineach section.• Many Problems employ real-world data andoccasionally may introduce applications beyondthose treated in the chapter.

Review Problems are specifically designed tohelp students prepare for examinations.• Review Problems often test comprehension

of concepts from more than one sectionwithin the chapter.• Review Problems often take a guided

approach by posing series of questions thatbuild on each other.

Problems and Review Problems are marked bylevel of difficulty:• Those without an asterisk are the most

common and require very little manipulationof existing equations; or they may merelyrequire retracing the steps of a workedExample.• Problems marked with one asterisk (*) are of

medium difficulty and may require use ofseveral concepts and manipulation more thanone equation to isolate and solve for theunknown variable.• Problems marked with two asterisks (**) are

challenging, demand considerable thought,may require significant mathematical skill, andare the least common.

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xxx OWNER’S MANUAL

Onl ine Concept Tu tor ia l swww.wwnorton.com/physics

1 Unit Conversion 1.5, 1.6

2 Significant Digits 1.6

3 Acceleration 2.4, 2.5, 2.6

4 Vector Addition and Vector Components 3.1, 3.2, 3.3

5 Projectile Motion 4.4

6 Forces 5.4

7 “Free-Body” Diagrams 5.3, 5.5, 5.6

8 Friction 6.1

9 Work of a Variable Force 7.1, 7.2, 7.4

10 Conservation of Energy 8.1, 8.2, 8.3

11 Circular Orbits 9.1, 9.3

12 Kepler’s Laws 9.4

13 Momentum in Collisions 11.1, 11.3

14 Elastic and Inelastic Collisions 11.2, 11.3

15 Rotation about a Fixed Axis 12.2

16 Oscillations and Simple Harmonic Motion 15.1

17 Simple Pendulum 15.4

18 Wave Superposition 16.3, 16.4

19 Doppler Effect 17.4

20 Fluid Flow 18.1, 18.2, 18.6

21 Ideal-Gas Law 19.1

22 Specific Heat and Changes of State 20.1, 20.4

23 Heat Engines 21.2

24 Coulomb’s Law 22.2

25 Electric Charge 22.1, 22.5

26 Electric Force Superposition 22.3

27 Electric Field 23.1, 23.3

28 Electric Flux 24.1

29 Gauss’ Law 24.2, 24.3

30 Electrostatic Potential 25.1, 25.2, 25.4

31 Superconductors 27.3

32 DC Circuits 28.1, 28.2, 28.3, 28.4, 28.7

33 Motion in a Uniform Magnetic Field 30.1

34 Electromagnetic Induction 31.2, 31.3

35 AC Circuits 32.1, 32.2, 32.3, 32.5

36 Polarization 33.3

37 Huygens’ Construction 34.1, 34.2, 34.3

38 Geometric Optics and Lenses 34.4, 34.5

39 Interference and Diffraction 35.3, 35.5

40 X-ray Diffraction 35.4

41 Special Relativity 36.1, 36.2

42 Implications of Special Relativity 36.2, 36.3

43 Bohr Model of the Atom 38.1, 38.2, 38.4

44 Quantum Numbers 39.1, 39.2

45 Radioactive Decay 40.4

Many Tutorials contain online experiments—in this case, determining how the kinetic frictionforce varies with the normal force and with thechoice of materials.

The online experiments allow students to changeindependent variables—in this case, mass and material.• Students may collect and display data in a built-in lab

notebook.• Each Tutorial includes an interactive self-quiz.

The Online Concept Tutorials listed here indicate eachtextbook section supported by the tutorial (in paratheses).

A n Online Concept Tutorial accompanies

many central topics in this textbook. When a

Tutorial is available, its numbered icon appears

at section heading within the chapter and a

dagger footnote appears in the end-of-chapter

Problems section as reminder. These Tutorials

are digitally delivered, either via the Internet or

via a CD-ROM for those without Internet

access.

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About the Support Resources xxxi

SmartWork is a subscription-based online homework-management system that makes

it easy for instructors to assign, collect and grade end-of-chapter problems from Physics

for Engineers and Scientists. Built-in hinting and feedback address common misper-

ceptions and help students get the maximum benefit from these assignments.

SmartWork is available as a stand-alone purchase, or with an integratedebook version of Physics for Engineersand Scientists.• Where appropriate, SmartWork

prompts students to review relevantsections of the text.• Links to the ebook make it easy for

students consult the text whileworking through problems.

Simple Feedback Problems antici-pate common misconceptions andoffer prompts at just the rightmoment to help students reach thecorrect solution.

Guided Tutorial Problems addresschallenging topics.• If a student solves one of these

problems incorrectly, she is presentedwith a series of discrete tutorial stepsthat lead to a general solution.• Each step includes hinting and

feedback. After working throughthese remedial steps, the studentreturns to a restatement of theoriginal problem, ready to apply thisnewly obtained knowledge.


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xxxii OWNER’S MANUAL

For each chapter of the textbook, the StudentActivity Workbook offers Activities designed to

break down a physical condition into constituent

parts.

• The Activities are unique to the Workbook and

not found in the textbook.

• The Activities are pencil and paper exercises well

suited to either individual or small group collabo-

ration.

• The Activities include both conceptual and quan-

titative questions.

Student Ac t iv i ty Workbook

Student So lu t ion Manual

S tudySpace

STUDY PLANS

The Student Solutions Manual contains worked solutions

for about 50% of the odd-numbered Problems and Review

Problems in the text.

• Appendix 11 in the back of the textbook contains only the

final answer for odd-numbered problems in the chapters,

not the intermediate steps of the solutions.

• 45 Online Concept Tutorials—at no additional cost.

• 41 Study Plans, one for each chapter—at no additional cost.

• Smartwork online homework system—a subscription service.

• ebook links to textbook chapters—as part of subscription service.

The Student Activity Workbook is

available in two paperback volumes:

Volume 1 comprises Chapters 1–21

and Volume 2 comprises Chapters

22–41.

T he Student Solution Manual is

available in two paperback volumes:

Volume 1 comprises Chapters 1–21

and Volume 2 comprises Chapters

22–41.

The StudySpace website is the free and open portal through which

students access the resources that accompany this text.

www.wwnor ton.com/phys ics

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xxxv

Physics is the study of matter. In a quite literal sense, physics

is the greatest of all natural sciences: it encompasses the small-

est particles, such as electrons and quarks; and it also encom-

passes the largest bodies, such as galaxies and the entire Universe.

The smallest particles and the largest bodies differ in size by a

factor of more than ten thousand billion billion billion billion! In

the pictures on the following pages we will survey the world of

physics and attempt to develop some rough feeling for the sizes

of things in this world. This preliminary survey sets the stage

for our explanations of the mechanisms that make things behave

in the way they do. Such explanations are at the heart of physics,

and they are the concern of the later chapters of this book.

Since the numbers we will be dealing with in this prelude

and in the later chapters are often very large or very small, we

will find it convenient to employ the scientific notation for these

numbers. In this notation, numbers are written with powers of

10; thus, hundred is written as 102, thousand is written as 103,

ten thousand is written as 104, and so on. A tenth is written as

10�1, a hundredth is written as 10�2, a thousandth is written as

10�3, and so on.The following table lists some powers of ten:

10 � 101 0.1 � 1/10 � 10�1

100 � 102 0.01 � 1/100 �10�2

1000 � 103 0.001 � 1/1000 � 10�3

10000 � 104 0.0001 � 1/10000 � 10�4

100000 � 105 0.00001 � 1/100000 � 10�5

1000000 � 106 0.000001 � 1/1000000 � 10�6 etc.

Note that the power of 10, or the exponent on the 10,

simply tells us how many zeros follow the 1 in the number

(if the power of 10 is positive) or how many zeros follow the

1 in the denominator of the fraction (if the power of 10 is

negative).

In scientific notation, a number that does not coincide

with one of the powers of 10 is written as a product of a dec-

imal number and a power of 10. For example, in this nota-

tion, 1500000000 is written as 1.5 � 109. Alternatively, this

number could be written as 15 � 108 or as 0.15 � 1010; but

in scientific notation it is customary to place the decimal point

immediately after the first nonzero digit.The same rule applies

to numbers smaller than 1; thus, 0.000 015 is written as

1.5 � 10�5.

The pictures on the following pages fall into two sequences.

In the first sequence we zoom out: we begin with a picture of

a woman’s face and proceed step by step to pictures of the

entire Earth, the Solar System, the Galaxy, and the Universe.

This ascending sequence contains 27 pictures, with the scale

decreasing in steps of factors of 10.

Most of our pictures are photographs. Many of these have

become available only in recent years; they were taken by

high-flying aircraft, Landsat satellites, astronauts, or sophis-

ticated electron microscopes. For some of our pictures no

photographs are available and we have to rely, instead, on

drawings.

PreludeThe World of Physics

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xxxvi Prelude

Fig. P1 SCALE 1:1.5 This is Erin, an intelligent biped of the planetEarth, Solar System, Orion Spiral Arm, Milky Way Galaxy, LocalGroup, Local Supercluster. Erin belongs to the phylum Chordata, classMammalia, order Primates, family Hominidae, genus Homo, species sapiens. She is made of 5.4 � 1027 atoms, with 1.9 � 1028 electrons, thesame number of protons, and 1.5 � 1028 neutrons.

Fig. P2 SCALE 1:1.5 � 10 Erin has a height of 1.7 meters and a mass of57 kilograms. Her chemical composition (by mass) is 65% oxygen, 18.5%carbon, 9.5% hydrogen, 3.3% nitrogen, 1.5% calcium, 1% phosphorus, and1.2% other elements.

The matter in Erin’s body and the matter in her immediate environ-ment occur in three states of aggregation: solid, liquid, and gas. All theseforms of matter are made of atoms and molecules, but solid, liquid, andgas are qualitatively different because the arrangements of the atomic andmolecular building blocks are different.

In a solid, each building block occupies a definite place. When a solidis assembled out of molecular or atomic building blocks, these blocks arelocked in place once and for all, and they cannot move or drift about exceptwith great difficulty. This rigidity of the arrangement is what makes theaggregate hard—it makes the solid “solid.” In a liquid, the molecular oratomic building blocks are not rigidly connected. They are thrown togetherat random and they move about fairly freely, but there is enough adhesionbetween neighboring blocks to prevent the liquid from dispersing. Finally,in a gas, the molecules or atoms are almost completely independent of oneanother. They are distributed at random over the volume of the gas and areseparated by appreciable distances, coming in touch only occasionallyduring collisions. A gas will disperse spontaneously if it is not held in con-finement by a container or by some restraining force.

PART I : THE LARGE-SCALE WORLD

0 10−1 m0.5 × 10−1

0 100 m0.5 × 100

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Fig. P3 SCALE 1:1.5 � 102 The building behind Erin is the NewYork Public Library, one of the largest libraries on Earth. This libraryholds 1.4 � 1010 volumes, containing roughly 10% of the total accumu-lated knowledge of our terrestrial civilization.

Fig. P4 SCALE 1:1.5 � 103 The New York Public Library is locatedat the corner of Fifth Avenue and 42nd Street, in the middle of NewYork City, with Bryant Park immediately behind it.

Prelude xxxvii

0 101 m0.5 × 101

0 102 m0.5 × 102

Fig. P5 SCALE 1:1.5 � 104 This aerial photograph shows an area of 1 kilometer � 1 kilometer in the vicinity of the New York PublicLibrary. The streets in this part of the city are laid out in a regular rec-tangular pattern. The library is the building in the park in the middleof the picture. The photograph was taken early in the morning, andthe high buildings typical of New York cast long shadows.

The photograph was taken from an airplane flying at an altitudeof a few thousand meters. North is at the top of the photograph.

0 103 m0.5 × 103

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xxxviii Prelude

Fig. P6 SCALE 1:1.5 � 105 This photograph shows a large portion ofNew York City. We can barely recognize the library and its park as asmall rectangular patch slightly above the center of the picture. Thecentral mass of land is the island of Manhattan, with the Hudson Riveron the left and the East River on the right.

This photograph and the next two were taken by satellites orbitingthe Earth at an altitude of about 700 kilometers.

Fig. P7 SCALE 1:1.5 � 106 In this photograph, Manhattan is in theupper middle. On this scale, we can no longer distinguish the pattern ofstreets in the city. The vast expanse of water in the lower right of thepicture is part of the Atlantic Ocean. The mass of land in the upperright is Long Island. Parallel to the south shore of Long Island we cansee a string of very narrow islands; they almost look man-made. Theseare barrier islands; they are heaps of sand piled up by ocean waves in thecourse of thousands of years.

Fig. P8 SCALE 1:1.5 � 107 Here we see the eastern coast of theUnited States, from Cape Cod to Cape Fear. Cape Cod is the hook nearthe northern end of the coastline, and Cape Fear is the promontory nearthe southern end of the coastine. Note that on this scale no signs ofhuman habitation are visible. However, at night the lights of large citieswould stand out clearly.

This photograph was taken in the fall, when leaves had brilliantcolors. Streaks of orange trace out the spine of the Appalachian moun-tains.

0 104 m0.5 × 104

0 105 m0.5 × 105

0 106 m0.5 × 106

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Prelude xxxix

Fig. P9 SCALE 1:1.5 � 108 In this photograph, taken by theApollo 16 astronauts during their trip to the Moon, we see a large partof the Earth. Through the gap in the clouds in the lower middle of thepicture, we can see the coast of California and Mexico. We can recog-nize the peninsula of Baja California and the Gulf of California. Erin’slocation, the East Coast of the United States, is covered by a bigsystem of swirling clouds on the right of the photograph.

Note that a large part of the area visible in this photograph isocean. About 71% of the surface of the Earth is ocean; only 29% island. The atmosphere covering this surface is about 100 kilometersthick; on the scale of this photograph, its thickness is about 0.7 mil-limeter. Seen from a large distance, the predominant colors of theplanet Earth are blue (oceans) and white (clouds).

0 107 m0.5 × 107

Fig. P10 SCALE 1:1.5 � 109 This photograph of the Earth was takenby the Apollo 16 astronauts standing on the surface of the Moon.Sunlight is striking the Earth from the top of the picture.

As is obvious from this and from the preceding photograph, theEarth is a sphere. Its radius is 6.37 � 106 meters and its mass is 5.98 �1024 kilograms.

0 108 m0.5 × 108

Jan. 1,

2000

Jan

. 25

Jan. 2

1

Jan. 17

Jan. 13

Jan.

9

Jan. 5

perigee

apogee

0 109 m0.5 × 109

Fig. P11 SCALE 1:1.5 � 1010 In this drawing, the dot at the centerrepresents the Earth, and the solid line indicates the orbit of the Moonaround the Earth (many of the pictures on the following pages are alsodrawings). As in the preceding picture, the Sun is far below the bottomof the picture. The position of the Moon is that of January 1, 2000.

The orbit of the Moon around the Earth is an ellipse, but anellipse that is very close to a circle. The solid red curve in the drawingis the orbit of the Moon, and the dashed green curve is a circle; bycomparing these two curves we can see how little the ellipse deviatesfrom a circle centered on the Earth. The point on the ellipse closest tothe Earth is called the perigee, and the point farthest from the Earth iscalled the apogee. The distance between the Moon and the Earth isroughly 30 times the diameter of the Earth. The Moon takes 27.3 daysto travel once around the Earth.

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xl Prelude

Fig. P14 SCALE 1:1.5 � 1013 This drawing shows the positionsof the Sun and the inner planets: Mercury, Venus, Earth, andMars. The positions of the planets are those of January 1, 2000.The orbits of all these planets are ellipses, but they are close tocircles. The point of the orbit nearest to the Sun is called the per-ihelion and the point farthest from the Sun is called the aphe-lion. The Earth reaches perihelion about January 3 and aphelionabout July 6 of each year.

All the planets travel around their orbits in the samedirection: counterclockwise in our picture. The marks along theorbit of the Earth indicate the successive positions at intervals of10 days.

Beyond the orbit of Mars, a large number of asteroids orbitaround the Sun; these have been omitted to prevent excessiveclutter. Furthermore, a large number of comets orbit around theSun. Most of these have pronounced elliptical orbits. The cometHalley has been included in our drawing.

Fig. P12 SCALE 1:1.5 � 1011 This picture shows the Earth, theMoon, and portions of their orbits around the Sun. On thisscale, both the Earth and the Moon look like small dots. Again,the Sun is far below the bottom of the picture. In the middle, wesee the Earth and the Moon in their positions for January 1,2000. On the right and on the left we see, respectively, theirpositions for 1 day before and 1 day after this date.

Note that the net motion of the Moon consists of thecombination of two simultaneous motions: the Moon orbitsaround the Earth, which in turn orbits around the Sun.

Fig. P13 SCALE 1:1.5 � 1012 Here we see the orbits of theEarth and of Venus. However, Venus itself is beyond the edge ofthe picture. The small circle is the orbit of the Moon. The dotrepresenting the Earth is much larger than what it should be,although the artist has drawn it as minuscule as possible. On thisscale, even the Sun is quite small; if it were included in this pic-ture, it would be only 1 millimeter across.

Jan. 2Jan. 1, 2000

Dec. 31

0 1010 m0.5 × 1010

0 0.5 × 1011 1011 m

0 0.5 × 1012 1012 m

aph

elion

Earth

Mars

Mercury

Venus

Halley

perih

elion

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Prelude xli

0.5 × 1013 1013 m 0

Halley

Pluto

Uranus

SaturnJupiter

Neptune

Fig. P15 SCALE 1:1.5 � 1014 This picture shows the positionsof the outer planets of the Solar System: Jupiter, Saturn, Uranus,Neptune, and Pluto. On this scale, the orbits of the inner planetsare barely visible. As in our other pictures, the positions of theplanets are those of January 1, 2000.

The outer planets move slowly and their orbits are very large;thus they take a long time to go once around their orbit. Theextreme case is that of Pluto, which takes 248 years to completeone orbit.

Uranus, Neptune, and Pluto are so far away and so faint thattheir discovery became possible only through the use of tele-scopes. Uranus was discovered in 1781, Neptune in 1846, andthe tiny Pluto in 1930. Pluto is now known as one of severaldwarf planets.

The Sun is a sphere of radius 6.96 � 108 meters. On the scaleof the picture, the Sun looks like a very small dot, even smallerthan the dot drawn here. The mass of the Sun is 1.99 � 1030

kilograms.The matter in the Sun is in the plasma state, sometimes called

the fourth state of matter. Plasma is a very hot gas in which violentcollisions between the atoms in their random thermal motion havefragmented the atoms, ripping electrons off them. An atom thathas lost one or more electrons is called an ion.Thus, plasma con-sists of a mixture of electrons and ions engaging in frequent colli-sions.These collisions are accompanied by the emission of light,making the plasma luminous.

0.5 × 1014 1014 m 0

Fig. P16 SCALE 1:1.5 � 1015 We now see that the SolarSystem is surrounded by a vast expanse of space. Although thisspace is shown empty in the picture, the Solar System isencircled by a large cloud of millions of comets whose orbitscrisscross the sky in all directions. Furthermore, the interstellarspace in this picture and in the succeeding pictures containstraces of gas and of dust. The interstellar gas is mainly hydrogen;its density is typically 1 atom per cubic centimeter.

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Fig. P17 SCALE 1:1.5 � 1016 More interstellar space. Thesmall circle is the orbit of Pluto.

Fig. P18 SCALE 1:1.5 � 1017 And more interstellar space. Onthis scale, the Solar System looks like a minuscule dot,0.1 millimeter across.

Fig. P19 SCALE 1:1.5 � 1018 Here, at last, we see the starsnearest to the Sun. The picture shows all the stars within acubical box 1017 meters � 1017 meters � 1017 meters centeredon the Sun: Alpha Centauri A, Alpha Centauri B, andProxima Centauri. All three are in the constellation Centaurus,in the southern sky.

The star closest to the Sun is Proxima Centauri. This is avery faint, reddish star (a “red dwarf ”), at a distance of 4.0 �1016 meters from the Sun. Astronomers like to express stellardistances in light-years: Proxima Centauri is 4.2 light-yearsfrom the Sun, which means light takes 4.2 years to travel fromthis star to the Sun.

xlii Prelude

0.5 × 10151015 m 0

0 1016 m0.5 × 1016

Pro

xim

a C

enta

uri

α C

enta

uri

A a

nd

B

Sun

0 1017 m0.5 × 1017

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Prelude xliii

Sun

Pro

cyon

Alt

air

Fom

alh

aut

Poll

ux

Cas

tor

Veg

a

Cap

ella

Den

ebola

Arc

turu

sM

enken

t

Sirius

Cent.

Caph

Alderamin

α

Aln

air

0 1018 m0.5 × 1018

Fig. P20 SCALE 1:1.5 � 1019 This picture displays the bright-est stars within a cubical box 1018 meters � 1018 meters � 1018

meters centered on the Sun. There are many more stars in thisbox besides those shown—the total number of stars in this boxis close to 2000.

Sirius is the brightest of all the stars in the night sky. If itwere at the same distance from the Earth as the Sun, it wouldbe 28 times brighter than the Sun.

COMA

PERSEUS

PLEIADES

HYADES

Bellatrix

Achernar

CanopusMiaplacidusSchaulaMimosa

Spica

AliothRegulus

Alk

aid

Dubh

eE

ltan

inE

l N

ath

An

tare

sA

crux

Had

ar

Mirfak

Sun

0 1019 m0.5 × 1019

Fig. P21 SCALE 1:1.5 � 1020 Here we expand our box to 1019

meters � 1019 meters � 1019 meters, again showing only thebrightest stars and omitting many others. The total number ofstars within this box is about 2 million. We recognize severalclusters of stars in this picture: the Pleiades Cluster, the HyadesCluster, the Coma Berenices Cluster, and the Perseus Cluster.Each of these has hundreds of stars crowded into a fairly smallpatch of sky. In this diagram, Starbursts signify single stars, cir-cles with starbursts indicate star clusters, and a circle with asingle star indicate a star cluster with its brightest star.

0 1020 m0.5 × 1020

Fig. P22 SCALE 1:1.5 � 1021 This photograph shows a viewof the Milky Way in the direction of the constellationSagittarius. Now there are so many stars in our field of viewthat they appear to form clouds of stars. There are about a mil-lion stars in this photograph, and there are many more stars toofaint to show up distinctly. Although this photograph is notcentered on the Sun, it is similar to what we would see if wecould look toward the Solar System from very far away.

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xliv Prelude

Fig. P23 SCALE 1:1.5 � 1022 This is the spiral galaxy NGC2997. Its clouds of stars are arranged in spiral arms woundaround a central bulge. The bright central bulge is the nucleusof the galaxy; it has a more or less spherical shape. Thesurrounding region, with the spiral arms, is the disk of thegalaxy. This disk is quite thin; it has a thickness of only about3% of its diameter. The stars making up the disk circle aroundthe galactic center in a clockwise direction.

Our Sun is in a spiral galaxy of roughly similar shape andsize: the Milky Way Galaxy. The total number of stars in thisgalaxy is about 1011. The Sun is in one of the spiral arms,roughly one-third inward from the edge of the disk toward thecenter.

Fig. P24 SCALE 1:1.5 � 1023 Galaxies are often found inclusters of several galaxies. Some of these clusters consist of justa few galaxies, others of hundreds or even thousands. Thephotograph shows a cluster, or group, of galaxies beyond theconstellation Fornax. The group contains an elliptical galaxylike a luminous yellow egg (center), three large spiral galaxies(left), and a spiral with a bar (bottom left).

Our Galaxy is part of a modest cluster, the Local Group,consisting of our own Galaxy, the great Andromeda Galaxy, theTriangulum Galaxy, the Large Magellanic Cloud, plus 16 othersmall galaxies.

According to recent investigations, the dark, apparentlyempty, space near galaxies contains some form of distributedmatter, with a total mass 20 or 30 times as large as the mass inthe luminous, visible galaxies. But the composition of this invis-ible, extragalactic dark matter is not known.

Fig. P25 SCALE 1:1.5 � 1024 The Local Group lies on thefringes of a very large cluster of galaxies, called the LocalSupercluster. This is a cluster of clusters of galaxies. At thecenter of the Local Supercluster is the Virgo Cluster withseveral thousand galaxies. Seen from a large distance, our super-cluster would present a view comparable with this photograph,which shows a multitude of galaxies beyond the constellationFornax, all at a very large distance from us. The photograph wastaken with the Hubble Space Telescope coupled to two very sensitive cameras using an exposure time of almost 300 hours.

All these distant galaxies are moving away from us and awayfrom each other. The very distant galaxies in the photo aremoving away from us at speeds almost equal to the speed oflight. This motion of recession of the galaxies is analogous tothe outward motion of, say, the fragments of a grenade after itsexplosion. The motion of the galaxies suggests that theUniverse began with a big explosion, the Big Bang, thatlaunched the galaxies away from each other.

0 1021 m0.5 × 1021

0 1022 m0.5 × 1022

0 1023 m0.5 × 1023

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0 1024 m0.5 × 1024

Fig. P26 SCALE 1:1.5 � 1025 On this scale a galaxy equal insize to our own Galaxy would look like a fuzzy dot, 0.1 mil-limeter across. Thus, the galaxies are too small to show upclearly on a photograph. Instead we must rely on a plot of thepositions of the galaxies. The plot shows the positions of about200 galaxies. The dense cluster of galaxies in the lower half ofthe plot is the Virgo Cluster.

Since we are looking into a volume of space, some of thegalaxies are in the foreground, some are in the background; butour plot takes no account of perspective.

The luminous stars in the galaxies constitute only a smallfraction of the total mass of the Universe. The space aroundthe galaxies and the clusters of galaxies contains dark matter,and the space between the clusters contains dark energy, astrange form of matter that causes an acceleration of theexpansion of the Universe.

Fig. P27 SCALE 1:1.5 � 1026 This plot shows the positionsof about 100,000 galaxies in a patch of the sky at distances ofup to 1 � 109 light years from the Earth. The false color in thisimage indicates the distance–red for shorter distances, blue forlarger distances.

The visible galaxies plotted here contribute only about 5%of the total mass in the universe. The dark matter near thegalaxies contribute another 25%. The remaining 70% of thetotal mass in the universe is in the form of dark energy, whichis uniformly distributed over the vast reaches of intergalacticspace.

This is the last of our pictures in the ascending series. Wehave reached the limits of our zoom out. If we wanted to drawanother picture, 10 times larger than this, we would need toknow the shape and size of the entire Universe. We do not yetknow that.

0 1025 m0.5 × 1025

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PART I I : THE SMALL-SCALE WORLD

Fig. P28 SCALE 1:1.5 We now return to Erin and zoom inon her eye.The surface of her skin appears smooth and firm. Butthis is an illusion. Matter appears continuous because thenumber of atoms in each cubic centimeter is extremely large. Ina cubic centimeter of human tissue there are about 1023 atoms.This large number creates the illusion that matter is continu-ously distributed—we see only the forest and not the individualtrees. The solidity of matter is also an illusion. The atoms in ourbodies are mostly vacuum. As we will discover in the followingpictures, within each atom the volume actually occupied by sub-atomic particles is only a very small fraction of the total volume.

Fig. P29 SCALE 1:1.5 � 10�1 Our eyes are very sophisticatedsense organs; they collect more information than all our othersense organs taken together. The photograph shows the pupiland the iris of Erin’s eye. Annular muscles in the iris changethe size of the pupil and thereby control the amount of lightthat enters the eye. In strong light the pupil automaticallyshrinks to about 2 millimeters; in very weak light it expands toas much as 7 millimeters.

0 10−1 m0.5 × 10−1

MAGNIFICATION 0.67 ×

0 10−2 m0.5 × 10−2

MAGNIFICATION 6.7 ×

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Fig. P30 SCALE 1:1.5 � 10�2 This false-color photographshows the delicate network of blood vessels on the front sur-face of the retina, the light-sensitive membrane lining the inte-rior of the eyeball. The rear surface of the retina is denselypacked with two kinds of cells that sense light: cone cells androd cells. In a human retina there are about 6 million cone cellsand 120 million rod cells. The cone cells distinguish colors; therod cells distinguish only brightness and darkness, but they aremore sensitive than the cone cells and therefore give us visionin faint light (“night vision”).

This and the following photographs were made withvarious kinds of electron microscopes. An ordinary micro-scope uses a beam of light to illuminate the object; an electronmicroscope uses a beam of electrons. Electron microscopes canachieve much sharper contrast and much higher magnificationthan ordinary microscopes.

Fig. P31 SCALE 1:1.5 � 10�3 Here we have a false-color photograph of rod cells prepared with a scanning electronmicroscope (SEM). For this photograph, the retina was cutapart and the microscope was aimed at the edge of the cut. Inthe top half of the picture we see tightly packed rods. Each rodis connected to the main body of a cell containing the nucleus.In the bottom part of the picture we can distinguish tightlypacked cell bodies of the cell.

Fig. P32 SCALE 1:1.5 � 10�4 This is a close-up view of a fewrods cells. The upper portions of the rods contain a special pig-ment—visual purple—which is very sensitive to light. Theabsorption of light by this pigment initiates a chain of chemi-cal reactions that finally trigger nerve pulses from the eye tothe brain.

0 10−3 m0.5 × 10−3

MAGNIFICATION 6.7 × 10 ×

0 10−4 m0.5 × 10−4


0 10−5 m0.5 × 10−5


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Fig. P33 SCALE 1:1.5 � 10�5 These are strands of DNA, ordeoxyribonucleic acid, as seen with a transmission electronmicroscope (TEM) at very high magnification. DNA is foundin the nuclei of cells. It is a long molecule made by stringingtogether a large number of nitrogenous base molecules on abackbone of sugar and phosphate molecules. The base mole-cules are of four kinds, the same in all living organisms. Butthe sequence in which they are strung together varies from oneorganism to another. This sequence spells out a message—thebase molecules are the “letters” in this message. The messagecontains all the genetic instructions governing the metabolism,growth, and reproduction of the cell.

The strands of DNA in the photograph are encrusted witha variety of small protein molecules. At intervals, the strands ofDNA are wrapped around larger protein molecules that formlumps looking like the beads of a necklace.

Fig. P34 SCALE 1:1.5 � 10�6 The highest magnifications areattained by a newer kind of electron microscope, the scanningtunneling microscope (STM). This picture was prepared withsuch a microscope. The picture shows strands of DNAdeposited on a substrate of graphite. In contrast to the strandsof the preceding picture, these strands are uncoated; that is,they are without protein encrustations.

Fig. P35 SCALE 1:1.5 � 10�7 This close-up picture ofstrands of DNA reveals the helical structure of this molecule.The strand consists of a pair of helical coils wrapped aroundeach other. This picture was generated by a computer fromdata obtained by illuminating DNA samples with X rays (X-ray scattering).

0 10−6 m0.5 × 10−6


0 10−7 m0.5 × 10−7


0 10−8 m0.5 × 10−8


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Fig. P36 SCALE 1:1.5 � 10�8 This picture shows a layer ofpalladium atoms on surface of graphite as seen with an STM.Here we have visual evidence of the atomic structure of matter.The palladium atoms are arranged in a symmetric, repetitivehexagonal pattern. Materials with such regular arrangementsof atoms are called crystals.

Each of the palladium atoms is approximately a sphere,about 3 � 10�10 meter across. However, the atom does nothave a sharply defined boundary; its surface is somewhat fuzzy.Atoms of other elements are also approximately spheres, withsizes that range from 2 � 10�10 to 4 � 10�10 meter across.

At present we know of more than 100 kinds of atoms orchemical elements. The lightest atom is hydrogen, with a massof 1.67 � 10�27 kilogram; the heaviest is element 114, unun-quadium, with a mass about 289 times as large.

Fig. P37 SCALE 1:1.5 � 10�9 The drawing shows the inte-rior of an atom of neon. This atom consists of 10 electrons orbit-ing around a nucleus. In the drawing, the electrons have beenindicated by small dots, and the nucleus by a slightly larger dotat the center of the picture. These dots have been drawn as smallas possible, but even so the size of these dots does not give a cor-rect impression of the actual sizes of the electrons and of thenucleus. The electron is smaller than any other particle weknow; maybe the electron is truly pointlike and has no size at all.The nucleus has a finite size, but this size is much too small toshow up on the drawing. Note that the electrons tend to clusternear the center of the atom. However, the overall size of theatom depends on the distance to the outermost electron; thiselectron defines the outer edge of the atom.

The electrons move around the nucleus in a very compli-cated motion, and so the resulting electron distribution resem-bles a fuzzy cloud, similar to the STM image of the previouspicture. This drawing, however, shows the electrons as theywould be seen at one instant of time with a hypotheticalmicroscope that employs gamma rays instead of light rays toilluminate an object; no such microscope has yet been built.

The mass of each electron is 9.11 � 10�31 kilogram, butmost of the mass of the atom is in the nucleus; the 10 electronsof the neon atom have only 0.03% of the total mass of theatom.

0 10−9 m0.5 × 10−9


0 10–10 m0.5 × 10–10


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l Prelude

Fig. P39 SCALE 1:1.5 � 10�11 In this drawing we finallysee the nucleus in its true size. At this magnification, thenucleus of the neon atom looks like a small dot, 0.5 millimeterin diameter. Since the nucleus is extremely small and yet con-tains most of the mass of the atom, the density of the nuclearmaterial is enormous. If we could assemble a drop of purenuclear material of a volume of 1 cubic centimeter, it wouldhave a mass of 2.3 � 1011 kilograms, or 230 million metrictons!

Our drawings show clearly that most of the volume withinthe atom is empty space. The nucleus occupies only a verysmall fraction of this volume.

0 10–12 m0.5 × 10–12


0 10–13 m0.5 × 10–13

MAGNIFICATION 6.7 × 1011 × Fig. P40 SCALE 1:1.5 � 10�12 We can now begin to distin-guish the nuclear structure. The nucleus has a nearly sphericalshape, but its surface is slightly fuzzy.

0 10–11 m0.5 × 10–11

MAGNIFICATION 6.7 × 109 × Fig. P38 SCALE 1:1.5 � 10�10 Here we are closing in on thenucleus. We are seeing the central part of the atom. Only twoelectrons are in our field of view; the others are beyond themargin of the drawing. The size of the nucleus is still muchsmaller than the size of the dot at the center of the drawing.

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Fig. P42 SCALE 1:1.5 � 10�14 This final picture shows threepointlike bodies within a proton. These pointlike bodies arequarks—each proton and each neutron is made of threequarks. Recent experiments have told us that the quarks aremuch smaller than protons, but we do not yet know theirprecise size. Hence the dots in the drawing probably do notgive a fair description of the size of the quarks. The quarkswithin protons and neutrons are of two kinds, called up anddown. The proton consists of two up quarks and one downquark joined together; the neutron consists of one up quarkand two down quarks joined together.

This final picture takes us to the limits of our knowledge ofthe subatomic world. As a next step we would like to zoom inon the quarks and show what they are made of. According to aspeculative theory, they are made of small snippets or loops ofstrings, 10�35m long. But we do not yet have any evidence forthis theory.

Fig. P41 SCALE 1:1.5 � 10�13 At this extreme magnificationwe can see the details of the nuclear structure. The nucleus ofthe neon atom is made up of 10 protons (white balls) and10 neutrons (red balls). Each proton and each neutron is asphere with a diameter of about 2 � 10�15 meter, and a massof 1.67 � 10�27 kilogram. In the nucleus, these protons andneutrons are tightly packed together, so tightly that theyalmost touch. The protons and neutrons move around thevolume of the nucleus at high speed in a complicated motion.

Magnification 6.7 � 1012 �

0 0.5 � 10–14 10–14 m

Magnification 6.7 � 1013 �

0 0.5 � 10–15 10–15 m

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Relativity,Quanta, andParticles

C O N T E N T S

36 The Theory of Special Relativity

37 Quanta of Light

38 Spectral Lines, Bohr’s Theory, and Quantum Mechanics

39 Quantum Structure of Atoms, Molecules, and Solids

40 Nuclei

41 Elementary Particles and Cosmology

P A R T

6

The solar cells in the 73-meter

long photovoltaic arrays on the

International Space Station

convert solar energy into elec-

trical power. The arrays are

fitted with gimbals that angle

the arrays toward the Sun at

all times so as to maximize the

power supplied to the Space

Station.

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C O N C E P T S I N C O N T E X T

Determination of latitude and longitude by means of radio signals from

Global Positioning System (GPS) satellites, such as the one shown here,

requires measurement of the travel time from several satellites to the rele-

vant point on the ground. GPS satellites are in relative motion with respect

to the Earth’s surface, but the speed of their radio signals is not affected by

this motion.

In our study of the theory of Special Relativity we will study the prop-

agation of light in different reference frames in relative motion, and we

can then ask:

? How are distances calculated from GPS signals, and how do we

know that the speed of light is unaffected by the motion of the

satellite or by the translational motion of the Earth? (Section 36.1,

page 1219)

The Theory ofSpecial Relativity36

36.1 The Speed of Light; the Ether

36.2 Einstein’s Principle ofRelativity

36.3 Time Dilation

36.4 Length Contraction

36.5 The Lorentz Transformationsand the Combination ofVelocities

36.6 Relativistic Momentum andEnergy

36.7 Mass and Energy

C H A P T E R

1216

Conceptsin

Context

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1217

? Relative to clocks on the surface of the Earth, clocks on GPS satellites are in

motion at a somewhat high speed. How does this affect the rate of the clocks?

(Example 2, page 1227)

? How does the motion of a GPS satellite relative to the Earth affect the frequency

of the radio signal? (Example 3, page 1228)

A s we saw in Chapter 5, Newton’s laws of motion are equally valid in every iner-

tial reference frame. All inertial reference frames are unaccelerated, but they can

differ in their uniform translational motion. Since Newton’s laws make no distinction

between different inertial reference frames, no mechanical experiment can detect a

uniform translational motion of one inertial reference frame by itself. Such motion

can be detected only as a relative motion of one reference frame with respect to another

reference frame. This is the Newtonian principle of relativity. For a concrete illustra-

tion of this principle, consider the reference frame of a cruise ship moving steadily

away from the shore, without acceleration, and consider the reference frame of the

shore. Both of these reference frames are inertial, and the behavior of a ball used in a

game of tennis on the ship is not different from the behavior of a similar ball on

shore—balls on the ship and on the shore accelerate in the same way when subjected

to forces, and they obey the same laws of conservation of energy, conservation of

momentum, etc.1 Thus, experiments with such balls aboard the ship will not reveal

the uniform motion of the ship relative to the shore. To detect this motion, the crew

of the ship must take sightings of points on the shore or use some other navigational

technique that fixes the position and velocity of the ship relative to the shore. Hence,

in regard to mechanical experiments, uniform translational motion of our inertial

reference frame is always relative motion—it can be detected only as motion of our ref-

erence frame with respect to another reference frame. There is no such thing as absolute

motion.

The question naturally arises whether the relativity of motion indicated by mechan-

ical experiments also applies to electric, magnetic, optical, and other experiments. Do

any of these experiments permit us to detect an absolute motion or our reference frame?

In 1905, Albert Einstein proposed that no experiment of any kind should ever permit

us to detect such motion. He postulated a principle of relativity for all the laws of

physics.This postulate serves as the foundation for Einstein’s theory of Special Relativity,

widely regarded as one of the greatest achievements of twentieth-century physics.The

theory of relativity requires a drastic revision of our concepts of space and time, and it

also requires a drastic revision of Newton’s laws. At high speeds—near the speed of light—

particles obey new, relativistic laws which are quite different from Newton’s laws. However,

at low speeds—small compared with the speed of light—the differences between

Einstein’s and Newton’s theories are usually undetectable. Hence Newton’s laws are

adequate for describing the behavior of particles and of other bodies at the relatively

low speeds we encounter in our everyday experience.

Before we deal with the details of Einstein’s theory of relativity, we will briefly

describe why nonmechanical experiments—and, especially, experiments with light—

might be expected to detect absolute motion, which mechanical experiments cannot

detect.

ALBERT EINSTEIN (1879–1955)German (and Swiss, and American) theoreti-

cal physicist, professor at Zurich, at Berlin,

and at the Institute for Advanced Study at

Princeton. Einstein was the most celebrated

physicist of the twentieth century. He formu-

lated the theory of Special Relativity in 1905

and the theory of General Relativity in 1916.

Einstein also made incisive contributions to

modern quantum theory, for which he received

the Nobel Prize in 1921. Einstein spent the

last years of his life in an unsuccessful quest for

a unified theory of forces that was supposed to

incorporate gravity and electricity in a single

set of equations.

1 This assumes the ship moves steadily. If the ship lurches forward, or pitches, or rolls, it ceases to be an

inertial reference frame, and ball will behave in a manner “inconsistent” with Newton’s laws.


36.1 THE SPEED OF L IGHT; THE ETHER

Since the laws of mechanics are the same in all inertial reference frames, it might seem

quite natural to assume that the laws of electricity, magnetism, and optics are also the

same in all inertial reference frames. But this assumption immediately leads to a par-

adox concerning the speed of light. As we know from Chapter 33, light is an oscillat-

ing electric and magnetic disturbance propagating through space, and Maxwell’s

equations permit us to deduce that the speed of propagation of this disturbance must

always be2 3.00 � 108 m/s. The trouble with this deduction is that, according to the

Galilean addition rule for velocity [Eq. (4.53)], the speed of light ought not to be the

same in all reference frames. For instance, imagine that an alien spaceship approach-

ing the Earth with a speed of, say, 1.00 � 108 m/s flashes a light signal toward the

Earth. If this signal has a speed of in the reference frame of the

spaceship, then the Galilean addition rule tells us that it ought to have a speed of

in the reference frame of the Earth (see Fig. 36.1).

To resolve this paradox, either we must give up the notion that the laws of electricity

and magnetism (and the values of the speed of light) are the same in all inertial refer-

ence frames, or else we must give up the Galilean addition rule for velocities. Both

alternatives are unpleasant: the former means that we must abandon all hope for a

principle of relativity embracing electricity and magnetism, and the latter means that

together with the Galilean addition rule we must abandon the transformation rule for

position vectors measured in different reference frames [Eq. (4.50)] as well as the intu-

itively “obvious” notions of absolute time and length from which these rules are derived.

Since the failure of a relativity principle embracing electricity and magnetism might

seem to be the lesser of two evils, let us first explore this alternative. Let us assume

that there exists a preferred inertial reference frame in which the laws of electricity

and magnetism take their simplest form, that is, the form expressed in Maxwell’s equa-

tions, Eqs. (33.4)–(33.7). In this reference frame, the speed of light has its standard

value of whereas in any other reference frame it is larger or smaller

according to the Galilean addition rule. The propagation of light is then analogous to

the propagation of sound. There exists a preferred reference frame in which the equa-

tions for the propagation of sound waves in, say, air take their simplest form: the ref-

erence frame in which the air is at rest. In this reference frame, sound has its standard

speed of 331 m/s. In any other reference frame, the equations for the propagation of

sound waves are more complicated, but the speed of propagation can always be obtained

directly from the Galilean addition rule. For instance, if a wind of 40 m/s (a hurri-

cane) is blowing over the surface of the Earth, then sound waves have a speed of 331

m/s relative to the air, but their speed relative to the ground depends on direction—

downwind the speed is 371 m/s, whereas upwind it is 291 m/s.

This analogy between the propagation of light and of sound suggests that there

might exist some pervasive medium whose oscillations bring about the propagation

of light, just as the oscillations of air bring about the propagation of sound. Presumably

this ghostly medium fills all of space, even the interplanetary and interstellar space,

which is normally regarded as a vacuum. The physicists of the nineteenth century

called this hypothetical medium the ether, and they attempted to describe light waves

as oscillations of the ether, analogous to sound waves as oscillations of the air.The pre-

ferred reference frame in which light has its standard speed is then the reference frame

in which the ether is at rest. The existence of such a preferred reference frame would

c � 3.00 � 108 m/s,

4.00 � 108 m/s

3.00 � 108 m/s

1218 CHAPTER 36 The Theory of Special Relativity

2The exact value of the speed of light is but throughout this chapter we will round

this off to 3.00 � 108 m/s.

2.997 924 58 � 108 m/s,

(a)

(b)

3�108 m/s

4�108 m/s?

1�108 m/s

Flash signal travels withspeed of light c relative to spaceship.

When spacecraftspeeds toward Earth…

…Galilean addition rule for velocity implies a faster speed of light relative to Earth.

FIGURE 36.1 Velocities according to the

Galilean addition rule in (a) reference frame

of a spaceship and (b) reference frame of

the Earth. A speed of light different from

3 � 108 m/s is at odds with Maxwell’s

equations.

ether

41Online

ConceptTutorial


http://www.wwnorton.com/college/physics/om/_tutorials/chap36/special_relativity/interface_new.swf

36.1 The Speed of Light; the Ether 1219

imply that velocity is absolute—the ether frame

would set an absolute standard of rest, and the veloc-

ity of any body could always be referred to this frame.

For instance, instead of describing the velocity of the

Earth relative to some other material body, such as

the Sun, we could always describe its velocity relative

to the ether.

Presumably the Earth has some nonzero veloc-

ity relative to the ether. Even if the Earth were at

rest in the ether at one instant, this condition could

not last, since the Earth continually changes its

motion as it orbits around the Sun.The motion of the

ether past the Earth was called the ether wind by the

nineteenth-century physicists (Fig. 36.2). If the Sun is at rest in the ether, then the

ether wind would have a velocity opposite to the velocity of the Earth around the

Sun—about 30 km/s; if the Sun is in (steady) motion, then the ether wind would vary

with the seasons—it would reach a maximum when the orbital motion of the Earth is

parallel to the motion of the Sun, and a minimum when antiparallel.

Experimenters attempted to detect this ether wind by its effects on the propaga-

tion of light. A light wave in a laboratory on the Earth would have a greater speed

when moving downwind and a smaller speed when moving upwind or across the wind.

If the speed of the ether wind “blowing” through the laboratory is V, then the speed

of light in this laboratory would be for a light signal with downwind motion,

for upwind motion, and for motion perpendicular to the wind (see

Fig. 36.3). With a value of 300 000 km/s for c and a value of approximately 30 km/s

for V, the increase or decrease of the speed of light amounts to only about 1 part in

10 000, and a very sensitive apparatus is required for the detection of this small change.

In a famous experiment first performed in 1881 and often repeated thereafter, A.

A. Michelson and E. W. Morley attempted to detect small changes in the speed of

light by means of an interferometer. The results of their experiment were negative. As

discussed in Section 35.2, the sensitivity of the original experiment of Michelson and

Morley was such that an ether wind of 5 km/s would have produced a detectable effect.

Since the expected wind is about 30 km/s, the experimental result contradicts the ether

theory of the propagation of light. Later, more refined versions of the experiment

established that if there were an ether wind, its speed would certainly have to be less

than 3 m/s. The experimental evidence therefore establishes conclusively that the

motion of the Earth has no effect on the propagation of light. As the Earth moves

around the Sun, its velocity is first in one direction, then in another, and another; and

the Earth is first in one inertial reference frame, then in another, and another. But all

these inertial reference frames appear to be completely equivalent in regard to the

propagation of light. There is no preferred reference frame. There is no ether.

2c2 � V 2c � V

c � V

(a)

ether

V

Motion of the Earth relative to a hypotheticalmedium, the ether,…

(b)

�V

etherwind

…corresponds to flow of the ether past Earth, forming the ether wind.

FIGURE 36.2 (a) Motion of Earth in ref-

erence frame of ether. (b) Motion of ether in

reference frame of Earth.

(a)

c V

c � V � c � V

Vc

c � V � c � V

(b) (c)V

c

c � V � c 2 � V 2

Light movingdownwind throughthe ether…

…would have a greater speedon Earth…

…and lightmoving upwind…

…or across the wind would havea lower speed.

FIGURE 36.3 The velocity of

light is c relative to the ether, and

the velocity of the ether is V relative

to the laboratory. The velocity of

light relative to the laboratory is

then the vector sum c � V. (a) If c

and V are parallel, the magnitude of

the vector sum is (b) If c and

V are antiparallel, the magnitude of

the vector sum is (c) If c and

V form this right triangle, the mag-

nitude of the vector sum is

2c 2 � V 2.

c � V.

c � V.

Conceptsin

Context

Michelson-Morley experiment


Today, we take it for granted that the speed of light is unaffected by the motion of

the Earth. Many experiments and instruments rely on the fact that the travel time for

an electromagnetic signal between an emitter and a receiver depends only on the

distance between the two; the speed of light, and thus the travel time, does not depend

on the motion of the emitter or of the receiver. For example, accurate determination

of a position on the ground using the Global Positioning System (GPS) is achieved by

precisely measuring the travel time of signals from several satellites. When a GPS

receiver calculates the distances to several satellites, it assumes that the speed of light

is unaffected by the motion of the satellites and by the translational motion of the

Earth.

Checkup 36.1

QUESTION 1: You are in an open automobile (a convertible) traveling at 80 km/h.

Does the speed of sound waves (relative to you) depend on direction?

QUESTION 2: At the equator of the Earth, the rotational speed is 0.46 km/s. Was the

original Michelson–Morley experiment capable of detecting the corresponding ether

wind?

QUESTION 3: If Michelson and Morley had detected an ether wind of 30 km/s oppo-

site to the motion of the Earth around the Sun, what could they have concluded about

the absolute motion of the Earth? The Sun?

QUESTION 4: In 1887, after observations lasting a few days, a null result was obtained

in the Michelson–Morley experiment with 5 km/s sensitivity. To be sure that there

was no ether, Michelson and Morley had to repeat the experiment

(A) With a different color of light (B) At 1 km/s resolution

(C) At a different time of year

36.2 E INSTE IN’S PR INCIPLE OF RELAT IV ITY

Neither the laws of mechanics nor the laws for the propagation of light reveal any

intrinsic distinction between different inertial reference frames.This motivated Einstein

to take a bold step and to propose a general hypothesis concerning all the laws of

physics. This hypothesis is the Principle of Relativity:

All the laws of physics are the same in all inertial reference frames.

In addition, Einstein proposed the Principle of the Universality of the Speed of

Light:

The speed of light (in vacuum) is the same in all inertial reference frames; it always

has the value c � 3.00 � 108 m/s.

These deceptively simple principles form the foundation of the theory of Special

Relativity. As we pointed out in the preceding section, the universality of the speed of

light conflicts with the Galilean addition rule for velocity. We will therefore have to dis-

✔


Principle of Relativity

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card this rule, and we will also have to discard the transformation rule for position

vectors or coordinates on which it is based (see Section 4.6).

The universality of the speed of light also requires that we give up some of our

intuitive, everyday notions of space and time. Obviously, the fact that a light signal

always has a speed of regardless of how hard we try to move toward

it or away from it in a fast aircraft or spaceship, does violence to our intuition. This

strange behavior of light is only possible because of a strange behavior of length and

time in relativistic physics. As we will see later in this chapter, neither length nor time

is absolute—they both depend on the reference frame in which they are measured,

and they suffer contraction or dilation when the reference frame changes.

Before we can inquire into the consequences of Einstein’s two principles, we must

carefully describe the construction of reference frames and the synchronization of their

clocks. A reference frame is a coordinate grid erected around some given origin and a

set of clocks (see Fig. 36.4), which can be used to determine the space and time coor-

dinates of any event. In relativity, as in everyday life, an event is an occurrence that

happens at one point of space at one point of time (for example, a climber stepping

onto the summit of Mt. McKinley). An event is therefore represented by one point of

space and time.The space coordinates of the event are directly given by the coordinate

grid intersections at the event. The time coordinate of the event is the time registered

by the clock at the event. Different choices of reference frame result in different space

coordinates and different time coordinates for the event, and in Section 36.5 we will

see how these different coordinates in different reference frames are related.

The clocks of any chosen reference frame must be synchronized with each other

and with the master clock sitting at its origin of coordinates. Einstein proposed that

this synchronization can be accomplished by sending out a flash of light from a point

exactly midway between the clock at the origin and the other clock (see Fig. 36.5).

The two clocks are synchronized if both show exactly the same time when the light from

the midpoint reaches them. Note that this synchronization procedure hinges on the uni-

versality of the speed of light. If the speed of light were not a universal constant, but

were dependent on the reference frame and on the direction of propagation (say, faster

toward the right in Fig. 36.5 and slower toward the left), then we could not achieve syn-

chronization by the simple procedure with a flash of light from the midpoint.

3.00 � 108 m/s,


y

xO

A reference frame consists of a coordinate grid anda set of clocks.

In any chosen referenceframe, all clocks mustbe synchronized.

FIGURE 36.4 A reference frame.

event

y

wave front

x

A flash of light is sentout from a point midwaybetween two clocks.

Synchronized clocks showexactly the same time when lightfrom midpoint reaches them.

FIGURE 36.5 Synchronization

procedure for a pair of clocks. The

leading wave front reaches the clocks

at the lower left corner and the upper

right corner simultaneously.


One immediate consequence of our synchronization procedure is that simultaneity

is relative, that is, the simultaneity of two events depends on the reference frame. Einstein illus-

trated this with the following concrete example. Suppose that a train is traveling at high

speed along a straight track and two bolts of lightning strike the front end and the rear

end of the train, leaving scorch marks on the train and on the ground (the existence of

these marks helps us to remember exactly where the lightning struck in each reference

frame). Suppose that these two strokes of lightning are exactly simultaneous in the ref-

erence frame of the Earth. Then in the reference frame of the train, the two strokes of

lightning will not be simultaneous—as judged by the clocks on the train, the stroke at the

front end of the train occurs slightly earlier than the stroke at the rear end.

To see how this difference between the two reference frames comes

about, let us apply our procedure for testing simultaneity. Suppose that in

the reference frame of the Earth, an observer stands near the track at the

midpoint between the two scorch marks the lightning made on the ground

(see Fig. 36.6); she will then receive flashes of light from the lightning at

her left and her right at the same instant. Thus, this observer will confirm

that in the reference frame of the Earth, the lightning was simultaneous.

In the reference frame of the train, an observer can likewise test for

simultaneity by placing himself exactly at the midpoint between the scorch

marks the lightning made at the front and rear ends of the train (see Fig.

36.7) and waiting for the arrival of the flashes of light. Will he receive the

flashes of light from the front and the rear of the train at the same instant?

We can answer this by examining the motion of this observer and the

propagation of the flashes of light in the reference frame of the ground

(see Fig. 36.8).This observer is traveling toward the flash of light approach-

ing him from the front end of the train, and he is traveling away from the

flash of light trying to catch up with him from the back end of the train.

Thus, this observer will encounter the flash of light from the front end

before the flash of light from the rear end can catch up. In the reference

frame of the ground, this delay between the flashes of light seen by the

observer on the train is attributed to his motion toward one flash and away

from the other. But in the reference frame of the train, the observer cannot

attribute the delay to such a difference in motion—the light flashes from

the front and the rear of the train originated at exactly equal distances

from him, and, according to the Principle of Universality of the Speed of

Light, they traveled at equal speeds. Hence this observer will conclude

that the stroke of lightning at the front end of the train occurred earlier and

the stroke of lightning at the rear end occurred later.

Although this qualitative argument shows that simultaneity depends

on the reference frame, it does not tell us by how much. A quantitative cal-

culation shows that for a train traveling at ordinary speed the delay is insignif-

icant, or less. But the delay increases with the speed and it also

increases with the distance between the flashes of lightning. For instance, con-

sider a fast spaceship traveling by the Earth at 90% of the speed of light,

and consider two flashes of lightning at two points on the Earth separated

by a fairly large distance, say, Boston and New York, separated by a distance

of 300 km. If these flashes are simultaneous in the reference frame of the

Earth, they will differ by 0.001 s in the reference frame of the spaceship.

If simultaneity is relative, then the synchronization of clocks is also rela-

tive. In the reference frame of the Earth, all the clocks of this reference

frame are synchronized, that is, the hands of these clocks reach the, say, noon

position simultaneously. But when observed from the reference frame of

10�13 s


…an observer on ground at midpoint of strikes will receive twolight flashes at the same instant.

Lightning strikes train and groundat front and rear. When strikes on ground are simultaneous, …

FIGURE 36.6 Observer on the ground at the midpoint

between the scorch marks on the track watches for the

arrival of flashes of light.

For an observer in train at midpoint, lightning flashes originate at equaldistances and travel at equal speeds.

FIGURE 36.7 Observer in the train at the midpoint

between the front and rear ends, where lightning has made

scorch marks.

…and flash catching up from rearlater. On train, observer concludesthat flash from front occurred earlier.

In reference frame of ground, moving observer on train encountersflash approaching from front first…

FIGURE 36.8 Observer on a train moving toward

the right.


the train or the spaceship, the clocks in the Earth’s reference frame that are further

ahead in the direction of clock motion show earlier times. Consequently, they reach the

noon position later than those on the right side—they are “late” in the same way as

the lightning of the left side is late. This can be seen in Fig. 36.9, which shows the

clocks belonging to the reference frame of the Earth as observed at one instant of time

from a fast spaceship traveling in the direction from Boston to New York.

The effect is symmetric. In the reference frame of the spaceship, all the clocks onboard

are synchronized. But, as observed from the reference frame of the Earth, the clocks on

the front part of the spaceship are late. Figure 36.10 shows the clocks belonging to the

reference frame of the spaceship as observed at one instant from the Earth. Note that

in Fig. 36.9 we are viewing the reference frame of the Earth moving past the spaceship,

and in Fig. 36.10 we are viewing the reference frame of the spaceship moving past the

Earth. In either case, the clocks on the leading edge of the reference frame are late.

The relativity of synchronization is a direct consequence of the universality of the

speed of light, since our procedure for testing simultaneity depends crucially on the

speed of light. The failure of an absolute synchronization valid for all reference frames

implies that there exists no absolute time. Each reference frame has its own way of

reckoning time—time is relative. Instead of the single absolute time coordinate t we

used in Newtonian physics, we now have to use a separate time coordinate for each

individual reference frame.

Checkup 36.2

QUESTION 1: A spaceship approaches the Earth at and sends a light

signal toward the Earth, as in Fig. 36.1. According to Einstein, what is the speed of this

light signal relative to the Earth?

QUESTION 2: Figure 36.10 shows the clocks of the reference frame of a spaceship in

motion relative to the Earth.The clocks at the front of the spaceship are late. Does this

mean that the crew of the spaceship find that their clocks are out of synchronization?

QUESTION 3: An earthquake occurs in San Francisco, and simultaneously (in Earth time)

another earthquake occurs in New York.These earthquakes are not simultaneous as seen

in the reference frame of a fast spaceship traveling westward in the direction from New

York to San Francisco. Which is late?

QUESTION 4: A satellite, moving away from you at speed v, emits a pulse of radio

waves in your direction. The speed of the waves relative to you is

(A) c � v (B) c � v (C) (D) c (E) v2c2 � v2

1.00 � 108 m/s

✔


O

Boston New York

y

x

Moving Earth clocks as viewed from spaceship are not synchronized.

Clocks further aheadin direction of clock motion are “later.”

FIGURE 36.9 Clocks of the

reference frame of the Earth as

observed at one instant of

spaceship time.

O

y'

x'

Clocks on moving spaceshipas viewed from Earthare not synchronized.

Clocks further aheadin direction of clock motion are “later.”

FIGURE 36.10 Clocks of the reference

frame of the spaceship as observed at one

instant of Earth time.


36.3 T IME DILAT ION

The relativity of time shows up not only in the synchronization of clocks, but also in

the rate of clocks. When experimenters on Earth observe a clock onboard the moving

spaceship, they find that it suffers time dilation relative to the clocks on Earth: the

rate of the moving clock is slow compared with the rate of identically manufactured

clocks at rest on Earth.To see how this comes about, imagine that experimenters in the

spaceship set up a racetrack of length L perpendicular to the direction of motion of

the spaceship (see Fig. 36.11). If the experimenters in the spaceship use one of their

clocks to measure the time of flight of a light signal that goes from one end of the

track to the other and returns to its starting point, they will then find that the light

signal takes a time of to complete the round trip.

But experimenters on the Earth see that the light signal has concurrent vertical

and horizontal motions (see Fig. 36.12). For the experimenters on Earth, the light

signal travels a total distance larger than 2L. Since the speed must still be the standard

speed of light c, they will find that according to their clocks the light signal now takes

a time longer than 2L/c to complete the trip. Thus, a given time interval regis-

tered by a clock on the spaceship is registered as a longer time interval by the clocks

on the Earth. This means that the clock on the spaceship runs slow when judged by

the clocks on the Earth. Note that the experiment involves one clock on the spaceship

(the clock at the starting point of the track), but several clocks on the Earth, because

the light signal does not return to the point at which it started on Earth, and the

observers on Earth will have to use one (stationary) clock at the starting point and

another (stationary) clock at the end point to measure the time of flight.

For a quantitative evaluation of the time dilation, we note that in Fig. 36.12 the

upward portion of the path of the light signal is the hypotenuse of a right triangle of

sides L and V where V is the speed of the spaceship relative to the Earth. The

total length of the path that the light signal has to cover in the reference frame of the

Earth is therefore

(36.1)

The time taken to cover this distance is

(36.2)

If we square both sides of this equation, we obtain

(36.3)

which we can solve for and then for

and

(36.4)¢t �2L�c

21 � V 2�c2

(¢t)2 �4L2�c2

1 � V 2�c2

¢t :(¢t)2

(¢t)2 �4L2 � V 2(¢t)2

c2

¢t �2 � 2L2 � (V¢t�2)2

c

2 � 2L2 � (V¢t�2)2

¢t�2,

¢t

¢t�¢t

¢t� � 2L�c

¢t�


L

In reference frame of spaceship, light makes a round trip, travelinga distance 2L in time �t' � 2L/c.

FIGURE 36.11 Spaceship with a “race-

track” for a light pulse.

Boston New York

V �t/2

L

In reference frame of the Earth, light making a round trip travelsa distance greater than 2L.

Total distance traveled is

2 � L2 � (V �t/2)2 .

FIGURE 36.12 The trajectory of the light

pulse as observed from the Earth. The light

pulse has both a vertical motion (up or

down) and also a horizontal motion (toward

the right).

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Since, in the reference frame of the spaceship, this gives us

(36.5)

This is the time-dilation formula. It shows that the time registered by the clocks on the

Earth is longer than the time registered by the clock on the spaceship by a factor of

that is, the clock on the spaceship runs slow when measured with the

clocks on the Earth. Figure 36.13 is a plot of the time-dilation factor

for speeds in the range from 0 to c. At low speeds the time-dilation effect is insignif-

icant, but at speeds near c, it becomes quite large.

The slowing down of the rate of lapse of time applies to all physical processes—

atomic, nuclear, biological, etc. Thus, the astronauts on the spaceship will perform all

their tasks in “slow motion,” and they will age slower than normal when measured

with the clocks on the Earth. But they themselves will be unaware of this. From their

point of view, they are in an inertial reference frame in which the usual laws of physics

are valid, and the physical processes in their reference frame proceed at the normal

rate, without any indication of anything unusual.

The time-dilation effect is symmetric: as measured by the clocks on the spaceship,

a clock on the Earth runs slow by the same factor:

(36.6)

The derivation of Eq. (36.6) can be based on an argument similar to that given

above, with a racetrack for light at rest on the Earth. Incidentally: in these arguments

we have implicitly assumed that the length of the racetrack is unaffected by the motion

of the spaceship or the Earth, that is, we have assumed that the length is absolute. As

we will see in the next section, this is true for lengths perpendicular to the direction of

motion, although it is not true for lengths along the direction of motion.

Very drastic time-dilation effects have been observed in the

decay of short-lived elementary particles. For instance, a muon

particle (see Chapter 41) usually decays in about but if it is moving

at high speed through the laboratory, then the internal processes that produce the

decay will slow down as judged by the clocks in our laboratory, and the muon lives

a longer time. In accurate experiments performed at the European Organization

for Nuclear Research (CERN) accelerator near Geneva, muons with a speed of

99.94% of the speed of light were found to have a lifetime 29 times as long as

the lifetime of muons at rest. Is this dilation of the lifetime in agreement with

Eq. (36.5)?

SOLUTION: We can regard the muon as a clock at rest in the reference frame of

an (imaginary) spaceship with a speed of V � 0.9994 c. Over the lifetime of the

muon, this clock registers a time interval of However, the

experimenters in the laboratory see the spaceship moving at and over

the lifetime of the muon they see their own clocks register a larger time than¢t

V � 0.9994c,

¢t� � 2.2 � 10�6 s.

2.2 � 10�6 s;

EXAMPLE 1

e clock at rest on Earth registers ¢t

clocks in spaceship reference frame measure ¢t�¢t� �

¢t

21 � V 2�c2

1�21 � V 2�c21�21 � V 2�c2,

e clock at rest in spaceship registers ¢t�

clocks in Earth reference frame measure ¢t¢t �

¢t�

21 � V 2�c2

2L�c � ¢t�,


time dilation

1/

V

1–V 2/c2

5.0

4.0

3.0

2.0

1.0

0.2c0 0.4c 0.6c 0.8c 1.0c

Time-dilationeffect is insignificantat low speeds…

…but becomesoverwhelming atspeeds close to c.

FIGURE 36.13 Time-dilation factor as a

function of speed V.


the time registered by the moving clock. According to Eq. (36.5), the times

and differ by the time-dilation factor that is,

This time-dilation factor is in agreement with the experimental result.

At everyday speeds, the time-dilation effect is extremely small. For example, con-

sider a clock aboard an airplane traveling at 300 m/s over the ground.The time-dilation

factor is then

(36.7)

Evaluation of this gives 1.000 000 000 000 5, which means that a clock in the airplane

will slow down by only 5 parts in 1013! However, detection of such a small change is

not beyond the reach of modern atomic clocks. In a notable experiment, scientists

from the National Institute of Standards and Technology placed portable atomic clocks

aboard a commercial airliner and kept them flying for several days, making a complete

round trip around the world. Before and after the trip, the clocks were compared with

an identical clock that was kept on the ground. The flying clocks were found to have

lost time—in one instance, the total time lost because of the motion of the clock was

about

The time-dilation effect leads to the famous twin paradox, which we can state as

follows: a pair of identical twins, Terra and Stella, celebrate their, say, twentieth birth-

day on Earth. Then Stella boards a spaceship that carries her at a speed of

to Proxima Centauri, 4 light-years away; the spaceship immediately turns around and

brings Stella back to Earth. According to the clocks on Earth, this trip takes about 4

years each way, so Terra’s age will be 28 years when the twins meet again. But Stella has

benefited from time dilation—relative to the reference frame of the Earth, the space-

ship clocks run slow by a factor

(36.8)

Hence 8 years of travel registered by the Earth clocks amount to only 8 � 0.14 � 1.1

years according to the spaceship clocks, and Stella’s biological age on return will be

only 21 years. Stella will be younger than Terra.

1

21 � V 2�c2�

1

21 � (0.99)2�

1

0.14

V � 0.99c

10�7 s.

1

21 � V 2�c2�

1

21 � (300)2�(3.00 � 108)2�

1

21 � 10�12

1

21 � V 2�c2�

1

21 � (0.9994c)2�c2�

1

21 � (0.9994)2�

1

0.035� 29

1>21 � V 2�c2,¢t�¢t

¢t�


twin paradox

(a)

In reference frame of the Earth, the spaceship moves, and itsclocks experience time dilation.

(b)

In reference frame of thespaceship, the Earth moves, andits clocks experience time dilation.

FIGURE 36.14 Time-dilation effect symmetry.


The paradox arises when we examine the elapsed times from the point of view of

the reference frame of the spaceship. In this reference frame, the Earth is moving away

from the spacecraft (see Fig. 36.14). Hence in this reference frame, the Earth clocks

run slow—and Terra should be younger than Stella.

The resolution of this paradox hinges on the fact that our time-dilation formula is

valid only if the time of a moving clock is measured from the point of view of an iner-

tial reference frame. The reference frame of the Earth is (approximately) inertial, and

therefore our calculation of the time dilation of the spaceship clocks is valid. But the

reference frame of the spaceship is not inertial—the spaceship must decelerate when

it reaches Proxima Centauri, stop, and then accelerate toward the Earth. If the refer-

ence frame is not inertial, the Principle of Relativity does not apply. Therefore, we

cannot use the simple time-dilation formula to find the time dilation of the Earth

clocks from the point of view of the spaceship reference frame. The “paradox” results

from the misuse of this formula.

A detailed analysis of the behavior of the Earth clocks from the point of view of

the spaceship reference frame establishes that the Earth clocks indeed do also run slow

as long as the spaceship is moving with uniform velocity, but that the Earth clocks run

fast when the spaceship is undergoing its acceleration to turn around at Proxima

Centauri. The time that the Earth clocks gain during the accelerated portions of the

trip more than compensates for the time they lose during the other portions of the

trip. This confirms that Stella will be younger than Terra, even from the point of view

of the spaceship reference frame.

The speed of a GPS satellite relative to a point on the Earth’s

surface is typically Assume that the clock

on a GPS satellite is synchronized with the clocks of the Earth’s reference frame at

one instant. By how much do they differ 1.0 hour later? What is the corresponding

distance error for a radio signal? In this calculation, ignore the rotation of the Earth

and treat the satellite motion the same as motion with uniform velocity.

SOLUTION: A time interval measured on Earth will differ from the interval

measured on the GPS satellite by the time-dilation factor:

We wish to find the difference when Since the speed

of the satellite is much less than the speed of light, we can simplify the time-

dilation factor using the expansion for small x. Here, with

and we have

(36.9)

Inserting this into the time-dilation expression, we have

or

(36.10)¢t � ¢t� � ¢t� �1

2 V

2

c 2

¢t � ¢t� � a1 �1

2 V

2

c 2b

1

21 � V 2�c

2� a1 �

V 2

c 2b�1�2

� 1 �1

2 V

2

c 2

x � �V 2�c2,n � �1

2

(1 � x)n � 1 � nx

¢t� � 1.0 hour.¢t � ¢t�

¢t �¢t�

21 � V 2�c2

¢t�

¢t

V � 3.9 � 103 m/s.EXAMPLE 2


Conceptsin

Context


With and we find that the clocks will

differ by

If such a timing error were uncorrected and the satellite clock were used to

determine the distance of a point on the Earth by light travel time, then the error

in this distance would be

(36.11)

Since the accuracy of GPS positioning is required to be much better than that,

corrections for the time dilation must be applied, in addition to many more mun-

dane corrections, for example, those due to refraction and the decrease in the speed

of radio waves in the atmosphere.

Finally, we point out that the time dilation of relativistic physics affects the Doppler

shift of the frequency of light waves whenever the emitter is in motion relative to the

receiver. Since we want to find the frequency detected by the receiver, we consider the

reference frame of the receiver, and we pretend that this reference frame serves as the

“medium” in which the light propagates. In Newtonian physics the Doppler shift for

an emitter moving at speed V through a medium in which the emitted wave has a

speed c is simply where f is the frequency radiated by the emitter,

is the frequency detected by the receiver, and the positive sign applies if the emit-

ter is receding, the negative sign if approaching [see the derivation of Eq. (17.17) for

the case of sound waves]. In relativistic physics, the Doppler shift must also include the

time dilation of the emitter. The time dilation of the period of the waves corresponds

to a reduction in the detected frequency by the additional factor Including

this factor, we have

(36.12)

where now V is to be interpreted as the speed of the emitter relative to the receiver.

Since Eq. (36.12) can be simplified by appropriate

cancellations:

for receding emitter

for approaching emitter

(36.13)

A GPS satellite transmits radio (microwave) signals at a fre-

quency of 1.575 GHz. Assume for simplicity that a GPS satel-

lite is directly approaching your location on the Earth’s surface with a relative speed

of By what factor is the frequency that you detect on Earth increased3.9 � 103 m/s.

EXAMPLE 3

f � � B1 � V�c

1 � V�c f

f � � B1 � V�c

1 � V�c f

1 � V 2�c

2 � (1 � V�c)(1 � V�c),

f � �21 � V

2�c 2

1 ; V�c f

21 � V 2�c2.

f �

f � � f �(1 ; V�c),

¢x � c � (¢t � ¢t�) � 3.00 � 108 m/s � 3.0 � 10�7 s � 90 m

¢t � ¢t� � 3600 s �1

2� a 3.9 � 103 m/s

3.00 � 108 m/sb 2

� 3.0 � 10�7 s

V � 3.9 � 103 m/s,¢t� � 1.0 h � 3600 s


Conceptsin

Context

relativistic Doppler shift


due to the ordinary Doppler shift of Newtonian physics? Due to the time dilation

of relativistic physics?

SOLUTION: For an approaching emitter, the Newtonian result for the upward-

shifted frequency that you receive is given solely by the denominator in Eq. (36.12)

[see also Eq. (17.17)]:

Taking the ratio of received to emitted frequencies and inserting the values, we

obtain

This is an upward shift of 13 parts per million, or of about (1.3 � 10�5) � (1.575

GHz) � 20 kHz.

The time dilation factor is

which shifts the received frequency downward by a factor that differs from unity

by

or less than a tenth of a part per billion. Thus for such a “low” speed, the time dila-

tion contribution to the frequency shift is completely negligible compared with

the ordinary Doppler effect. We already used this fact when we examined police

radar guns in Example 17.7.

Checkup 36.3

QUESTION 1: Distinguish between the relativity of the synchronization of clocks and

the relativity of the rates of clocks.

QUESTION 2: Consider the plot of the time-dilation factor given in Fig. 36.13. If we

increase the speed of a clock by a factor of 2, do we increase the time-dilation factor

by a factor of 2, or by less, or by more?

QUESTION 3: An astronaut aims a beam of light from a green laser pointer at you. If

you approach the astronaut at do you detect blue light or yellow light?

QUESTION 4: A spaceship moves away from Earth at high speed. How do experi-

menters on the Earth measure a clock in the spaceship to be running? How do those

in the spaceship measure a clock on the Earth to be running?

(A) Slow; fast (B) Fast; slow (C) Slow; slow (D) Fast; fast

V � 0.10c,

✔

1

2 V

2

c 2

�1

2�

(3.9 � 103 m/s)2

(3.00 � 108 m/s)2� 8.4 � 10�11

B1 �V 2

c2� 1 �

1

2 V

2

c 2

f �

f�

1

1 � (3.9 � 103 m/s)�(3.00 � 108 m/s)� 1.000 013

f � �1

1 � V�c f



36.4 LENGTH CONTRACTION

In the preceding sections we have seen that time is relative—both the synchronization

of clocks and the rate of clocks depend on the reference frame. Now we will see that length

is also relative. A measuring rod, or any other body, onboard the spaceship suffers length

contraction along the direction of motion. The length of the moving measuring rod

will be short when compared with the length of an identically manufactured measuring

rod at rest on the Earth.The reason for this is that the length measurement of a moving

body depends on simultaneity, and since simultaneity is relative, so is length.

Suppose that the spaceship, traveling from Boston toward New York, carries a

measuring rod that has a length of, say, 300 km in the reference frame of the spaceship

(see Fig. 36.15). To measure the length of this rod in the reference frame of the Earth,

we station observers in the vicinity of New York and Boston with instructions to ascer-

tain the positions of the front and the rear ends of the measuring rod at one instant of

time, say, at noon.

But when the observers on the Earth do this, the observers on the spaceship will

claim that the position measurements were not done simultaneously, and that the

observers in the vicinity of Boston measured the position of the rear end at a later

time. In the extra time, the rear end moves an extra distance to the right, and hence the

distance between the positions measured for the rear and the front ends will be reduced.

From the point of view of the observers on the spaceship, it is therefore immediately

obvious that the length measured by the observers on the Earth will be short. Figures

36.16 and 36.17 show the reference frame of the spaceship moving past the Earth and

the reference frame of the Earth moving past the spaceship, respectively. In these fig-

ures the length contraction has been included (it was left out in Figs. 36.9 and 36.10).

As illustrated in these figures, the length-contraction effect is symmetric: a body at

rest in the spaceship will suffer contraction when measured in the reference frame of

the Earth, and a body at rest on the Earth will suffer contraction when measured in the

reference frame of the spaceship.

We can obtain a formula for the length contraction by exploiting the formula for

the time dilation. Consider a rod of length at rest in the spaceship. According to the

spaceship clocks, an observer on the Earth takes a time to travel from

one end of the rod to the other. Taking time dilation into account, we then conclude

that the observer on the Earth will judge that this takes a shorter time of only

(36.14)

However, the observer on the Earth cannot attribute this reduction of the travel time

to a slowing of his clock—in his own reference frame his clock runs at a normal rate.

Instead, he must attribute the reduction of travel time to a contraction of length of

the rod. Measured in the reference frame of the Earth, the rod in the spaceship has

some contracted length L, and it moves at speed V. Hence, the rod passes by a point

on the Earth in a time L�V. This time must agree with the time calculated in Eq.

(36.14), so

(36.15)

or

L

V� B1 �

V 2

c 2 L�

V

¢t

¢t � B1 �V 2

c 2 ¢t� � B1 �

V 2

c 2 L�

V

¢t� � L��V

L�


Boston New York

Measuring rod is at rest inreference frame of spaceship.

FIGURE 36.15 Spaceship with a measur-

ing rod oriented along the direction of its

motion.

O

y'

x'

Lengths on spaceship alongdirection of motion measuredby observers on Earth are short.

FIGURE 36.16 Reference frame of the

spaceship as observed at one instant of

Earth time (including length contraction;

this length contraction was not included in

Fig. 36.10).

O

Boston New York

y

x

Lengths on the Earthalong direction of motion measured by observerson spaceship are short.

FIGURE 36.17 Reference frame of the

Earth as observed at one instant of space-

ship time (including length contraction;

this length contraction was not included in

Fig. 36.9).

PhysicsPagesCh36_Ac.qxd 08/02/2007 2:53 AM Page 1230 PMAC-291 PMAC-291:Books:GK-Jobs:GK023/Ohanion: TechBooks [PPG -QUARK]

for a rod at rest in spaceship (36.16)

This is the formula for length contraction. According to this formula, the length of a

rod or any body in motion relative to a reference frame is shortened by a factor of

Figure 36.18 is a plot of the length-contraction factor as

a function of speed.

As already mentioned, this contraction effect is symetric: if the rod is at rest on

the Earth and is measured in the reference frame of the spaceship, the formula for the

length contraction is

for a rod L at rest on Earth (36.17)

The length contraction has not been tested directly by experiment. There is no

practical method for a high-precision measurement of the length of a fast-moving

body. Our best bet might be high-speed photography, but this is nowhere near accu-

rate enough, since the contraction is extremely small even at the highest speeds that we

can impart to a macroscopic body. Note, however, that the experimental evidence for

time dilation can be regarded as indirect evidence for length contraction, since, as we

saw above, the former implies the latter.

The contraction effect applies only to lengths along the direction of motion of the

body. Lengths perpendicular to the direction of motion are unaffected. The proof of

this is by contradiction: imagine that we have two identically manufactured pieces of

pipe, one at rest on the Earth, one at rest on the spaceship (see Fig. 36.19). If the

motion of the spaceship relative to Earth were to bring about a transverse contraction

of the spaceship pipe, then, by symmetry, the motion of the Earth relative to the space-

ship would bring about a contraction of the Earth pipe. These contraction effects are

contradictory, since in one case the spaceship pipe would fit inside the Earth pipe, and

in the other case it would fit outside.

A proton is passing by the Earth at a speed of 0.50c. In the

reference frame of the proton, what is the length of the

diameter of the Earth in a direction parallel to that of the motion of the proton?

By how much is this shorter than the diameter in the reference frame of the

Earth?

SOLUTION: We can regard the reference frame of the proton as the reference

frame of a spaceship, and we can regard the diameter of the Earth as a rod at rest

in the reference frame of the Earth. Relative to the proton, or the spaceship, the Earth

has a speed In the reference frame of the Earth, the diameter has the

familiar value But in the reference frame of the proton, or the

spaceship, the Earth has a speed of and the rod is observed to have a

shorter length According to Eq. (36.16),the lengths L and differ by the

length contraction factor that is,

21 � V 2�c

2 � 21 � (0.50c)2�c 2 � 21 � 0.25 � 0.87

21 � V 2�c

2,

L�L�.

V � 0.50c,

L � 1.3 � 107 m.

V � 0.50c.

EXAMPLE 4

L� � B1 �V 2

c2 L

21 � V 2�c

221 � V 2�c

2.

L�L � B1 �V 2

c2 L�

36.4 Length Contraction 1231

length contraction

V

1–V 2/c2

1.0

0.8

0.6

0.4

0.2

0.2c0 0.4c 0.6c 0.8c 1.0c

Length-contractioneffect is insignificantat low speeds…

…but becomes severeat speeds close to c.

FIGURE 36.18 Length-contraction

factor as a function of speed V.

…and this one is at rest in referenceframe of spaceship.

This piece of pipe isat rest in referenceframe of the Earth…

Only length alongdirection of motionis contracted.

FIGURE 36.19 Two identical pieces of

pipe.


Since the diameter of the Earth in its own reference frame is the

length in the reference frame of the proton is

This is shorter than by or 2000 km!

COMMENTS: The dimensions of the Earth perpendicular to the direction of

motion do not contract. This implies that, in the reference frame of the proton,

the Earth is not a sphere, but an ellipsoid.

From the length contraction of a three-dimensional body we can deduce the volume

contraction. The volume of the Earth, which is calculated by taking a product of the

dimension parallel to the motion and the two dimensions perpendicular to the motion,

will be contracted by just one factor of that is, a factor of 0.87 in the

case of Example 4.

Checkup 36.4

QUESTION 1: A cannonball is perfectly round in its own reference frame. Describe

the shape of the cannonball in a reference frame relative to which it is moving at a

high speed, say, 0.5c.

QUESTION 2: Could we use the argument based on the two identical pieces of pipe

(see Fig. 36.19) to prove that lengths along the direction of motion are not affected?

Why, or Why not?

QUESTION 3: In view of the length contraction, does the density of a material depend

on its speed?

QUESTION 4: A track is 100 m long. A particle moves parallel to the track with speed

V such that In the reference frame of the particle, the length of

the track is

(A) 25 m (B) 50 m (C) 100 m (D) 200 m (E) 400 m

36.5 THE LORENTZ TRANSFORMATIONSAND THE COMBINATION OF VELOCIT IES

Suppose that we measure the space and time coordinates of an event—such as the

impact of lightning on a point on the ground—in the reference frame of the Earth

and also in the reference frame of a moving spaceship. We will then obtain different

values of these coordinates in the Earth and in the spaceship reference frames, but

these different values of the coordinates are related by transformation formulas. In

Einstein’s physics, the transformation formulas for the coordinates are fairly compli-

cated, because they are designed so as to keep the speed of light the same in all refer-

ence frames, and they incorporate the length contraction and the time dilation. Before

dealing with these complicated formulas, let us examine the much simpler transformation

formulas for coordinates in Newton’s physics, where there is no length contraction and

no time dilation.

1 � V 2�c

2 � 0.25.

✔

21 � V 2�c

2,

2 � 106 m,1.3 � 107 m

L� � 0.87 � L � 0.87 � 1.3 � 107 m � 1.1 � 107 m

L�

L � 1.3 � 107 m,



Figure 36.20 shows the coordinate grids x–y and of the first (Earth) and the

second (spaceship) reference frames.The second reference frame is moving with veloc-

ity V along the x axis of the first reference frame. We assume that at time the

origins of the two coordinate grids coincide; at time t, the distance between the ori-

gins is then Vt. The coordinates of the point P are x, y in the first reference frame and

in the second reference frame. By inspection of Fig. 36.20, we see that the dis-

tance x equals the sum of the distances and Vt :

(36.18)

Hence

(36.19)

Furthermore, the distance y equals the distance

(36.20)

Equations (36.19) and (36.20) are the transformation equations for the x and y coor-

dinates in Newton’s physics.These two equations are merely the x and y components of

the general vector equation r� � r � Vt for the transformation of the position vector

we found in Chapter 4 [see Eq. (4.50)]. We could have obtained our equations for the

transformation of the x and y coordinates from the general vector equation; but it is just

as easy to re-derive these results by inspection of Fig. 36.20. Note that although Fig.

36.20 makes the equations for the transformation of the x and y coordinates seem self-

evident, these equations hinge on the absolute character of length in Newton’s physics.

Absolute length means that the observers in the two reference frames agree on the

measurement of any length or any distance between two points. If the observers dis-

agreed on the values of the distances x or —for example, if one observer claimed

that the distance x was 3.0 m and the other observer claimed that this distance x was

contracted to 2.5 m—then Eq. (36.18) would not be valid.The left side of Eq. (36.18)

is a distance defined at one instant in the first reference frame, whereas the right side

is a sum of a distance defined at one instant in the second reference frame and a

distance (Vt) defined at one instant in the first reference frame, and such a sum makes

no sense unless the observers agree on the values of these distances.

x�

x�

y� � y

y�

x� � x � Vt

x � x� � Vt

x�

y�x�,

t � 0,

x��y�

36.5 The Lorentz Transformations and the Combination of Velocities 1233

y

y

y'

Vt

x x'

V

P

OO'

xx'

y'

The x' –y' grid moveswith speed V along the x axis of the x –y grid.

FIGURE 36.20 The coordinate grid x–y belonging to the reference frame of the Earth

and the coordinate grid belonging to the reference frame of the spaceship. The coor-

dinates of the point P are x, y in the first grid and x�, y� in the second grid.

x��y�


Furthermore, in Newton’s physics time is absolute. This means that the times reg-

istered by the clocks in the two reference frames are always equal,

(36.21)

Taken together, Eqs. (36.19)–(36.21) are called the Galilean transformation equations;

they relate the space and time coordinates in one reference frame to those in the other.

From these equations we can deduce the Galilean addition rule for the compo-

nents of the velocity. For instance, if the x coordinate changes by dx in a time dt, then

Eqs. (36.19) and (36.21) give us

(36.22)

(36.23)

Dividing these two equations side by side, we obtain

(36.24)

Here, is the x velocity of the particle, light signal, or whatever, measured in the

second reference frame; and dx�dt is the x velocity measured in the first reference

frame. Hence Eq. (36.24) says

(36.25)

This, of course, is simply the Galilean addition rule for the x component of the velocity

[see Eq. (4.53)].

In Einstein’s physics, the Galilean formulas for the transformation of the coordi-

nates and for the addition of velocities must be replaced by more complicated formu-

las, designed in such a way as to keep the speed of light the same in all reference frames.

The transformation equations that accomplish this trick are called the Lorentz trans-

formations. If the new reference frame moves, again, with velocity V along the x axis

of the first reference frame, and if the origins coincide at time the Lorentz trans-

formations take the form

(36.26)

(36.27)

(36.28)

These equations cannot be obtained by simple inspection of Fig. 36.20, because the dis-

tances displayed in this figure are not absolute in Einstein’s physics, and they cannot

simply be added as in Newton’s physics.

We will not bother with a formal derivation of the Lorentz transformation equa-

tions, because we can achieve a clear understanding of the various terms and factors in

these equations by comparing them with the Galilean transformation equations.

Equation (36.26) differs from the Galilean equation (36.19) only by the factor

this factor represents the length contraction. Equation (36.27) is iden-

tical to the Galilean equation, because lengths perpendicular to the direction of motion

1�21 � V 2�c

2;

t� �t � Vx�c2

21 � V 2�c

2

y� � y

x� �x � Vt

21 � V 2�c

2

t � 0,

v�x � vx � V

dx��dt�

dx�

dt��

dx

dt� V

dt� � dt

dx� � dx � V dt

t� � t


Lorentz transformations

Galilean transformation equations


remain unchanged. And Eq. (36.28) differs from the Galilean equation in two ways:

it contains an extra factor representing the time dilation, and it con-

tains an extra term representing the relativity of synchroniza-

tion discussed in Section 36.2. We already gave quantitative treatments of the time

dilation and the length contraction, and we therefore do not need to reexamine these

here. But we did not yet give a quantitative treatment of the relativity of synchronization,

and we will now examine this, to justify the presence of the extra term in Eq. (36.28).

Suppose that observers in the Earth frame send a light signal in the positive x direc-

tion from the origin O (with coordinate to some point P (with coordinate

The signal leaves the origin O at time and arrives at the point P at time t. According

to the observers in the Earth frame, the arrival time is since the light signal

travels at speed c. We want to know the arrival time as seen by the observers in the

spaceship frame. For these observers, the light signal moves in the positive direction

at speed c, while simultaneously the Earth frame and the point P move in the negative

direction at speed V. Hence the “closing speed” of the light signal and the point P

is (this closing speed is larger than c, but that is not objectionable; it merely

reflects the fact that if the target and light signal are both moving toward each other,

they will meet sooner than if the target is at rest).To calculate the time of arrival of the

light signal at the point P, the spaceship observers have to divide the length OP by the

closing speed But since the length OP is a moving length, these observers

must take the length contraction into account: the length between O and P is not x, but is

Accordingly, the observers in the spaceship frame find that the arrival

time of the light signal is

Before we compare this with the value of given by the Lorentz transformation

(36.28), let us multiply the numerator and denominator by and rearrange

the result:

(36.29)

Here, the term x�c in the numerator is simply the time t that the light signal takes to

arrive according to observers in the Earth frame. The term �V x�c2 in the numerator

represents the relativity of synchronization. Comparing Eqs. (36.28) and (36.29), (with

we see they agree exactly—and this agreement provides the justification of

the extra term in Eq. (36.28).

A measuring rod of length is at rest along the axis of the

spaceship frame, which is moving in the positive direction

with speed V relative to the Earth frame. What is the length of the measuring rod

in the Earth frame according to the Lorentz transformation equations?

x�

x�¢x�EXAMPLE 5

x�c � t),

�x�c � Vx�c

2

21 � V 2�c

2

�x (1 � V�c)(1 � V�c)

c (1 � V�c)21 � V 2�c

2

t� �x21 � V

2�c 2

c � V�21 � V

2�c 2

21 � V 2�c

2�

x (1 � V 2�c

2)

c (1 � V�c)21 � V 2�c

2

21 � V 2�c

2t�

t� �x21 � V

2�c 2

c � V

x21 � V 2�c

2.

c � V.

c � V

x�

x�

t�,

t � x�c,

t � 0

x 0).x � 0)

�(Vx�c 2)�21 � V

2�c 2

1�21 � V 2�c

2


HENDRIK ANTOON LORENTZ(1853–1928) Dutch theoretical physicist,

professor at Leiden. He investigated the rela-

tionship between electricity, magnetism, and

mechanics. In order to explain the observed

effect of magnetic fields on emitters of light

(Zeeman effect), he postulated the existence of

electric charges in the atom, for which he was

awarded the Nobel Prize in 1902. He derived

the Lorentz transformation equations by

examining Maxwell ’s equations, but he was

not aware that this leads to a new concept of

space and time.


SOLUTION: We begin with Eq. (36.26) written as an equation for differences:

(36.30)

In the Earth frame, the length of the rod is measured at one instant t of time, so

Hence Eq. (36.30) reduces to

which gives

This is the expected length-contraction formula (36.16).

A clock is at rest in the Earth frame. If a time elapses as

shown by this clock, how much time elapses according to the

clocks of a spaceship moving at speed V relative to the Earth frame?

SOLUTION: Again, we begin by writing the Lorentz transformation equation

(36.28) in terms of differences:

(36.31)

For the clock at rest in the Earth frame, and therefore

which is the expected time-dilation formula, Eq. (36.6).

COMMENT: Note that both the Lorentz transformation equations (36.26) and

(36.28) have factors of even though the length contraction has a

factor of not a factor of The reason becomes clear

from inspection of these two examples: the Lorentz transformation equations

(36.26) and (36.28) incorporate the time dilation for a clock at rest on Earth, but

Eq. (36.26) provides the length of a rod at rest in the spaceship frame. If we want the

length contraction for a rod at rest in the Earth frame, we would need to use the

inverse Lorentz transformation equations, that is, the equations for x, t in terms

of These can be obtained by solving Eqs. (36.26) and (36.28) for x and t,

that is, by pretending x and t are unknowns, to be evaluated by combining these

equations to obtain each of them in terms of and t � (see Problem 33). The

resulting equations have exactly the same form as Eqs. 36.26 and 36.28 with

primed and unprimed space and time coordinates exchanged and with V replaced

by �V.

x�

x�, t�.

1�21 � V 2�c

2.21 � V 2�c

2,

1�21 � V 2�c

2,

¢t� �¢t

21 � V 2�c

2

¢x � 0,

¢t� �¢t � V ¢x�c2

21 � V 2�c

2

¢tEXAMPLE 6

¢x � ¢x� � B1 �V 2

c2

¢x� �¢x

21 � V 2�c

2

¢t � 0.

¢x� �¢x � V¢t

21 � V 2�c

2


inverse Lorentz transformations


Note that if the relative velocity between the two reference frames is small com-

pared with the speed of light, then V�c in Eqs. (36.26) and (36.28) is small, and any

term involving this quantity can be omitted in the equations. The Lorentz transfor-

mations then reduce to

Thus, for low speeds, the Lorentz transformations reduce to the Galilean transfor-

mations (36.19)–(36.21).

The crucial feature of the Lorentz transformation equations is that they leave the

speed of light unchanged. To verify this, we need to find the relativistic combination

rule for velocity. If the x coordinate changes by dx in a time dt, then the Lorentz trans-

formation equations tell us that

(36.32)

(36.33)

and dividing these two equations side by side, we find

(36.34)

On the right side we can divide both the numerator and the denominator by dt, with

the result

(36.35)

In this expression, dx�dt is the x component of the velocity of a light signal or parti-

cle measured in the first reference frame and is the x component of the veloc-

ity measured in the second reference frame. Hence Eq. (36.35) may be written

(36.36)

This is the relativistic combination law for the x components of the velocity (there are

somewhat different formulas for the combination of the y and z components of the veloc-

ity; see Problem 41).

It is instructive to compare the relativistic combination rule for velocities with the

Galilean addition rule

(36.37)

It is the denominator in Eq. (36.36) that makes all the difference. For instance, sup-

pose that is the velocity of a light signal propagating along the x axis of the first

reference frame. Then and Eq. (36.36) yields

(36.38)v�x �c � V

1 � cV�c2�

c (1 � V�c)

1 � V�c� c

vx � c,

vx

v�x � vx � V

v�x �vx � V

1 � vxV�c2

dx��dt�

dx�

dt��

dx�dt � V

1 � V (dx�dt)�c2

dx�

dt��

dx � Vdt

dt � Vdx�c2

dt� �dt � Vdx�c2

21 � V 2�c2

dx� �dx � Vdt

21 � V 2�c2

t� � t

y� � y

x� � x � Vt


relativistic velocity combination


Thus, as required by the Principle of Universality of the Speed of Light, the velocity

of the light signal in the second reference frame has exactly the same magnitude as in

the first reference frame (see Fig. 36.21).

The relativistic combination rule for light velocities has been explicitly tested in

an experiment at CERN, on the French-Swiss border; involving a beam of very fast

pions. These particles decay spontaneously by a reaction that emits a flash of very

intense, very energetic light (gamma rays). Hence, such a beam of pions can be regarded

as a high-speed light source. In the experiment, the velocity of the pions relative to

the laboratory was V � 0.999 75c. The Galilean addition law for velocity would have

predicted laboratory velocities of 1.999 75c for light emitted in the forward direction

and 0.000 25c for light emitted in the backward direction. But the experiments con-

firmed the relativistic combination rule—the laboratory velocity of the light had the

same magnitude c in all directions.

An alien spaceship approaching the Earth at a speed of 0.40c

fires a rocket at the Earth (see Fig. 36.22). If the velocity of the

rocket is 0.80c in the reference frame of the spaceship, what is its velocity in the ref-

erence frame of the Earth?

SOLUTION: The equation for the combination of velocities is easiest to use if

is treated as known and as unknown. Accordingly, we take to be the veloc-

ity in the reference frame of the spaceship, and we take v�x to be the velocity in the

reference frame of the Earth. The x axis is directed from the spaceship toward the

Earth, and the velocity of the rocket in the reference frame of the spaceship is

The velocity V must be taken to be that of the Earth relative to the

spaceship; this velocity is negative, Then Eq. (36.36) gives

Checkup 36.5

QUESTION 1: Suppose that the new reference frame moves in the direction of the neg-

ative x axis of the first reference frame. What are the Lorentz transformation equations

in this case?

QUESTION 2: How do we know that the Lorentz transformation equations are con-

sistent with the requirement that the speed of light is left unchanged?

QUESTION 3: If the spaceship in Example 7 is moving away from the Earth instead of

approaching the Earth, how does this change the answer for

QUESTION 4: A radioactive nucleus approaches Earth at and emits an elec-

tron toward the Earth at relative to the nucleus. What is the electron’s speed

relative to the Earth?

(A) c�4 (B) 4c�5 (C)

(D) c (E) 4c�3

23�4c

v � c�2v � c�2

v�x?

✔

v�x �vx � V

1 � vxV�c2�

0.80c � (�0.40c)

1 � (0.80c)(�0.40c)�c 2

� 0.91c

V � �0.40c.

vx � 0.80c.

vxv�x

vx

EXAMPLE 7


0.80c

0.40c

Spaceship approaches theEarth at speed of 0.40c.

Velocity of rocketis 0.80c in referenceframe of spaceship.

What is velocity of rocket inreference frame of the Earth?

FIGURE 36.22 A spaceship launches a

rocket toward Earth.

(a)

(b)

3 � 108 m/s

3 � 108 m/s

Light signal has speed 3 � 108 m/sin reference frame of spaceship speeding toward the Earth.

Light signal has the samespeed of 3 � 108 m/s inreference frame of the Earth.

FIGURE 36.21 Addition of velocities

according to the relativistic combination

rule. A spaceship speeding toward the Earth

emits a light signal toward the Earth.

(a) Reference frame of the spaceship.

(b) Reference frame of the Earth.


36.6 RELAT IV IST IC MOMENTUM AND ENERGY

The drastic revision that the theory of Special Relativity imposes on Newton’s concepts

of space and time implies a corresponding revision of the concepts of momentum and

energy.The formulas for momentum and energy and the equations expressing their con-

servation are intimately tied to the transformation equations of the space and time coor-

dinates. To see that this is so, we briefly examine the Newtonian (nonrelativistic) case.

In Newton’s physics, the momentum of a particle of mass m and velocity v is

(36.39)

To find the momentum of this particle in a new reference frame, we note that the

Galilean transformation equation for the velocity vector is

v� � v � V (36.40)

Multiplying this by the mass, we find the transformation equation for the momentum:

p� � mv� � mv � mV (36.41)

From this we see that the momentum p� in the new reference frame differs from the

momentum p in the old reference frame by only the constant quantity mV (a quantity

independent of the velocity v of the particle). Hence, if the total momentum of a

system of colliding particles is conserved in one reference frame, it will also be conserved

in the other reference frame—and the Law of Conservation of Momentum obeys the

Principle of Relativity.This shows that the nonrelativistic formula for momentum and

the nonrelativistic Galilean formula for the addition of velocities match in just the

right way.

According to the relativistic physics of Einstein, we must replace the Galilean addi-

tion formula for velocities by the relativistic combination rule. If the Law of Conservation

of Momentum is to obey the Principle of Relativity, we must then design a new rela-

tivistic formula for momentum that matches the new relativistic combination rule for

velocities. The required relativistic formula for momentum is

(36.42)

We will not give a proof of this formula.

If the speed of the particle is small compared with the speed of light, then

and Eq. (36.42) becomes approximately

p � mv (36.43)

This shows that for low speeds, the relativistic and the Newtonian formulas for the

momentum agree. We can therefore regard the Newtonian formula for the momen-

tum as a simple and useful approximation for low speeds. This approximation is quite

adequate for the description of all the phenomena we encounter in everyday life and

(almost) all the phenomena we encounter in the realm of engineering, such as the phe-

nomena we dealt with in the earlier chapters of this book. But at high speeds, the for-

mulas differ drastically. We must then abandon the Newtonian formula, and rely

entirely on the relativistic formula. Note that the relativistic momentum becomes infi-

nite as the speed of the particle approaches the speed of light. Figure 36.23 is a plot of

the magnitude of the momentum as a function of the speed.

21 � v 2�c

2 � 1

p �mv

21 � v 2�c

2

p � mv


relativistic momentum


An electron in the beam of a TV tube has a speed of 1.0 � 108

m/s. What is the magnitude of the momentum of this electron?

SOLUTION: For this electron,

According to Eq. (36.42), the magnitude of the momentum is then

COMMENTS: Note that if we had calculated the momentum according to the

nonrelativistic formula we would have obtained and

we would have been in error by about 6%.

We also need a new relativistic formula for kinetic energy. This formula is

(36.44)

For low speeds, this relativistic formula for kinetic energy can be shown to agree approx-

imately with the nonrelativistic formula

The relativistic kinetic energy becomes infinite as the speed of the particle approaches

the speed of light. This indicates that, for any particle with mass (and for any body),

the speed of light is unattainable, since it is impossible to supply a particle with an

infinite amount of energy. Figure 36.24 is a plot of the kinetic energy vs. the speed.

K � 12 mv2.

K �mc2

21 � v2�c2� mc2

9.1 � 10�23 kgm/s,p � mv,

p �mv

21 � v2�c2�

9.11 � 10�31 kg � 1.0 � 108 m/s

21 � (0.33)2� 9.7 � 10�23 kgm/s

v�c � (1.0 � 108 m/s)�(3.0 � 108 m/s) � 0.33.

EXAMPLE 8


relativistic kinetic energy

v

K

0.2c0 0.4c 0.6c 0.8c 1.0c

4mc2

3mc2

2mc2

mc2

…but increases much more quickly with vat speeds close to c.

Kinetic energyK ≈ mv 2

is proportional tosquare of velocityat low speeds…

21

FIGURE 36.24 Kinetic energy of a

particle as function of speed v.

v

p

0.2c0 0.4c 0.6c 0.8c 1.0c

4mc

3mc

2mc

mc

Momentum p ≈ mvis proportional to velocity at low speeds…


FIGURE 36.23 Momentum of a particle

as a function of speed v.


The maximum speed that electrons achieve in the Stanford

Linear Accelerator (SLAC) is 0.999 999 999 67c. What is the

kinetic energy of an electron moving at this speed?

SOLUTION: The relativistic formula (36.44) contains a factor Since

v�c for these electrons is extremely close to 1, most calculators are unable to eval-

uate To get around this difficulty, we write

(36.45)

and we evaluate “by hand,”

1�v�c � 1 � 0.999 999 999 67 � 3.3 � 10�10

The rest of the calculation is within the reach of our calculator:

Although the theory of Special Relativity requires a revision of the basic equations

of mechanics, it does not require any revision of the basic equations of electricity and

magnetism. Maxwell’s equations are already relativistic, that is, they match the relativistic

behavior of length and of time in just the right way.This concordance between Maxwell’s

equations and the requirements of relativity is no accident. Maxwell’s equations incor-

porate a universal speed of light—they imply in every reference frame.

Einstein’s search for a theory of relativity was motivated by his faith in Maxwell’s equa-

tions and his recognition that if Maxwell’s equations were right then the Galilean

coordinate transformations had to be wrong.

Checkup 36.6

QUESTION 1: For a given speed, is the relativistic value of the momentum always larger

than the Newtonian value? By what factor?

QUESTION 2: Is the relativistic momentum always in the direction of the velocity of

the particle?

QUESTION 3: In science fiction stories, spaceships routinely reach speeds equal to or

in excess of the speed of light. What is wrong with this?

QUESTION 4: According to the relativistic formula (36.44) for the kinetic energy, what

is the kinetic energy of a particle of zero speed?

(A) 0 (B) mc2 (C) �mc2

(D) 2mc2 (E) Infinite

✔

c � 1�1m0�0

� 3.2 � 10�9 J

� 9.11 � 10�31 kg � (3.00 � 108 m/s)2 � a 1

2223.3 � 10�10� 1 b

K � mc2 a 1

21 � v2�c2� 1 b � mc2 a 1

2221 � v�c� 1 b

1 � v�c

B1 �V 2

c2� B1 �

V

cB1 �v

c� 22B1 �

v

c

21 � v2�c2.

21 � v2�c2.

EXAMPLE 9



36.7 MASS AND ENERGY

One of the great discoveries that emerged from relativity is that energy can be trans-

formed into mass, and mass can be transformed into energy. Thus, mass is a form of

energy. The amount of energy contained in an amount m of mass at rest is given by

Einstein’s famous formula

(36.46)

The quantity is called the rest-mass energy.3 The formula (36.46) can be derived

from the theory of relativity, but, as with some other equations in this chapter, we will

not give the derivation.

The most spectacular demonstration of Einstein’s mass–energy formula is found

in the annihilation of matter and antimatter (see Chapter 41). If a proton collides with

an antiproton, or an electron with an antielectron, the two colliding particles react vio-

lently and they annihilate each other in an explosion that generates an intense flash

of very energetic light. According to Eq. (36.46), the annihilation of just 1000 kg of

matter and antimatter (500 kg of each) would release an amount of energy

(36.47)

This is enough energy to satisfy the needs of the United States for a full year.

Unfortunately, antimatter is not readily available in large amounts. On Earth,

antiparticles can be obtained only from reactions induced by the impact of beams

of high-energy particles on a target. These collisions occasionally result in the cre-

ation of a particle–antiparticle pair. Such pair creation is the reverse of pair annihila-

tion. The creation process transforms some of the kinetic energy of the collision into

mass, and a subsequent annihilation merely gives back the original energy.

But the relationship between energy and mass in Eq. (36.46) also has another

aspect. Energy has mass. Whenever the internal energy stored in a body is changed, its

rest mass (and weight) is changed. The change in rest mass that accompanies a given

change of energy is

(36.48)

For instance, in the fission of uranium, the nuclear material loses energy, and corre-

spondingly its mass (and weight) decreases.The complete fission of 1.0 kg of uranium

releases an energy of and correspondingly the mass of the nuclear mate-

rial decreases by or about 0.1%.

The fact that energy has mass indicates that energy is a form of mass. Conversely,

as we have seen above, mass is a form of energy. Hence mass and energy must be

regarded as different aspects of essentially the same thing. The laws of conservation of

mass and conservation of energy are therefore not two independent laws—each implies

the other. For example, consider the fission reaction of uranium inside the reactor

vessel of a nuclear power plant (for details, see Chapter 40). The reaction conserves

¢m � (8.2 � 1013 J)�c2 � 0.000 91 kg,

8.2 � 1013 J,

¢m � ¢E�c2

E � mc2 � 1000 kg � (3.00 � 108 m/s)2 � 9.00 � 1019 J

mc2

E � mc2


3 Throughout this section, mass means the mass that a body has when at rest or nearly at rest; to emphasize

this, we use the term rest mass. The definition and measurement of mass for a body in motion at high (rela-

tivistic) speeds is rather tricky, because Newton’s equation fails and the direction of the acceleration

is not necessarily the direction of the force.The only kind of mass that is unambiguously defined in Einstein’s

physics is the mass that the body has when at rest, and this is the only kind of mass we will consider.

ma � F

rest mass energy

mass and energy changes


energy—it merely transforms nuclear energy into heat, light, and kinetic energy but does

not change the total amount of energy. The reaction also conserves mass—if the reac-

tor vessel is hermetically sealed and thermally insulated from its environment, then the

reaction does not change the mass of the contents of the vessel. However, if the vessel

has an opening that lets some of the heat and light escape, then the mass of the residues

will not match the mass of the original amount of uranium. The mass of the residues

will be about 0.1% smaller than the original mass of the uranium. This mass defect

represents the mass carried away by the energy that escapes. Thus, the nuclear fission

reaction merely transforms energy into new forms of energy and mass into new forms

of mass. In this regard, a nuclear reaction is not fundamentally different from a chem-

ical reaction. The mass of the residues in an exothermic chemical reaction is slightly

less than the original mass. The heat released in such a chemical reaction carries away

some mass, but, in contrast to a nuclear reaction, this amount of mass is so small as to

be quite immeasurable.

The total energy of a free particle in motion is the sum of its rest-mass energy

(36.46) and its kinetic energy (36.44):

(36.49)

This leads to a simple formula for the relativistic total energy of the particle:

(36.50)

It is easy to verify (see Problem 66) that the relativistic energy can be expressed as

follows in terms of the relativistic momentum:

(36.51)

For an ultra-relativistic particle, moving at a speed close to that of light, the first term

within the square root is much larger than the second term Hence, for such

a particle we can ignore the second term, and we then obtain the simple result

or

(ultra-relativistic particle) (36.52)

Thus, the momentum and the energy of an ultra-relativistic particle are directly pro-

portional.

Consider an electron of speed 0.999 999 999 67c, as in

Example 9. What is the momentum of such an electron?

SOLUTION: Such as electron is ultra-relativistic. Its kinetic energy is much larger

than its rest-mass energy, and the total energy is therefore approximately equal to

the kinetic energy, which we have already calculated in Example 9:

E � mc2 � K � K � 3.2 � 10�9 J

EXAMPLE 10

E � cp

E � 2c 2p2

(m2c4).(c2p2)

E � 2c2p2 � m2c4

E �mc2

21 � v2�c2

E � mc2 � K � mc2 �mc2

21 � v2�c2� mc2

36.7 Mass and Energy 1243

relativistic total energy


Hence Eq. (36.47) yields

Checkup 36.7

QUESTION 1: What is the rest-mass energy of a 1.0-kg piece of stone? Why can’t we

exploit this energy?

QUESTION 2: Does kinetic energy have mass? For instance, does the kinetic energy

of the particles of a gas contribute to the overall mass of the gas?

QUESTION 3: For an ultra-relativistic particle (of speed near c), the momentum and the

energy are proportional. Is this also true for a particle of lower speed, say, 0.9c or lower?

QUESTION 4: We must add energy to a hydrogen atom to ionize it and thus obtain a

proton and an electron. A neutron, on the other hand, will spontaneously decay to

provide a moving proton and a moving electron. From this information, which has a

greater mass, the hydrogen atom or the neutron? Or do they have the same mass?

(A) Hydrogen atom (B) Neutron (C) Both have same mass

✔

p �E

c�

3.2 � 10�9 J

3.0 � 108 m/s� 1.1 � 10�17 kgm/s


SUMMARY

PRINCIPLE OF RELATIVITY All the laws of physics are the same in

all inertial reference frames.

(36.5)¢t �¢t�

21 � V 2�c2

PRINCIPLE OF UNIVERSALITY OF SPEED OF LIGHT The speed of

light is the same in all inertial reference frames.

TIME DILATION registered by clock in its

own reference frame.)

(¢t� 1/

V

1–V 2/c2

5.0

4.0

3.0

2.0

1.0

0.2c0 0.4c 0.6c 0.8c 1.0c

for receding emitter

(36.13)

for approaching emitterf � � B1 � V�c

1 � V�c f

f � � B1 � V�c

1 � V�c f

RELATIVISTIC DOPPLER SHIFT


Summary 1245

(36.16)L � 21 � V 2�c2L�

(36.26)

(36.27)

(36.28) t� �t � Vx�c2

21 � V 2�c2

y� � y

x� �x � Vt

21 � V 2�c2

LENGTH CONTRACTION is length of body

in its own reference frame.)

(L�

LORENTZ TRANSFORMATIONS

RELATIVISTIC COMBINATION OF VELOCITIES

RELATIVISTIC MOMENTUM

(36.36)v�x �vx � V

1 � vxV�c2

(36.42)p �mv

21 � v2�c2

(36.44)K �mc2

21 � v2�c 2� mc2

(36.50)

E � 2c2p2 � m2c4

E �mc2

21 � v2�c2

(36.46)E � mc2

�m � �E�c2 (36.48)

RELATIVISTIC KINETIC ENERGY

RELATIVISTIC TOTAL ENERGY

REST-MASS ENERGY

MASS AND ENERGY CHANGES

O

y'

x'

Lengths on spaceship alongdirection of motion measuredby observers on Earth are short.

v

K

0.2c0 0.4c0.6c0.8c 1.0c

4mc2

3mc2

2mc2

mc2


Kinetic energyK ≈ mv 2

is proportional tosquare of velocityat low speeds…

21

v

p

0.2c0 0.4c 0.6c 0.8c 1.0c

4mc

3mc

2mc

mc

Momentum p ≈ mvis proportional to velocity at low speeds…




4 George Gamow, Mr. Tompkins in Wonderland.

10. According to the arguments of Section 36.3, a light signal trav-

eling along a track placed perpendicular to the direction of

motion of the spaceship (see Fig. 36.11) takes a longer time to

complete a round trip when measured by the clocks on the

Earth than when measured by the clocks on the spaceship.

Would the same be true for a light signal traveling along a track

placed parallel to the direction of motion? Explain qualitatively.

11. A cannonball is perfectly round in its own reference frame.

Describe the shape of this cannonball in a reference frame rel-

ative to which it has a speed of 0.95c. Is the volume of the

cannonball the same in both reference frames?

12. A rod at rest in the ground makes an angle of with the x

axis in the reference frame of the Earth. Will the angle be

larger or smaller in the reference frame of a spaceship moving

along the x axis?

13. In the charming tale “City Speed Limit” by George Gamow,4

the protagonist, Mr. Tompkins, finds himself riding a bicycle

in a city where the speed of light is very low, roughly 30 km/h.

What weird effects must Mr. Tompkins have noticed under

these circumstances?

14. A long spaceship is accelerating away from the Earth. In the

reference frame of the Earth, are the instantaneous speeds of

the nose and of the tail of the spaceship the same?

15. Suppose that a very fast runner holding a long horizontal pole

runs through a barn open at both ends. The length of the pole

(in its rest frame) is 6 m, and the length of the barn (in its rest

frame) is 5 m. In the reference frame of the barn, the pole will

suffer length contraction and, at one instant of time, all of the

pole will be inside the barn. However, in the reference frame

of the runner, the barn will suffer length contraction and all of

the pole will never be inside the barn at one instant of time. Is

this a contradiction?

16. Why can a spaceship not travel as fast as or faster than the

speed of light?

17. If the beam from a revolving searchlight is intercepted by a

distant cloud, the bright spot will move across the surface of

the cloud very quickly, with a speed that can easily exceed the

speed of light. Does this conflict with our conclusion of

Section 36.6, that the speed of light is unattainable?

30�

QUEST IONS FOR DISCUSSION

1. An astronaut is inside a closed space capsule coasting through

interstellar space. Is there any way the astronaut can measure

the speed of the capsule without looking outside?

2. Why did Michelson and Morley use two light beams, rather

than a single light beam, in their experiment?

3. When Einstein was a boy he wondered about the following

question: A runner holds a mirror at arm’s length in front of

his face. Can he see himself in the mirror if he runs at (almost)

the speed of light? Answer this question both according to the

ether theory and according to the theory of Special Relativity.

4. Consider the piece of paper on which one page of this book is

printed. Which of the following properties of the piece of

paper are absolute, that is, which are independent of whether

the paper is at rest or in motion relative to you? (a) The thick-

ness of the paper, (b) the mass of the paper, (c) the volume of

the paper, (d) the number of atoms in the paper, (e) the chemi-

cal composition of the paper, (f ) the speed of light reflected by

the paper, and (g) the color of the colored print on the paper.

5. Two streetlamps, one in Boston and the other in New York

City, are turned on at exactly 6:00 P.M. Eastern Standard

Time. Find a reference frame in which the streetlamp in New

York was turned on late.

6. According to the theory of Special Relativity, the time order of

events can be reversed under certain conditions. Does this

mean that a sparrow might fall from the sky before it leaves

the nest?

7. Because of the rotational motion of the Earth about its axis, a

point on the equator moves with a speed of 460 m/s relative to

a point on the North Pole. Does this mean that a clock placed

on the equator runs more slowly than a similar clock placed on

the pole?

8. According to Jacob Bronowski, author of The Ascent of Man,

the explanation of time dilation is as follows: If you are

moving away from a clock tower at a speed nearly equal to the

speed of light, you keep pace with the light that the face of

the clock sent out at, say, 11 o’clock. Hence, if you look toward

the clock tower, you always see its hands at 11 o’clock. Is this

explanation correct? If not, what is wrong with it?

9. Suppose you wanted to travel into the future and see what the

twenty-fifth century is like. In principle, how could you do

this? Could you ever return to the twenty-first century?


Problems 1247

†For help, see Online Concept Tutorial 41 at www.wwnorton.com/physics ††For help, see Online Concept Tutorial 42 at www.wwnorton.com/physics

PROBLEMS

†36.1 The Speed of L igh t ; the E ther

1. Consider the case where the Sun moves at a high speed v

through the hypothetical ether. What are the minimum and

maximum ether-wind speeds on Earth when the Sun moves

through the ether at (a) 30 km/s and (b) 60 km/s? Assume the

orbital speed of the Earth is 30 km/s.

*2. A Michelson–Morley interferometer determines the shift of

two waves traveling in perpendicular directions (see also

Section 35.2 and Fig. 35.9).

(a) Assume that one wave travels a distance L along the ether

wind with speed to a mirror, and back with speed

as in Figs. 36.3a–b. Show that the round-trip time

can be written

(b) Assume the other wave travels the same distance perpen-

dicular to the ether wind with speed (see Fig.

36.3c). Show that its round-trip time is

(c) Use the expansion for small x to show

that the difference in arrival times is

(d) What fraction of a full period is this shift for light with

Use the values and

*3. Ordinarily, the two arms of a Michelson–Morley interferome-

ter cannot be set exactly equal, and instead have two values,

and Insert these respective values into the results of

Problem 2a and b and obtain a new expression for (see

Problem 2c). Note that this result alone cannot be used to

determine the ether-wind speed, since the difference between

and is not accurately known. In an actual experiment,

the entire apparatus is rotated (thus interchanging and

Obtain an expression for the net shift by subtracting the

differences in arrival times for the two orientations.

†,††36.2 E ins te in ’s Pr inc ip le o f Re la t iv i ty

4. A spaceship traveling at speed c relative to the Earth ejects a

spacepod traveling in the forward direction at speed c relative

to the spaceship. The spacepod emits a light signal toward the

Earth at speed c relative to the spacepod. What is the speed of

the light signal relative to the spaceship? What is the speed of

the light signal relative to the Earth? Which observer (on the

spaceship or on the spacepod) determines that the light strikes

the Earth earlier?

14

12

L2).

L190�

L2L1

¢t

L2.

L1

V � 30 km/s.L � 11 ml � 500 nm?

¢t � t ‘ � t� �LV 2

c3

(1 � x)n � 1 � nx

t� �2L

c a1 �

V 2

c 2b�1�2

2c2 � V 2

t ‘ �2L

c a1 �

V 2

c 2b�1

c � V,

c � V

††36.3 T ime Di la t ion

5. If a moving clock is to have a time-dilation factor of 10, what

must be its speed?

6. Neutrons have an average lifetime of 15 minutes when at rest

in the laboratory. What is the average lifetime of neutrons of a

speed of 25% of the speed of light? 50%? 90%?

7. Consider an unstable particle, such as a pion, which has a life-

time of only when at rest in the laboratory.

What speed must you give such a particle to make its lifetime

twice as long as when at rest in the laboratory?

8. The speed of the Sun around the center of our Galaxy is 200

km/s. Clocks in the Solar System will therefore run slow as

compared with clocks at rest in the Galaxy. By what factor are

the Solar System clocks slow?

9. The orbital speed of the Earth around the Sun is 30 km/s. In

one year, how many seconds do the clocks on the Earth lose

with respect to the clocks of an inertial reference frame at rest

relative to the Sun? [Hint: If V�c is small, the approximation

is valid.]

10. In 1961, the cosmonaut G. S. Titov circled the Earth for 25 h

at a speed of 7.8 km/s. According to Eq. (36.5), what was the

time-dilation factor of his body clock relative to the clocks on

Earth? By how many seconds did his body clock fall behind

during the entire trip? (Hint: Use the approximation given in

Problem 9.)

11. At a speed V, the time-dilation factor has some value. Suppose

that at speed 2V, the time-dilation factor has twice the previ-

ous value. What is the speed V ?

12. An astronaut traveling at V � 0.80 c taps her foot 3.0 times per

second. What is the frequency of taps determined by an

observer on the Earth?

13. An atomic clock aboard a spaceship runs slow compared with

an Earth-based atomic clock at a rate of 1.0 second per day.

What is the speed of the spaceship?

14. A spaceship equipped with a chronometer is sent on a round-

trip to Alpha Centauri, 4.4 light-years away. The spaceship

travels at 0.10 c, and returns immediately.

(a) According to clocks on the Earth, how long does this trip

take?

(b) According to the chronometer on the spaceship, how long

does this trip take?

15. Consider the Doppler-shift formula for a receding source. By

what factor does the frequency decrease for For

For V � 0.90c?V � 0.70c?

V � 0.50c?

21 � (V�c)2 � 1 � 12 (V�c)2

2.6 � 10�8 s


16. The frequencies of light received from distant galaxies and

quasars are shifted due to the Doppler effect. Frequencies 5.0

times smaller than expected for a stationary source have been

detected from receding quasars. What is V�c for such a quasar?

*17. In a test of the relativistic time-dilation effect, physicists com-

pared the rates of vibration of nuclei of iron moving at differ-

ent speeds. One sample of iron nuclei was placed on the rim of

a high-speed rotor; another sample of similar nuclei was

placed at the center. The radius of the rotor was 10 cm, and it

rotated at 35 000 rev/min. Under these conditions, what was

the speed of the rim of the rotor relative to the center?

According to Eq. (36.5), what was the time-dilation factor of

the sample at the rim compared with the sample at the center?

(Hint: Use the approximation given in Problem 9.)

*18. If cosmonauts from the Earth wanted to travel to the

Andromeda galaxy in a time of no more than 10 years as reck-

oned by clocks aboard their spaceship, at what (constant)

speed would they have to travel? How much time would have

elapsed on Earth after 10 years of time on the spaceship? The

distance to the Andromeda galaxy is light-years.

*19. Because of the rotation of the Earth, a point on the equator

has a speed of 460 m/s relative to a point at the North Pole.

According to the time-dilation effect of Special Relativity, by

what factor do the rates of two clocks differ if one is located

on the equator and the other at the North Pole? After 1.00

year has elapsed, by how many seconds will the clocks differ?

Which clock will be ahead? (Although the special-relativistic

time dilation slows one clock at the equator, there is an addi-

tional gravitational time dilation that slows the other clock.

These two time-dilation effects balance, and the two clocks

actually run at the same rate.)

**20. The star Alpha Centauri is 4.4 light-years away from us.

Suppose that we send a spaceship on an expedition to this star.

Relative to the Earth, the spaceship accelerates at a constant

rate of 0.10g until it reaches the midpoint, 2.2 light-years from

Earth. The spaceship then decelerates at a constant rate of

0.10g until it reaches Alpha Centauri. The spaceship performs

the return trip in the same manner.

(a) What is the time required for the complete trip according

to the clocks on the Earth? Ignore the time that the

spaceship spends at its destination.

(b) What is the time required for the complete trip according

to the clocks on the spaceship? Assume that the instanta-

neous time-dilation factor is still even

though the speed V is a function of time.

36.4 Length Cont rac t ion

21. A meterstick is moving by an observer in a direction parallel to

its length. The speed of the meterstick is 0.50c. What is its

measured length in the reference frame of the observer?

22. According to the manufacturer’s specifications, a spaceship has

a length of 200 m. At what speed (relative to the Earth) will

21 � V 2�c 2

2.2 � 106

this spaceship have a length of 100 m in the reference frame of

the Earth?

23. A cannonball flies through our laboratory at a speed of 0.30c.

Measurement of the transverse diameter of the cannonball gives

a result of 0.20 m. What can you predict for the measurement

of the length, or the longitudinal diameter, of the cannonball?

24. What is the percent length contraction of an automobile traveling

at 96 km/h? (Hint: Use the approximation given in Problem 9.)

25. A hangar for housing spaceships is 100 m long. How fast must

a 200-m-long spaceship be traveling to (briefly) fit in the

hangar?

26. A right triangle of sheet metal with two angles lies in the

x–y plane, with one of its sides along the x axis (see Fig.

36.25). The length of each side is 0.20 m, and the length of

the hypotenuse is Suppose that this triangle is

observed from an reference frame moving at 0.80c along

the x axis. What are the lengths of the sides and of the

hypotenuse in this reference frame? What are the angles?

x�, y�

12 � 0.20 m.

45�


y

x

0.20 m

45°

45°

0.20 m

FIGURE 36.25 A triangle.

*27. Two identical spaceships are traveling in the same direction.

An observer on Earth measures the first to have speed 0.80c

and observes the second to be 1.50 times as long as the first

one. What is the speed of the second spaceship?

*28. Suppose that a meterstick at rest in the reference frame of the

Earth lies in the x–y plane and makes an angle of with the

x axis. Suppose that one end of the meterstick is at the origin.

At a fixed time t, what are the x and y components of the dis-

placement from this end of the meterstick to the other? At a

fixed time what are the and components of the dis-

placement from one end of the meterstick to the other in a

new reference frame moving with velocity in the

positive x direction? What is the angle the meterstick makes

with the axis of this new reference frame?

*29. Electric charge is uniformly distributed throughout a sphere;

the charge density is If this sphere is put in

motion relative to the laboratory at a speed of 0.80c, what will

be the charge density? Keep in mind that the total amount of

electric charge is unchanged by the motion of the sphere.

2.0 � 10�6 C/m3.

x�

V � 0.70c

y�x�t�,

30�


*30. It can be shown that when a point charge moves at uniform

velocity of relativistic magnitude, its pattern of electric field

lines is contracted by the usual length-contraction factor

in the longitudinal direction and is unchanged

in the transverse direction. Figure 36.26 shows the resulting

pattern of field lines for a speed Draw a similar

picture for a speed of 0.80c.

V � 0.60c.

21 � V 2�c 2

34. A spaceship has a length of 300 m, measured in its own refer-

ence frame. It is traveling in the positive x direction at a speed

of 0.80c relative to the Earth. A strobe light at the nose of the

spaceship sends a pulse of light toward the tail of the spaceship.

(a) As measured in the reference frame of the spaceship, how

long does this light pulse take to reach the tail?

(b) As measured in the reference frame of the Earth, how

long does this light pulse take to reach the tail?

35. A spaceship is moving at a speed of 0.60c toward the Earth. A

second spaceship, following the first one, is moving at a speed

of 0.90c. What is the speed of the second spaceship as

observed in the reference frame of the first?

36. Find the inverse of Eq. (36.36); that is, express in terms of

37. The captain of a spaceship traveling away from Earth in the x

direction at observes that a nova explosion occurs at

a point with spacetime coordinates

as measured in the

reference frame of the spaceship. He reports this event to the

Earth via radio without delay.

(a) What are the spacetime coordinates of the explosion in

the reference frame of the Earth? Assume that the master

clock of the spaceship coincides with the master clock of the

Earth at the instant when the midpoint of

the spaceship passes by the Earth, and that the origin of

the spaceship coordinates is at the midpoint of the

spaceship.

(b) Will the Earth receive the captain’s report before or after

astronomers on the Earth see the nova explosion in their

telescopes? No calculation is required for this question.

*38. Consider the situation described in Problem 37. Since light

takes some time to travel from the nova to the spaceship, the

space and time coordinates that the captain reports are not

directly measured but, rather, deduced from the time of arrival

and the direction of the nova light reaching the spaceship.

(a) At what time time) did the nova light reach the space-

ship?

(b) If the captain sends a report to Earth via radio as soon as

he sees the nova, at what time (t time) does the Earth

receive the report?

(c) At what time do Earth astronomers see the nova?

*39. At A.M. a boiler explodes in the basement of the

Museum of Modern Art in New York City. At

A.M. a similar boiler explodes in the basement of a soup fac-

tory in Camden, New Jersey, at a distance of 150 km from the

first explosion. Show that, in the reference frame of a space-

ship moving at a speed greater than from New York

toward Camden, the first explosion occurs after the second.

*40. A radioactive atom in a beam produced by an accelerator has a

speed 0.80c relative to the laboratory. The atom decays and

ejects an electron of speed 0.50c relative to itself. What is the

speed of the electron relative to the laboratory, if ejected in the

forward direction? If ejected in the backward direction?

V � 0.60c

11h0m0.0003s

11h0m0.0000s

(t�

x�, y�

t � t� � 0

y� � 1.2 � 1017 mx� � 1.9 � 1017 m,

t� � �6.0 � 108 s,

V � 0.80c

v�x.vx

Problems 1249

0.60c

FIGURE. 36.26 Electric field lines

of a charge moving at V � 0.60c.

v

FIGURE 36.27 A drive belt and two flywheels.

*31. A flexible drive belt runs over two flywheels whose axles are

mounted on a rigid base (see Fig. 36.27). In the reference

frame of the base, the horizontal portions of the belt have a

speed v and therefore are subject to length contraction, which

tightens the belt around the flywheels. However, in a reference

frame moving to the right with the upper portion of the belt,

the base is subject to length contraction, which ought to

loosen the belt around the flywheels. Resolve this paradox by a

qualitative argument. (Hint: Consider the lower portion of the

belt as seen in the reference frame of the upper portion.)

36.5 The Lorentz Trans format ions andthe Combinat ion of Ve loc i t ies

32. In the reference frame of the Earth, a firecracker is observed

to explode at at According to the

Lorentz transformation equations, what are the and coor-

dinates of this event as observed in the reference frame of a

spaceship traveling in the x direction at a speed of 0.50c?

According to the Galilean transformation equations?

33. Obtain the inverse Lorentz transformation equations by

solving Eqs. (36.26) and (36.28) for x and t, each in terms of

and t�.x�

t�x�

t � 4.0 s.x � 6.0 � 108 m,


*41. In a manner similar to the procedure of Eqs. (36.32)–(36.36),

show that relativistic combination formula for the y compo-

nent of the velocity is

*42. Consider two speeds and V, each of which is less than the

speed of light. Show that if these speeds are combined by the

relativistic combination formula, the result is always less than

the speed of light.

*43. The speed of light with respect to a medium is

where n is the index of refraction. Suppose that the medium,

say, flowing water, is moving past a stationary observer in the

same direction as the light with speed V. Show that the

observer measures the speed of light to be approximately

This effect was first observed by Fizeau in 1851.

**44. The acceleration of a particle in one reference frame is

where the particle has instantaneous velocity vx

in that frame. Consider a reference frame moving with speed

V parallel to the positive x axis of the first frame. Show that

the acceleration in the second frame is given by

36.6 Re la t iv i s t i c Momentum and Energy

45. Consider a particle of mass m moving at a speed of 0.10c.

What is its kinetic energy according to the relativistic formula?

What is its kinetic energy according to the Newtonian for-

mula? What is the percent deviation between these two results?

46. Suppose you want to give a rifle bullet of mass 0.010 kg a

speed of 1.0% of the speed of light. What kinetic energy must

you supply?

47. The yearly energy expenditure of the United Stated is about

Suppose that all of this energy could be converted

into kinetic energy of an automobile of mass 1000 kg. What

would be the speed of this automobile?

48. The speed of an electron in a hydrogen atom is

For this speed, does your calculator show any difference

between the kinetic energies calculated according to the rela-

tivistic formula and the Newtonian formula?

49. What is the speed of an electron if its kinetic energy is 1.6 �

50. What is the momentum and what is the kinetic energy of an

electron moving at a speed of one-half the speed of light?

51. Show that the momentum of a particle can be expressed in the

concise form

p �Ev

c 2

10�13 J?

2.6 � 106 m/s.

8 � 1019 J.

a�x �dv�x

dt�� ax

(1 � V 2�c 2 )3�2

(1 � vxV�c 2 )3

ax � dvx �dt,

vx �c

n� a1 �

1

n2bV

vx� � c�n,

vx

v�y �vy21 � V 2�c 2

1 � vxV2�c 2

*52. What is the percent difference between the Newtonian and

the relativistic values for the momentum of a meteoroid reach-

ing the Earth at a speed of 72 km/s?

*53. Consider three accelerators that produce high-energy parti-

cles: the Large Hadron Collider, which will soon produce pro-

tons with kinetic energy 7 TeV; the Stanford Linear

Accelerator, which produces electrons with kinetic energy 50

GeV; and the Relativistic Heavy Ion Collider, which produces

gold nuclei (mass 197 u) with kinetic energy 20 TeV. In each

case, calculate the difference c � v between the speed of light

and the speed of the particle.

*54. The most energetic cosmic-ray particles have energies of

about 50 J. Assume that such a cosmic ray consists of a

proton. By how much does the speed of such a proton

differ from the speed of light? Express your answer in

meters per second. [Hint: Use the approximation given in

Eq. (36.45)].

*55. Consider the electrons of a speed of 0.999 999 999 67c pro-

duced by the Stanford Linear Accelerator. What is the magni-

tude of the momentum of such an electron?

*56. At the Fermilab accelerator, protons are given kinetic energies

of By how many meters per second does the

speed of such a proton differ from the speed of light? What is

the magnitude of the momentum of such a proton?

*57. A mass M at rest decays into two particles of masses and

Use Eq. (36.51) to show that the magnitude of the

momentum of each of the two particles is

*58. A particle of mass is at rest. A second particle of mass

and kinetic energy K strikes the first particle and sticks to it, a

perfectly inelastic collision. Use Eq. (36.51) to show that the

mass M of the composite particle is

*59. Show that the velocity of a relativistic particle can be

expressed as follows:

**60. At the Brookhaven AGS accelerator, protons of kinetic energy

are made to collide with protons at rest.

(a) What is the speed of a moving proton in the laboratory

reference frame?

(b) What is the speed of a reference frame in which the two

colliding protons have the same speed (and are moving in

opposite directions)?

(c) What is the total energy of each proton in the latter refer-

ence frame?

5.3 � 10�9 J

v �cp

2m2c 2 � p2

M �B(m1 � m2)2 �

2m1K

c 2

m2m1

p �2M 2 � (m1 � m2)

2 2M 2 � (m1 � m2)2 c

2M

m2.

m1

1.6 � 10�7 J.



71. A spaceship travels in the positive x direction with speed

0.80c. A man on Earth makes these observations:

At a photon with is emitted at the

rear of the ship moving toward the front.

At a photon with is

emitted at the front of the spaceship moving toward the rear.

A woman on the spaceship observes the same events.

(a) What time interval does she measure between the two

events? Which happens earlier?

(b) What wavelengths does she measure for the two photons?

(c) What is the length of the ship (as determined by the

woman on it)?

*72. An observer on Earth sees one spaceship traveling away to the

west at speed 0.40c and a second spaceship, also traveling away

but to the east, at 0.70c. Each spaceship emits a signal in its

own reference frame at 2.00 GHz. What frequency does the

Earth observer measure for each signal? What frequency does

each spaceship measure for the signal from the other?

*73. Consider a cube measuring in its

own reference frame. If this cube moves relative to the Earth

at a speed of 0.60c, what are its dimensions in the reference

frame of the Earth? What are the areas of its faces? What is its

volume? Assume that the cube moves in a direction perpendi-

cular to one of its faces.

*74. A spaceship has a length of 200 m in its own reference frame.

It is traveling at 0.95c relative to the Earth. Suppose that the

tail of the spaceship emits a flash of light.

1.0 m � 1.0 m � 1.0 m

l � 600 nmx � 960 m,t � 1.00 ms,

l � 400 nmx � 0,t � 0,

Review Problems 1251

REVIEW PROBLEMS

*68. Muons are unstable particles which—if at rest in a labora-

tory—decay after a time of only Suppose that a

muon is created in a collision between a cosmic ray and an

oxygen nucleus at the top of the Earth’s atmosphere, at an alti-

tude of 20 km above sea level.

(a) If the muon has a downward speed of relative

to the Earth, at what altitude will it decay? Ignore gravity

in this calculation.

(b) Without time dilation, at what altitude would the muon

have decayed?

*69. Suppose that a special breed of cat (Felis einsteinii) lives for

exactly 7.0 years according to its own body clock. When such

a cat is born, we put it aboard a spaceship and send it off at

V � 0.80c toward the star Alpha Centauri. How far from the

Earth (reckoned in the reference frame of the Earth) will the

cat be when it dies? How long after the departure of the

spaceship will a radio signal announcing the death of the cat

reach us? The radio signal is sent out from the spaceship at the

instant the cat dies.

*70. Suppose that a proton speeds by the Earth at along

a line parallel to the axis of rotation of the Earth.

(a) In the reference frame of the proton, what is the polar

diameter of the Earth? The equatorial diameter?

(b) In the reference frame of the proton, how long does the

proton take to travel from the point of closest approach to

the North Pole to the point of closest approach to the

South Pole? In the reference frame of the Earth, how long

does this take?

v � 0.80c

v � 0.990c

2.2 � 10�6 s.

36.7 Mass and Energy

61. How much energy will be released by the annihilation of one

electron and one antielectron (both initially at rest)? Express

your answer in electron-volts.

62. The atomic bomb dropped on Hiroshima had an explosive

energy equivalent to that of 20 000 tons of TNT, or

How many kilograms of rest mass must have

been converted into energy in this explosion?

63. The mass of the Sun is The thermal energy in

the Sun is about How much does the thermal

energy contribute to the mass of the Sun? Express your answer

in percent.

64. Combustion of one gallon of gasoline releases of

energy. How much mass is converted to energy? Compare this

with 2.8 kg, the mass of one gallon of gasoline.

1.3 � 108 J

2 � 1041 J.

2.0 � 1030 kg.

8.4 � 1013 J.

65. The masses of the proton, electron, and neutron are 1. 672 623 �

10�27 kg, 9.11 � 10�31 kg, and

respectively. When a neutron decays into a proton and an elec-

tron, how much energy is released (other than the energy of

the rest mass of the proton and electron)? Compare this extra

energy with the energy of the rest mass of the electron.

66. From Eqs. (36.42) and (36.50) show that the relativistic

energy and the relativistic momentum are related by

**67. A particle at rest decays spontaneously into a particle

and a particle. What will be the speed of each of the latter?

The mass of the is and the masses of the

and particles are each.2.49 � 10�28 kgp�p�

8.87 � 10�28 kg,K0

p�

p�K0

E2 � c 2p2 � m2c4

1.674 929 � 10�27 kg,


(a) In the reference frame of the spaceship, how long does the

light take to reach the nose?

(b) In the reference frame of the Earth, how long does this

take? Calculate the time directly from the motions of the

spaceship and the flash of light; then compare it with the

results calculated by applying the Lorentz transformations

to the result obtained in (a).

75. Suppose that a spaceship is moving at a speed of

relative to the Earth and a meteoroid is moving at a speed of

relative to the Earth; in the same direction as the

spaceship. What is the speed of the meteoroid relative to

the spaceship according to Eq. (36.36)? What is the percent

difference between this relativistic result and the Galilean

result?

76. A collision between two gamma rays creates an electron and

an antielectron that travel away from the point of creation in

opposite directions, each with a speed of 0.95c in the labora-

tory. What is the speed of the antielectron in the reference

frame of the electron, and vice versa?

*77. A spaceship traveling at 0.70c away from the Earth launches a

projectile of muzzle speed 0.90c (relative to the spaceship).

What is the speed of the projectile relative to the Earth if it is

launched in the forward direction? In the backward direction?

*78. Three particles are moving along the positive x axis, in the

positive direction. The first particle has a speed of 0.60c rela-

tive to the second, the second has a speed of 0.80c relative to

the third, and the third has a speed of 0.50c relative to the lab-

oratory. What is the speed of the first particle relative to the

laboratory?

79. What is the kinetic energy of a spaceship of rest mass 50

metric tons moving at a speed of 0.50c? How many metric

tons of matter–antimatter mixture would have to be consumed

to make this much energy available?

80. A particle has a kinetic energy equal to its rest-mass energy.

What is the speed of this particle?

81. Suppose that a spaceship traveling at 0.80c through our Solar

System suffers a totally inelastic collision with a small mete-

oroid of mass 2.0 kg.

v�x

vx � 0.10c

V � 0.20c

(a) What is the kinetic energy of the meteoroid in the refer-

ence frame of the spaceship?

(b) In the collision all of this kinetic energy suddenly becomes

available for inelastic processes that damage the spaceship.

The effect on the spaceship is similar to an explosion.

How many tons of TNT will release the same explosive

energy? One ton of TNT releases

*82. At the Brookhaven AGS accelerator, protons of kinetic energy

are made to collide with protons at rest.

(a) What is the speed of one of these moving protons in the

laboratory reference frame?

(b) What is the magnitude of the momentum?

*83. At the SSC accelerator that was to be built in the United

States, protons would have been given kinetic energies of

What is the value of c � v for such a proton,

that is, by how many meters per second does the speed differ

from the speed of light?

*84. Free neutrons decay spontaneously into a proton, an electron,

and an antineutrino:

The neutron has a rest mass of the

proton, the electron,

and the antineutrino nearly zero. Assume that the neutron is

at rest. Other than the rest-mass energy of the proton and

electron, what is the energy released in this decay?

**85. A particle moving at a speed of 0.60c through the labora-

tory decays into a muon and an antimuon.

(a) In the reference frame of the what is the speed of each

muon? The mass of the is and the

masses of the muon and the antimuon are 1.88 � 10�28

kg each.

(b) Assume that the muon moves in a direction parallel to the

original direction of motion of the and that the

antimuon moves in the opposite direction. What are the

speeds of the muon and the antimuon with respect to the

laboratory?

K0

8.87 � 10�28 kg,K0

K0,

K0

9.11 � 10�31 kg;1.6726 � 10�27 kg;

1.6749 � 10�27 kg;

n S p � e � n

3.2 � 10�6 J.

5.3 � 10�9 J

4.2 � 109 J.


Answers to Checkups

Checkup 36.1

1. Yes. Sound waves propagate in a medium (air), and thus the

speed of sound waves relative to you depends on your speed

relative to the air. The speed is largest for sound waves travel-

ing opposite to your motion, from the front to the back of the

convertible automobile.

2. An ether wind due only to rotation means that the Earth

remains translationally at rest in the ether (the ether moves

with the Earth). The original experiment had a sensitivity of

about 5 km/s, and thus could not detect an ether wind due

only to the rotational speed of 0.46 km/s.

3. Such an observation would have led to the conclusion that the

Earth has an absolute motion of 30 km/s relative to the ether.

Since this is the same as the velocity of the Earth relative to

the Sun, such a result, if observed year round, would have also

implied that the Sun is at rest with respect to the ether.


3. Yes. Since the volume decreases with increasing speed, the

density, or mass per unit volume, increases with increasing

speed.

4. (B) 50 m. The 100-m track is contracted by the factor

that is, to a length of

Checkup 36.5

1. The Lorentz transformation equations will be the same as

Eqs. (36.26)–(36.28), but now with a negative value of V.

2. We know the Lorentz transformation equations are consistent

with a speed of light that is unchanged because they lead

directly to the relativistic velocity combination law, Eq.

(36.36), which gives when

3. This is the same as if the Earth is moving away from the

spaceship, that is, V is now positive, and so the relativistic

velocity combination rule gives

4. (B) 4c�5. As in Example 7, we can obtain the relative speed

from the velocity combination rule, Eq. (36.36): v�x �

[(c�2) � (�c�2)]�

Checkup 36.6

1. Yes; other than for speed the relativistic value of the

momentum, given by Eq. (36.42), is always larger than the

Newtonian value, by the factor

2. Yes, the momentum vector p is proportional to the velocity v,

and so p is always in the same direction as v.

3. For any mass, attaining the speed of light would require infi-

nite kinetic energy, and this is impossible.

4. (A) 0. With in the first term of Eq. (36.44), the two

terms in the realistic kinetic energy cancel.

Checkup 36.7

1. From we have

This energy can’t be converted to useful forms of

energy, except by annihilation, which would require an equal

amount of (unavailable) antimatter.

2. Yes. For example, the mass of a warm container of gas is

greater than the mass of a cold container of gas, due to the

additional kinetic energy

3. No. This is only true when For example, for

the energy becomes very nearly proportional to the momen-

tum.

4. (B) Neutron. Since the neutron produces a proton, an elec-

tron, and some kinetic energy, its total energy (its mass) must

be greater than the hydrogen atom, which requires added

energy just to separate the proton and electron.

v � 0.99c,v � c.

1016 J.

1.0 kg � (3.00 � 108 m/s)2 � 9.0 �E � mc2

v � 0

1�21 � v2�c2.

v � 0,

[1 � (1�2) � (�1�2)]� c�(5�4)� (4�5)c.

(1 � 0.80 � 0.40) � 0.59c.

v�x � (0.80c � 0.40c)�

vx � c.v�x � c

0.50 � 100 m � 50 m.

21 � V 2�c 2 � 10.25 � 0.50,

Answers to Checkups 1253

4. (C) At a different time of year. A single null result could have

implied that the Earth was (coincidentally) nearly at rest with

respect to the ether. Repeating the experiment when the

Earth’s velocity was in a different direction ensures that this

was not the case.

Checkup 36.2

1. The speed of light in vacuum always has the same value,

2. No; the clocks shown in Fig. 36.10 are as observed from the

Earth. For the crew or anyone in the reference frame of the

spaceship, the clocks are all synchronized.

3. Relative to a spaceship traveling westward, the Earth reference

frame is traveling eastward. Thus, New York is at the leading

edge of this reference frame, so the New York clocks, and the

New York earthquake, are late.

4. (D) c. The speed of an electromagnetic wave is the speed of

light; it does not depend on the speed of the emitter or

receiver.

Checkup 36.3

1. Relativity of synchronization means that the times indicated

by different clocks in a moving reference frame are different

but all these clocks run at the same rate; relativity of rates

means that the rate of the clocks in the moving reference

frame differs from that of the clocks on the ground.

2. At low speeds, a factor of 2 increase in velocity has a negligible

effect on the time-dilation factor, which remains nearly equal

to unity. At higher speeds, the time-dilation factor increases

more quickly; for example, an increase from 0.45c to 0.90c

increases the time-dilation factor from 1.1 to 2.3, more than a

factor of 2, and higher speeds result in larger-factor increases.

3. An approaching receiver is the same as an approaching emit-

ter, since only relative motion matters; and the lower relation

in Eq. (36.13) applies. Thus the frequency increases. So the

wavelength decreases, and you detect blue light.

4. (C) Slow; slow. The time-dilation effect is symmetric, so

observers in each reference frame measure a clock in the other

reference frame to be running slow.

Checkup 36.4

1. Like the shape of the Earth moving at high speed relative to

the reference frame of the proton in Example 4, the cannon-

ball is an ellipsoid, flattened along the direction of motion, in a

reference frame relative to which it is moving at 0.5c.

2. No; contraction along the direction of motion does not affect

which pipe fits inside which. Moreover, any apparent contra-

diction can be explained in terms of differences in simultaneity

at any two different positions along the direction of motion.

3.00 � 108 m/s.



The Sun emits thermal radiation consisting of electromagnetic waves of

many wavelengths. For the Sun and many hot bodies, there are universal fea-

tures in the distribution of the emitted wavelengths. The explanation of

the wavelength distribution of such thermal radiation requires the revolu-

tionary idea that electromagnetic waves are made up of small, indivisible

packets of energy, called quanta of light, or photons.

Our study of thermal radiation will enable us to ask:

? How do we know the temperature of the surface of the Sun?

(Section 37.1, page 1257)

? At what wavelength is the Sun’s thermal radiation maximum?

(Example 1, page 1261)

? How many photons are there in the sunlight that reaches the surface

of the Earth? (Example 4, page 1264)

Quanta of Light37

37.1 Blackbody Radiation

37.2 Energy Quanta

37.3 Photons and thePhotoelectric Effect

37.4 The Compton Effect

37.5 X Rays

37.6 Wave vs. Particle

C H A P T E R

1254

Conceptsin

Context

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I n Chapter 35 we examined the wave properties of light. We saw that light exhibits

interference and diffraction, in agreement with Maxwell’s theory, according to which

light is a wave consisting of oscillating electric and magnetic fields with a smooth,

continuous density of energy. In this chapter we will see that light has particle prop-

erties. We will discuss experimental evidence that establishes that a light beam consists

of a stream of discrete, particle-like energy packets. These energy packets are called

quanta of light or photons.

The discovery of the quantization of light by Max Planck in 1900 initiated the

modern era in physics. Physicists quickly came to recognize that quantization of energy

is a pervasive feature of the atomic and subatomic realm. The energies of the atoms and

the energies of the subatomic particles—electrons, protons, and neutrons—are quan-

tized. As we will discuss in the next chapters, such a quantization of energy is in con-

flict with Newton’s laws, and physicists had to find new laws that govern the behavior

of atoms and of subatomic particles.The new theory that governs the realm of the atom

is called quantum physics. In contrast, the old theory of Newton is called classical

physics.

The fact that light has the dual attributes of wave and of particle indicates that

neither the classical wave nor the classical particle concept we have used in earlier

chapters gives an adequate description of light. We have to think of light as a wave–

particle object, which sometimes behaves pretty much as a wave, sometimes pretty

much as a particle, and sometimes as a bit of both. Furthermore, we will see in the

next chapter that electrons, protons, neutrons, and all the other known “particles” also

exhibit such dual attributes of wave and of particle.

37.1 BLACKBODY RADIAT ION

The first hint of a failure of classical physics emerged around 1900 from the study of

thermal radiation. When we heat a body to high temperature, it gives off a glow. For

instance, when we heat a bar of iron to 1200 or 1300 K, it glows in a bright orange or

yellow color. This glow is thermal radiation (“radiant heat”). The color of the thermal

radiation depends on the temperature; an extremely hot iron bar glows in a nearly

white color (“white-hot”); at an intermediate temperature it glows yellow; and at a

lower temperatures it glows orange and then bright red to dull red (Fig. 37.1). You

can observe this change of color with temperature by turning on the heat-

ing coil of a kitchen range; the coil first glows dull red, and then orange, but

it never reaches white-hot. Bodies at room temperature also emit thermal

radiation, but the glow is infrared, and not visible to the eye (see the ther-

mogram of Fig. 37.2). The spectrum of thermal radiation is continuous—if

we analyze the light emitted by a glowing body with a prism, we find that

the energy of the light is smoothly distributed over all wavelengths.

For a quantitative description of the distribution of energy over differ-

ent wavelengths, we plot the energy flux (or the power per unit area) radi-

ated by the surface of the glowing body vs. the wavelength of the radiation.

We can think of such a plot as the intensity distribution seen in the spec-

trum that an (ideal) prism produces when we use it to analyze thermal radi-

ation. Measurements of the thermal radiation emitted by a glowing body

show that the energy flux at very long and at very short wavelengths is

quite small, and that the energy flux has a maximum at some intermediate

wavelength.The location of this maximum depends on the temperature. For

quantum physics vs. classical physics

FIGURE 37.1 A hot, glowing iron bar.

The tip is at the highest temperature

(“white-hot”), and the temperature gradu-

ally decreases along the bar.


example, Fig. 37.3 gives plots of the distribution of energy flux radiated by glowing

bodies at 1000 K, 1250 K, and 1450 K. By comparing these plots, we see that an increase

of temperature produces more radiation at all wavelengths (the 1450-K curve is every-

where higher than the other curves); and we also see that an increase of temperature

shifts the location of the maximum to shorter wavelengths (the peak of the 1450-K

curve is located at a shorter wavelength than the peaks of the other curves).

The thermal radiation emerging from the surface of a glowing body is generated

within the volume of the body by the random thermal motions of atoms and electrons.

Before the radiation reaches the surface and escapes, it is absorbed and re-emitted many

times and it attains thermal equilibrium with the atoms and electrons. This equilibra-

tion process distributes the radiation continuously over all wavelengths and shapes its

spectrum, completely washing out all of the original spectral features that the radiation

had when first emitted by the atoms in the body.

1256 CHAPTER 37 Quanta of Light

Exposed skin radiates morethan clothed parts of bodies.

20°C20°C

28°C

FIGURE 37.2 A thermogram. Each

“false” color represents a different wave-

length of infrared electromagnetic radiation;

unlike the visible spectrum, the brightest

colors here are hottest (white, then red,

yellow, green, blue, p).

energy flux(arbitrary units)

classical 1450 Kclassical 1450 K

1450 K

1250 K

1000 K

20000 4000 6000 nmwavelength

Higher temperature shifts location of peak to shorter wavelength.

Higher temperature produces moreintensity at all wavelengths.

FIGURE 37.3 Distribution of the energy flux

in the spectra of thermal radiation emitted by

glowing bodies at 1000 K, 1250 K, and 1450 K.

The maxima (peaks) of these curves lie in the

infrared region. The maximum is at 2900 nm for

1000 K, at 2300 nm for 1250 K, and at 2000 nm

for 1450 K. The dashed curve gives the predic-

tion of classical physics according to Rayleigh’s

calculation.



The flux of thermal radiation emerging from the surface of a glowing body depends

to some extent on the characteristics of the surface. The surface usually permits the

escape of only a fraction of the flux reaching it from the inside of the body.

Correspondingly, if the body is irradiated with an equal flux of thermal radiation from

the outside, the surface permits the ingress of only an equal fraction of this flux, reflect-

ing the rest. This equality of the emissive and absorptive characteristics of the surface

can be deduced by an argument based on thermodynamics. Thus we are led to a gen-

eral rule: a good absorber is a good emitter; and a poor absorber is a poor emitter. With this

rule we can understand how the silvered, mirrorlike glass walls of thermos bottles or

dewars provide such excellent thermal insulation. These bottles are constructed with

a double glass wall, and the space between these walls is evacuated (Fig. 37.4). Heat

cannot flow across the evacuated space by conduction or convection; it can only flow

by radiation. To inhibit radiation, the glass walls are silvered and thereby made into

mirrorlike reflecting surfaces; these highly reflective surfaces are very poor absorbers

and emitters of radiation. This keeps the heat transfer between the walls very small.

A body with a perfectly absorbing (and emitting) surface is called a blackbody;

when such a body is cold and emits no radiation of its own, it looks black because it

does not reflect any of the illumination reaching it from the outside. But when a black-

body is hot, its surface emits more thermal radiation than any other hot body at the same

temperature. A body covered with black soot is an approximate blackbody. Experimental

physicists prefer to achieve the characteristics of an ideal blackbody by a trick: take a

body with a cavity, such as a hollow cube, and drill a small hole in one side of the cube

(Fig. 37.5). The hole then acts as a blackbody—any radiation incident on the hole

from outside will be completely absorbed. Because of this equivalence between a black-

body and a hole in a cavity, the terms blackbody radiation and cavity radiation are used

interchangeably. The curves plotted in Fig. 37.3 are based on measurements of radi-

ation emerging from a small hole in a body with a cavity; thus, these curves represent

the spectra of blackbody radiation.

The Sun is an almost perfect blackbody radiator. A very small fraction of the Sun’s

radiation is due to specific chemical features of the Sun; the overwhelming majority of

sunlight makes up a spectrum of thermal radiation that precisely follows the shape

given in Fig. 37.3. As we have seen, a blackbody spectrum depends on the tempera-

ture of the blackbody, so we can tell the temperature of the surface of the Sun from

the spectrum of thermal radiation that the Sun sends to the Earth.

FIGURE 37.4 A thermos bottle.

Since any radiation entering a small hole suffers multiplereflections and is trapped,…

…the hole is therefore a perfect absorber.

FIGURE 37.5 A cavity with a small hole.

blackbody radiation or cavity radiation

blackbody

Conceptsin

Context

silveredglass

Vacuum layer preventsheat flow by conductionor convection.

Highly reflectingsurfaces are poorabsorbers of radiation.


The blackbody plays a special role in the study of thermal radiation because the

spectrum of its thermal radiation does not depend on the material of which it is made

or on any other characteristics of the body. By an argument based on thermodynamics,

it can be established that the spectrum depends only on the temperature of the blackbody. The

mathematical formula for the distribution of energy in this spectrum is therefore a

universal law, and in the last years of the nineteenth century, physicists engaged in an

intensive experimental and theoretical effort to find this universal law of blackbody

radiation.

Checkup 37.1

QUESTION 1: A wood-burning stove is made of cast iron. How would the perform-

ance of the stove change if it were made of polished stainless steel?

QUESTION 2: One house has a roof of dark-colored shingles; another has a roof of

light-colored shingles. Which roof absorbs more heat during the day, in sunlight?

QUESTION 3: For space “walks,” astronauts wear suits with a shiny, silvery surface layer.

What is the purpose of this surface layer?

QUESTION 4: Two ideal cavities emit thermal radiation. The spectrum emitted by the

first cavity is most intense at a wavelength of 600 nm, and that of the second at 500 nm.

Compared with the second cavity, the radiation emitted by the first cavity is

(A) More intense at all wavelengths

(B) More intense at short wavelengths and less intense at long wavelengths

(C) Less intense at short wavelengths and more intense at long wavelengths

(D) Less intense at all wavelengths

37.2 ENERGY QUANTA

Before 1900, physicists made several attempts at a theoretical explanation of the dis-

tribution of energy in the spectrum of blackbody radiation, but they met with disas-

ter. One of the best of these attempts was that of Lord Rayleigh. Since the energy flux

of the radiation emerging from the hole in a cavity is directly proportional to the energy

density of the radiation inside the cavity [compare Eqs. (33.17) and (33.22)], Rayleigh

decided to calculate the latter quantity. He began by noting that the radiation in a

cavity is made up of a large number of standing electromagnetic waves. Figure 37.6

shows some of these standing waves; they are analogous to the standing waves in an

organ pipe closed at both ends. Each of these standing waves can be regarded as a

mode of vibration of the cavity. Rayleigh then appealed to the equipartition theorem,

according to which, at thermal equilibrium, each mode of vibration has an average

thermal energy of kT, where k is Boltzmann’s constant (in Section 19.4 we stated a

special case of the equipartition theorem for free translational or rotational motion of

a gas molecule).Thus, each of the standing waves of Fig. 37.6 ought to have an energy

kT, and from this we can calculate the energy distribution in the spectrum of the radi-

ation (the dashed curve in Fig. 37.3 shows the energy distribution obtained from

Rayleigh’s calculation). Although this calculation gave reasonable results at the

long-wavelength end of the blackbody spectrum, it gave disastrous results at the short-

wavelength end: the number of possible standing-wave modes of very short wave-

length is infinitely large, and if each of these modes had an energy kT, the total energy

✔




ultraviolet catastrophe

x

x

Ey

x

y

(a)

0 0 0

Ey

x

(b)

Ey

(c)

L

L

L

L

Radiation in cavity is made upof a large number of standingelectromagnetic waves.

There are few modesof long wavelength…

…and an infinite number of possiblemodes of shorter wavelength.

FIGURE 37.6 Some of the possible standing

electromagnetic waves in a closed cavity. For the

sake of simplicity, only waves with a horizontal

direction of propagation are shown. The plots give

the electric fields as a function of x at one instant

of time.

in the cavity would be infinite! This disastrous failure of classical physics has been

called the ultraviolet catastrophe.

The correct formula for the distribution of energy in the spectrum of blackbody radi-

ation was finally obtained by Max Planck in 1900. By some inspired guesswork, based

on thermodynamics, Planck found a mathematical formula that gave a precise fit to the

experimental curves of blackbody radiation, such as those plotted in Fig. 37.3. He then

searched for a theoretical explanation for his formula. The search led Planck to a rev-

olutionary discovery: the quantization of energy. This discovery was to bring about

the overthrow of classical physics and the birth of quantum physics. For Planck, the pos-

tulate of the quantization of energy was “an act of desperation,” which he committed

because “a theoretical explanation had to be found at any cost, whatever the price.”

Planck’s derivation of the blackbody radiation formula involves some sophisticated

statistical mechanics which we will not reproduce. We will merely give a sketchy out-

line of this derivation. Planck began by making a theoretical model of the walls of the

cavity: he regarded the atoms in the walls as small harmonic oscillators of many dif-

ferent natural frequencies, that is, small masses (with electric charges) attached to

springs of many different spring constants. Although this is a rather crude model of the

atoms that make up the walls of the cavity, it was adequate for his purposes since, as

described in the preceding section, the radiation in a cavity is known to be completely

independent of the characteristics of the wall. The random thermal motions of the

oscillators result in the emission of electromagnetic radiation. This radiation fills the

cavity and acts back on the oscillators. When thermal equilibrium is attained, the aver-

age rate of emission of radiation energy by the oscillators matches their rate of absorp-

tion of radiation energy. Thus, the oscillators share their energy with the radiation in

the cavity, and Planck was able to show that, under equilibrium conditions, the aver-

age radiation energy at some frequency f (or at a wavelength is directly pro-

portional to the average energy of an oscillator of frequency f.

These steps of Planck’s calculation involved nothing but classical mechanics. But

in the next step of the calculation, Planck departed radically from classical physics.

l � c�f )


He postulated that the energy of the oscillators is quantized according to the follow-

ing rule: In an oscillator of frequency f, the only permitted values of the energy are

(37.1)

All other values of the energy are forbidden. The constant h in Eq. (37.1) is a new fun-

damental constant, called Planck’s constant. The value of this constant is

(37.2)

The energy hf is called an energy quantum; according to the quantization rule, the

energy of an oscillator is always some multiple of the basic energy quantum hf :

(37.3)

The integer n is called the quantum number of the oscillator.

With this quantization condition, Planck calculated the average energy of the

oscillators; and from that he derived his formula for the energy flux radiated by a black-

body per unit wavelength,

(37.4)

The energy flux per unit wavelength means that if is a small interval of wavelength

centered on a given wavelength then the energy flux dS of electromagnetic waves that

have wavelengths in the interval is

The distribution function (37.4) agrees with the experimentally measured distribu-

tions shown in Fig. 37.3. Although we cannot go into the details of this derivation,

we can achieve a rough understanding of how Planck’s calculation avoids the ultravi-

olet catastrophe. The thermal energy of the walls of the cavity is shared at random

among all the oscillators in these walls. Some of these oscillators have high frequen-

cies, some have low frequencies. For an oscillator of very high frequency, the energy quan-

tum hf is very large. If this oscillator is initially quiescent it cannot begin to

move unless it acquires one energy quantum; but since this energy quantum hf is very

large, the random thermal disturbances will be insufficient to provide it—the oscilla-

tor will remain quiescent. Thus, the quantization of energy tends to inhibit the ther-

mal excitation of the high-frequency oscillators. If the high-frequency oscillators remain

quiescent, they will not supply energy to the corresponding high-frequency standing

waves in the cavity.These waves will then not have the energy kT predicted by Rayleigh;

instead they will have no energy at all. This avoids the ultraviolet catastrophe.

Note that for an oscillator with a frequency of which is typical for atomic

vibrations, the energy quantum is

Since this is a very small amount of energy, quantization does not make itself felt at a

macroscopic level. But quantization plays a pervasive role at the atomic level.

Unfortunately, Planck could not offer any basic justification for his postulate of

quantization of energy. His postulate gave him a blackbody radiation law which was

in complete agreement with the experimentally measured distribution of energy over

wavelength (as displayed in Fig. 37.3), but his postulate brought him into conflict with

classical physics. Superficially, the quantization of energy is analogous to the quanti-

zation of electric charge—we know from Chapter 22 that the electric charge of any

� 6.6 � 10�19 J.hf � 6.6 � 10�34 J�s � 1015 Hz

f � 1015 Hz,

(n � 0),

dS � Sl dl

dl

l,

dl

Sl �2phc2

l5 � (e hc�lkT � 1)

n � 0, 1, 2, 3, …E � nhf

h � 6.63 � 10�34 J�s

E � 0, hf, 2hf, 3hf, …


Planck’s constant

energy quantum hf

energy of oscillator

quantum number n

MAX PLANCK (1858–1947) German

theoretical physicist, professor at Berlin and

President of the Kaiser Wilhelm Institute

(later renamed the Max Planck Institute).

Planck made important contributions to

thermodynamics before he became involved

with the problem of blackbody radiation. He

was deeply troubled by the quantization of

energy, because he recognized that this held

disastrous consequences for classical mechanics

and electromagnetism. He was awarded the

Nobel Prize in 1918 for his discovery of

energy quanta.


particle or body is always some multiple of the fundamental charge e. However, whereas

the quantization of charge is consistent with the laws of classical physics, the quanti-

zation of energy is inconsistent with these laws.The energy of any oscillator that obeys

Newton’s laws—such as a mass on a spring—can be changed by a small amount by

pushing on the oscillator with a very weak force, and it should therefore be possible to

change the energy of the oscillator continuously by any amount we please, not just in

discrete steps of one energy quantum. Thus, quantization of energy makes no sense

in classical physics. Planck could only justify his postulate by its consequences; but, in

theoretical physics, the end does not justify the means. A deeper explanation of the

quantization of energy emerged only much later, with the development of quantum

mechanics (see Chapter 38).

From Planck’s formula (37.4) for the distribution of energy in the spectrum of black-

body radiation one can establish that the energy flux has a maximum at a wavelength

(37.5)

where c is, as always, the speed of light, k is Boltzmann’s constant (see Section 19.1),

h is Planck’s constant, and T is the absolute temperature of the blackbody. Equation

(37.5) is called Wien’s displacement Law. If we insert the numerical values of h, c,

and k, Wien’s Law takes the simple form

(37.6)

Wien’s Law asserts that is inversely proportional to the temperature T.This means

that an increase of temperature shifts the location of the maximum to shorter wave-

lengths, in agreement with the experimental results presented in Fig. 37.3. If the tem-

perature is sufficiently high—6000 K or so—the maximum of the spectrum lies in the

visible region. For instance, the Sun, with a surface temperature of 5800 K, emits its

largest flux of thermal radiation in the visible region.

According to Wien’s Law, at what wavelength does the thermal

radiation emitted by the Sun have its maximum? At what wave-

length does the thermal radiation emitted by the tungsten filament in a lightbulb

have its maximum? Assume that the Sun and the tungsten filament are approxi-

mately blackbodies at temperatures of 5800 K and 3200 K, respectively.1

SOLUTION: With Wien’s Law gives

and with it gives

lmax �2.90 � 10�3 m�K

3200 K� 9.1 � 10�7 m � 910 nm

T � 3200 K,

lmax �2.90 � 10�3 m�K

5800 K� 5.0 � 10�7 m � 500 nm

T � 5800 K,

EXAMPLE 1

lmax

lmax �2.90 � 10�3 m�K

T

lmax �1

4.965 hc

k�

1

T


Wien’s displacement Law

Conceptsin

Context

1The blackbody approximation is fairly good for the Sun (except at sunspots; see the chapter photo). But it

is not good for an ordinary tungsten filament, because the tungsten surface is not a good absorber.The black-

body approximation becomes better if the filament is tightly coiled, like a solenoid, and if we examine the radi-

ation in the interior of this coil (a cavity).


Thus, the maximum of the thermal radiation from the Sun lies in the visible region

(in the green), but the maximum of the thermal radiation from the lightbulb lies

in the infrared.

Furthermore, from Planck’s formula (37.4) one can calculate the combined energy

flux for all wavelengths radiated from the surface of a blackbody by integration of the

intensity per unit wavelength, over all wavelengths.This gives a total intensity pro-

portional to the fourth power of the temperature,

(37.7)

This is called the Stefan–Boltzmann Law. The constant of proportionality in this

law can, again, be expressed in terms of h, c, and k. With the appropriate numerical

values, the Stefan–Boltzmann constant then has the value

(37.8)

Both the Wien and Stefan–Boltzmann laws had been discovered empirically many

years before Planck supplied their theoretical foundation. Note that the energy flux S

in Eq. (37.7) depends only on the temperature of the blackbody; it does not depend

on the material from which the body is made. Since S is the energy flux, or the power

per unit area, we can obtain the net power radiated from the body by multiplying S

by the surface area; thus, the net power depends on the size of the body.

On a clear night, the surface of the Earth loses heat by radia-

tion. Suppose that the temperature of the ground is and

that the ground radiates like a blackbody. What is the rate of loss of heat per square

meter?

SOLUTION: The temperature of the ground is so the absolute temperature

of the ground is 283 K. Hence, the Stefan–Boltzmann Law tells us that the radi-

ated flux, or power per unit area, is

COMMENTS: This large radiative heat loss of the ground explains the sharp drop

of temperature experienced during clear nights. The drop of temperature is much

less severe if there is an overcast sky. The clouds then reflect most of the radiation

back to the ground—they act like a blanket to keep the ground warm.

In a house, a room is heated by means of a radiator filled with

hot water at The radiator consists of a large vertical panel.

If you place your hand near the panel (see Fig. 37.7a), what is the rate at which

thermal radiation is incident on your hand? The area of one side of your hand is

SOLUTION: When your hand is very near the panel, the energy flux reaching your

hand is the same as the energy flux, or intensity, emitted by the radiator panel.The

power incident on your hand is the incident flux times the area A of your hand,

0.016 m2.

82�C.EXAMPLE 3

S � sT 4 � 5.67 � 10�8 W/(m2�K4) � (283 K)4 � 364 W/m2

10�C,

10�CEXAMPLE 2

s � 5.67 � 10�8 W�(m2�K4)

s

S � sT 4

Sl,


Stefan–Boltzmann Law


COMMENTS: The distance at which you place your hand from the panel is not cru-

cial. If your hand is somewhat farther away (see Fig. 37.7b), then some of the radi-

ation emitted by the portion of the panel directly facing your hand will miss your

hand, but an equal amount of radiation from a more distant portion of the panel

will now be able to reach your hand, and these amounts compensate. You can do

a simple experiment to verify this: place your hand very near a radiator panel (or

near a stove), and gradually move your hand away—you will hardly notice any dif-

ference for the first few centimeters of motion. For a radiator panel of finite size,

the energy flux incident on your hand will decrease once you move far enough

away. But if you are in a room all of whose walls, ceiling, and floor are lined with

radiator panels, a compensation similar to that depicted in Fig. 37.7b is valid

throughout the room—in such an environment, your hand receives the same flux

no matter where it is located or in what direction it faces.

Checkup 37.2

QUESTION 1: The caption for Fig. 37.3 gives the wavelengths of the maxima for dif-

ferent temperatures. What is the product of wavelength and temperature in each case?

Is this in agreement with Wien’s Law?

QUESTION 2: If we increase the temperature of a blackbody from 1000 K to 2000 K,

by what factor do we change the wavelength of the maximum of the spectrum? By

what factor do we change the total energy radiated from the surface?

QUESTION 3: An oscillator has a quantized energy hf. Is its kinetic energy quantized?

Is its potential energy quantized?

QUESTION 4: The pendulum of a clock can be regarded as an oscillator. Is the energy

of the pendulum quantized? Why don’t we notice the quantization?

QUESTION 5: Suppose that Planck’s constant was instead of

Would we notice the quantization of a pendulum?

QUESTION 6: Figure 37.8 shows photos of three stars.The light emitted by these stars

is thermal radiation. Which of these stars is the hottest? The coolest?

(A) Red; yellow (B) Red; blue (C) Yellow; blue

(D) Blue; red (E) Blue; yellow

6.6 � 10�34 J�s.

6.6 � 10�2 J�s

✔

� (82 � 273)4 K4 � 0.016 m2 � 14 W� 5.67 � 10�8 W/(m2�K4)

P � S A � sT 4 � A


(a)

(b)

Intensity reaching hand equalsthat emitted by radiator panel.

Some intensity from nearestpart of panel misses hand,…

Hand issomewhat awayfrom panel.

…but is mostly com-pensated by radiationfrom other parts.

Hand is verynear panel.

FIGURE 37.7 (a) Hand placed near a flat

radiator panel. (b) Hand placed farther away.

FIGURE 37.8 Betelgeuse, Bellatrix, and

Rigel (red, yellow, and blue, respectively) in

the constellation Orion.

Different colors of thermal radiation from stars indicate different surface temperatures.


37.3 PHOTONS AND THEPHOTOELECTRIC EFFECT

In 1905, Einstein showed that Planck’s formula could be understood much more simply

in terms of a direct quantization of the energy of the radiation. Planck had postulated

that the oscillators in the wall of the cavity have discrete quantized energies, but he

had treated the electromagnetic radiation as a smooth, continuous density of energy,

exactly as it is supposed to be according to classical electromagnetic theory.

In contrast, Einstein proposed that electromagnetic radiation consists of discrete

particle-like packets of energy. He regarded a wave of some given frequency f as a

stream of more or less localized energy packets, each with one quantum of energy hf

(see Fig. 37.9). The particle-like energy packets of magnitude hf are called photons.

The wave then has an energy hf if it contains only one photon, 2hf if it contains two

photons, and so on. The thermal radiation in a cavity, with waves traveling randomly

in all directions, can then be regarded as a gas of photons. Einstein applied statistical

mechanics to calculate the energy spectrum of this gas and he thereby obtained the

energy spectrum of the cavity radiation.

The essential difference between Planck’s and Einstein’s view of the cavity radia-

tion is that Planck quantized only the exchange of radiation with the walls of the cavity,

whereas Einstein quantized the radiation itself. Thus, in Einstein’s view, electromag-

netic radiation is always quantized, regardless of where or how it is produced. Not only

is the radiation quantized when it is produced by the oscillators in the walls of a cavity

(as in the case of thermal radiation), but also when it is produced outside of a cavity,

say, by the acceleration of electric charges on the antenna of a radio transmitter.


photon

Conceptsin

Context

FIGURE 37.9 (a) According to classical

theory, the energy is smoothly distributed

over an electromagnetic wave, although the

energy density has maxima wherever the wave

reaches maximum amplitude. (b) According

to Einstein, the energy is localized in small

energy packets, which move with the wave.

The energy flux of sunlight reaching the surface of the Earth is

at normal incidence. How many photons

reach the surface of the Earth per square meter per second? For the purposes of

this calculation assume that all the photons in sunlight have an average wavelength

of 500 nm.

SOLUTION: The energy of a photon of wavelength 500 nm is

(37.9)

The energy incident per square meter per second is To obtain the

number of photons, we must divide this by the energy per photon,

That is,

4.0 � 10�19 J.

1.0 � 103 J.

� 6.63 � 10�34 J�s �3.00 � 108 m/s

5.0 � 10�7 m� 4.0 � 10�19 J

E � hf � h c

l

1.0 � 103 W/m2EXAMPLE 4

(a) (b)

cc

Classically, an electromagnetic wave hasa continuously varying energy density.

In quantum theory, wave is madeup of a large number of photons, each with energy E � hf.


For a square meter, this gives

photons per square meter per second

COMMENTS: Because the number of photons in sunlight and other common

light sources is so large, at the macroscopic level we do not perceive the grainy

character of the energy distribution in light.

With this concept of light as a stream of photons, Einstein was also able to offer

an explanation of the photoelectric effect. In some early experiments on the produc-

tion of radio waves by sparks, Hertz had noticed that light shining on an electrode

tended to promote the formation of sparks. Subsequent careful experimental investi-

gations demonstrated that the impact of light on an electrode can eject electrons,

which trigger the sparks. The electrons emerge with a kinetic energy which increases

directly with the frequency of the light.

Figure 37.10 is a schematic diagram of the apparatus used in the investigation of

the photoelectric effect. Light from a lamp illuminates an electrode of metal (C )

enclosed in an evacuated tube. Electrons ejected from this electrode travel to the col-

lecting electrode (A ), and then flow around the external circuit. A galvanometer (G )

detects this flow of electrons. The kinetic energy of the ejected photoelectrons can be

determined by applying a potential difference V between the emitting and the col-

lecting electrodes by means of an adjustable source of emf. With the polarity shown in

the figure, the collector has a negative potential relative to the emitter, that is, the col-

lector (A) exerts a repulsive force on the photoelectrons. If an electron is to travel from

the emitter to the collector, the change in its potential energy between emittere � V

1.0 � 103 J�(s�m2)

4.0 � 10�19 J�photon� 2.5 � 1021

photons

second�

energy�second

energy�photon


G

�

�

VC

A

lamp

Light from lamp illuminates…

…a metal surface, the cathode, which emits electrons.

Electrons can reach a collectingelectrode, the anode,…

…and electron flowis measured by meter.

FIGURE 37.10 Schematic diagram

of the apparatus for the investigation

of the photoelectric effect.

photoelectric effect


and collector must be less than or equal to its initial kinetic energy. If the change in

potential energy exceeds the initial kinetic energy, then the electron will reverse its

motion and return to the emitter. The critical potential Vstop that stops the flow of

electrons from emitter to collector is called the stopping potential. The measured

value of this stopping potential gives us the initial kinetic energy of the electrons:

(37.10)

Experimentally, it is found that the kinetic energy determined in this way increases

directly with the frequency of the incident light. For example, Fig. 37.11 is a plot of

kinetic energy vs. frequency of the light for photoelectrons ejected from sodium. Note

that, according to this plot, if the frequency is below then the light is

incapable of ejecting electrons.

Einstein’s quantum theory of light accounts for these experimental observations

as follows. The electrons in the illuminated electrode absorb photons from the light,

one at a time. When an electron absorbs a photon, it acquires an energy hf. But before

this electron can emerge from the electrode, it must overcome the restraining forces that

bind it to the metal of the electrode.The energy required for this is called the work func-

tion of the metal, designated by The remaining energy of the electron is then

and this must be the kinetic energy of the emerging electron:

(37.11)

Some electrons suffer extra energy losses in collisions within the metal before they

emerge; thus, actually is the maximum possible kinetic energy with which elec-

trons can emerge.

Equation (37.11) is Einstein’s photoelectric equation. It shows that the kinetic energy

does indeed increase directly with the frequency, in agreement with the data of Fig. 37.11.

According to Eq. (37.11), a minimum frequency is required to achieve the ejection

of an electron. This minimum frequency, called the threshold frequency, corresponds

to the ejection of an electron of zero kinetic energy; such an electron is just barely

ejected. The threshold frequency is given by

or

(37.12)fthresh �f

h

0 � hfthresh � f

fthresh

hf � f

K � hf � f

hf � f,

f.

4.4 � 1014 Hz,

K � eVstop


stopping potential

K

f

3 eV

2

1

3 4 5 6 7 8 9 10 11 12�1014 Hz

Kinetic energy increaseslinearly with frequency of light.

A minimum photon energy, the work function, is requiredto emit an electron.

FIGURE 37.11 Kinetic energies

(in electron-volts) of photoelectrons

ejected from sodium by light of dif-

ferent frequencies.

threshold frequency

Einstein’s photoelectric equation

work function �


When an electron absorbs a photon of this frequency, all of the energy of the photon

is used to overcome the restraining forces that bind the electron to the metal, and no

kinetic energy is left over for the ejected electron.

According to Fig. 37.11, the threshold frequency for sodium

is (photons of a frequency below this are inca-

pable of ejecting electrons). What is the work function for sodium? Express the

answer in eV.

SOLUTION: With Eq. (37.12) gives

The work function for platinum is 6.2 eV. If ultraviolet light

of frequency illuminates a platinum electrode,

what is the maximum kinetic energy of the ejected electrons? What is the stop-

ping potential?

S O L U T I O N : With we

find from Eq. (37.11)

The stopping potential is, from Eq. (37.10),

(37.13)

Einstein’s photoelectric equation was verified in detail by a long series of meticulous

experiments by R. A. Millikan (the data in Fig. 37.11 are due to him). In order to obtain

reliable results, Millikan found it necessary to take extreme precautions to avoid con-

tamination of the surface of the photosensitive electrode. Since the surfaces of metals

exposed to air quickly accumulate a layer of oxide, he developed a technique for shav-

ing the surfaces of his metals in a vacuum, by means of a magnetically operated knife.

The results of these experiments gave strong support to the quantum theory of

light. This success of Einstein’s theory was all the more remarkable in view of the fail-

ure of the classical wave theory of light to account for the features of the photoelec-

tric effect. According to classical theory, an electromagnetic wave acts on the electron

by means of its electric field, which exerts a force on the electron. Therefore the cru-

cial parameter that determines the ejection of a photoelectron should be the intensity

of light, since this determines the strength of the electric field in the wave. If an intense

wave strikes an electron, it should be able to jolt it loose from the metal, regardless of

the frequency of the wave. Furthermore, the kinetic energy of the ejected electron

should depend on the intensity of the wave. The observational evidence contradicts

these predictions of the classical theory: A wave with a frequency below the threshold

frequency never ejects an electron, regardless of its intensity. And, furthermore, the

Vstop �K

e�

2.3 � 10�18 J

1.6 � 10�19 C� 14 V

� 2.3 � 10�18 J

K � hf � f � (6.63 � 10�34 J�s � 5.0 � 1015 Hz) � 9.9 � 10�19 J

f � 6.2 eV � 6.2 � 1.6 � 10�19 J � 9.9 � 10�19 J,

5.0 � 1015 HzEXAMPLE 6

� 2.9 � 10�19 J �1 eV

1.6 � 10�19 J� 1.8 eV

f � hfthresh � 6.63 � 10�34 J�s � 4.4 � 1014 Hz � 2.9 � 10�19 J

fthresh � 4.4 � 1014 Hz,

4.4 � 1014 HzEXAMPLE 5



kinetic energy depends on the frequency [as specified by Eq. (37.11)] and not on the

intensity. High-intensity light ejects more photoelectrons, but does not give the indi-

vidual electrons more kinetic energy.

Checkup 37.3

QUESTION 1: The colors of light range from red to violet. What color has the most ener-

getic photons?

QUESTION 2: Platinum has a larger work function than sodium. Qualitatively, how

does the K vs. f plot for platinum differ from that for sodium (Fig. 37.11)?

✔


PHYSICS IN PRACTICE PHOTOMULTIPL IER

FIGURE 1 Schematic diagram of a photomultiplier tube. The

curved electrodes are called dynodes. For the purpose of this dia-

gram, it has been assumed that each electron impact on a dynode

releases two electrons. The arrows show an avalanche of electrons.

photon

electron

�100 V

�300 V

�500 V

�400 V

�200 V

dynode

FIGURE 2 Photomultiplier tubes.

The photoelectric effect finds many practical applications in

sensitive electronic devices for the detection of light. For

instance, in a photomultiplier tube, an incident photon ejects

an electron from an electrode at the faceplate of the tube

(Fig. 1).To convert this single electron into a measurable pulse

of current, electric fields within the tube accelerate this single

electron toward a second electrode (called a dynode; see the

figure), where its impact ejects several secondary electrons.

These, in turn, are accelerated toward a third electrode, where

their impact ejects tertiary electrons, and so on. Thus, the

single electron from the first electrode generates an avalanche

of electrons. In a high-gain photomultiplier tube, a pulse of

electrons emerges from the last electrode, delivering a

pulse of current to an external circuit. In this way, the photo-

multiplier tube can detect the arrival of individual photons.

A similar solid-state device known as an avalanche photodi-

ode is also used for photon counting. Some sensitive video

cameras rely on the same multiplier principle to convert the

arrival of a photon at a photosensitive faceplate into a pulse

of current.This permits these cameras to take pictures in faint

light, where there are few photons.

109


QUESTION 3: Figure 37.12 shows a plot of current vs. applied potential for the photo-

electric current emitted by the surface of metal illuminated with light of a given wave-

length. Qualitatively, why is the current zero if V �Vstop? Why does the current

level off for large positive V ? Why do the curves differ for different intensities of light?

QUESTION 4: Sodium has a work function of 1.8 eV; for platinum, the value is 6.2 eV.

Compared with the plot of the maximum kinetic energy of ejected electrons as a func-

tion of the frequency of the incident light for sodium (Fig. 37.11), a similar plot for plat-

inum has

(A) A larger slope (B) A smaller slope (C) The same slope

37.4 THE COMPTON EFFECT

Very clear experimental evidence for the particle-like behavior of photons was uncov-

ered by Arthur Holly Compton in 1922. Compton had been investigating the scattering,

or the deflection, of X rays by a target of graphite (see Fig. 37.13). According to

Maxwell’s theory, X rays are merely light waves of extremely high frequency. But accord-

ing to quantum theory, they ought to consist of photons; and since their frequency is

much higher than that of ordinary light, the energy of the X-ray photons ought to be

much larger than that of photons of ordinary light (X rays will be discussed in more

detail in the next section).

When Compton bombarded the graphite with X rays of one selected wavelength,

he found that the scattered X rays had wavelengths somewhat larger than that of the

incident X rays. Classical theory cannot explain such a change of wavelength. When

a classical electromagnetic wave is incident on a graphite target, its oscillating electric

fields exert forces on the electrons in the carbon atoms, and this causes the electrons

to oscillate at the same frequency as the wave. The oscillating, accelerated electrons

then radiate a new electromagnetic wave, which spreads outward in all directions.This

radiated wave is the scattered wave. Its frequency is necessarily the same as that of the

oscillating electrons, which is the same as that of the incident wave.

Compton soon recognized that the change of wavelength could be understood in

terms of collisions of photons with electrons, collisions in which the photons behave

like particles. In such a collision the electron of a carbon atom can be regarded as free,

because the force binding the electron to the atom is insignificant compared with the

force exerted by the incident photon. When the photon bounces off the electron, the

electron recoils and thereby picks up some of the photon’s energy—the deflected photon


�

target

detector

incidentX rays

scatteredX rays

A thin beam of X rays approaches target…

…and is deflected (scattered) in several directions by target.

Detector measures X raysemerging at an angle �.

FIGURE 37.13 Scattering of X rays by a target of graphite.

ARTHUR HOLLY COMPTON (1892–1962) American experimental physicist. For

his discovery of the Compton effect, he received

the Nobel Prize in 1927.

V

�Vstop

I

(a)

(b)

For each intensity of light, current is zerofor V �Vstop…

…and levels offfor high V.

Current is smaller forlower intensity of light.

FIGURE 37.12 I vs. V for two different

values of the intensity of light: (a) high

intensity; (b) low intensity.


is left with a reduced energy. Since the energy of the photon is a reduc-

tion of the energy of the photon implies an increase of wavelength. Qualitatively, we

expect that photons that suffer the most violent collisions will be deflected through

the largest angles and lose the most energy; and therefore such photons should emerge

with the longest wavelength. And this is just what Compton found in his experiments.

For a quantitative discussion of the photon–electron collision, we need an expres-

sion for the momentum of the photon. We can derive this expression from the relativistic

formula for the momentum of a particle. Since the speed of the photon is the speed of

light, the photon must be regarded as an ultra-relativistic particle. For such a particle,

Eq. (36.52) tells us that

(37.14)

[Note that according to Eq. (36.50), the energy of a particle with a speed equal to the

speed of light can be finite only if the rest mass m is zero; thus, photons must be

regarded as particles of zero rest mass.] Since the energy of the photon is

the expression for the momentum can be written as

(37.15)

With the expression (37.15) for the momentum of the photon, Compton calcu-

lated the change of energy and the change of wavelength of a photon in an elastic col-

lision with an electron, initially at rest. For a photon that emerges from the collision

at an angle (the deflection angle; see Fig. 37.14), he calculated from the laws of con-

servation of energy and momentum that the change of wavelength is

(37.16)

This change of wavelength is called the Compton shift. The change of wavelength is

always small; it is always less than 0.005 nm (see the next example). Such changes of

wavelength are difficult to detect unless the wavelength of the X rays is itself quite

small, so is an appreciable fraction of the wavelength.¢l

¢l �h

mec (1 � cos u)

u

p �hl

E � hf � hc�l,

p �E

c



Compton wavelength shift

momentum of photon

photon

photon

electron

�

Incident photon collideswith initially stationaryelectron.

Deflection angle � is angle betweenphoton directions of propagationbefore and after collision.

Electron emergesfrom collision.

FIGURE 37.14 Directions of propagation, or directions

of the momentum, of a photon before and after collision

with an electron.


Evaluate the Compton shift for X rays scattered at and for

X rays scattered at

SOLUTION: If the photon emerges from the collision at an angle of then

and the wavelength shift of Eq. (37.16) is

� 0.002 43 nm

If the photon emerges at in a direction opposite to that of its initial direc-

tion of motion, then and the wavelength shift is

This is the maximum wavelength shift that can be produced by the Compton effect.

The formula (37.16) for the wavelength shift can be deduced

by examining the conservation of energy and momentum in

the photon–electron collision. As a simple special case of such a collision, consider

a photon that recoils back in the direction it came from [that is, in Eq.

(37.16)] while the electron, initially at rest, acquires a motion in the forward direc-

tion. For such a straight-line motion of photon and electron, write down the laws

of conservation of energy and momentum, and deduce the wavelength change of

the photon.

SOLUTION: We will use the relativistic expressions for momentum and energy,

because, in a collision with a high-energy photon, the electron will emerge with an

energy large enough to require such a relativistic treatment. In the notation famil-

iar from Chapter 11, we designate the initial momentum of the photon by the

final momentum by likewise, we designate the initial momentum of the elec-

tron by (which is zero, since the electron is initially at rest), the final momen-

tum by Conservation of momentum then tells us

(37.17)

and conservation of energy tells us

(37.18)

where is the initial energy of the electron (the rest-mass energy). To solve

this pair of equations for the final photon energy, we need to express the momenta

in Eq. (37.17) in terms of the energies. For this purpose, we first rewrite Eq. (37.17)

as and square both sides:

(37.19)

Now we substitute the relativistic relation for the electron

[compare Eq. (36.51)] and the relations and for the photon:

aE2cb 2

� m2ec

2 � aE1

c�

E1cb 2

p1 � �E1�cp1 � E1�c

(p2)2 � (E2�c)2 � m2

ec 2

( p2)2 � ( p1 � p1)

2

p2 � p1 � p1

mec 2

E1 � mec 2 � E1 � E2

p1 � p1 � p2

p2.

p2

p1;

p1,

u � 180�

EXAMPLE 8

¢l �h

mec� [1 � (�1)] �

2hmec

� 4.85 � 10�12 m � 0.004 85 nm

cos u � cos 180� � �1,

180�,

¢l �h

mec�

6.63 � 10�34 J�s

9.11 � 10�31 kg � 3.00 � 108 m/s� 2.43 � 10�12 m

cos u � cos 90� � 0,

90�,

180�.

90�EXAMPLE 7



By means of Eq. (37.18) we can now eliminate the energy of the electron and

obtain an equation that relates the initial and final energies of the photon:

When we expand the squares in this equation, several terms cancel, leaving us with

the expression

Multiplying both sides by c�(2me E1E1) gives

(37.20)

But from Eq. (37.15), so Eq. (37.20) is equivalent to

that is,

(37.21)

This result agrees with our general formula (37.16) when we set and

Checkup 37.4

QUESTION 1: The colors of light range from red to violet. What color has photons of

the largest momentum?

QUESTION 2: Does the photon gas in a cavity in a hot body exert a pressure on the

surrounding walls?

QUESTION 3: Consider X rays of wavelengths 0.2 nm and 0.6 nm. If these suffer Compton

collisions with electrons and emerge at the same scattering angle, which have the larger

wavelength shift? Which have the larger percentage change of wavelength?

QUESTION 4: Can the Compton effect occur with visible light? Would it be observable?

QUESTION 5: For Compton scattering, it is impossible that the wavelength shift

is

(A) Smaller than the wavelength of the incident photon

(B) Larger than the wavelength of the incident photon

(C) Less than 0.001 nm

(D) Greater than 0.005 nm

(E) Larger for larger scattering angles

¢l

✔

cos u � �1.

u � 180�

¢l �2hmec

l

h�l

h�

2

mec

E1�c � p1 � h�l

1

E1�c�

1

E1�c�

2

mec

2mec aE1

c�

E1cb � 4

E1

c E1c

aE1

c� mec �

E1cb 2

� m2ec2 � aE1

c�

E1cb 2



37.5 X RAYS

X rays were discovered in 1895 by W. K. Röntgen in experiments

with beams of energetic electrons. Röntgen found that when such

a beam of electrons moving in an evacuated glass tube struck the

wall of the tube, some invisible, mysterious rays were emitted that

caused a faint luminosity (fluorescence) on a nearby sheet of paper

impregnated with chemicals. He also found that these rays could

affect covered and wrapped photographic plates—the rays could

penetrate through the wrapping.These rays proved capable of pen-

etrating though thick layers of opaque materials, and Röntgen imme-

diately recognized the possible medical applications of these rays

for making images of the tissues inside the human body, especially

bones (see Fig. 37.15).

The distance that X rays can penetrate through a material depends on the density

of the material. When X rays pass through atoms, they tend to be absorbed by the

atomic electrons. Therefore materials of high density, such as lead, with a concomi-

tant high density of electrons, strongly absorb and block X rays. In X-ray photographs

of parts of the human body, bones throw sharp shadows because their density is higher

than that of the surrounding tissues. Organs made of soft tissues of low density, such

as the gastrointestinal tract, do not throw sharp shadows, and to enhance the contrast

of the X-ray photograph it is advantageous to fill the organ with a barium solution, a

high-density material that blocks the X rays (see Fig. 37.16).

Röntgen had named his rays “X rays” because their nature was unknown. Although

he and his contemporaries suspected that they might be electromagnetic waves of

extremely short wavelength, the conclusive experimental proof of this conjecture was

not obtained until 1912, when Max von Laue argued that if X rays are waves, they

should display interference effects when passing through crystals.The distances between

the rows of atoms in a crystal, such as rock salt, are of the same order of magnitude as

the wavelengths of X rays, and von Laue proposed that the crystal therefore can play

the role of a “grating” for X rays, analogous to a multiple-slit grating used for interference

experiments with light. Figure 37.17 shows a photograph of the interference pattern

produced by X rays incident on a crystal. In this photograph the interference maxima

show up as dark spots. The wavelength of the X rays can be deduced from the angu-

lar positions of the interference maxima and the known size of the spacings in the

crystal. The wavelengths of X rays range from about 0.001 nm to 10 nm.

The spots in such photographs of X-ray interference patterns are called Laue spots.

The beautiful symmetry of the pattern of Laue spots reflects the symmetry of the

arrangement of the atoms in the crystal. In modern crystallography laboratories, the

patterns of Laue spots produced by X rays incident on a crystal are often used to inves-

tigate the structure of the crystal and that of the molecules in it. For instance, such X-

ray interference experiments played a large role in the determination of the structure

of DNA.

Since X rays are electromagnetic waves, their generation by the impact of ener-

getic electrons on some sort of obstacle can be understood by the familiar mechanism

of the emission of radiation by acceleration of electric charges, which we discussed in

Chapter 33. When the electrons in the beam collide with the atoms in the obstacle, they

will suffer sudden decelerations and radiate intense electromagnetic waves of short

wavelength. This kind of radiation is called Bremsstrahlung (German for braking

radiation). Figure 37.18 shows an X-ray tube in which electrons emerging from a hot

filament are accelerated through a potential difference of several kilovolts and then

37.5 X Rays 1273

FIGURE 37.15 One of the first X-ray

photographs prepared by Röntgen. It shows

the hands of his wife; note the sharp image

of the ring.

FIGURE 37.16 X-ray photograph of a

stomach with barium shadowing.

FIGURE 37.17 Interference pattern pro-

duced by X rays incident on a crystal.

Positions of interference maxima(Laue spots) indicate symmetryand spacing of atoms in a crystal.


strike an obstacle, or target, made of a heavy metal, such as tungsten or molybdenum.

The X rays produced in the decelerations of the electrons escape through the side of

the tube. Such X-ray tubes are widely used in medical X-ray machines.

Figure 37.19 is a plot of the distribution of energy of the X rays generated by elec-

trons of 35 kilovolts striking a molybdenum target. Note that the X-ray energy is

smoothly distributed over a wide range of wavelengths, but there also are two con-

spicuous spikes in the energy distribution. The smooth portion of the X-ray spectrum

is due to Bremsstrahlung, whereas the discrete spikes are generated in the interior of

the atoms of molybdenum, in much the same way as spectral lines of visible light are

generated by the atoms (see Chapter 38). In the plot of the energy distribution, the

shape and the location of the broad peak of the Bremsstrahlung portion of the spec-

trum depend on the energy of the incident electrons. But the locations of the spikes

do not depend on the electron energy; instead they depend on the material of the

target. The spikes are called the characteristic spectrum; each kind of target atom

has its own distinctive characteristic spectrum, just as each kind of atom has its own

distinctive spectrum of visible light (for more on the characteristic spectra of atoms, see

the next chapter).

Let us ignore the spikes for now, and concentrate on the smooth Bremsstrahlung

spectrum. Note that below a certain wavelength—for instance, below about 0.03 nm

in Fig. 37.19—there is no Bremsstrahlung. The minimum wavelength emitted by the

electrons is called the cutoff wavelength. Experimentally, the cutoff wavelength is

found to be inversely proportional to the kinetic energy of the electrons. This cutoff

makes no sense from the point of view of classical theory—if an electron suffers a suf-

ficiently violent collision with an atom, it ought to be able to radiate waves of arbi-

trarily short wavelength (although the intensity of short-wavelength waves is expected

to be low). But the cutoff is readily explained by the quantum theory of light, accord-


X rays

target

electrons

�

�

When electrons areaccelerated through apotential difference andthen strike a metal target…

…the sudden decelerationof electrons produces intenseelectromagnetic waves ofshort wavelength.

FIGURE 37.18 An X-ray tube.

intensity

�0.04 0.06 0.08 nm

0.036 nm

Smooth part of X-ray spectrum(Bremsstrahlung) is due to electron decelerations…

Minimum wavelength (maximumenergy) corresponds to an electron giving all its energy to one photon.

…and discrete line(characteristic) spectrumis due to specific kindof target atoms.

FIGURE 37.19 Distribution of energy in the spectrum

of X rays produced by electrons of 35 kilovolts incident

on a molybdenum target.


ing to which the decelerating electron emits photons. Obviously, the maximum energy

that a decelerating electron can give to a photon is all its energy. In this case, the elec-

tron emits only one single photon, with a maximum energy and a maximum frequency,

or cutoff frequency, The energy of this single photon of maximum energy

then equals the kinetic energy K of the electron:

(37.22)

from which

The wavelength is related to the frequency by so the cutoff wavelength is

(37.23)

If an X-ray tube is operated at a potential of 35 kilovolts, the

electron energy is 35 keV. What is the cutoff wavelength for

the X rays generated by electrons of this energy?

SOLUTION: In joules, the electron energy is

Equation (37.23) then gives

This is in agreement with the cutoff wavelength indicated in Fig. 37.19.

Checkup 37.5

QUESTION 1: A fine slit cut in a sheet of lead produces noticeable diffraction effects

for light, but not for X rays. Why not?

QUESTION 2: Figure 37.19 shows the energy distribution for X rays produced by elec-

trons of 35 keV incident on a molybdenum target. If we reduce the electron energy to

17.5 keV, qualitatively, what will change in this figure?

QUESTION 3: Sometimes an X-ray photograph of the intestines reveals bubbles of gas.

Keeping in mind that X-ray photographs are usually negatives (they turn dark where

the X rays strike, as in Fig. 37.15), would you expect a bubble of gas to look light or

dark?

QUESTION 4: When electrons are accelerated through an electrostatic potential

and strike a target, the frequencies f of emitted X rays obey the relation

(A) (B) (C)

(D) (E) f � eV0 �hcf � eV0 �hf � eV0 �hf � V0 �hf � V0 �h

V0

✔

� 3.6 � 10�11 m � 0.036 nm

lcutoff �hc

K�

6.63 � 10�34 J�s � 3.00 � 108 m/s

5.6 � 10�15 J

K � 35 keV � 1.60 � 10�19 J/eV � 5.6 � 10�15 J

EXAMPLE 9

lcutoff �hc

K

l � c�f,

fcutoff �K

h

hfcutoff � K

hfcutofffcutoff.

37.5 X Rays 1275

cut-off wavelength


37.6 WAVE VS. PART ICLE

From the photoelectric effect and the Compton effect, we learned that photons have

particle properties. On the other hand, we know that light displays interference and dif-

fraction phenomena, which prove that photons also have wave properties. Thus pho-

tons are neither classical particles nor classical waves.They are some new kind of object,

unknown to classical physics, with a subtle combination of both wave and particle

properties. Arthur Eddington coined the name wavicle for this new kind of object.2

It is difficult to achieve a clear understanding of the character of a wavicle because

these objects are very remote from our everyday experience. We all have an intuitive grasp

of the concept of classical particles and classical waves from our experience with, say,

billiard balls and water waves, but we have no such experience with wavicles.

We can gain some insight into the interplay between the particle behavior and the

wave behavior of a photon by a new, detailed examination of a simple diffraction exper-

iment. Figure 37.20 shows a light beam striking a plate with a narrow slit. In the usual

diffraction experiments described in Chapter 35, we installed a screen or a photo-

graphic film at the far right, and with this we recorded the intensity of the light in the

diffraction pattern. Instead, in our new arrangement we will install the faceplate of a

very sensitive video camera in place of the customary screen. With this, we can detect

the individual photons of the light in the diffraction pattern.

If the incident light beam has a very low intensity, so there is only one photon

passing through the slit at a time, we can watch the photons arriving one by one at

the faceplate of our video camera. Figure 37.21a shows a typical pattern of impacts of

30 photons. The pattern seems quite random. If the photons behaved like classical

particles, they would travel along a straight line and they would reach only those points

on the faceplate that are within the geometric shadow of the slit. The widely scattered

impacts prove that the photons are certainly not traveling along such straight lines.

Figure 37.21b shows the pattern of accumulated impacts for 300 photons, and Fig.

37.21c shows it for 3000 photons. In these figures we can recognize a tendency of the

photons to cluster in bandlike zones. These zones correspond to the maxima of the

diffraction pattern predicted by the wave theory of light. Finally, Fig. 37.21d shows

the pattern of accumulated impacts for a very large number of photons; this is simply

the familiar intensity pattern of light diffracted by a slit (see also Fig. 35.33).


2Other names that have been proposed are quanticle and quon. Most physicists prefer to avoid all such neolo-

gisms; if anything, they favor the descriptive phrase quantum-mechanical particle or wave-mechanical particle.

wavicle

P

�

video camera

Light beam is incidentfrom left on a platewith a narrow slit.

Sensitive camera detects arrival of individual photons.

FIGURE 37.20 A diffraction experiment.

FIGURE 37.21 Patterns of impacts of

photons on the video camera in Fig. 37.20.

(a) 30 photons; (b) 300 photons; (c) 3000

photons; (d) a very large number of photons.

The first three pictures are computer simula-

tions of accumulated photon impacts; the

last picture is a diffraction pattern obtained

with a laser beam.

(a)

(b)

(c)

(d)

For a few photon impacts, patternis not a shadow of slit, but is spreadout and seems random…

…but after many impacts, photondistribution resembles diffractionpattern of wave theory of light.


From this diffraction experiment we learn that photons display both a wave aspect

and a particle aspect. They behave like waves while passing through the slit; but they

behave like particles when they strike the faceplate of the video camera. Whether the

wave aspect or the particle aspect predominates depends on the experimental equip-

ment with which the photon is interacting. The slit brings out the wave aspect of the

photon; the faceplate of the video camera brings out the particle aspect.

The behavior of the photons is governed by a probabilistic law.The point of impact

of an individual photon on the faceplate is unpredictable. Only the average distribu-

tion of impacts of a large number of photons is predictable: the distribution of photons

matches the intensity distribution calculated from the wave theory of light. Thus, the

probability that a photon arrives at a given point on the faceplate is proportional to

the intensity of the wave at that point. Since the intensity of an electromagnetic wave

is proportional to the square of its electric field, we can write the proportionality of

probability and intensity as

(37.24)

Here we have a connection between the wave and the particle aspects of the photon:

the intensity of the photon wave at some point determines the probability that there is a

photon particle at that point. This probability interpretation of the intensity of the wave

was discovered by Max Born.

Our single-slit experiment can also teach us something about the limitations

that quantum theory imposes on the ultimate precision of measurement of the posi-

tion of a wavicle. Suppose we have a light wave consisting of one photon and we

want to measure the position of this photon. The position has x, y, and z compo-

nents; we will concentrate on the y component, perpendicular to the direction of

propagation. Figure 37.22 shows the wave propagating in the horizontal direction;

the y direction is vertical. To determine this vertical position of the photon, we use

a narrow slit placed in the path of the wave. If the photon succeeds in passing through

this slit, then we will have achieved a determination of the vertical position to within

an uncertainty

(37.25)

where a is the width of the slit. If the photon fails to pass through this slit, then our

measurement is inconclusive and will have to be repeated.

By making the slit very narrow, we can make the uncertainty of our determina-

tion of the y coordinate very small. But this has a surprising consequence for the y

component of the momentum of the photon: if we make the uncertainty in the y coor-

dinate small, we will make the uncertainty in the y component of the momentum large.

To see how this comes about, let us recall that according to our preceding discussion

of the single-slit experiment, the photon suffers diffraction by the slit and emerges at

some angle (Fig. 37.22).This angle is unpredictable; all we can say about the photon

after it emerges from the slit is that it will be heading toward some point within the

diffraction pattern.Thus, the direction of motion of the photon is uncertain, and there-

fore the y component of its momentum is uncertain. If we make the slit very narrow,

the diffraction pattern will become very wide, and the uncertainty in the direction of

motion and the uncertainty in the y component of the momentum will become very

large.

As a rough quantitative measure of the magnitude of the uncertainty in direction,

we can take the angular width of the central diffraction maximum (most of the inten-

sity of the photon wave is gathered within the region of this central maximum, and

uu

¢y � a

[probability for photon at a point] r E 2

37.6 Wave vs. Particle 1277

y

a

x�p

A photon incident fromleft on a slit…

…will emerge at someunknown angle � withinthe diffraction pattern.

FIGURE 37.22 Photon passes through a

narrow slit and emerges at an angle u.

MAX BORN (1882–1970) German, and

later British, theoretical physicist. He was

awarded the Nobel Prize somewhat tardily in

1954 for his discovery of the probabilistic

interpretation of quantum waves in 1926.

probability interpretation of wave intensity


hence the photon is most likely to be found in this region).This estimate of the uncer-

tainty of the angle gives us

(37.26)

The y component of the momentum is (see Fig. 37.22); since we are con-

cerned with a small angle, we can use the approximation and therefore

The uncertainty in is then

(37.27)

But, according to Eq. (37.15), and therefore

(37.28)

Thus, the uncertainty in the y component of the momentum is inversely proportional

to the width of the slit.

Comparing Eqs. (37.25) and (37.28), we find that the product of the uncertainty

of position and the uncertainty of momentum is

(37.29)

This equation states that and cannot both be small; if one is small then the

other must be large, so that their product equals Planck’s constant.

Although we have obtained Eq. (37.29) by examining the special case of a position

measurement by means of a slit, it turns out that this relation is actually of general valid-

ity for any kind of position measurement—the product of the uncertainty in the posi-

tion and the uncertainty in the momentum always equals or exceeds Planck’s constant.

A more rigorous derivation uses the rms uncertainty—the square root of the mean of

the squares of the deviations from the mean value; for example,

Careful consideration of a variety of arrangements for the simultaneous measurement

of position and momentum shows that the product of rms uncertainties rigorously

obeys the inequality

(37.30)

Equation (37.30) is one example of a Heisenberg uncertainty relation.There are cor-

responding relations for the other components of position and momentum. The

Heisenberg uncertainty relations tell us that there exist ultimate, insuperable limita-

tions in the precision of our measurements. At the macroscopic level, the quantum

uncertainties in our measurements can be neglected. But at the atomic level, these

quantum uncertainties are often so large that it is completely meaningless to speak of

the position or momentum of a wavicle.

Suppose we measure the vertical position of a photon by means

of a narrow horizontal slit of width With what

uncertainty in vertical momentum does the photon emerge from this slit?

SOLUTION: Equation (37.28) tells us that for such a position measurement

¢py �ha

�6.63 � 10�34 J�s

1.0 � 10�5 m� 6.6 � 10�29 kg�m/s

1.0 � 10�5 m.EXAMPLE 10

¢y ¢py �h

4p

¢x � [(x � x)2]12.

¢py¢y

¢y ¢py � h

¢py � h�a

p � h�l,

¢u � pl�a¢py � p

pypy � p u.

sin u � upy � p sin u

¢u � l�a


WERNER HEISENBERG (1901–1976)German theoretical physicist. He was one of

the founders of the new quantum mechanics,

and received the Nobel Prize in 1932.

Heisenberg uncertainty relation


Checkup 37.6

QUESTION 1: If photons were classical particles, what pattern of impact points would

we find on the faceplate placed beyond the narrow slit in Fig. 37.22?

QUESTION 2: Suppose that the amplitude of the electric field at some point in the dif-

fraction pattern displayed in Fig. 37.21c is one-half the amplitude at the center. By

what factor is the probability for a photon smaller than at the center?

QUESTION 3: Suppose that an electron moving in a TV tube has an uncertainty of

in position. Can this electron be free of uncertainty in momentum? In veloc-

ity? In energy?

QUESTION 4: Given that an electron has zero uncertainty in its y position

can this electron have zero uncertainty in the x component of the momentum

The z component of the momentum

QUESTION 5: Consider a double-slit experiment, with slit width and slit spacing sev-

eral times the wavelength of the incident light. However, now the intensity is greatly

reduced, so that only individual photons are incident on the slits. After a long time, the

transmitted light recorded by a sensitive camera would show

(A) A double-slit interference pattern

(B) A single-slit diffraction pattern

(C) A random pattern of uniform intensity

(D) Two sharp shadows of the slits without diffraction effects

(¢pz � 0)?(¢px � 0)?

(¢y � 0),

10�6 m

✔

Summary 1279

SUMMARY

ENERGY QUANTIZATION OF OSCILLATOR

WIEN’S DISPLACEMENT LAW

STEFAN–BOLTZMANN LAW where S is the

energy flux radiated by a blackbody

(37.3)n � 0, 1, 2, …E � n hf

(37.6)lmax � (2.90 � 10�3 m�K)�T

(37.7)S � sT 4

PHYSICS IN PRACTICE Photomultiplier (page 1268)

h � 6.63 � 10�34 J�s (37.2)PLANCK’S CONSTANT

STEFAN–BOLTZMANN CONSTANT (37.8)s � 5.67 � 10�8 W/(m2�K4)

ENERGY AND MOMENTUM OF A PHOTON and (37.15)p � hf�c � h�lE � hf

EINSTEIN’S PHOTOELECTRIC EQUATION

where � is the work function of the metal

(37.11)K � hf � f

energy flux(arbitrary units)

classical 1450 Kclassical 1450 K

1450 K

1250 K

1000 K

20000 4000 6000 nmwavelength



COMPTON WAVELENGTH SHIFT OF PHOTON

PROBABILITY INTERPRETATION OF WAVE

INTENSITY

(37.16)¢l �h

mec (1 � cos u)

(37.24)[probability for presence of photon] r [intensity of wave] r E 2

(37.30)¢y ¢py �h

4p

HEISENBERG UNCERTAINTY RELATION

FOR y AND py

�cutoff � (37.23)hc

K

CUTOFF WAVELENGTH FOR X RAYS

12. Fluorescent paints achieve their exceptionally bright orange or

red color by converting short-wavelength photons into long-

wavelength (red) photons. Why can we not make such a paint

in blue or violet colors?

13. When light of a given wavelength ejects photoelectrons from

the surface of a metal, why is it that not all of the photoelec-

trons emerge with the same kinetic enegy?

14. Can a particle of mass zero ever be at rest?

15. According to Eq. (37.16), a photon suffers a maximum change

of wavelength in a collision with an electron if it emerges at an

angle and a minimum change of wavelength (no

change) if it emerges at an angle Is this reasonable?

16. Suppose that a photon and an electron have the same momen-

tum. Which has the larger energy, taking into account both

the rest-mass energy and the kinetic energy?

17. Can the Compton effect occur with visible light? Would it be

observable?

18. Photons of short wavelength are more particle-like than

photons of long wavelength. Why?

19. Give an example of an experiment in which photons behave

like waves. Give an example of an experiment in which they

behave like particles.

20. What happens to the y momentum that a photon gains or

loses in the diffraction experiment described in Fig. 37.22?

21. According to Eq. (37.24), we can predict only the probability

that a photon will be found at some given point. Does this

mean that quantum physics is not deterministic?

u � 0�.

u � 180�,


1. Is the light emitted by a neon tube thermal radiation? The

light emitted by an ordinary incandescent lightbulb?

2. Does your body emit thermal radiation?

3. In Example 2 we calculated the radiative heat loss from the

ground during a clear night. Qualitatively, how would a cover

of snow on the ground affect the result?

4. The insulation used in the walls of homes consists of a thick

blanket of fiberglass covered on one side by a thin aluminum

foil. What is the purpose of these two layers?

5. Black velvet looks much blacker than black paint. Why?

6. For protection against the heat of sunlight, parts of the Lunar

Lander (and some other spacecraft) were wrapped in shiny

aluminum foil. Why is shiny foil useful for this purpose?

7. If you look into a kiln containing pottery heated to a tempera-

ture equal to that of the walls of the kiln, you can scarcely see

the pottery. Explain.

8. According to Fig. 37.3, at what wavelength is the energy flux

maximum for a body at 1450 K? At 1250 K? At 1000 K? Do

these wavelengths satisfy Wien’s Law?

9. The quantization of electric charge is consistent with classical

physics, but the quantization of energy is not. Does this make

sense?

10. Suppose that Planck’s constant were much larger than it is,

say, 1034 times larger. What strange behavior would you notice

in a simple harmonic oscillator consisting of a mass hanging

on a spring?

11. Why do we not notice the discrete quanta of light when we

look at a lightbulb?

photon

photon

electron

�

intensity

�0.04 0.06 0.08 nm

Minimum wavelength (maximumenergy) corresponds to an electron giving all its energy to one photon.


Problems 1281

PROBLEMS

37.1 B lackbody Radia t ion37.2 Energy Quanta

1. Consider a seconds pendulum, that is, a pendulum that has a

period of 2 seconds. What is the magnitude of one energy

quantum for such a pendulum? Would you expect that quan-

tum effects are noticeable in such a pendulum?

2. In molecules, the atoms can vibrate about their equilibrium

positions. For instance, in the molecule, the hydrogen

atoms vibrate about their equilibrium positions with a fre-

quency of What is the magnitude of the

energy quantum for this oscillating system? Note that

although there are two masses in this system, they must be

regarded as a single oscillator because the vibrational motions

of the two masses are always equal.

3. In a solid, each atom is held in its position by elastic forces,

which permit the atom to oscillate about its equilibrium point.

The frequency of such oscillations of an aluminum atom in

solid aluminum is The energy of this oscilla-

tory motion is quantized. What is the magnitude of an energy

quantum?

4. If we take Planck’s model of the walls of a blackbody cavity

seriously, we will have to assume that the oscillating masses are

electrons. Imagine an electron oscillating with a frequency of

under the influence of a springlike force. What

is the amplitude of oscillation of this electron if its energy of

oscillation is one energy quantum? Two energy quanta?

5. An oscillator of frequency consists of a mass

of attached to a spring. What is the ampli-

tude of oscillation of this oscillator if its energy of oscillation is

one energy quantum? Two energy quanta?

6. Suppose that two stars have the same size but the temperature

of one is twice that of the other. By what factor will the ther-

mal power radiated by the hotter star be larger than that radi-

ated by the cooler star?

7. The theoretical expression for the constant in the

Stefan–Boltzmann Law is

Verify that the numerical value implied by this theoretical

expression agrees with Eq. (37.8).

8. The flux of thermal radiation from the star Procyon B is

observed to have a maximum at a wavelength of 440 nm.

Assuming the star radiates like a blackbody, what temperature

can you deduce for the surface of this star?

9. Melting iron has a temperature of 1808 K. At what wave-

length does the iron radiate a maximum flux? Assume that the

iron acts like a blackbody.

s �2p5k4

15h3c 2

s

1.67 � 10�27 kg

3.00 � 1013 Hz

2.0 � 1015 Hz

8.0 � 1011 Hz.

1.31 � 1014 Hz.

H2

10. At what wavelength does your skin radiate a maximum flux of

thermal radiation? Assume that the temperature of your skin

is

11. Nanomechanical oscillators with a frequency of 5.0 GHz and

masses as small as can be fabricated. What is

the amplitude of oscillation of such an oscillator if its energy

of oscillation is one energy quantum?

12. Thermonuclear reactions near the core of the Sun maintain a

temperature of Assume that the distribution of

this radiation is thermal. At what wavelength does the spectrum

of this radiation have a maximum energy flux? What is this part

of the electromagnetic spectrum called? What is the energy per

second emitted per square meter near the Sun’s core due to this

radiation?

13. Low temperatures are commonly attained using liquid helium,

which has a latent heat such that 1.0 watt will boil approxi-

mately 1.0 liter of liquid helium per hour. Consider a region

with a surface area of What is the rate of energy

transfer into the region due to thermal radiation if it is

surrounded by a room-temperature (295 K) environment?

If surrounded by a liquid nitrogen (77 K) environment?

Which is preferable in order to conserve liquid helium?

14. The nichrome heater wire in a toaster has a radius of 0.20 mm

and a total length of 1.0 m. Assume that the wire acts as a

blackbody and emits 1200 W of thermal radiation. What is

the surface temperature of the wire? At what wavelength is the

thermal radiation maximum?

15. A black hole radiates with a thermal spectrum due to quantum

effects. A black hole with a radius of 30 km radiates a total of

of such thermal radiation, or Hawking radia-

tion. What is the temperature of this blackbody?

*16. The tungsten filament of a lightbulb is a wire of diameter

0.080 mm and length 5.0 cm. The filament is at a temperature

of 3200 K. Calculate the power radiated by the filament.

Assume the filament acts like a blackbody.

*17. The temperature of the surface of the Sun is 5800 K and the

Sun’s radius is If the Sun radiates like a black-

body, what can you predict for the thermal energy flux it

emits? What can you predict for the energy flux of sunlight

arriving at the Earth, at a distance of Compare

with the measured value,

*18. The star Procyon B is at a distance of 11 light-years from

Earth. The flux of its starlight reaching us is 1.7 � 10�12

W/m2, and the surface temperature of the star is 6600 K.

Calculate the size of the star.

*19. Consider the integral over wavelengths of Planck’s expression,

Eq. (37.4), for the distribution of blackbody radiation

S � �dS � �Sldl � �q

0

2phc2

l5 � (e hc�lkT � 1)

dl

1.35 � 103 W/m2.

1.5 � 1011 m?

6.96 � 108 m.

8.8 � 10�31 W

0.10 m2.

5.0 � 107 K.

2.0 � 10�20 kg

33�C.


Show that the total intensity emitted by a blackbody is pro-

portional to the fourth power of the absolute temperature, as

required by the Stefan–Boltzmann Law. (Hint: You need not

evaluate the integral; you need only use the substitution

to make the integral dimensionless.)

*20. At the Earth, the flux of sunlight per unit area facing the Sun is

The Earth absorbs heat from the sunlight

and reradiates heat as thermal infrared radiation. For equilibrium,

the power arriving from the Sun must equal the average power

radiated by the surface of the Earth.This permits us to make a

rough prediction for the average temperature of the Earth.

(a) Assume that the Earth absorbs all of the sunlight striking

it. What is the power of the sunlight absorbed? (Hint: The

relevant area is the cross-sectional area of the Earth.)

(b) Assume that the surface of the Earth radiates like a black-

body. If the temperature of the surface is T, what is the

expression for the power radiated? (Hint: The relevant area

is the total surface area If this power is to match

the power calculated in (a), what must be the value of T ?

**21. Deduce the surface temperature of Pluto by the method

described in the preceding problem. Pluto is 39 times as far

away from the Sun as the Earth, and hence the flux of sunlight

at Pluto is (39)2 times smaller than the flux at the Earth.

**22. Maximize Planck’s expression for the intensity of blackbody

radiation per unit wavelength, Eq. (37.4), and thus obtain Wien’s

Law, Eq. (37.5). (Hint: Once you obtain the condition for a

maximum, consider the dimensionless variable

Find a solution for x iteratively, for example, using a calculator.)

**23. In Problem 37 of Chapter 24, you will find a description of

the Thomson model of the hydrogen atom.

(a) What is the frequency of oscillation of the electron in

this atom?

(b) The electron can be regarded as an oscillator. Show that if

the energy of the electron is one quantum, the amplitude of

oscillation exceeds the radius of the atom.

37.3 Photons and the Photoelectr ic Ef fec t

24. Photons of green light have a wavelength of 550 nm. What is

the energy and what is the momentum of one of these photons?

25. You are lying on a beach, tanning in the sun. Roughly how

many photons strike your skin in one hour? Assume that the

energy flux of sunlight is as described in Example 4.

26. For each of the following kinds of electromagnetic waves, find

the energy of a photon: FM radio wave of wavelength 3.0 m,

infrared light of visible light of

ultraviolet light of X rays of

27. A radio transmitter radiates 10 kW at a frequency of

How many photons does the transmitter radi-

ate per second?

28. The energy flux in the starlight reaching us from the bright

star Capella is If you are looking at the1.2 � 10�8 W/m2.

8.0 � 105 Hz.

1.0 � 10�10 m.1.0 � 10�7 m,

5.0 � 10�7 m,1.0 � 10�5 m,

(5 � 10�11 m)

x � hc�lkT.

4pR2.)

pR2

1.35 � 103 W/m2.

x � hc�lkT

star, how many photons per second enter your eyes? The

diameter of your pupil is 0.70 cm. Assume that the average

wavelength of the light is 500 nm.

29. A laser emits a light beam with a power of 1.0 W. The wave-

length of the light is 630 nm. How many photons per second

does this laser emit?

30. The energy density in a radio wave of wavelength 3.0 m is

What is the corresponding density of

photons?

31. Calculate the range of energies of photons of visible light,

with wavelengths from 700 nm to 400 nm. Express your

answers in units of electron-volts.

32. For light-sensing applications that do not require the extreme

sensitivity of a photomultiplier (see Physics in Practice:

Photomultiplier), photodiodes are often used. A typical pho-

todiode can measure an incident photon flux provided the

total power incident is at least otherwise, the

signal is lost in the electrical noise of the photodiode. Laser

light of wavelength 633 nm is to be detected. What is the

minimum number of photons per second required?

33. Light of wavelength 486 nm is incident on a material with a

work function of 2.26 eV. What is the maximum kinetic

energy of ejected electrons? When light of wavelength 434 nm

is used, what stopping potential is necessary?

34. Show that if we express the energy of a photon in keV and the

wavelength in nanometers, then

35. According to Fig. 37.11, what is the work function of sodium?

Express your answer in electron-volts.

36. The work function of potassium is 2.26 eV. What is the

threshold frequency for the photoelectric effect in potassium?

37. The work functions of K, Cr, Zn, and W are 2.26, 4.37, 4.24,

and 4.49 eV, respectively. Which of these metals will emit pho-

toelectrons when illuminated with red light

Blue light Ultraviolet light

38. The photons emitted by a sodium atom have a wavelength of

589 nm when at rest. If this atom is moving away from you at

a speed of what is the energy that you measure

for one of these photons?

*39. By inspection of Fig. 37.11, find the slope of the line in

eV/Hz. Convert these units into J. s, and verify that the slope

is the same as Planck’s constant.

*40. The binding energy of an electron in a hydrogen atom is 13.6

eV. Suppose that a photon of wavelength 40 nm strikes the

atom and gives up all of its energy to the electron. With what

kinetic energy will the electron be ejected from the atom?

37.4 The Compton E f fec t

41. X rays emitted by molybdenum have a wavelength of 0.072

nm. What are the energy and the momentum of one of the

photons of these X rays?

2.0 � 106 m/s,

(l � 280 nm)?(l � 400 nm)?

(l � 700 nm)?

E � 1.24>l

1.0 � 10�14 W;

2.0 � 10�13 J/m3.



42. In a collision with an initially stationary electron, a photon

suffers a wavelength increase of 0.0022 nm. What must have

been the deflection angle of the photon?

43. For an experiment on the Compton effect, you want the X

rays emerging at from the incident direction to suffer an

increase of wavelength by a factor of 2. What wavelength do

you need for your incident X rays?

44. A photon of wavelength 0.030 nm collides with a free electron

at rest. Calculate the wavelength if the photon emerges from

this collision with a deflection of

45. X rays of wavelength 0.030 nm are incident on a graphite

target. Calculate the wavelength of the X rays that emerge

from this target with a deflection of Calculate the wave-

length of the X rays that emerge from this target with a

deflection of

46. X-ray photons of wavelength 0.154 nm are produced by a

copper source. Suppose that of these photons are

absorbed by a target each second.

(a) What is the total momentum p transferred to the target

each second?

(b) What is the total energy E of the photons absorbed by the

target each second?

(c) For these values, verify that the force on the target is related

to the rate of energy transfer by

47. Calculate the percentage change in wavelength,

when an X-ray photon of wavelength

� � 0.0010 nm is Compton-scattered through an angle of

(a) (b) and (c)

48. Suppose that a photon is “Compton-scattered” from a proton

instead of an electron. What is the maximum wavelength shift

in this case?

*49. In a Compton scattering experiment, X rays of wavelength

are incident. The maximum energy transferred to an electron

is Find an expression for in terms of h, and c.

What is the value of for

*50. Derive the general Compton-shift formula (37.16). [Hint:

Repeat the calculation of Example 8, but replace Eq. (37.19)

by the relation obtained from

inspection of the diagram representing the vector sum of

momentum vectors. Note that here the p’s represent magni-

tudes of momentum vectors and they are positive, whereas in

Example 8 the p’s are positive or negative depending on the

direction of motion.]

*51. What is the maximum energy that a free electron (initially sta-

tionary) can acquire in a collision with a photon of energy

*52. A photon of initial wavelength 0.040 nm suffers two succes-

sive collisions with two electrons. The deflection in the first

collision is and in the second collision it is What is

the final wavelength of the photon?

60�.90�

4.0 � 103 eV?

p21 � 2p1p1 cos u � p2

1 � p22 ,

Ee � 25 keV?l

me,Ee,lEe.

l

180�.30�,10�,

(¢l�l) � 100%,

dp�dt � (1�c)(dE�dt).

1.00 � 1018

120�.

60�.

30�.

90�

37.5 X Rays

53. What is the energy of the photons in X rays of a wavelength

of 0.050 nm?

54. An X-ray tube is being operated with electrons of energy 25

keV. What is the cutoff wavelength of the emitted X rays?

55. We want to use an X-ray tube to generate X rays of wave-

length 1.00 nm. What is the minimum potential difference we

must use to accelerate the electrons in the tube?

56. The tube in a medical X-ray machine can be operated at

potential differences in the range from 25 kV to 150 kV. What

is the cutoff wavelength of the X rays emitted when the tube is

operated at 25 kV? At 150 kV?

57. For studies of the structure of materials, a characteristic spec-

tral line of copper, with a wavelength of 0.1542 nm, is often

used. For such X rays to be emitted, what is the minimum

potential difference that could be used to accelerate electrons

toward a copper target?

58. Tungsten targets are often used in X-ray tubes to produce an

intense, broad spectrum of Bremsstrahlung. To study the spac-

ing of different planes of atoms in a particular crystal, X-ray

wavelengths throughout the range 0.10–40 nm are desired.

What is the corresponding minimum accelerating potential

for electrons?

37.6 Wave vs . Par t i c le

59. A photon passes through a horizontal slit of width

What uncertainty in the vertical position will

this photon have as it emerges from the slit? What uncertainty

in the vertical momentum?

60. A photon traveling in the z direction passes through a square

hole on a side. What is the uncertainty in each

of the x and y components of the momentum? Assuming the

uncertainty in the z component of the momentum is negligi-

ble, what is the uncertainty in the total momentum?

*61. A particular semiconductor laser emits photons with an uncer-

tainty in position of 1.5 cm along the direction of propagation,

known as the longitudinal coherence length. What is the cor-

responding uncertainty in the momentum of the photons?

The photons have an average wavelength of 678 nm. What is

the uncertainty in this wavelength?

*62. Consider a radio wave in the form of a pulse lasting 0.0010 s.

The pulse then has a length of

Since an individual photon of this radio wave can be anywhere

within this pulse, the uncertainty in the position of the photon

is along the direction of propagation.

(a) According to Heisenberg’s relation, what is the correspon-

ding uncertainty in the momentum of the photon?

(b) What is the uncertainty in the frequency of the photon?

¢x � 3.0 � 105 m

0.0010 s � c � 3.0 � 105 m.

2.0 � 10�6 m

5.0 � 10�6 m.

Problems 1283



REVIEW PROBLEMS

*63. Consider a mass of 1.0 g attached to a spring of spring con-

stant What is the frequency of oscillation of

this system? What is the magnitude of an energy quantum?

What is the amplitude of oscillation if the energy of the

system equals one quantum?

*64. We know from Chapter 15 that at small amplitudes a pendu-

lum behaves like a simple harmonic oscillator. Suppose that a

pendulum consists of a mass of 0.10 kg attached to a (mass-

less) string of length 1.0 m.

(a) Taking into account the quantization of energy, what is

the least (nonzero) amount of energy that this pendulum

can have?

(b) What is the amplitude of oscillation of the pendulum

with this least amount of energy?

*65. Interplanetary and interstellar space is filled with thermal radi-

ation of a temperature of 2.7 K left over from the Big Bang.

(a) At what wavelength is the flux of this radiation maximum?

What is this part of the electromagnetic spectrum called?

(b) What is the power incident on the surface of the Earth

due to this radiation?

*66. If you stand naked in a room, your skin and the walls of the

room will exchange heat by radiation. Suppose the tempera-

ture of your skin is the total area of your skin is

The temperature of the walls is Assume your skin and

the walls behave like blackbodies.

(a) What is the rate at which your skin radiates heat?

(b) What is the rate at which your skin absorbs heat? What is

your net rate of loss of heat?

67. An incandescent bulb radiates 40 W of thermal radiation from

a filament of temperature 3200 K. Estimate the number of

photons radiated per second; assume that the photons have an

average wavelength equal to the given by Wien’s Law.lmax

15�C.

1.5 m2.33�C;

3.0 � 10�2 N/m.

68. In order for a photoelectron to be emitted from a nickel sur-

face, the wavelength of the incident photon must be no longer

than 247.5 nm. What is the work function for nickel? When a

photon of wavelength 200 nm is used, what stopping potential

is required?

69. A photon loses 5.0% of its initial energy when a collision with

an electron at rest deflects it by What is the wavelength of

the incident photon? With what speed does the electron

emerge from the collision?

70. The energy density of starlight in intergalactic space is

What is the corresponding density of

photons? Assume the average wavelength of the photons is

500 nm.

*71. If you want to make a very faint light beam that delivers 1

photon per square meter per second, what must be the ampli-

tude of the electric field in this light beam? The wavelength of

the light is 500 nm.

*72. A photon has an energy of 5.0 eV in the reference frame of

the laboratory. What is the energy of this photon in the refer-

ence frame of a proton moving through the laboratory at a

speed of c in the same direction as the photon? (Hint: Use

the Doppler-shift formula given in Section 36.3.)

*73. In a collision with a free electron, a photon of energy

is deflected by What energy does the elec-

tron acquire in this collision?

*74. A photon of energy collides elastically with a

proton initially at rest. The photon is deflected by What is

its new energy?

*75. Figure 37.19 shows two of the discrete spectral lines of molyb-

denum. According to this figure, what are the wavelengths of

these spectral lines? Would these spectral lines appear if

instead of using electrons of 35 keV we were to use electrons

of 19 keV in the X-ray tube?

45�.

1.6 � 108 eV

90�.2.0 � 103 eV

12

1.0 � 10�15 J/m3.

90�.

Answers to Checkups

Checkup 37.1

1. If the stove were polished instead of black, it would not be as

good an emitter of thermal radiation, since a black surface

radiates heat most efficiently, and a shiny surface does not. A

good absorber is a good emitter, and a poor absorber is a poor

emitter.

2. The roof color provides direct evidence of the light absorbed or

emitted: a roof of dark-colored shingles absorbs more sunlight,

and a roof of light-colored shingles reflects more sunlight.

3. The shiny surface layer of the suit reflects sunlight, and thus

prevents overheating.

4. (D) Less intense at all wavelengths. As discussed in Section

37.1 and shown in Fig. 37.3, a blackbody with a peak at a

longer wavelength also produces less radiation at all wave-

lengths (and also has a lower temperature).

Checkup 37.2

1. For the first peak,

in agreement with Wien’s Law; likewise for the

other two peaks.

10�3 m�K,

lmaxT � 1450 K � 2000 nm � 2.9 �



2. Wien’s Law asserts that the wavelength at maximum intensity

varies inversely with the absolute temperature, so doubling the

temperature changes the wavelength of the maximum of the

spectrum by a factor of The Stefan–Boltzmann Law indi-

cates that the total energy radiated varies as the fourth power

of temperature, so doubling the temperature increases the total

energy radiated from the surface by a factor of 16.

3. No to both—only the total energy of the oscillator is quan-

tized, that is, the sum of kinetic and potential energies is

quantized.

4. Yes, the energies are quantized; but for frequencies around 1

Hz, the energy quanta are about

and thus individual increments in the motion are much too

small to be noticed.

5. For such a large value of Planck’s constant, the energy quanta

would be nearly one-tenth of one joule, and thus the ampli-

tude of oscillation of a pendulum would jump by a noticeable

amount when one quantum of energy is added or removed.

6. (D) Blue; red. By Wien’s Law, the hottest star emits more

strongly at shorter wavelengths (the blue) and the coolest at

longer wavelengths (the red). See Fig. 33.23 for the relation

between the colors of the spectrum and wavelength.

Checkup 37.3

1. Violet light has the shortest wavelength, and thus the highest

frequency and the most energetic photons. See Fig. 33.23 for

the relation between the colors of the spectrum and wavelength.

2. For a larger work function, the threshold frequency is larger,

that is, the straight line intercepts the horizontal frequency

axis at a larger value of f.

3. For repulsive voltages more negative than electrons do

not have enough kinetic energy to reach the collector. The cur-

rent levels off for sufficiently high positive voltages because then

all ejected electrons have enough energy to reach the collector

and, for a given intensity, no more are available. The current

decreases for lower intensity since the number of photons, and

thus the rate of ejection of electrons, is proportional to intensity.

4. (C) The same slope. By Eq. (37.11), the slope of all such plots

is Planck’s constant h.

Checkup 37.4

1. Violet light has the shortest wavelength and, since

the largest momentum. See Fig. 33.23 for the relation

between the colors of the spectrum and wavelength.

2. Yes, when a photon bounces off the wall, its momentum

reverses, and the wall experiences an impulsive force.

3. At any given scattering angle, the wavelength shift is the

same for all incident wavelengths. Since the wavelength shift

is fixed, the percentage change is largest for the smaller wave-

length, 0.2 nm.

¢l

p � h>l,

�Vstop,

E � hf � 6.63 � 10�34 J,

12.

4. Yes, visible photons scatter in the same way as X rays and

experience a Compton shift. The effect is extremely difficult

to observe, since the change is so small, only a few picometers

out of hundreds of nanometers.

5. (D) Greater than 0.005 nm. A wavelength shift greater than

0.005 nm is impossible; as in Example 7, the maximum possi-

ble wavelength shift is 2h�mec � 0.004 85 nm.

Checkup 37.5

1. X-ray wavelengths are much shorter than those of visible

light; diffraction effects increase with the ratio of the wave-

length to the slit width, and so are much smaller for X rays.

2. Since [Eq. (37.23)], halving the kinetic energy

will double the cutoff wavelength to (compare

Example 9). The characteristic spikes will remain at the

same wavelengths but disappear for molybdenum since

is too large; their locations are determined

by the target material.

3. A bubble would allow more X rays to pass than the surround-

ing tissue, and so would appear dark in a negative.

4. (D) The maximum energy of the photon is the

kinetic energy of the electron,

Checkup 37.6

1. We would find a narrow line of the same width as the narrow

slit; classically, the slit would merely create an ordinary

shadow.

2. Since the probability that a photon arrives at a given point is

proportional to the intensity of the wave at that point, and

since the intensity is proportional to the square of the electric

field amplitude, the point where the amplitude is decreased by

a factor of 2 has a photon arrival probability smaller by a factor

of 4.

3. No; the uncertainty principle dictates that there must be

uncertainty in the momentum p. The electron is thus also not

free from uncertainty in the velocity or in the kinetic

energy

4. Yes in both cases. The uncertainty relation applies to the same

component of the position and momentum. Additionally, Eq.

(37.29) implies that if the uncertainty in, say, pz is zero, then

the uncertainty in z must be infinite.

5. (A) A double-slit interference pattern. Like the case of the

single slit, the distribution of photons matches the intensity

distribution calculated from the wave theory of light; that is,

the probability that a photon arrives at a given point is propor-

tional to the intensity of a classical wave at that point.

K � p2�2m.

v � p�m,

E � hf � K � eV0.

f � eV0 �h.

lcutoff � 0.072 nm

2 � 0.036 nm

lcutoff � hc�K



When a sample of a chemical element is heated, often by means of an

electric discharge, it emits characteristic electromagnetic radiaton of dis-

crete wavelengths. The visible light in such a discrete-line spectrum for

hydrogen gas is shown in the upper frame of the figure above; the lower

frame includes additional lines in the ultraviolet spectrum of hydrogen.

In our study of the spectra of hydrogen, we will be able to ask:

? The spacings between these lines decrease systematically from right to

left, from long wavelengths to short wavelengths. This systematic pat-

tern suggests that the wavelengths form a series, described by a simple

mathematical formula. What is this formula? (Section 38.2, page 1291)

? Are there other series of spectral lines in the spectrum of hydrogen?

What are they, and what is the shortest wavelength in any of these

series? (Section 38.2, and Example 1, page 1292)

Spectral Lines,Bohr’s Theory, andQuantum Mechanics38

38.1 Spectral Lines

38.2 Spectral Series of Hydrogen

38.3 The Nuclear Atom

38.4 Bohr’s Theory

38.5 Quantum Mechanics; theSchrödinger Equation

C H A P T E R

1286

Conceptsin

Context

350 nm 660 nm

410.3 nm 434.2 nm 486.3 nm 656.5 nm410.3 nm 434.2 nm 486.3 nm 656.5 nm

H� H� H� H� H�



? How can we calculate the wavelengths of the series from the quantum processes

within the hydrogen atom? (Example 2, page 1300)

? What series of wavelengths will hydrogen absorb, when illuminated with light

containing a wide range of wavelengths? (Example 3, page 1301)

T he photograph shown in Fig. 38.1 gives convincing visual evidence that solids, liq-

uids, and gases are made of atoms, small grains of matter with a diameter of about

This photograph was prepared with a powerful electron microscope of a spe-

cial design. Unfortunately, not even this microscope is sufficiently powerful to reveal

the inside of the atom. For the exploration of the internal structure of the atom, we

still have to rely on the technique developed by Ernest Rutherford and his associates

around 1910: bombard the atom with a beam of particles and use this beam as a probe

to “feel” the interior of the atom.

By 1910, most physicists had come to believe that atoms are made of some com-

bination of positive and negative electric charges, and that the attractions and repul-

sions between these electric charges are the basis for all the chemical and physical

phenomena observed in solids, liquids, and gases. Since electrons were known to be

present in all of these forms of matter, it seemed reasonable to suppose that each atom

consists of a combination of electrons and positive charge. The vibrational motions of

the electrons within the atom would then result in the radiation of electromagnetic

waves; this was supposed to account for the emission of light by the atom. However,

both the arrangement of the electric charges within the atom and the mechanism that

accounts for the characteristic colors of the emitted light remained mysteries until

Rutherford’s discovery of the nucleus and Niels Bohr’s discovery of the quantization

of atomic states. In this chapter we will look at these two momentous discoveries.

The exploration of the internal structure of the atom led to the inescapable con-

clusion that in the atomic realm Newton’s laws of motion are not valid. Electrons and

other subatomic particles obey new equations of motion that are drastically different

from the old equations of motion obeyed by planets, billiard balls, or tennis balls. The

new equations of motion incorporate the quantization of energy and the wave–

particle behaviors we discussed in the preceding chapter, and they extend quantiza-

tion to the realm of the atom.The new theory of motion that rules the realm of the atom

is called quantum mechanics. The discovery of an entirely new set of laws of motion

was the greatest scientific revolution of the twentieth century.

38.1 SPECTRAL L INES

The earliest attempts at a theory of atomic structure ended in failure—these early the-

ories were not able to explain the characteristic colors of the light emitted by atoms.

These colors show up very distinctly when a small sample of gas is made to emit light

by the application of heat or of an electric current. For instance, if we put a few grains

of ordinary salt into a flame, the sodium vapor released by the salt will glow with a

characteristic yellow color. If we put neon gas into an evacuated glass tube and connect

the ends of the tube to a high-voltage generator (Fig. 38.2), the gas will glow with the

familiar orange red color of neon signs (Fig. 38.3).

The light emitted by an atom can be precisely analyzed with a prism (Fig. 38.4);

this breaks the light up into its component colors. In the arrangement shown in Fig.

38.4, each discrete color generates a bright line, called a spectral line. Each kind of

atom has its own discrete spectral lines. The color print on page 1289 shows the

spectral lines of hydrogen, helium, lithium, mercury, and sodium; the numbers above

10�10 m.

Scanning tunneling microscopy image resolves individual atoms.

Surface of this crystal has a defect, a missing atom, knownas a vacancy.

FIGURE 38.1 A scanning tunneling

microsope image of the surface of a semi-

conductor, gallium arsenide (GaAs).

spectral line

quantum mechanics

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http://www.wwnorton.com/college/physics/om/_tutorials/chap38/bohr_model/interface_new.swf

the visible spectral lines give the wavelengths in nanometers. Hydrogen has four spec-

tral lines in the visible region (already mentioned in Chapter 34; see Fig. 34.29) and

many ultraviolet and infrared lines not visible to the human eye. From the color print

we see that the set of spectral lines, or spectrum, belonging to hydrogen is unmistak-

ably different from the spectra of helium, mercury, lithium, and sodium—the spec-

trum of an atom can serve as a fingerprint for its identification.

Spectroscopy provides us with a useful alternative to the traditional “wet” analy-

sis familiar to students of chemistry. By heating a small sample of an unknown sub-

stance—preferably by means of an electric discharge—and inspecting the spectral

lines, we can identify the kinds of atoms in the sample. It so happens that an atom

capable of emitting light of a given wavelength is also capable of absorbing light of that

wavelength. In spectroscopy laboratories, scientists often take advantage of this to per-

form the quantitative analysis of samples of atoms with absorption lines rather than emis-

sion lines. When we illuminate a sample of atoms with white light (a mixture containing

all colors or wavelengths), the atoms will absorb light of their characteristic wave-

length, and upon analyzing the remaining light with a prism, we find dark absorption

lines in the continuous background generated by the white light.The last picture in the

color print on page 1289 shows such an absorption spectrum for sodium vapor.The dark

lines of this absorption spectrum coincide with the bright lines in the emission spec-

trum (see the next-to-last picture on page 1289).

One advantage of spectroscopy over “wet” chemistry is that the analysis can be

performed even on minuscule amounts of material. What is more, atoms can be iden-

tified at a distance. For example, we can identify the atoms on the surface of the Sun

by careful analysis of the distribution of colors in sunlight— we do not need to pluck

a sample of atoms from the Sun. The power of this technique is best illustrated by the

story of the discovery of helium (the “Sun element”—helios is Greek for the Sun). In

1868, this gas was yet unknown to chemists when astronomers discovered it on the

Sun by means of its spectral lines; 30 years later chemists finally found traces of helium

in minerals on the Earth. By spectroscopic techniques, astronomers can identify atoms

in remote stars, clouds of interstellar gas, galaxies, and quasars. For example, Fig. 38.5

shows the spectrum of the star Caph in the constellation Cassiopeia; the spectral

absorption lines indicate the presence of hydrogen, calcium, iron, manganese, chromium,

etc.; these lines are formed in the layers of gas on the surface of the star. Using spec-

troscopy, astronomers can perform a “chemical analysis” of the material on this star,

even though it is many light-years away.

Note that apart from these discrete spectral lines the bulk of the starlight is white

light, a continuous and more or less uniform mixture of all colors. This is the thermal

radiation emitted from the stellar surface. As we know from the preceding chapter, this

1288 CHAPTER 38 Spectral Lines, Bohr's Theory, and Quantum Mechanics

FIGURE 38.3 A neon sign.

prism

screen

spectral lines

slit

dischargetube

Slit provides narrowbeam of light fromdischarge tube.

…and reveals a setof separate spectrallines on screen.

Prism disperses differentcolors of light in beam…

FIGURE 38.4 Analysis of light by means of a prism.

spectrum

4000 V

When terminals of a tube of low-pressure gas are connected to a high-voltage generator, …

…an electric currentflows through the gasand makes it glow.

FIGURE 38.2 An electric discharge tube.



400.0 nm 500.0 nm 600.0 nm 700.0 nm

656.5 nm486.3 nm434.2 nm410.3 nm

402.7 nm 447.3 nm 471.4 nm 501.7 nm 587.7 nm 668.0 nm

671.0 nm610.5 nm497.3 nm460.4 nm413.3 nm

404.8 nm 577.1 nm407.9 nm 436.0 nm 491.7 nm 546.2 nm 579.2 nm

589.8 nm589.2 nm

sodium, absorption spectrum

sodium

mercury

lithium

helium

hydrogen

spectrum of sunlight with main Fraunhofer lines

spectrum of white light

G F E D2 C BD1

(All wavelengths are measured in vacuum.)

492.3 nm438.9 nm


kind of light does not retain the fingerprint of the atoms that produced it. Light origi-

nating in the stellar interior cannot escape directly, but is first tossed back and forth (scat-

tered) many times by the restless atoms of the hot stellar gas. The random motion of

these atoms communicates random changes of wavelength to the light, and what finally

emerges from the stellar interior is a continuous mixture of a wide range of wavelengths.

Figure 38.6 shows a portion of the spectrum of the white light from our Sun.

White light is a continuous and nearly uniform mixture of all colors. However, the

high-resolution spectrum in Fig. 38.6 displays many dark lines in the spectrum of the

Sun, caused by absorption in the layers of gas on the surface of the Sun.These dark lines

are called the Fraunhofer lines.


Ca Ca H H

390 400 410 420 430 440 450 nm

Strong absorption lines on leftare due to ionized calcium…

…and other two stronglines are due to hydrogen.

FIGURE 38.5 A portion of the spectrum

of light from the star Caph Cassiopeiae),

from 390 nm to 450 nm. All the spectral

lines displayed here are absorption lines.

(�

Ca Ca H Ca Fe H

390 400 410 420 430 440 450 nm

Sun’s strong calciumand hydrogen absorptionlines are similar to thosefrom star in Fig. 38.5, …

…but there are alsostrong lines of iron, andmany other somewhatweaker lines.

Fraunhofer lines

FIGURE 38.6 A portion of the spectrum of

light from the Sun, from 390 nm to 450 nm.

Checkup 38.1

QUESTION 1: If, instead of spectral lines, you want the prism in Fig. 38.4 to produce spec-

tral dots (one dot for each color),how do you have to change the experimental arrangement?

QUESTION 2: The flame of the gas burner in a kitchen stove is bluish. Is this color

that of the continuum thermal radiation, or the color of one (or several) spectral lines?

QUESTION 3: Sodium lamps are widely used to illuminate streets and parking lots.According

to the information on the color print on page 1289, what is the wavelength of the light?

QUESTION 4: When the otherwise continuous spectrum of radiation from a small

opening to a hot cavity exhibits reduced intensity at a few particular wavelengths, such

“dark lines” are most likely due to

(A) Shadows caused by the shape of the cavity opening.

(B) Absorption of light by atoms in the cavity.

(C) Destructive interference from multiple sources within the cavity.

(D) Diffraction minima related to the size of the cavity opening.

✔


38.2 SPECTRAL SERIES OF HYDROGEN

Careful examination of the spectral lines produced by an element reveals certain

systematic regularities in the spacings of the lines. These regularities sometimes

become especially striking if, instead of examining only the visible region of the

spectrum, we examine both the visible region and the adjacent ultraviolet and

infrared regions. Figure 38.7 again shows the spectrum of hydrogen in the visible

and the near-ultraviolet regions. We can notice immediately that the spacings of

the lines decrease systematically as we look at shorter and shorter wavelengths. The

hydrogen lines in Fig. 38.7 are said to form a spectral series. In the spectra of other

elements we find similar series; however, the spectra usually contain several over-

lapping series, and this makes it somewhat harder to perceive the regularities in the

spacings.

The systematic pattern in the spacing of the spectral lines of hydrogen shown

in Fig. 38.7 suggests that the wavelengths of these lines should be described by

some simple mathematical formula. Table 38.1 lists the wavelengths of many of

these spectral lines; the spacing between the lines is smaller at shorter wavelengths,

where very many spectral lines are crowded together. In 1855, Johann Balmer scru-

tinized the numbers in such a table and discovered that the wavelengths accurately

fit the formula

(38.1)

with etc. This infinite series of spectral lines is called the Balmer series.

Note that if n approaches infinity, the wavelength approaches the ultimate value

This value is called the series limit, because there

are no spectral lines of shorter wavelength beyond this value.

Balmer’s formula is usually written compactly as

(38.2)

where R is called the Rydberg constant,

(38.3)

Balmer’s formula was purely descriptive, or phenomenological; it did not explain

the atomic mechanism responsible for the production of the spectral lines. Nevertheless,

it proved very fruitful because it led to more general formulas describing other series

R �1

91.176 nm� 1.096 78 107 m�1

1

l� R a 1

22�

1

n2b

l � 4 91.176 nm � 364.70 nm.

n � 3, 4, 5,

1

l�

1

91.176 nm a 1

4�

1

n2b

38.2 Spectral Series of Hydrogen 1291

Spacing of lines decreasessystematically from right…

…to left, and such lines form a spectral series.

397.1

nm

410.3

nm

434.2

nm

486.3

nm

656.5

nm

H� H� H� H� H�

FIGURE 38.7 Spectrum of hydrogen in

the visible and near-ultraviolet regions. The

spectral lines of this spectral series are con-

secutively labeled and so on.

The numbers give the wavelengths of the

spectral lines.

H�, H�, Hg,

WAVELENGTHa

656.47 nm 486.27 nm 434.17 nm 410.29 nm 397.12 nm 389.02 nm 383.65 nm 379.90 nm etc.

a Wavelengths are measured in vacuum.

l

THE BALMER SERIES IN THE HYDROGEN SPECTRUMTABLE 38.1

Balmer series

series limit

Rydberg constant

Conceptsin

Context

43Online

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of spectral lines. Balmer proposed that there might be other series in the hydrogen

spectrum, with the in the parentheses in Eq. (38.2) replaced by or or etc.

This yields the series of wavelengths

(38.4)

(38.5)

(38.6)

These series of spectral lines were actually discovered some years after Balmer proposed

them; they are called, respectively, the Lyman, the Paschen, and the Brackett series.

The first of these series lies in the ultraviolet region of the spectrum; the other series

lie in the infrared.

We can combine all the formulas for all the spectral series of hydrogen into a single

general formula

(38.7)

where and are positive integers and is larger than

According to Eq. (38.7), what is the shortest wavelength of

light that a hydrogen atom will emit or absorb?

SOLUTION: To find the shortest wavelength, we must chose and in Eq.

(38.7) so as to obtain the largest possible value for the right side. This means that

the positive term must be as large as possible, and the negative term as small as

possible, which demands and and gives

(38.8)

and

Since this spectral line belongs to the Lyman series—it is the series limit

for the Lyman series.

Checkup 38.2

QUESTION 1: Figure 38.7 shows that spectral lines of short wavelength in the Balmer

series are tightly spaced. Is this also true for the other spectral series of hydrogen?

QUESTION 2: Consider the series limits for the Lyman, Balmer, and Paschen series.

Which has the longest wavelength? The shortest?

✔

n1 � 1,

l � 1R � 91.176 nm

1

l� R a 1

12�

1

q2b � R

n1 � qn2 � 1

n2n1

EXAMPLE 1

n2.n1n2n1

1

l� R a 1

n22

�1

n21

b

n � 5, 6, 7, …1

l� R a 1

42�

1

n2b

n � 4, 5, 6, …1

l� R a 1

32�

1

n2b

n � 2, 3, 4, …1

l� R a 1

12�

1

n2b

42,32,12,22


Lyman, Paschen, and Brackett series

wavelengths of spectral series of hydrogen

Conceptsin

Context

Conceptsin

Context


QUESTION 3: To find the longest wavelength in the Lyman series, what value of

must you select? In the Balmer series? In the Paschen series? In the Brackett series?

QUESTION 4: In Example 1, we found that the shortest wavelength emitted by hydro-

gen, is in the Lyman series. We previously saw that the Bolmer series

limit is � � 364.70 nm. One of the spectral lines of hydrogen has In

which series is this line?

(A) Lyman (B) Balmer (C) Paschen (D) Brackett

38.3 THE NUCLEAR ATOM

The regularity in the series of spectral lines of the atom must be due to an underlying

regularity in the structure of the atom. We may think of an atom as analogous to a

musical instrument, such as a flute. The atom can emit only a discrete set of spectral

lines, just as the flute can emit only a discrete set of tones which make up a musical scale.

The regularity in the spacing of tones in this musical scale is due to an underlying reg-

ularity in the structure of the flute—the tube of the instrument has regularly spaced tone-

holes that determine what kind of standing waves can build up within the tube and

what kind of waves will be radiated.

J. J.Thomson, the discoverer of the electron, made one of the first attempts at explain-

ing the emission of light in terms of the structure of the atom. Having established that elec-

trons are a ubiquitous component of matter, Thomson proposed the following picture:

An atom consists of a number of electrons, say, Z electrons, embedded in a cloud of pos-

itive charge.The cloud is heavy, carrying almost all of the mass of the atom.The positive

charge in the cloud is so it exactly neutralizes the negative charge of the elec-

trons. In an undisturbed atom, the electrons will sit at their equilibrium positions, where

the attraction of the cloud on the electrons balances their mutual repulsion (Fig. 38.8).

But if the electrons are disturbed by, say, a collision, then they will vibrate around their

equilibrium positions, and this accelerated motion will cause the emission of electro-

magnetic radiation, that is, the emission of light.This model of the atom, called the “plum-

pudding” model, does yield frequencies of vibration of the same order of magnitude as

the frequency of light, but it does not yield the observed spectral series; for instance, on

the basis of this model, hydrogen should have only one single spectral line, in the far ultra-

violet. And in 1910, experiments by Ernest Rutherford and his collaborators established

conclusively that most of the mass of the atom is not spread out over a cloud—instead, the

mass is concentrated in a small kernel, or nucleus, at the center of the atom.

Rutherford had been studying the emission of alpha particles from radioactive

substances. These alpha particles carry a positive charge and they have a mass of

about 4 times the mass of a proton (alpha particles are nuclei of

helium atoms; that is, they are completely ionized helium atoms, with all electrons

removed; see Section 40.3). Some radioactive substances, such as radioactive polo-

nium and radioactive bismuth, spontaneously emit alpha particles with kinetic ener-

gies of several million electron-volts.These energetic alpha particles readily pass through

thin foils of metal, thin sheets of glass, or other materials.

Rutherford was much impressed by the penetrating power of these alpha parti-

cles, and it occurred to him that a beam of these particles can serve as a probe to “feel”

the interior of the atom. When a beam of alpha particles strikes a foil of metal, the

alpha particles penetrate the atoms and are deflected by collisions with the sub-

atomic structures; the magnitude of these deflections gives a clue about the sub-

atomic structures. For example, if the interior of the atom had the “plum-pudding”

6.64 10�27 kg,

�2e,

�Ze�Ze,

l � 121.57 nm.

l � 91.176 nm,

n1

38.3 The Nuclear Atom 1293

In the (incorrect) “plum-pudding”model, an atom consists of a heavycloud of positive charge…

…and electrons sittingat equilibrium positions.

FIGURE 38.8 The lithium atom accord-

ing to the “plum-pudding” model.

“plum-pudding” model

nucleus


structure proposed by J. J. Thomson, then the alpha particles would suffer only very

small deflections, since neither the electrons, with their small masses, nor the diffuse

cloud of positive charge would be able to disturb the motion of a massive and ener-

getic alpha particle.

The crucial experiments were performed by H. Geiger and E. Marsden working

under Rutherford’s direction. They used thin foils of gold and of silver as targets

and bombarded these with a beam of alpha particles from a radioactive source. After

the alpha particles passed through the foil, they were detected on a fluorescent screen,

which registers the impact of each particle by a faint scintillation (Fig. 38.9). To

Rutherford’s amazement, some of the alpha particles were deflected by such a large

angle that they came out backward. In Rutherford’s own words, “It was quite the

most incredible event that has ever happened to me in my life. It was almost as

incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back

and hit you.”

Rutherford immediately recognized that the large deflection must be produced by

a close encounter between the alpha particle and a very small but very massive kernel

inside the atom. He therefore proposed the following picture: An atom consists of a small

nucleus of charge containing almost all of the mass of the atom; this nucleus is sur-

rounded by a swarm of Z electrons, of charge Thus, the atom is like a solar system—

the nucleus plays the role of Sun and the electrons play the role of planets.

On the basis of this nuclear model of the atom, Rutherford calculated what frac-

tion of the beam of alpha particles should be deflected through what angle.The deflec-

tions are caused by the repulsive electric force between the positive alpha particle

(charge and the positive nucleus (charge If an alpha particle passes close

to the nucleus, the electric force will be large, and it will be deflected by a large angle;

if it passes farther from the nucleus, the electric force will be smaller, and it will be

deflected by a smaller angle. Figure 38.10 shows the trajectories of several alpha par-

ticles approaching a nucleus; these trajectories are hyperbolas. The perpendicular dis-

tance between the nucleus and the original (undeflected) line of motion is called the

impact parameter. In order to suffer a large deflection, the alpha particle must hit an

atom with a very small impact parameter, or less; since the alpha particles in

the beam strike the foil of metal at random, only very few of them will score such a close

hit, and only very few will be deflected by a large angle.

10�13 m

�Ze).�2e)

�Ze.

�Ze


evacuatedbox

foil microscope

fluorescent screen

source of� particles

Alpha particles are emittedby a radioactive source…

…and their deflection by atoms in a thin metal foil is detectedupon impact with a screen.

FIGURE 38.9 Rutherford’s apparatus.

b

nucleus

Perpendicular distancebetween original line ofmotion and nucleus isthe impact parameter b.

Particles with smallerimpact parameters sufferlarger deflections.

FIGURE 38.10 Four hyperbolic orbits

of different impact parameters.

impact parameter

SIR ERNEST RUTHERFORD (1871–1937) British experimental physicist and

director of the Cavendish Laboratory at

Cambridge. Rutherford identified alpha and

beta rays. He founded nuclear physics with his

discoveries of the nucleus and of transmutation

of elements by radioactive decay; he also pro-

duced the first artificial nuclear reaction. He

was awarded the Nobel Prize in chemistry in

1908.


Checkup 38.3

QUESTION 1: The hydrogen atom has one single electron. What does this atom look

like according to the “plum-pudding” model? If the cloud of positive electric charge is

spherically symmetric, what is the equilibrium position of the electron?

QUESTION 2: If an alpha particle approaches a nucleus with an impact parameter of

zero, what will happen to the alpha particle?

QUESTION 3: Is the distance of closest approach between a nucleus and an alpha par-

ticle in a hyperbolic orbit (such as shown in Fig. 38.10) larger or smaller than the

impact parameter?

QUESTION 4: If the alpha particles in Fig. 38.10 had a negative electric charge

instead of their positive charge how would this affect the deflections illustrated

in this figure?

QUESTION 5: Rutherford’s experiments used foils of silver (atomic number 47) or gold

(atomic number 79) as targets. For the same impact parameter, an alpha particle of a

given energy incident on a silver target is deflected

(A) Through a smaller angle than when incident on a gold target.

(B) Through the same angle as when incident on a gold target.

(C) Through a larger angle than when incident on a gold target.

38.4 BOHR’S THEORY

Rutherford’s experiments did reveal the gross arrangement of the electrons in the atom,

but not the details of their motion. Since the electrons make up the outer layers of an

atom, their arrangement and motion determine the chemical bonds of the atom and the emis-

sion of light, that is, they determine all the chemical and spectroscopic properties. But when

physicists tried to calculate the electron motion according to the laws of classical

mechanics and electromagnetism, they immediately ran into trouble.

To gain some insight into the source of this trouble, consider the case of the hydro-

gen atom. Suppose that the single electron of this atom is moving around the nucleus,

according to the laws of classical mechanics, in a circular orbit of atomic size, that is,

with a radius of about . In such an orbit, the electron would have a centripetal

acceleration due to the attractive electric force of the nucleus, which would be very

large, about Because of this acceleration, the electron would emit high-fre-

quency electromagnetic radiation, that is, it would emit light (see Chapter 33). The

energy carried away by the light would have to be supplied by the electron. Hence the

emission process would have the same effect on the electron as a friction force—it

would remove energy from the electron. This kind of friction would cause the elec-

tron to leave its circular orbit and gradually spiral in toward the nucleus, in just the

way the residual atmospheric friction on an artificial satellite in a low-altitude orbit

around the Earth causes it to spiral down toward the ground. A calculation using the

laws of classical mechanics and electricity shows that the rate of emission of light by

the orbiting electron in a hydrogen atom would be quite large. Correspondingly, the

rate of energy loss of the electron would be large—the electron would spiral inward

and collide with the nucleus within a time as short as s!

Thus, our classical calculation leads us to the incongruous conclusion that hydro-

gen atoms, and other atoms, ought to be unstable—all the electrons ought to collapse

into the nucleus almost instantaneously. Furthermore, the light that the electron emit-

10�10

1023 m/s2.

10�10 m

�2e,

�2e

✔


NIELS BOHR (1885–1962) Danish

theoretical physicist. He worked under J. J.

Thompson and Rutherford in England and

then became director of the Institute of

Theoretical Physics in Copenhagen, for the

foundation of which he was largely responsible.

After formulating the quantum theory of the

atom, he played a leading role in the further

development of the new quantum mechanics.

He received the Nobel Prize in 1922.

43Online

ConceptTutorial



ted during the spiraling motion would have to be a wave of continually increasing

amplitude and increasing frequency (in musical terminology, crescendo and glissando);

this is so because when the electron came closer to the nucleus, the electric force would

become larger, leading to a larger centripetal acceleration and a higher frequency of

the orbital motion. Hydrogen atoms do not behave as this classical calculation pre-

dicts. Hydrogen atoms are stable, and when they do emit light, they emit discrete fre-

quencies (spectral lines) instead of a continuum of frequencies.

These irreconcilable disagreements between the observed properties of atoms and

the calculated properties gave evidence of a serious breakdown of the classical mechan-

ics of Newton and the classical theory of electromagnetism. Although these theories

had proved very successful on a macroscopic scale, they were in need of some drastic

modifications on an atomic scale.

In 1913, Niels Bohr took a bold step toward resolving these difficulties. He made

the radical proposal that, at the atomic level, the laws of classical mechanics and of classi-

cal electromagnetism must be replaced or supplemented by other laws. Bohr expressed these

new laws of atomic mechanics in the form of several postulates:

1. The orbits and the energies of the electrons in an atom are quantized, that is,

only certain discrete orbits and energies are permitted. When an electron is in one

of the quantized orbits, it does not emit any electromagnetic radiation; thus, the

electron is said to be in a stationary state. The electron can make a discontinuous

transition, or quantum jump, from one stationary state to another. During such a

transition from one stationary state to another stationary state of lower energy, the

electron does emit radiation.

2. The laws of classical mechanics apply to the orbital motion of the electrons in a

stationary state, but these laws do not apply during the transition from one state

to another.

3. When an electron makes a transition from one stationary state to another, the excess

energy �E is released as a single photon of frequency f � �E�h.

4. The permitted orbits are characterized by quantized values of the orbital angu-

lar momentum L.This angular momentum is always an integer multiple of h�2:

(38.9)

The number n is called the angular-momentum quantum number.

Let us now see how to calculate the stationary states and the spectrum of the hydro-

gen atom on the basis of these postulates. For the sake of simplicity, we will assume that

the electron moves in a circular orbit around the proton, which remains at rest (Fig.

38.11). The centripetal acceleration of the electron is This centripetal accelera-

tion is produced by the force of attraction between the electron and the proton, that

is, the Coulomb force Thus, the equation of motion for the

electron is

(38.10)

The orbital angular momentum for a circular orbit is [see Eq. (13.34)].

According to Bohr’s postulate, this orbital angular momentum must be multi-

plied by an integer n:

(38.11)mevr � n h

2p

h2pL � mevr

me

v2

r�

1

4p�0

e

2

r 2

mea � Fe 2(4p�0r

2).

v2r.

n � 1, 2, 3, …L � n h

2p


stationary state

quantization of angular momentum

Bohr’s quantum postulates

r

v

a

Electron moving in assumed circular orbit experiences a centripetal acceleration…

…produced by attractive Coulomb force of nucleus.

FIGURE 38.11 Electron in circular

orbit around a proton.


It is convenient to rewrite this as

(38.12)

where (pronounced “h-bar”) is Planck’s constant h divided by

(38.13)

From Eq. (38.12), we obtain

(38.14)

and when we substitute this into Eq. (38.10), we obtain an equation for the radius of

the orbit,

(38.15)

Solving this for the radius, we find

(38.16)

From Eq. (38.16) we see that the radii of the permitted orbits are proportional to

The smallest value of n is and this leads to the radius of the smallest permit-

ted orbit, which is called the Bohr radius and designated by

(38.17)

The radii of the other permitted orbits are multiples of the Bohr radius:

(38.18)

For this gives respectively. Figure 38.12 shows

the permitted circular orbits, drawn to scale.

r � 4a0, 9a0, 16a0, …,n � 2, 3, 4, …,

r �4p�0n2U2

mee 2

� n2a0

a0 �4p�0U2

mee 2

� 0.529 10�10 m � 0.0529 nm

a0:

n � 1,

n2.

r �4p�0n2U2

mee 2

me a nUmerb 2

1

r�

1

4p�0

e

2

r 2

v �nUmer

U �h

2p�

6.63 10�34 J�s

2p� 1.05 10�34 J�s

2p:U

mevr � nU


Bohr radius

n � 1

n � 2 n � 3

n � 4

0 2 4 6 8 10

10�10 m

Orbits permitted by Bohr theory are circular, …

…with radii r � n2a0 that aresquare-integer multiples ofBohr radius a0 � 0.0529 nm.

FIGURE 38.12 The possible Bohr

orbits of an electron in the hydrogen atom.

h (h-bar)


The energy of the electron in one of these orbits is a sum of kinetic and potential

energies.The kinetic energy is and the potential energy is the electrostatic

potential energy [see Eq. (25.11)]. Hence

the total energy is

(38.19)

Next, we substitute from Eq. (38.14) and then r from Eq. (38.16), and we obtain

(38.20)

(38.21)

or

(38.22)

This energy depends on the quantum number n, and as a reminder of this depend-

ence, we label this energy with a subscript n:

(38.23)

According to this equation, the energies of all the stationary states are negative. The

stationary state with the least energy, or the most negative energy, is the state with

for which

(38.24)

And, by Eq. (28.23), the energies of the other stationary states are fractions of this

energy:

(38.25)

Thus, and so on.

Figure 38.13 displays these quantized energies in an energy-level diagram. Each

horizontal line represents one of the energies given by Eq. (38.25). According to Bohr’s

E3 � �(13.69) eV,E2 � �(13.64) eV,E1 � �13.6 eV,

En � �

13.6 eV

n2

� �2.18 10�18 J � �13.6 eV

� �

9.11 10�31 kg (1.602 10�19 C)4

2 (4p 8.85 10�12 F/m)2 (1.055 10�34 J�s)2

E1 � �

mee4

2(4p�0)2U2

n � 1,

En � �

mee4

2(4p�0)2U2

1

n2

E � �

mee4

2(4p�0)2U2

1

n2

�mee

4

2(4p�0)2U2

1

n2�

mee4

(4p�0)2U2

1

n2

�1

2 me

n2U2

m2e

a mee 2

4p�0n2U2b 2

�e

2

4p�0

a mee 2

4p�0n2U2b

E �1

2 me

n2U2

m2er

2�

1

4p�0

e2

r

v2

E �1

2 mev

2 �1

4p�0

e

2

r

E � K � U

U � (�e) (14p�0)er � �(14p�0)e 2r

K � 12mev

2,


energies of stationary states

energy-level diagram


assumptions, the electron emits a photon when it makes a transition, or a quantum

jump, from one stationary state to a lower stationary state. Such quantum jumps have

been indicated by arrows in Fig. 38.13. The stationary state of lowest energy

is called the ground state, the next state is called the first excited state, the next

state is called the second excited state, and so on.

Ordinarily, the electron of the hydrogen atom is in the ground state, that is, the

circular orbit with radius and energy This is the configuration of least

energy, and it is the configuration into which the atom tends to settle when it is left

undisturbed. As long as the atom remains in the ground state, it does not emit light.

To bring about the emission of light, we must first kick the electron into one of the

excited states, that is, a circular orbit of larger radius and higher energy. We can do

this by heating a sample of atoms or by passing an electric current through the sample.

Collisions between the atoms will then disturb the electronic motions and occasion-

ally kick an electron into a larger orbit. From there, the electron will spontaneously

jump into a smaller orbit, emitting a photon. Note that the quantum jumps indicated

by colored arrows in Fig. 38.13 form several series: one series consists of all those jumps

(indicated by blue arrows) that end in the ground state, another series consists of all those

jumps (indicated by red arrows) that end in the first excited state, etc. These series of

jumps give rise to the series of spectral lines: the Lyman series, the Balmer series, etc.

From our formula for the energies of the states of the hydrogen atom we can cal-

culate the frequency of the light emitted in a quantum jump from some initial state

to a final state, as in the following example.

�13.6 eV.a0

(n � 3)

(n � 2)

(n � 1)


ground state and excited states

E

–13.6

eV

–13

–14

–12

–11

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

0 6543

n

2

1Lyman series

Balmer series

Pfund series

Brackett seriesPaschen series

In an energy-leveldiagram, each horizontalline represents one of theallowed energies…

…and a colored arrow shows anelectron’s possiblequantum jump.

Lowest energy state iscalled the ground state.

FIGURE 38.13 Energy-level diagram for

the hydrogen atom. Jumps (blue) that end in

the ground state (n = 1) give rise to the

Lyman series; jumps (red) that end in the

first excited state give rise to the

Balmer series; etc.

(n � 2)


wavelength of emitted photon

frequency of emitted photon

Calculate the frequency and the wavelength of the light emitted

by the electron in a quantum jump from the second excited

state to the first excited state

SOLUTION: The energy of the initial state is and the energy

of the final state is Hence the electron releases an energy

According to Bohr’s postulates, this energy is radiated as a single photon of fre-

quency

and wavelength

This wavelength agrees with wavelength of the first spectral line of the Balmer

series (see Table 38.1), except for a round-off error.

More generally, we can calculate the frequency of the light emitted in a quantum

jump from some initial state of quantum number to some final state of quantum

number nf . The initial energy of the electrons is and the final energy is Thus,

the electron releases an energy

(38.26)

This energy is radiated as a single photon of frequency

(38.27)

From the frequency of the light, we can calculate the wavelength. Since the wave-

length is inversely proportional to the frequency, Eq. (38.27) implies the

following expression for the wavelength of the emitted light:

(38.28)

or, with Eq. (38.26),

(38.29)

Comparison of Eqs. (38.7) and (38.29) yields the following theoretical formula for

the Rydberg constant:

1

l�

mee4

4p(4p�0)2U3c

a 1

n2f

�1

n2i

b

1

l�

Ei � Ef

hc�

Ei � Ef

2pU c

l � cf,

f �¢E

h�

Ei � Ef

h

Ei � Ef �mee

4

2(4p�0)2U2

a 1

n2f

�1

n2i

b

Ei � Ef ,

Ef .Ei

ni

l �c

f�

3.00 108 m/s

4.56 1014 Hz� 6.58 10�7 m � 658 nm

f �¢E

h�

1.89 eV

h�

1.89 eV 1.60 10�19 J/eV

6.63 10�34 J�s� 4.56 1014 Hz

¢E � E3 � E2 � �

13.6

9 eV � a�

13.6

4 eV b � 1.89 eV

E2 � �(13.64) eV.

E3 � �(13.69) eV,

(n � 2).(n � 3)

EXAMPLE 2


Conceptsin

Context


(38.30)

Upon insertion of the accurate values of the fundamental constants given in Appendix

6, we obtain

(38.31)

This theoretical value of R agrees quite well with the experimental value quoted in

Eq. (38.3).1

Suppose that the atoms in a sample of hydrogen gas are ini-

tially in the ground state. If we illuminate these atoms with

light (from some kind of lamp), what frequencies will the atoms absorb?

SOLUTION: Absorption of light is the reverse of emission. When an electron in

an atom absorbs a photon (supplied by the lamp), it jumps from the initial state to

a state of higher energy. The energy of the photon must match the energy differ-

ence between the states. Thus, the frequencies of the photons that the electrons

can absorb when jumping upward from the ground state are exactly those fre-

quencies that they emit when jumping downward into the ground state, that is,

the frequencies of the Lyman series (see Fig. 38.13).

What is the ionization energy of the hydrogen atom, that is,

what energy must we supply to remove the electron from the

atom when it is initially in the ground state?

SOLUTION: To remove the electron from the atom, we must lift it into an orbit

of infinite radius, that is, an orbit of quantum number The energy required

to accomplish this transition from to is

This is the ionization energy.

With his theory Bohr attained the goal of explaining the regularities in the spectrum

of hydrogen in terms of the regularities of the structure of the atom. By showing that

this structure is based on a simple numerical sequence, he fulfilled the ancient dream of

Pythagoras of a universe based on simple numerical ratios, a dream that arose from an

analogy with musical instruments. Bohr’s theory tells us how the atom plays its tune.

E1 � Eq � �13.6 eV a 1

12�

1

q2b � 13.6 eV

n � qn � 1

n � q.

EXAMPLE 4

EXAMPLE 3

� 1.097 37 107 m�1

R �mee

4

4p(4p�0)2U3c


1The small disagreement between the theoretical value of R given in Eq. (38.31) and the experimental value

given in Eq. (38.3) is due to the motion of the nucleus of the hydrogen atom, which we have neglected in

our calculation. A careful calculation that takes into account the motion of electron and proton about their

common center of mass requires the substitution of the so-called reduced mass for

the electron mass, and eliminates the disagreement. See also Problem 31.

m � memp (me � mp)

Conceptsin

Context

R �9.109 53 10�31 kg (1.602 189 10�19 C)4

4p(4p 8.854 178 10�12 F/m)2 (1.054 589 10�34 J�s)3 2.997 925 108 m/s


Checkup 38.4

QUESTION 1: Why is Bohr’s postulate of stationary states in direct contradiction with

classical mechanics and electromagnetism?

QUESTION 2: If an electron in a hydrogen atom makes a transition from some state to

a lower state, does its kinetic energy increase or decrease? Its potential energy? Its

orbital angular momentum?

QUESTION 3: Suppose that an electron is initially in the state. It then makes

three quantum jumps in succession, such that the first jump produces a photon in the

Paschen series, the second a photon in the Balmer series, and the third a photon in

the Lyman series. In the energy-level diagram of Fig. 38.13, find the arrows that indi-

cate these transitions.

QUESTION 4: In the energy-level diagram of Fig. 38.13, what transition would pro-

duce the series limit of the Balmer series?

(A) to (B) to (C) to

(D) to (E) to

38.5 QUANTUM MECHANICS; THE SCHR

.O

.DINGER EQUATION

Bohr’s theory is a hybrid. It relies on some basic classical features (orbits) and grafts onto

these some quantum features (quantum jumps, quanta of light). In the 1920s the coop-

erative efforts of several brilliant physicists—L. de Broglie, E. Schrödinger,

W. Heisenberg, M. Born, P. Jordan, P. A. M. Dirac—established that the remaining

classical features had to be eradicated from the theory of the atom. Bohr’s semiclassical

theory had to be replaced by a new quantum mechanics with an entirely different equa-

tion of motion.

The basis of the new quantum mechanics was laid by the discovery that electrons—

as well as protons, neutrons, and all the other “particles” found in nature—have not only

particle properties but also wave properties. When a beam of electrons is made to pass

through an extremely narrow slit, the electrons exhibit diffraction.This means that elec-

trons are neither classical particles nor classical waves. Electrons, just like photons, are a

new kind of object with a subtle combination of particle and wave properties. Electrons are

wavicles. As in the case of photons [see Eq. (37.15)], the wavelength associated with an

electron or some other wavicle is inversely proportional to its momentum:

(38.32)

This formula was postulated by de Broglie, and it is called the de Broglie wavelength.

This wavelength is quite small, even for electrons of the lowest energies attainable in

experiments with beams of electrons.

What is the de Broglie wavelength of an electron of kinetic

energy 1.0 eV, which is about the lowest energy that can be

attained in experiments with beams of electrons?

EXAMPLE 5

l �hp

n � 2n � 11n � 2n � qn � 1n � qn � 2n � 3n � 1n � 2

n � 5

✔


de Broglie wavelength


SOLUTION: In joules, a kinetic energy of 1.0 eV is This kinetic

energy is related to the momentum as follows:

Hence

According to the de Broglie equation, the wavelength of the electron is then

The wave properties of electrons were confirmed experimentally by C. J. Davisson

and L. Germer, who observed interference effects with beams of electrons scattered

by a crystal. In these experiments the crystal acts as a grating for electron waves, in

much the same way as it acts as a grating for X rays in X-ray interference experiments

(see the discussion of X-ray interference and the production of Laue spots in Section

37.5). Davisson and Germer found that constructive interference of the electron waves

scattered by the rows of atoms in the crystal produced strong interference maxima in

selected directions, and they were able to confirm that these directions agreed with

calculations based on the de Broglie formula (38.32).

In the new quantum mechanics, or wave mechanics, the motion of an electron is

described by a wave equation, the Schrödinger equation. This Schrödinger equation

plays the same role for electrons as the wave equation derived from the Maxwell equa-

tions plays for photons [see Eq. (33.33)]. As in the case of photons, the behavior of

an electron is governed by a probabilistic law. The electron is represented by an elec-

tron wavefunction (Greek letter psi) calculated from the Schrödinger equation, and

the intensity of this electron wave at some point determines the probability that there

is an electron particle at that point [compare Eq. (37.24)]:

(38.33)

Furthermore, as a consequence of their wave properties, electrons obey the Heisenberg

uncertainty relations for position and momentum [see Eq. (37.30)],

(38.34)

These quantum uncertainties are of crucial importance for the behavior of an elec-

tron inside an atom. For such an electron, the uncertainty in the position is very large—

about as large as the size of the atom. This implies that the electron follows no definite

orbit. It is therefore not surprising that the Bohr theory should have failed in all

attempts at calculating the electron motions in the helium atom and in other atoms with

several electrons; what is surprising is that this theory should have succeeded as well

as it did in the case of the hydrogen atom.

¢y ¢py �U2

[probability for an electron at a point] r �c�2

c

� 1.2 10�9 m � 1.2 nm

l �hp

�6.63 10�34 J�s

5.4 10�25 kg�m/s

� 5.4 10�25 kg�m/s

p � 22meE � 22 9.11 10�31 kg 1.6 10�19 J

E �p2

2me

1.6 10�19 J.

38.5 Quantum Mechanics; the Schrödinger Equation 1303

Schrödinger equation

probability interpretation of wavefunction

LOUIS VICTOR, PRINCE DE BROGLIE(DE BROY) (1885–1962) French

theoretical physicist. He discovered the de

Broglie wavelength by reasoning that if waves

have particle properties, then maybe particles

have wave properties. For his discovery of the

wave properties of matter he was awarded the

Nobel Prize in 1929, after the existence of

these wave properties was confirmed experi-

mentally.


Consider an electron in the ground state of hydrogen. Show

that a well-defined orbit is inconsistent with the Heisenberg

uncertainty relations.

SOLUTION: If the electron is to follow a well-defined orbit, the uncertainty in its

momentum (in any direction) must be much smaller than the magnitude of the

momentum. According to Eq. (38.14), the speed of the electron in the smallest

circular orbit is and the magnitude of the momentum is

For a well-defined orbit, the uncertainty of the momentum must

be much smaller than this magnitude of the momentum:

(38.35)

Furthermore, for a well-defined orbit, we must require that the uncertainty in the

position must be much smaller than the size of the orbit:

V r (38.36)

In view of the inequalities (38.35) and (38.36), the product �y �py must be smaller

than :

V (for a well-defined orbit) (38.37)

This is inconsistent with the Heisenberg uncertainty relation (38.34).

In wave mechanics, the quantization of the energy in the hydrogen atom and other

atoms is an automatic consequence of the wave properties of the electron.The attractive

electric force of the nucleus confines the electron wave to some region near the nucleus

and causes the wave to reflect back and forth across the region, forming a standing

wave. The different stationary states of the atom correspond to different standing-wave

modes. As in the case of standing waves on a string, the standing electron waves in the

atom have a discrete set of eigenfrequencies. For electrons and other particles, just as

for photons, the energy is related to the frequency by

(38.38)

Thus a discrete set of frequencies for electron standing waves corresponds to a dis-

crete set of energies. The ground state of the atom is analogous to the fundamental

mode of the string, the first excited state is analogous to the first overtone, and so on.

However, whereas the determination of the eigenfrequencies of standing waves on a

string is a quite trivial mathematical exercise, the determination of the eigenfrequen-

cies of the electron waves in an atom is a formidable mathematical problem, which

requires an investigation of the solutions of the Schrödinger wave equation.

Although we will not deal here with the mathematical complexities of the

Schrödinger wave equation in three dimensions, we can gain some insight into how elec-

tron waves determine the discrete energies in the hydrogen atom by means of the fol-

lowing simple calculation. Let us assume that the electron travels around the nucleus

along an orbit of radius r, but instead of thinking of the electron as a particle, as in the

Bohr theory, let us think of it as a wave. Figure 38.14 shows a “snapshot” of such an

electron wave at one instant of time. If the wave is to have a well-defined amplitude

at all points, it must repeat whenever we go once around the circumference—if it did

E � hf

U¢y ¢py

U

¢y

¢py V Ur

p � mev � U r.

v � U mer,(n � 1)

EXAMPLE 6


energy of stationary state in terms of frequency

ERWIN SCHRÖDINGER (1887–1961)Austrian theoretical physicist. Another of the

founders of the new quantum mechanics, he

received the Nobel Prize in 1933 for the dis-

covery of his wave equation.


not, then the wave amplitude would have two different values at a single point, which

makes no sense. Hence, the wave is subject to the condition that some integer number

of wavelengths must fit around the circumference:

(38.39)

Although this equation looks like the condition for a standing wave on a string of

length we are here dealing with a traveling wave, for which the entire wave pat-

tern in Fig. 38.14 rotates rigidly around the center. With the de Broglie relation

Eq. (38.39) becomes

or

(38.40)

Since rp is the orbital angular momentum, this equation coincides with Bohr’s quan-

tization condition for the angular momentum, Eq. (38.9). Thus, the wave picture of the

electron implies the quantization of the angular momentum and, therefore, the quantiza-

tion of the energy. But we must not take this calculation too seriously—its legitimacy

is questionable, since it relies in part on the wave picture and in part on the particle pic-

ture. Furthermore, analysis of the Schrödinger equation shows that it is not enough

to consider the behavior of the wave around the circumference; we must also consider

the behavior of the wave outward along each radius.Thus, this simple calculation pro-

vides no more than a crude qualitative sketch of the role of the wave properties of elec-

trons in the atom.

There are some applications where a simple one-dimensional calculation gives

accurate and meaningful results, such as electron reflection from a barrier, electron

transmission through a barrier (known as “tunneling”; see Physics in Practice:

Ultramicroscopes), and confinement of electrons in one direction in some modern lay-

ered semiconductor devices. In one dimension, each allowed stationary state is described

by an electron wavefunction that is a solution to the time-independent

Schrödinger equation:

(38.41)

When the potential energy U(x) in this equation is some known function of position,

then solutions may be found for the allowed values of the total energy E and the cor-

responding allowed position-dependent wavefunctions Since the terms in

Eq. (38.41) with U and E involve the potential and total energies, respectively, it is

not surprising that the first term represents the kinetic energy Thus the

Schrödinger equation incorporates a wave-mechanical version of the the energy rela-

tion of classical mechanics.

Consider an electron in one dimension.The electron is confined

to a region of length L, say, to the interval (see Fig.

38.15). Inside this region, the potential energy and outside

This is sometimes referred to as a “particle in a box” or the “infinite potential

U (x) � q.U (x) � 0,

0 � x � LEXAMPLE 7

K � U � E

K � p22me.

c � c(x).

�

U2

2me

d2

dx2 c(x) � U(x) c(x) � E c(x)

c � c(x)

rp �nh2p

2pr �nhp

l � hp,

2pr,

2pr � nl


If we think of an electronorbiting the nucleus as a wave…

…then some integer numberof wavelengths must fitaround the circumference.

FIGURE 38.14 In this example, six wave-

lengths of the electron wave fit around the

circumference of the orbit.

time-independent Schrödinger equation


well.” What electron wavefunctions are solutions to the time-independent one-

dimensional Schrödinger equation? What are the corresponding allowed energies?

SOLUTION: Since the potential energy U(x) is infinite outside the box, the only

possible solution for the Schrödinger equation (38.41) outside the box is

We thus know that both at and at Inside the box, we have

and the Schrödinger equation becomes

We can rearrange this in the more familar form

(38.42)

We encountered similar equations, where the second derivative of a function is

proportional to the negative of the function, in our study of oscillations in Chapter

15. By analogy with Eq. (15.15), we know solutions to Eq. (38.42) are of the form

where the wave vector of the sinusoidal function is given by

(38.43)

Since we know that we require

(38.44)

The amplitude A must be nonzero, since the probability of finding the electron

cannot be zero everywhere. Equation (38.44) thus determines that Hence

our wavefunctions are of the form

(38.45)

Similarly, since we know that we require

which has solutions when the argument of the sine function is an integer multiple

of Hence the solutions correspond to discrete values of the wave vector k, which

we label with a quantum number n:

for (38.46)

Substituting these values into Eq. (38.45), we see that the corresponding allowed

wavefunctions are

(38.47)cn(x) � A sin an px

Lb

n � 1, 2, 3, …kn � n p

L

p.

c(L) � A sin (kL) � 0

c(L) � 0,

c(x) � A sin (kx)

d � 0.

c(0) � A sin (0 � d) � A sin d � 0

c(0) � 0,

k2 �2me E

U2

k � 2pl

c(x) � A sin (kx � d)

d 2

dx2 c(x) � �

2meE

U2 c(x)

�

U2

2me

d2

dx2 c(x) � E c(x)

U(x) � 0,

x � L.x � 0c(x) � 0

c(x) � 0.


L

x

0

3(x)

L

x

0

2(x)

L

x

0

1(x)

Wave functionsresemble standingwaves on a string:an integer numberof half wavelengthsmust fit intolength L.

FIGURE 38.16 Wavefunction (x) for the

first few states of the infinite potential well.

U � 0

positionx

U

L0

Since potential energy U(x) isinfinite for x � 0 and x � L, …

…electron is confined tothis region, where U(x) � 0.

FIGURE 38.15 The one-dimensional

infinite potential well. The electron is con-

fined to the region 0 � x � L.


probability of finding electron in a region

The wavefunctions corresponding to and 3 are shown in Fig. 38.16; in

all cases, an integer number of half wavelengths fits into the length L of the box.

This condition on the wavelength is analogous to the condition for standing waves

on a string.

From Eq. (38.43) we can express the energy in terms of the wave vector k,

(38.48)

or, inserting the allowed values of we obtain the allowed energies

(38.49)

The allowed energies are thus square-integer multiples of the ground-state energy

for (38.50)

where the ground-state energy is

(38.51)

The energies of the first five wavefunctions are shown in the energy-level diagram

of Fig. 38.17.

COMMENTS: Since inside the box, the energy of an electron confined

to a region of length L is all kinetic energy. Note that in the ground state,

this energy is nonzero, and is known as zero-point energy; even at absolute zero

temperature, the energy of the electron cannot be reduced below this value. This

zero-point energy of the infinite potential well is inversely proportional to the square

of the length L of the confinement region.Thus confinement costs energy. Note also

that the higher kinetic energy of an excited state corresponds to a shorter wavelength.

This is also evident from the de Broglie relation, since

The probabilistic information contained in the wavefunction was mentioned in Eq.

(38.33): the probability of finding the electron at a given point in space is proportional to

the square of the magnitude of the wavefunction at that point.2 The probability of find-

ing an electron at a single point is infinitesimal, so the quantity in Eq. (38.33) repre-

sents the probability per unit volume for finding an electron in a small volume around

some point. Similarly, in one dimension the quantity represents a probability per

unit length, that is, is the probability that the electron is in a region of width dx

centered at the point x.Thus in practice, the probabilistic relation (38.33) means that the

probability P of finding the electron in the region between two points a and b is given by

(38.52)P � �b

a

�c(x)�2dx

�c(x)�2�c(x)�2

�c�2

c

K � p22m � h22ml2.

n � 1

U (x) � 0

E1 �U2p2

2meL2

n � 1, 2, 3, …En � n2E1

E1,

En �U2

2me

an p

Lb 2

kn,

E �U2k2

2me

n � 1, 2,


2We use the square of the magnitude of to allow for cases where is a complex function. Although in prac-

tice the wavefunction is often complex, in this text we will only encounter examples where is a real function.c

cc


the one-dimensional infinite potential well.

25E1

16E1

9E1

4E1

= E1U 2 22meL

2

E

n � 5

n � 4

n � 3

n � 2

n � 1

Allowed energies are square-integer multiples of the ground-state energy, En = n2E1.

zero-point energy


normalization condition


Since the sum of the probabilities of all possible locations of the electron must be 1,

that is, the electron must be somewhere, we must also require that the wavefunction

satisfy

(38.53)

This requirement is known as the normalization condition. In addition to satisfying

the Schrödinger equation and being continuous and smooth at boundaries, a meaningful

wavefunction must satisfy the normalization condition.

In classical mechanics, an electron with energy E in a one-

dimensional box of length L simply bounces back and forth

between the walls at constant speed. A classical electron thus has equal probabil-

ity of being anywhere in the box; the probability per unit length for an electron to

be at any point is then For example, classical mechanics predicts that

an electron would be found in the middle half of the box with a probability of

exactly or 50% of the time. What are the normalized wavefunctions for an elec-

tron in the one-dimensional box of Example 7? In the quantum-mechanical ground

state, what is the probability per unit length that the electron is at the center of the

box? What is the probability of finding the electron in the middle half of the box?

SOLUTION: In order to use the wavefunctions from Eq.

(38.47) to calculate probabilities, we must determine the overall multiplicative

constant A from the normalization condition (38.53). Since outside the

box, we need only consider the region inside the box. For any quantum number n,

inserting the solutions (38.47) into the normalization condition (38.53) gives

(38.54)

where we have used the fact that for any integer number of half wavelengths, the

average value of the square of the sine function is and so the integral over the

length L is Solving for A, we obtain

(38.55)

independent of the quantum number n. So the normalized wavefunctions are

(38.56)

For the ground state, we use the wavefunction. The probability per unit

length that the electron is at the center of the box is obtained by evaluating

at

Thus, compared with classical mechanics, where quantum mechan-

ics predicts that the electron is twice as likely to be at the center of the box.

Pclassical � 1L,

`c1a L

2b ` 2 � `B

2

L sin apL2

Lb ` 2 �

2

L sin2

ap2b �

2

L

x � L2:

�c1(x)�2n � 1

cn(x) � A2

L sin a npx

Lb

A � A2

L

12L.

12,

1 � �L

0

�cn(x)�2dx � �L

0

A2 sin2

an px

Lb dx � 1

2 LA2

c(x) � 0

cn(x) � A sin (npxL)

12,

Pclassical � 1L.

EXAMPLE 8

��q

�q�c(x)�2dx � 1


The middle half of the box extends from to so the probabil-

ity of finding the electron between these points is

With a change of variable this becomes

From our table of integrals (Appendix 4), Hence

Thus quantum mechanics indicates that the electron in the ground state is much

more likely to be in the middle half of the box than classical physics would predict:

approximately 82% compared with 50%.

COMMENT: The probability distribution depends on the state. For the ground

state, the electron is most likely to be found at the center of the box, but for the first

excited state, it is least likely to be found there (see Fig. 38.18).

The above examples of an electron in a box demonstrate that we cannot be cer-

tain about the position of the electron. Since the electron may be traveling to the right

or to the left, its momentum is also uncertain. Within our small confinement region

or, similarly, within the atom, electrons always behave very much like waves. But out-

side the atom, they will sometimes behave pretty much like classical particles. Roughly,

we can say that classical mechanics will be a good approximation whenever the quan-

tum uncertainties are small compared with the relevant magnitudes of positions and

momenta. For instance, for the electrons in the beam of a TV tube, the quantum uncer-

tainty in the momentum is negligible compared with the magnitude of the momen-

tum. Under these conditions, classical mechanics gives an adequate description of the

motion of the electrons.

What we have said about the wave mechanics of electrons also applies to other par-

ticles, or wavicles, found in nature—they all have wave properties and they all have quan-

tum uncertainties in their position and momentum. Strictly speaking, even large

macroscopic bodies have wave properties. For example, an automobile is a wavicle and

it has some quantum uncertainty in its position. However, it turns out that the quan-

tum uncertainties are very small whenever the mass of the body is large compared with

atomic masses—the quantum uncertainty in the position of an automobile is typically no

more than about a number that can be ignored for all practical purposes. Hence,

for automobiles and other macroscopic bodies, quantum effects are completely insignif-

icant and classical mechanics gives an excellent description of the motion of these bodies.

10�18 m,

�1

2�

1

p� 0.818

P �2

p a 1

2 u �

1

4 sin 2ub ` 3p4

p4�

2

pe c 3

8 p � a�

1

4b d � a 1

8 p �

1

4b f

�sin2 u du � 1

2 u � 14 sin 2u.

P �2

L L

p �3p4

p4sin2

u du

u � pxL,

P � �3L4

L4�c1(x)�2dx � �

3L4

L4

2

L sin2

apx

Lb dx

x � 34L,x � 1

4L


L

x

0

3(x) 2

— —

L

x

0

2(x) 2

— —

L

x

0

1(x) 2

L2

L2

L2

— —Probability of finding an electronis zero at edges and nodes…

…and has the same maximumvalue for any n(x).

FIGURE 38.18 Probability per unit

length, as a function of position for

the first few states of the infinite potential

well.

�c(x)�2,



PHYSICS IN PRACTICE ULTRAMICROSCOPES

The wave properties of electrons find practical application in

the transmission electron microscope (TEM), which

employs electron waves to form a highly magnified image of

an object in the same way as an ordinary microscope employs

light waves.The maximum magnification attainable by ordi-

nary microscopes is limited by the wavelength of light. If we

attempt to observe an object as small as or smaller than the

wavelength of light, the image becomes indistinct, because

the light waves suffer pronounced diffraction when passing

through and around such a small object, and the resulting

diffraction fringes blur the image. The electron waves used

in typical electron microscopes have wavelengths of 0.05 nm,

which is 10 000 times shorter than the wavelength of light;

the electron waves are therefore much less susceptible to dif-

fraction than light.

The main “optical” elements in a transmission electron

microscope are the same as in an ordinary microscope, an

objective lens and an ocular, or projector, lens (see Fig. 1).

The “lenses” are not made of glass, but of magnetic fields,

carefully shaped so as to provide deflections similar to those

experienced by light in a glass lens. The electron rays emerg-

ing from the projector lens are intercepted by a detector array,

which records the image digitally for viewing and analysis.

Another kind of electron microscope is the scanning elec-

tron microscope (SEM). The principle of operation of this

microscope is quite different, and bears no resemblance to

the operation of ordinary light microscopes. The scanning

electron microscope relies on the particle properties of elec-

trons as well as their wave properties. Instead of forming an

image by means of refracted electron waves, the scanning

electron microscope uses a diffraction-limited fine beam of

electrons to bombard the object.The beam “scans” across the

object line by line in a sweep pattern, like the sweep pattern

of the electron beam in an ordinary TV tube. A detector picks

up the current of electrons that recoil from the object and of

secondary electrons knocked out of the object by the inci-

dent primary electrons (Fig. 2).This detected current is ampli-

fied and fed into a video monitor, which displays a picture.

Scanning electron microscopes produce very crisp pictures

with an exceptional depth of field and strong shadows that

give a vivid three-dimensional impression, but they do not

attain the extremely high magnifications of transmission elec-

tron microscopes.

FIGURE 1 (a) Schematic diagram of a

transmission electron microscope (TEM).

(b) Picture of a human chromosome pre-

pared with a TEM.

FIGURE 2 (a) Schematic diagram of a

scanning electron microscope (SEM).

(b) Picture of human chromosomes

prepared with an SEM. Figure 1b is a

close-up view of one such chromosome.

(a) (b)

electron gun

specimen

objectivelens

projectorlens

detectorarray

to vacuumpump

electron gun

electronbeam

condenserlens

objectivelens

scanningcoil

specimen

detector

to vacuumpump

amplifier

videomonitor

(a)

(b)



A newer and more powerful kind of electron microscope

is the scanning tunneling microscope (STM). This micro-

scope scans the surface of the object with a fine tungsten

needle whose motion is precisely controlled by a delicate sus-

pension system (see Fig. 3a). A potential difference is applied

between the needle and the surface of the object. The needle

is not actually in contact with the surface; it merely sweeps

across the surface line by line. The gap between the needle

and the surface is effectively an insulator, which tends to block

the motion of the electrons from the needle to the surface of

the object. However, the wave properties of electrons permit

them to spread for some distance into this gap, and leak across

to the surface. This kind of leakage of an electron across a

gap where its motion is classically forbidden is called tun-

neling. The probability for an electron to succeed in tunnel-

ing across the gap depends drastically on the size of the gap.

Whenever the needle, during its sweep, comes near any peak

in the surface—such as a bulging atom—the tunneling prob-

ability starts to increase drastically, and so does the current

from the needle to the surface. This incipient increase of the

current is detected by the electronic circuit connected to the

needle, and an amplified feedback signal is sent to the sus-

pension system to lift the needle, so as to keep it at constant

height from the peak. Thus, the needle skims over the peaks

and valleys of the surface, maintaining a relatively constant

height above the “terrain.” The amplified signal used to con-

trol the suspension is also sent to a computer, where it is

processed and then fed into a video monitor for display.

Modern scanning tunneling microscopes attain magnifica-

tions of up to and they permit us to “see” individual atoms

(Fig. 3b).

This technique for constructing a picture of the surface by

sweeping a needle across it is also exploited in the atomic-

force microscope (AFM). But the needle of this kind of

microscope is placed in direct contact with the surface—it is

pressed against the surface with a force of, say 10�8 N, and it

is lifted or lowered by the suspension system so as to keep the

force constant during the sweep. Thus, the needle actually

explores the shape of the surface by directly feeling it, just as

you might explore the shape of a surface by feeling it with the

tip of your finger. An alternative AFM technique is “tapping

mode” (see Example 2 of Chapter 15), where the needle oscil-

lates and taps the surface. While tapping the surface during

scanning, the needle is lifted or lowered to keep the oscilla-

tion amplitude constant. Magnifications attained with atomic-

force microscopes are nearly the same as those attained with

scanning tunneling microscopes. The picture displayed in

Fig. 4 was prepared with an atomic-force microscope.

108,

FIGURE 4 Picture of a carbon nanotube lying across ultra thin

stripes of gold, prepared with an atomic-force microscope (AFM).

FIGURE 3 (a) Schematic diagram of a scan-

ning tunneling electron microscope (STM).

(b) Picture of iron atoms prepared with an

STM. The 48 iron atoms arranged in a circle

on the surface of a copper crystal form a con-

fining potential known as a “quantum

corral,” and the ripples in the image show the

wave nature of the surface copper electrons.

tunnelingcurrent

preamppiezoelectricscanner

specimenstage

needle

horizontalsweep motioncontrol

specimen

vertical motioncontrol

steppermotor

(a)

(b)



SUMMARY

PHYSICS IN PRACTICE Ultramicroscopes (page 1310)

SPECTRAL SERIES OF HYDROGEN

QUANTIZATION OF ANGULAR MOMENTUM

(38. 7)R � 1.096 78 107 m�11

l� R a 1

n22

�1

n21

b

(38.9)n � 1, 2, 3, …L � nU

(38.13)U �h

2p� 1.05 10�34 J�s

(38.17)a0 �4p�0U2

mee 2

� 0.0529 nm

PLANCK’S CONSTANT (“h-bar”)

BOHR RADIUS

Checkup 38.5

QUESTION 1: One electron has a kinetic energy of 1 eV; another has a kinetic energy

of 100 eV. Which has the shorter de Broglie wavelength? By what factor?

QUESTION 2: Suppose that an electron has a de Broglie wavelength of 20 nm. If we

increase the speed of this electron by a factor of 4, what will be its new de Broglie

wavelength?

QUESTION 3: An electron and a proton have the same kinetic energy. Which has the

longer de Broglie wavelength?

QUESTION 4: What is the probability per unit length that an electron in the first

excited state of a one-dimensional box is at the center of the box? An elec-

tron in the second excited state (See Fig. 38.18.)

QUESTION 5: Bohr’s theory of the hydrogen atom assumes that the electron remains

in a given orbital plane, say, the x–y plane, and that it has no motion in the z direc-

tion. What are and in this case? Is this consistent with the uncertainty prin-

ciple?

QUESTION 6: Take the proton-to-electron mass ratio to be For the

ground states of the respective one-dimensional boxes, what is the ratio of the energy

of a proton in a region of width to that of an electron in a region of width

(A) 0.02 (B) 50 (C) 2000 (D) 5 106 (E) 2 1013

10�10 m?

10�15 m

mp me � 2000.

¢pz¢z

(n � 3)?

(n � 2)

✔

(38.23, 38.25)En � �

mee4

2(4p�0)2U2

1

n2� �

13.6 eV

n2

ENERGY OF STATIONARY STATES OF HYDROGEN


4. How can Rutherford’s experiment tell us something about the

size of the nucleus?

5. At low temperatures, the absorption spectrum of hydrogen

displays only the spectral lines of the Lyman series. At higher

temperatures, it also displays other series. Explain.

6. The planets move around the Sun in circular orbits. Is their

orbital angular momentum quantized?

7. In a muonic atom, a muon orbits around the nucleus. The

mass of the muon is 207 times the mass of the electron. What

is the Bohr radius for a muonic atom with a hydrogen nucleus?

Questions for Discussion 1313


1. Do the spectral lines seen in a stellar spectrum (for example,

Fig. 38.5) tell us anything about the chemical composition of

the stellar interior?

2. The lower spectrum of hydrogen shown in the chapter photo

displays all of the spectral lines simultaneously. Since a hydro-

gen atom emits only one spectral line at a time, how can all

the lines be visible simultaneously?

3. The target used in Rutherford’s scattering experiment was a

very thin foil of metal. What is the advantage of a thin foil

over a thick foil in this experiment?

and (38.27)1

l�

f

c�

Ei � Ef

hcf �

Ei � Ef

h

(38.32)l � hp

(38.41)�

U2

2me

d2

dx2 c(x) � U(x) c(x) � E c(x)

E � hf (38.38)

is a solution to the Schrödinger equation is everywhere continuous

(normalization) is everywhere continuous [except if U(x) is infinite]dc(x)dx��q

�q�c(x)�2dx � 1

c(x)c(x)

En � n2E1 � n2 U2p2

2meL2

cn(x) � B2

L sin a npx

Lb

FREQUENCY AND WAVELENGTH OF PHOTON

EMITTED IN TRANSITION

DE BROGLIE WAVELENGTH OF WAVICLE

ENERGY OF STATIONARY STATE IN TERMS OF

FREQUENCY

TIME-INDEPENDENT SCHRÖDINGER EQUATION

One-dimension:

CONDITIONS FOR WAVEFUNCTION

One-dimension:

ELECTRON IN A BOX One-dimension, 0 � x � L:

∂ n � 1, 2, 3, …

(38.50, 38.51)

Wavefunctions:

Energies:

(38.52)P � �b

a

�c(x)�2dxPROBABILITY FOR FINDING ELECTRON IN A

REGION a � x � b

U � 0

positionx

U

L0



PROBLEMS

†38.2 Spec t ra l Ser ies o f Hydrogen

1. Use Eq. (38.4) to calculate the wavelengths of the first four

lines of the Lyman series.

2. Show that the spectral lines of the Balmer series all have a

higher frequency than the spectral lines of the Paschen series.

Do the spectral lines of the Paschen series all have a higher

frequency than those of the Brackett series?

3. A hydrogen atom at rest emits a photon. By Eq. (37.15),

the photon of highest momentum has the shortest

wavelength, corresponding to the Lyman series limit. What is

the recoil velocity of the hydrogen atom when this photon is

emitted?

4. Which of the spectral lines of the Brackett series is closest in

wavelength to the first spectral line of the

Pfund series? By how much do the wavelengths differ?

*5. When astronomers examine the light of a distant galaxy, they

find that all the wavelengths of the spectral lines of the atoms

are longer than those of the atoms here on Earth by a common

multiplicative factor. This is the red shift of light; it is a Doppler

shift caused by the motion of recession of the galaxy, away from

Earth. In the light of a galaxy beyond the constellation Virgo,

astronomers find spectral lines of wavelengths 411.7 nm and

435.7 nm.

(a) Assume that these are two spectral lines of hydrogen, with

the wavelengths multiplied by some factor. Identify these

lines. What is the factor by which these wavelengths are

longer than the normal wavelengths of the two spectral

lines? What is the factor by which the frequencies are lower?

(n2 � 5; n1 � 6)

p � hl,

(b) If the speed of recession is low compared with the speed

of light, the Doppler shift of light obeys a formula similar

to that for the Doppler shift of sound [see Eq. (17.17)].

Calculate the speed of recession of the galaxy.

*6. One of the series of spectral lines of the lithium atom is the

principal series, with the following wavelengths, measured in

vacuum: 617.0 nm, 323.4 nm, 274.2 nm, 256.3 nm, 247.6 nm.

These wavelengths approximately fit the formula

where is the Rydberg constant for

lithium, and where s and p are constants characteristic of the

series. Find the values of these constants. (Hint: Write the

equations for 1�� for two different wavelengths, eliminate the

unknown s by subtracting these formulas, and then solve for p

by trial and error.)

*7. Another of the spectral series of the lithium atom is the dif-

fuse series, with the following wavelengths, measured in

vacuum: 610.5 nm, 460.4 nm, 413.4 nm, 391.6 nm, 379.6 nm.

These wavelengths approximately fit the formula

where, as in the preceding problem,

and where p and d are constants.

(a) Find the values of these constants.

(b) The principal series (see Problem 6) and the diffuse series

of lithium are analogous to two spectral series of hydrogen.

Which two series? (Hint: See the preceding problem.)

R � 1.097 29 107 m�1

n � 3, 4, 5, …1

l� R c 1

(2 � p)2�

1

(n � d ) 2d

R � 1.097 29 107 m�1

n � 2, 3, 4, …1

l� R c 1

(1 � s)2�

1

(n � p)2d

8. Given that the orbital angular momentum of an atom is quan-

tized, can we conclude that the orbital magnetic moment is

also quantized?

9. Bohr’s theory of the hydrogen atom can be adapted to the

singly ionized helium atom, that is, the helium atom with one

missing electron. To what other ionized atoms can Bohr’s

theory be adapted?

10. According to the Complementarity Principle, formulated by

Bohr, a wavicle has both wave properties and particle proper-

ties, but these properties are never exhibited simultaneously; if

the wavicle exhibits wave properties in an experiment, then it

will not exhibit particle properties, and conversely. Give some

examples of experiments in which wave or particle properties

(but not both simultaneously) are exhibited.

11. Show that photons obey the de Broglie relation.

12. If the de Broglie wavelengths of two electrons differ by a

factor of 2, by what factor must the kinetic energies differ?

13. According to the de Broglie relation, the wavelength of an

electron of very small momentum is very large. Could we take

advantage of this to design an experiment that makes the wave

properties of the electron obvious?

14. Describe the interference pattern expected for an electron

wave incident on a plate with two very narrow parallel slits

separated by a small distance.

15. Electron microscopes achieve high resolution because they

use electron waves of very short wavelength, usually less than

0.05 nm. Why can we not build a microscope that uses photons

of equally short wavelength?

†For help, see Online Concept Tutorial 43 at www.wwnorton.com/physics


15. What is the angular momentum of a vinyl record (a uniform

disk) rotating at revolutions per minute? The moment of

inertia of the record is Express the answer

as a multiple of

16. What are the angular momentum, the kinetic energy, the

potential energy, and the net energy of an electron in the

smallest Bohr orbit in the hydrogen atom? Express

your answers in SI units.

17. What is the speed of an electron in the smallest Bohr

orbit? Express your answer as a fraction of the speed of light.

Is it justified to ignore relativistic effects in the Bohr model?

18. What is the centripetal acceleration of an electron in the

smallest Bohr orbit in the hydrogen atom?

19. Hydrogen atoms in highly excited states with a quantum

number as large as have been detected in interstellar

space by radio astronomers. What is the orbital radius of the

electron in such an atom? What is the energy of the electron?

20. If a hydrogen atom is initially in the first excited state, what is

the longest wavelength of light it will absorb? What is the

shortest wavelength of light it will absorb?

21. If you bombard hydrogen atoms in their ground state with a

beam of particles, the collisions will (sometimes) kick atoms

into one of their excited states. What must be the minimum

kinetic energy of the bombarding particles if they are to

achieve such an excitation?

22. Suppose that the electron in a hydrogen atom is initially in the

second excited state What wavelength will the atom

emit if the electron jumps directly to the ground state? What

two wavelengths will the atom emit if the electron jumps to

the first excited state and then to the ground state?

23. A hydrogen atom emits a photon of wavelength 102.6 nm.

From what stationary state to what lower stationary state did

the electron jump?

*24. Show that the speed of an electron in a Bohr orbit with quan-

tum number n is given by

Also, find an expression for the centripetal acceleration of an

electron in the nth Bohr orbit.

*25. The quantity is called the Compton wavelength. The

quantity is called the “classical electron radius.”

Show that the Bohr radius, the Compton wavelength, and

the classical electron radius are in the ratios where

The quantity � is called the fine-structure

constant. What is the numerical value of this constant?

*26. A hydrogen atom is initially in the ground state. In a collision

with an argon atom, the electron of the hydrogen atom absorbs

an energy of 15.0 eV. With what speed will the electron be

ejected from the hydrogen atom?

*27. Suppose that a sample of hydrogen atoms, initially all in the

ground state, is under bombardment by a beam of electrons of

� � e2(4p�0U c).

1:�:�2,

e2(4p�0mec 2)

U (mec)

v �1

n

e2

4p�0U

(n � 3).

n � 732

(n � 1)

(n � 1)

(n � 1)

U .

1.3 10�2 kg�m2.

3313

Problems 1315

38.3 The Nuc lear Atom

*8. What is the distance of closest approach for a 5.5-MeV alpha

particle in a head-on collision with a gold nucleus? With an

aluminum nucleus?

*9. The nucleus of platinum has a radius of and

an electric charge of 78e. What must be the minimum energy of

an alpha particle in a head-on collision if it is to just barely reach

the nuclear surface? Assume the alpha particle is pointlike.

*10. Consider a model of the hydrogen atom with a pointlike nucleus,

but assume that the electron charge is uniformly distributed

over a sphere of radius The atom is in equilib-

rium when the nucleus is at the center of the electron charge

distribution. Find the frequency of oscillation when the electron

sphere is displaced from equilibrium by a distance

What is the wavelength of a photon with this frequency?

*11. The probability for a scattering event to occur is equal to the

fraction of the area of the target for which that event will

occur. The area per nucleus for an event is the scattering cross

section This area is related to the impact parameter b for

that event by The impact parameter for a deflection

of more than (backscattering) is for alpha

particles of energy 5.7 MeV scattering from gold nuclei. What

is the scattering cross section for backscattering? Consider an

incident beam of alpha particles with a beam radius of 2.0

millimeters and a target with gold nuclei in the

beam area. What is the probability for backscattering?

*12. An alpha particle of energy 5.5 MeV is incident on a silver

nucleus with an impact parameter The dis-

tance of closest approach of the particle is

Find the speed at the point of closest approach. (Hint: The

angular momentum is conserved.)

**13. A foil of gold, thick, is being bombarded by

alpha particles of energy 7.7 MeV. The particles strike at

random over an area of of the foil of gold.

(a) How many atoms are there within the volume

under bombardment? The

density of gold is and the mass of one atom is

(b) It can be shown that to suffer a deflection of more than

an alpha particle must strike within of

the center of a gold nucleus. What is the probability for

this to happen? (Hint: See Problem 11.)

(c) If alpha particles impact on the foil, how many

will suffer deflections of more than 30�?

†38.4 Bohr ’s Theor y

14. If you wanted to give an apple an angular momentum of at

what rate would you have to spin it about its axis? Treat the

apple as a uniform sphere of mass 0.20 kg and radius of 4.0 cm.

U ,

1.0 1010

5.5 10�14 m30�,

3.27 10�25 kg.

19.3 g/cm3,

1.0 cm2 2.1 10�5 cm

1.0 cm2

2.1 10�5 cm

2.7 10�14 m.

8.0 10�15 m.

5.0 1017

1.0 10�13 m90�

s � pb2.

s.

x � R.

R � 0.053 nm.

6.96 10�15 m



kinetic energy 12.2 eV. In an inelastic collision between a

hydrogen atom and one of the incident electrons, the hydro-

gen atom will occasionally absorb all, or almost all, the kinetic

energy of the electron and make a transition from the ground

state to an excited state. If so, what excited state will the

hydrogen atom attain? What are the possible spectral lines

the atom can emit subsequently?

*28. The singly ionized helium atom (usually designated HeII or

He�) has one electron in orbit around a nucleus of charge 2e.

(a) Apply Bohr’s theory to this atom and find the energies of

the stationary states. What is the value of the ionization

energy, that is, the energy that you must supply to remove

the electron from the atom when it is in the ground state?

Express the answer in electron-volts.

(b) Show that for every spectral line of the hydrogen atom,

the ionized helium atom has a spectral line of identical

wavelength.

*29. Assume that, as proposed by J. J. Thomson, the hydrogen

atom consists of a cloud of positive charge e, uniformly dis-

tributed over a sphere of radius R. However, instead of placing

the electron in static equilibrium at the center of the sphere,

assume that the electron orbits around the center with uni-

form circular motion under the influence of the electric cen-

tripetal force If the angular momentum of

this orbiting electron is quantized according to Bohr’s theory

(so that what are the radii and the energies of the

quantized orbits? What are the frequencies of the photons

emitted in transitions from one quantized orbit to another?

What must be the value of R if at least two orbits are to fit

inside this atom?

**30. In principle, Bohr’s theory also applies to the motion of the

Earth around the Sun. The Earth plays the role of the electron,

the Sun that of the nucleus, and the gravitational force that of

the electric force.

(a) Find a formula analogous to Eq. (38.18) for the radii of

the permitted circular orbits of the Earth around the Sun.

(b) The actual radius of the Earth’s orbit is

What value of the quantum number n does this corre-

spond to?

(c) What is the radial distance between the Earth’s actual

orbit and the next larger orbit?

**31. In our calculation of the energies of the stationary states of

hydrogen we pretended that the proton remains at rest. Actually,

both the electron and the proton orbit about their common

center of mass. Show that the energies of the stationary states,

taking into account this motion of the proton, are given by

where

m �memp

me � mp

En � �

me4

2(4p�0)2U2

1

n2

1.50 1011 m.

L � nU ),

(e24p�0)(rR3).

(Hint: The electron and the proton move in circles of radii

and

where r is the distance between the electron and the proton.

According to Bohr’s theory, the net angular momentum of this

system of two particles is quantized,

**32. The atom of positronium consists of an electron and a

positron (or antielectron) orbiting about their common center

of mass. According to Bohr’s theory, the net angular momen-

tum of this system is quantized, What is the radius

of the smallest possible circular orbit of this system? What is

the wavelength of the photon released in the transition from

to (Hint: the electron and the positron have the

same mass. See also Problem 31.)

38.5 Quantum Mechanics ; the Schrödinger Equat ion

33. What must be the energy of an electron if its wavelength is to

equal the wavelength of visible light, about 550 nm?

34. Find the de Broglie wavelength for each of the following

electrons with the specified kinetic energy: electron of 20 keV

in a TV tube, conduction electron of 5.4 eV in a metal, orbit-

ing electron of 13.6 eV in a hydrogen atom, orbiting electron

of 91 keV in a lead atom.

35. What is the de Broglie wavelength of an electron in the

ground state of hydrogen? In the first excited state?

36. What is the de Broglie wavelength of a tennis ball of mass

0.060 kg moving at a speed of 1.0 m/s?

37. An electron microscope operates with electrons of kinetic

energy 40 keV. What is the wavelength of such electrons? By

what factor is this wavelength smaller than that of visible light?

38. A photon and an electron each have an energy of

What are their wavelengths?

39. If the de Broglie wavelengths of two electrons differ by a

factor of 4, by what factor must their kinetic energies differ?

40. The “thermal” neutrons in a nuclear reactor typically have a

kinetic energy of about 0.050 eV. What is the de Broglie

wavelength of such a neutron?

41. Suppose that the velocity of an electron has been measured to

within an uncertainty of — 1.0 cm/s. What minimum uncer-

tainty in the position of the electron does this imply?

42. The nucleus of the aluminum atom has a diameter of

Consider one of the protons in this nucleus.

The uncertainty in the position of this proton is necessarily

less than What is the minimum uncertainty in

its momentum and velocity?

43. If the position of a parked automobile of mass is

uncertain by what is the corresponding

uncertainty in its velocity?

_1.0 10�18 m,

2.0 103 kg

7.2 10�15 m.

7.2 10�15 m.

6.0 103 eV.

n � 1?n � 2

L � nU .

L � nU .)

rp � r me

mp � me

re � r mp

mp � me



44. For an electron confined to a one-dimensional box of length

what is the kinetic energy in the ground state?

What is the corresponding speed? Is it justified to ignore rela-

tivistic effects for such an electron?

45. Suppose that a particle confined to a one-dimensional box of

length L is in its first excited state. What is the probability per

unit length that the particle is in a small interval near

Near Near

46. Consider the possibility that a nucleus might contain an

electron. What would be the energy of an electron in the

ground state of a one-dimensional box of nuclear size, say

Since nuclear energies are typically only on

the order of 1 MeV, do you expect that an electron might be

confined within the nucleus?

47. For a particle in the state of quantum number n of a one-

dimensional box of length L, at how many points within the

box is the probability per unit length of finding the particle

zero? At how many points is it maximum? In each case,

describe the locations of those points.

48. An electron in a one-dimensional box of length

makes a transition from the first excited state to the ground

state. What is the wavelength of the emitted photon?

49. Use symmetry and the result of Example 8 to determine the

probability that an electron in the ground state of a one-

dimensional box is in the region

*50. Neutrinos of energy 10 MeV are emitted by the sun. What is

the de Broglie wavelength of such a neutrino? (Hint: Since the

neutrino is a particle of very small mass, its momentum at

such high kinetic energy must be calculated according to the

formula for the momentum of an ultra-relativistic particle.)

*51. What is the de Broglie wavelength of an electron if its kinetic

energy is equal to its rest-mass energy? (Hint: For a relativistic

particle the momentum must be calculated according to the

relativistic formula.)

*52. What is the de Broglie wavelength of the Earth in its motion

around the Sun?

0 � x � 14 L.

L � 0.30 nm

1.0 10�14 m?

x � 34 L?x � 1

2 L?

x � 14 L?

L � 0.10 nm,

*53. Show that the de Broglie wavelength of an electron can be cal-

culated from the formula

where the wavelength is expressed in nm and the kinetic

energy K is expressed in eV.

*54. Consider the solutions to the Schrödinger equation for a

particle in a one-dimensional box. For n large, show that the

probability per unit length, averaged over a small but suitable

interval, yields the classical result

*55. Obtain the energies [Eqs. (38.50–38.51)] of an electron con-

fined to an interval from to along the x axis by a

simple de Broglie wavelength calculation:

(a) What are the de Broglie wavelengths of standing electron

waves in this interval? Assume that, as for a standing wave

on a string, the amplitude of oscillation is zero at

and so that an integer number of half wavelengths

fit into the length L.

(b) What are the momenta corresponding to these wave-

lengths?

(c) What are the energies of the stationary states?

(d) Evaluate the energies of the ground state and the first

three excited states numerically for

**56. The harmonic oscillator potential is a parti-

cle of mass m in this potential oscillates with frequency �0.

The ground-state wavefunction for a particle in the harmonic

oscillator potential has the form

(a) By substituting U(x) and into the one-dimensional,

time-independent Schrödinger equation [Eq. (38.41)],

find expressions for the ground-state energy E and the

constant a in terms of m, and

(b) Apply the normalization condition to determine the con-

stant A in terms of m, and v0.U ,

v0.U ,

c(x)

c(x) � Ae�ax2

U(x) � 12mv2

0x2;

L � 0.10 nm.

x � L,

x � 0

x � Lx � 0

Pclassical � 1L.

l

l � 1.231K


REVIEW PROBLEMS

57. The series limit for the Balmer series is 364.7 nm. What are

the series limits for the Lyman, Paschen, and Brackett series?

58. Are the wavelengths of all the spectral lines of the Lyman

series shorter than the series limit of the Balmer series? Are

the wavelengths of all the spectral lines of the Balmer series

shorter than the series limit of the Paschen series? Are the

wavelengths of all the spectral lines of the Paschen series

shorter than the series limit of the Pfund (n2 � 5) series?

59. If a hydrogen atom is in the ground state, what is the longest

wavelength it will absorb?

60. What is the frequency of the orbital motion for an electron in

the smallest Bohr orbit? In the next Bohr

orbit? Do either of these frequencies coincide with the fre-

quency of the light emitted during the transition from

to

61. Find the orbital radius, the speed, the angular momentum,

and the centripetal acceleration for an electron in the

Bohr orbit of hydrogen.

62. When a hydrogen atom is initially in the ground state, the

ionization energy (or the energy for removal of the electron) is

n � 2

n � 1?

n � 2

(n � 2)(n � 1)



4. (B) Absorption of light by atoms in the cavity. Before emis-

sion from the opening, thermal radiation can strike atoms in

the cavity; the atoms will absorb light at their characteristic

wavelengths, resulting in dark absorption lines.

Checkup 38.2

1. Yes. In any series, the separation of successive spectral lines is

given by the difference of the inverse square of successive inte-

gers, which approaches zero as the integers approach infinity.

2. Similar to the result of Example 1, the series limits are given

by For the series given, the largest wavelength

corresponds to for the Paschen series, and the shortest

wavelength to for the Lyman series.n2 � 1

n2 � 3

1l � Rn22.

Answers to Checkups

Checkup 38.1

1. The original spectra are in the form of lines because of the

presence of the slit between the light source and the prism; to

produce spectral dots, you would have to replace the slit by a

small hole.

2. For the moderately hot temperature of a normal flame, we saw

in Chapter 37 that the continuum thermal radiation is peaked

in the infrared, not the blue; thus the bluish color is due to

spectral lines.

3. According to the color print on page 1289, this yellow light is

due to two nearly equal wavelengths (a “doublet”) with values

589.2 nm and 589.8 nm.

13.6 eV. What is the ionization energy for a hydrogen atom

initially in the first excited state?

63. For an electron in the first excited state of a one-dimensional

box, what is the probability for finding the electron in the

interval Compare this with the classical result,

64. An electron makes a transition from the state to the

state of a one-dimensional box, emitting a photon of

wavelength 450 nm. What is the length of the box?

*65. An alpha particle of energy 5.5 MeV is incident on a copper

nucleus with an impact parameter of Find the

distance of closest approach of the particle, and find the speed

at the point of closest approach. (Hint: Both the energy and

the angular momentum of the alpha particle are conserved.)

*66. The muon (or mu meson) is a particle somewhat similar to an

electron; it has a charge and a mass 206.8 times as large as

the mass of the electron. When this particle orbits around a

proton, the two particles form a muonic hydrogen atom, similar

to an ordinary hydrogen atom, but with the muon playing the

role of the electron. Calculate the Bohr radius of this muonic

atom and calculate the energies of the stationary states. What

is the energy of the photon emitted when the muon makes a

transition from the to the state?

*67. An electron in a hydrogen atom is initially in the ground state.

The electron absorbs a photon from an external light source and

thereby makes a transition to the state. What must have

been the energy and the wavelength of the absorbed photon? If

the electron now jumps spontaneously to the state, what

are the energy and the wavelength of the emitted photon?

n � 3

n � 4

n � 1n � 2

�e

5.0 10�15 m.

n � 1

n � 3

P � 18.

0 � x � 18 L?

*68. The doubly ionized lithium atom (usually designated LiIII or

has one electron in orbit around a nucleus of charge 3e.

What is the radius of the smallest Bohr orbit in doubly ion-

ized lithium? What is the energy of this orbit?

*69. Consider a helium atom in interstellar space in a circular orbit

around a meteoroid of mass 4.0 kg and radius 10 cm under the

influence of the gravitational force. We can apply Bohr’s

theory to this system; the meteoroid plays the role of the

nucleus, the atom that of the electron, and the gravitational

force that of the electric force. Find formulas analogous to

Eqs. (38.18) and (38.23) for the orbital radii and the energies

of the permitted circular orbits. Because of the finite size of

the meteoroid, the smallest feasible orbit has a radius of 10 cm.

What is the quantum number and what is the energy (in eV)

of a helium atom in this orbit?

*70. What is the de Broglie wavelength of a nitrogen molecule

in air at room temperature Assume that the mol-

ecule is moving with the rms speed of molecules at this tem-

perature.

*71. Interferometric methods permit us to measure the position of

a macroscopic body to within Suppose we

perform a position measurement of such a precision on a body

of mass 0.050 kg. What uncertainty in momentum is implied

by the Heisenberg relation? What uncertainty in velocity?

*72. Consider an electron in a circular orbit of quantum number n

in a hydrogen atom. The orbit is well defined provided that

and Show that if these require-

ments are not in conflict with the Heisenberg uncertainty rela-

tions. Thus, large orbits in the hydrogen atom are well defined.

n W 1,¢y V r.¢p V p

; 1.0 10�12 m.

(20�C)?(N2)

Li2�)



3. In each case, the longest wavelength is given by the smallest

difference in the two terms of Eq. (38.7); this corresponds to

choosing Since for the

Lyman, Balmer, Paschen, and Brackett series, respectively, the

corresponding values of are 2, 3, 4, and 5.

4. (A) Lyman. Similar to Example 1, the shortest wavelength of

each series is given by Thus

the series with = 2, 3, and 4 have wavelengths too long to

include the given wavelength, so the given wavelength must

belong to the or Lyman series.

Checkup 38.3

1. According to the “plum-pudding” model, the positive charge

is spread over a cloud of atomic size, approximately

and the electron is a pointlike particle in this cloud. If the

cloud of positive charge is spherically symmetric, the electron

is in equilibrium when at rest at the center of the cloud, where

the electric field is zero.

2. An impact parameter of zero means that the alpha particle

approaches the nucleus head-on; thus, the alpha particle

bounces back, along its initial line of motion (or if it has suffi-

cient energy, it penetrates into the nucleus).

3. Since the impact parameter is the perpendicular distance

between the nucleus and the undeflected original line of

motion, the distance of closest approach in the repulsively

deflected hyperbolic orbit is larger than the impact parameter.

4. If the incident particles were negatively charged, the deflections

would be attractive; thus, the orbits would go partially around

the nucleus, and the alpha particle would emerge in a some-

what downward direction, instead of being deflected upward.

5. (A) Through a smaller angle than when incident on a gold

target. For the smaller charge on the silver nucleus, the

repulsive electric force is smaller, and so the deflection is

smaller.

Checkup 38.4

1. According to classical mechanics and electromagnetism, an

electron in a circular orbit is accelerating and thus must radi-

ate, lose energy, and spiral inward toward the nucleus.

2. The kinetic energy is the first term in Eqs. (38.19)–(38.21),

which is positive and inversely proportional to the square

of the quantum number n. For a transition to a lower state,

n decreases, so the kinetic energy increases. The potential

energy, the second term in the same equations, is also inversely

proportional to the square of n, but is negative and so

�Ze

10�10 m,

n2 � 1,

n2

l � n22R � n2

2 91.176 nm.

n1

n2 � 1, 2, 3, and 4n1 � n2 � 1.

decreases as n decreases. According to Bohr’s postulate [see

Eq. (38.9)], the orbital angular momentum is proportional to

n and thus decreases when the hydrogen atom makes a transi-

tion to a lower state.

3. Transitions of the Paschen series all end at so the first

transition is from to Transitions of the Balmer

series all end at so the second transition is from

to the final transition must then be from to

Transitions that end at are in the Lyman series.

4. (D) to Transitions of the Balmer series all end

at the transition that produces the series limit has the

largest energy difference, and so begins at

Checkup 38.5

1. Higher energy corresponds to shorter wavelength. Since the

kinetic energy is proportional to the square of the momentum

(that is, and the de Broglie wavelength is

the wavelength varies inversely with the square root

of the energy. So the 100-eV electron has a factor of 10

shorter wavelength than the 1-eV electron.

2. The de Broglie wavelength is inversely proportional to the

momentum and, since also to the speed. A factor of 4

increase in speed thus results in a factor of 4 decrease in wave-

length to 5 nm.

3. Since the de Broglie wavelength is the kinetic

energy is Equality of energies thus

implies that the wavelength must be longer for the electron,

because of its smaller mass.

4. Since the state has a node at the center of the box

the probability per unit length that an electron

is at the center is zero. The state has a maximum at the

center. In Example 8, we found that all the stationary states of

the one-dimensional box have the same wavefunction ampli-

tude, and thus the probability per unit length at the

maximum is 2�L.

5. If an electron remains within the x–y plane, then it has

If there is no motion in the z direction, then the elec-

tron has These are inconsistent with the uncertainty

relation, which requires

6. (D) The energy varies inversely with the mass and

inversely with the square of the length, so compared with the

electron, the proton energy will be both decreased by a factor

of 2000 and increased by a factor of for an overall

increase by a factor of The given lengths correspond

to the sizes of nuclear and atomic regions; atomic energies are

typically of order eV, and nuclear energies, MeV.

5 106.

1010,

5 106.

¢z ¢pz � U 2.

¢pz � 0.

¢z � 0.

12L,

n � 3

[c(L2) � 0],

n � 2

K � p22m � h22ml2.

p � hl,

p � mv,

l � hp,

K � 12 mv2 � p22m)

n � q.

n � 2;

n � 2.n � qn � 1n � 1.

n � 2n � 2;

n � 3n � 2,

n � 3.n � 5

n � 3,


C O N C E P T S I N C O N T E X TThis scanning tunneling microscope (STM) image depicts the electron

densities associated with individual atoms on the surface of gallium arsenide,

a modern semiconductor preferred in some applications over silicon, the

material predominantly used in the fabrication of transistors and inte-

grated circuits. The atoms are arranged in a periodic structure, known as

a lattice.

As we learn about the quantum structure of materials, we will con-

sider such questions as:

? The nearly free electrons in a metal interact with the atoms in a lat-

tice, and this restricts the permitted energies of the electrons. How

do these restrictions arise? (Section 39.4, page 1336)

? The bright spot in the photo of the crystal lattice shows an indium

atom, an impurity. Such impurities are intentionally added to a

Quantum Structureof Atoms, Molecules,and Solids39

39.1 Principal, Orbital, andMagnetic QuantumNumbers; Spin

39.2 The Exclusion Principle andthe Structure of Atoms

39.3 Energy Levels in Molecules

39.4 Energy Bands in Solids

39.5 Semiconductor Devices

C H A P T E R

1320

Conceptsin

Context


39.1 Principal, Orbital, and Magnetic Quantum Numbers; Spin 1321

semiconductor to modify the conductivity. How does the presence of impurity

atoms affect the conductivity? (Section 39.4, page 1339)

? How are regions with different impurities combined to make useful devices?


The physical and chemical properties of atoms and of molecules depend on the quantum

behavior of their electrons. These electrons occupy most of the volume of the atom,

and their arrangement in different quantum states determines the size and the shape

of the atom, the chemical bonds that the atom forms with other atoms, the energy

required for ionization, the spectrum of light emitted and absorbed, and so on. Likewise,

the physical properties of a solid—such as diamond, silicon, silver, copper—depend

on the quantum behavior of the electrons in the solid. The spacing of the crystal lat-

tice of the solid, the electric and thermal conductivities, the magnetic properties, and

the mechanical properties (such as elasticity and hardness) all hinge on the arrange-

ment of the electrons. For instance, we will see in Section 39.4 how the arrangement

of the electrons determines the ability of the solid to conduct an electric current, and

we will see how the differences among conductors, semiconductors, and insulators

depend on what stationary quantum states are available in the solid and which of these

are occupied by electrons.

The arrangement of the electrons in the stationary states of an atom or a solid is

subject to an important restriction: no more than two electrons can occupy the same

orbital state. This is called the Pauli Exclusion Principle. In Section 39.1, we will

become acquainted with the spin, or the intrinsic angular momentum, of the electron,

and in Section 39.2, we will see that the Exclusion Principle is intimately linked to

the spin. In later sections we will examine the implications of the Exclusion Principle

for the arrangement of the electrons in atoms and in solids.

39.1 PRINCIPAL, ORBITAL , ANDMAGNETIC QUANTUM NUMBERS; SP IN

In the preceding chapter, we saw that Bohr’s theory characterizes the stationary states

of the hydrogen atom and their energies by a single quantum number n. However,

Bohr’s simple theory deals only with circular orbits. We know from the study of plan-

etary orbits (Chapter 9) that the general orbit of a particle moving under the influ-

ence of an inverse-square force is an ellipse with one focus at the center of attraction.

For an electron in such an elliptical orbit around a nucleus, the quantum number n, now

called the principal quantum number, characterizes the overall size of the ellipse,

that is, its major axis; this quantum number also characterizes the energy of the elec-

tron [the formula for the quantized energies of the elliptical orbits is the same as that

for the circular orbits, Eq. (38.25)]. But for the complete definition of the ellipse we

need two extra quantum numbers l and m that characterize the elongation of the ellipse

and the inclination of the ellipse, that is, its orientation in space.These two extra quan-

tum numbers l and m are called, respectively, the orbital quantum number and the

magnetic quantum number.

For a classical orbit, the elongation of the ellipse is closely related to the magnitude

of the angular momentum of the electron. Among ellipses of equal sizes (equal major

axes), the most elongated has the least angular momentum. Thus, the orbital quan-

tum number l characterizes both the elongation of the ellipse and the magnitude of its

angular momentum.

principal quantum number

orbital quantum number

magnetic quantum number

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magnitude of angular momentum

The inclination of the ellipse is closely related to the direction of the angular

momentum.To understand this relationship, we have to think of the angular momen-

tum as a vector, with a direction perpendicular to the plane of the orbit (see Fig. 39.1).

The orientation of the orbit in space is then specified by the direction of the angular-

momentum vector; for instance, if the orbit is inclined at an angle of relative to

the x–y plane, as in Fig. 39.1, then the angular momentum vector is inclined at an

angle of relative to the z axis. As we will see below, the magnetic quantum number

m specifies the z component of the angular-momentum vector, and it thereby charac-

terizes both the direction of the angular-momentum vector and the inclination of the

orbit. A large value of the magnetic quantum number m corresponds to a large value

of this z component and a small inclination of the orbit; a small value of the magnetic

quantum number m corresponds to a small value of this z component and a large incli-

nation of the orbit, near (The quantum number m is called “magnetic” because it

acquires a special significance when the atom is placed in a magnetic field; the energy

of the orbit then depends not only on the size of the orbit and on the principal quan-

tum number n, but also on the orientation of the orbit and, therefore, on the “mag-

netic” quantum number m.)

Although the two extra quantum numbers l and m were originally introduced on

the basis of semiclassical considerations involving elliptical orbits, a later, more rigor-

ous analysis based on wave mechanics and the Schrödinger equation confirmed that

these quantum numbers are, indeed, required for the description of the stationary

quantum states. In wave mechanics a stationary state corresponds to a three-dimensional

standing wave, and the quantum numbers n, l, and m characterize the “shape” of this stand-

ing wave, and they also characterize the energy, the magnitude of the angular momentum,

and the z component of the angular momentum.

According to wave mechanics, the magnitude of the angular momentum is quan-

tized. However, this quantization differs from that of the simple Bohr theory in that

the quantum number for angular momentum is the orbital quantum number l, not the

principal quantum number n. Furthermore, the formula for the quantization condition

is somewhat more complicated. From the study of the mathematical properties of the

Schrödinger equation, it can be demonstrated that the magnitude L of the angular

momentum obeys the quantization condition

(39.1)

where the orbital quantum number l is restricted to integer values from 0 to

that is,

l � 0, 1, 2, # # #, n � 1 (39.2)

For example, if then the possible values of the orbital quantum number are

and and the corresponding magnitudes of the angular momentum are

and

Note that the smallest possible magnitude of the angular momentum is actually zero,

in contrast to what was claimed by the Bohr theory, where the smallest magnitude of

the angular momentum was U .

L � 12 � (2 � 1)U � 16U

L � 11 � (1 � 1)U � 12U

L � 10 � (0 � 1)U � 0

l � 2;l � 1,

l � 0,n � 3,

n � 1,

L � 1l (l � 1)U

90�.

35�

35�

FIGURE 39.1 This elliptical orbit is

inclined at an angle of relative to the x–y

plane. The angular-momentum vector L is

perpendicular to the plane of the orbit; this

vector is therefore inclined at an angle of

relative to the z axis. The z component of

the angular-momentum vector is Lz.

35�

35�

1322 CHAPTER 39 Quantum Structure of Atoms, Molecules, and Solids

p

r

L

Lz

z

y

x

35�

35�

Angular momentum vector Lis normal to plane of orbit.

…L is tilted 35�from the z axis.

Since plane oforbit is tilted35� from x –yplane…


z component of angular momentum


Since the angular momentum is a vector quantity, it has not only magnitude, but

also a direction. However, according to wave mechanics, the direction of the angular-

momentum vector is only partially determined. The direction of the angular-momen-

tum vector always has substantial quantum uncertainties, and, correspondingly, the x,

y, and z components of this vector have substantial uncertainties. It can be demon-

strated that at most one of the components is well determined—the other two com-

ponents are completely uncertain. If we take the well-determined component to be

the z component, then the quantization condition for this component is

(39.3)

Here, the magnetic quantum number m is restricted to integer values from �l to � l ; that is,

m � �l , �l � 1, . . . , 0, . . . , l � l , l (39.4)

Note that, for a given value of l, there are possible values of m. Each of these

possible values of m corresponds to a possible direction of the angular-momentum

vector relative to the z axis. For instance, if then there are five possible values of

m, namely, and thus, there are five

possible choices for the direction of the angular-momentum vector relative to the z

axis. Figure 39.2 shows these possible directions.

From Eq. (39.3) we see that the z component of the angular momentum has a

maximum possible value of this corresponds to the best attainable alignment of

the angular momentum with the z axis. If we compare this maximum possible value

of the z component with the magnitude of a nonzero angular momentum, we see that

the former is always smaller than the latter, This means that the

angular-momentum vector is never perfectly aligned with the z axis. The uppermost

vector drawn in Fig. 39.2 indicates the best attainable alignment for

For the angular-momentum vector corresponding to a quan-

tum number calculate the angle between the vector

and the z axis for the cases and

SOLUTION: The directions of the angular-momentum vector for these cases are

illustrated in Fig. 39.2. The magnitude of the angular-momentum vector is

In the case the z compo-

nent is and the angle is given by

With a calculator, we find that the angle with this cosine is

Likewise, in the case the z component is and

which gives us an angle of

Finally, in the case the z component is which gives us an angle

of This angular-momentum vector lies in the x–y plane.

Each possible set of values of n, l, and m corresponds to one kind of three-dimen-

sional standing wave, often called an orbital.The simplest of these waves is that for the

90�.

Lz � 0,m � 0,

66�.

cos u �U16U

� 0.41

Lz � U ,m � 1,

35�.

cos u �Lz

L�

2U16U

�2

16� 0.82

uLz � mU � 2U ,

m � 2,L � 1l (l � 1)U � 12 � (2 � 1)U � 16U .

m � 0.m � 1,m � 2,

ul � 2,EXAMPLE 1

l � 2.

lU � 1l (l � 1)U .

lU ;

m � �2;m � �2, m � �1, m � 0, m � �1,

l � 2,

2l � 1

Lz � mU

z

y

x

�

L

m � 2, Lz � 2�

m � 1, Lz � �

m � 0, Lz � 0

m � �1, Lz � ��

m � �2, Lz � �2�

z component of angularmomentum for a givenl may range from �l�…

…to �l� in integermultiples of �.

z component is always smallerthan magnitude so nonzero Lis always tilted away from z axis.

FIGURE 39.2 Possible directions of

the angular-momentum vector for the case

In this case, there are five possible

values of characterized by

and �2.m � �2, �1, 0, �1,

Lz,

l � 2.

orbital


ground state, with and Figure 39.3a shows a picture of the inten-

sity of this wave, denoted the square of the amplitude of the wavefunction. In this

picture, generated from the theoretical formulas of wave mechanics by a computer,

the density of the dots indicates the intensity of the wave; thus, the wave is strongest

near the center (near the nucleus of the atom), and gradually fades with increasing

distance from the center. The intensity of the wave at any point is proportional to the

probability for finding the electron at this point; thus, in the ground state of the hydro-

gen atom, the most probable position for the electron is right at the nucleus. Because

of this probabilistic interpretation, the pictures of wave intensity shown in Fig. 39.3 are

often called probability clouds.

Waves with other values of n, l, and m correspond to the excited states; all these

waves are more complicated than the wave for the ground state, especially if n, l, and

m are large. Figures 39.3b–d give some examples of waves corresponding to some

excited states.

In addition to these quantum numbers n, l, and m, one more quantum number is

needed for the complete characterization of the stationary states of an electron in the

hydrogen atom. This is the spin quantum number that characterizes the spin, or

intrinsic angular momentum, of the electron. The quantum number was origi-

nally proposed by Wolfgang Pauli in an attempt to describe the “hyperfine” structure

of the spectral lines: when the spectral lines of hydrogen and of other atoms are examined

with a spectroscope of high resolving power, they are often found to consist of pairs,

or doublets, of very closely spaced lines. This implies that one or both of the atomic

energy levels involved in the transition must be a closely spaced pair of energy levels,

and it implies that, besides n, l, and m, there must be another quantum number that dis-

tinguishes between the two energy levels in the pair. Pauli assigned the values

and to the two energy levels in the pair, but offered no explanation of the

physical significance of this new quantum number.

ms � �12

ms � �12

ms

ms

�c�2,m � 0.l � 0,n � 1,


spin quantum number ms

spin, or intrinsic angular momentum

(a) (b) (c) (d)

For l = 0, probability of findingelectron is spherically symmetricand is highest at nucleus.

For many excited states, pro-bability is zero at nucleus andhas complicated patterns.

FIGURE 39.3 Intensities of possible standing waves, or orbitals, in the hydrogen atom. The nucleus

is at the center of each picture. The density of the dots is proportional to the intensity of the wave, and

thus to the probability of finding the electron, �c�2.

Ground state,

m � 0.

l � 0,n � 1, First excited state,

m � 0.

l � 1,n � 2, First excited state,

or �1.m � �1

l � 1,n � 2, Second excited state,

or �1.m � �1l � 2,

n � 3,



Shortly thereafter, S. Goudsmit and G. Uhlenbeck suggested that the electron has

an intrinsic spin angular momentum, and that the two values of correspond to different

orientations of the axis of spin (see Fig. 39.4). They imagined the electron as a small

ball of charge spinning about its axis, like the Earth spinning about its axis, with an angu-

lar momentum of magnitude which means that the intrinsic angular

momentum quantum number, or the spin quantum number, is If we use the same rule

for the possible directions of the spin as for the possible directions of the orbital angu-

lar momentum, we find that there are possible directions for the spin.

One of these directions is characterized by a magnetic quantum number and the

other by a magnetic quantum number these two possible spin directions are called

spin up and spin down, respectively.

Goudsmit and Uhlenbeck’s simple picture of the spin as due to a rotation of the elec-

tron about its axis proved untenable. A consistent, relativistic picture of the spin was later

contrived by P. A. M. Dirac. Modern wave mechanics tells us that the spin is an angu-

lar momentum generated by a circulating energy flow in the electron wave. The orbital

angular momentum is also generated by such a circulating energy flow in the electron wave,

but the energy flow that generates the spin is distinguished in that it persists even in the

electron wave of an isolated, free electron at rest (that is, an electron outside of the hydro-

gen atom). Although the simple picture of the electron as a rotating ball of charge is

false, it can sometimes serve as a convenient crutch for our imagination. According to this

simple picture, we expect that the electron has a magnetic moment, since a piece of

charge rotating about an axis amounts to a current loop (see Fig. 39.5).The energy dif-

ference between the two energy levels that gives rise to closely-spaced pairs of spectral

lines (“doublets”, such as seen in the sodium and mercury spectra in the color print on

page 1289) is due to this magnetic interaction of the magnetic moment with the mag-

netic field generated by the motion of the nuclear charge seen in the reference frame of

the electron. As already mentioned in Section 30.4, the magnetic moment of the elec-

tron also plays an important role in the behavior of ferromagnetic materials.

ms � �12;

ms � �12,

2 � 12 � 1 � 2

12.

212 (1

2 � 1)U ,

ms

WOLFGANG PAULI (1900–1958)Austrian and later Swiss theoretical physicist.

For his discovery of the Exclusion Principle, he

was awarded the Nobel Prize in 1945. Pauli

made an important contribution to the theory

of beta decay by proposing that the emission of

the beta particle is always accompanied by the

emission of a neutrino (see Chapter 40).

(a) (b)

spin up spin down

In a simple picture, weimagine electron spinning.

Sense of rotation…

…determines direction of in-trinsic spin angular momentum.

FIGURE 39.4 Simple picture of the electron as a ball of

charge spinning about its axis. (a) The electron can spin in

the counterclockwise direction, seen from above, so the

direction of the angular momentum is up. (b) Alternatively,

the electron can spin in the clockwise direction, so the

direction of the angular momentum is down.

FIGURE 39.5 A ball of negative charge

spinning about an axis.

rotation

spin

currents

magneticmoment

For a negative charge,rotation around axis in one direction…

…is equivalent to loopsof current around axis in opposite direction.

Current loops implyelectron has a magnetic moment.


Many other elementary particles besides electrons have spin. For example, the proton

and the neutron have spin angular momenta of magnitude (spin quantum

number the photon has a spin angular momentum of magnitude (spin

quantum number 1), etc.

The quantum numbers n, l, m, and provide a complete characterization of the

stationary states of the hydrogen atom. The energies of the stationary state depend

mainly on the principal quantum number n. According to Eq. (38.25), the energies

are

(39.5)

Thus, stationary states of the same value of n have (almost) the same energy, regard-

less of the values of the other quantum numbers. Classically, this means that the energy

depends on the overall size of the elliptical orbit, but not on its elongation, or on its ori-

entation in space. However, a wave-mechanics calculation of the energies of the sta-

tionary states of the hydrogen atom shows that the energies depend slightly on the

orbital quantum number l and on the orientation of the spin relative to the orbital

angular momentum.This means that Eq. (39.5) for the energies of the stationary states

of the hydrogen atom is not quite accurate. But the deviations from Eq. (39.5) are very

small, and we can often ignore them.

Table 39.1 lists the permitted quantum numbers for the stationary states of the

hydrogen atom.1 These quantum numbers can also be used to characterize the stationary

states of atoms other than hydrogen, but the energies of other atoms are not given by the

simple formula (39.5).

E � �13.6 eV

n2

ms

11(1 � 1)U12),

212 (1

2 � 1)U


1Strictly, these quantum numbers are appropriate for a hydrogen atom placed in a magnetic field. For a

hydrogen atom by itself, the appropriate quantum numbers are n and some intricate combinations of l, m,

and ms. For the sake of simplicity, we will ignore this complication.

PAUL ADRIEN MAURICE DIRAC(1902–1984) British theoretical physicist,

professor at Cambridge. He formulated the

relativistic wave equation for the electron,

incorporating spin, and he used this equation

to predict the existence of antielectrons. He

received the Nobel Prize in 1933, when anti-

electrons were confirmed experimentally.

QUANTUM NUMBERS OF ELECTRONIC STATES

QUANTUM NUMBER SYMBOL VALUES PHYSICAL QUANTITY

principal n 1, 2, 3, energy

orbital l magnitude of orbital angular momentum,

magnetic m z component of orbital angular momentum,

spin z component of spin angular momentum, Sz � ; 12 U�12, �

12ms

Lz � mU�l, �l � 1, p , 0, p , l � 1, l

L � 1l (l � 1)U0, 1, 2, p , n � 1

p

TABLE 39.1

(a) What are the permitted quantum numbers for the case

(b) What are the possible quantum numbers for the

case

SOLUTION: (a) For Table 39.1 tells us that the only permitted value of l

is 0. Furthermore, if then the only permitted value of m is also 0. Thel � 0,

n � 1,

n � 2?

n � 1?EXAMPLE 2



permitted values of are and regardless of the values of the other quan-

tum numbers. Hence, for the list of permitted quantum numbers is

(b) For Table 39.1 tells us that the permitted values of l are 0 and 1. If

then the only permitted value of m is 0, and we find the same list as in part (a):

However, if then the permitted values of m are �1, 0, and �1, and we find

the following list:

Each alternative listed here represents a possible electronic state.

Checkup 39.1

QUESTION 1: Consider the ground state of the hydrogen atom. According to wave

mechanics, what is the magnitude of the orbital angular momentum? According to

Bohr’s theory? According to wave mechanics, what is the value of the quantum number

n? According to Bohr’s theory?

QUESTION 2: If the orbital quantum number is what is the magnitude of the

angular momentum?

QUESTION 3: If the principal quantum number is what are the possible values

of the orbital angular-momentum quantum number?

QUESTION 4: Figures 39.3b, c, and d give the probability clouds for some excited states

of the hydrogen atom. According to these figures, what are the probabilities for find-

ing the electron at the center (at the nucleus)?

QUESTION 5: If the orbital angular-momentum quantum number is what are

all the possible values of the magnetic quantum number m?

(A) 0 (B) 0 or 1 (C) 0, 1, or 2 (D) or (E) �1, 0, or �1�12�1

2

l � 1,

n � 3,

l � 3,

✔

or n � 2; l � 1; m � �1; ms � �12

n � 2; l � 1; m � �1; ms � �12

n � 2; l � 1; m � 0; ms � �12

n � 2; l � 1; m � 0; ms � �12

n � 2; l � 1; m � �1; ms � �12

n � 2; l � 1; m � �1; ms � �12

l � 1,

or n � 2; l � 0; m � 0; ms � �12

n � 2; l � 0; m � 0; ms � �12

l � 0,

n � 2,

or n � 1; l � 0; m � 0; ms � �12

n � 1; l � 0; m � 0; ms � �12

n � 1,

�12,�1

2ms


39.2 THE EXCLUSION PRINCIPLE ANDTHE STRUCTURE OF ATOMS

The paramount question in atomic structure is the determination of the detailed

arrangement of the electrons in their orbital states around the nucleus. An atom of

atomic number Z has Z electrons in orbitals around a nucleus of positive charge Ze.

The electron arrangement, or configuration, determines all the physical and chemical prop-

erties of the atom—if the electron configuration is known, all the properties of the atom

can be deduced by theoretical considerations. For instance, the observed similarities

of chemical properties among select groups of elements are due to similarities in

their electronic configurations. Chemists list similar elements in columns in the peri-

odic table of the elements (see Table 39.2). Thus, the elements helium, neon, argon,

krypton, etc., are listed in the last column; these are the noble gases, or inert gases, which

are practically unable to react chemically with anything at all. The elements fluorine,

chlorine, bromine, iodine, etc., are listed in the column adjacent to the noble gases; these

are the halogens, irritating, corrosive substances, all with quite noticeable and distinc-

tive colors (pale yellow, greenish yellow, red, and blue violet, respectively). And the ele-

ments lithium, sodium, potassium, rubidium, etc., are listed in the first column; these

are the alkalis, silvery white metals, which are extremely reactive. The pattern of the ele-

ments displayed in the periodic table can be explained by a study of the electron configurations.

For the case of the hydrogen atom, the determination of the electron configuration

is trivial: the single electron of this atom is in one or another of the stationary states

characterized by the quantum numbers n, l, m, and If the atom is in the ground

state, the values of the quantum numbers of the electron configuration are

and thus, everything is fixed, except the direction of the spin,

which can be up or down.

But for atoms with several electrons, the determination of the electron configura-

tion is not so trivial. It might be tempting to suppose that the ground state of the atom

(the state of least energy) is attained by placing all the electrons in the lowest station-

ary state, with as for the hydrogen atom. But this

would imply that all the atoms ought to have a spectral series similar to that of hydro-

gen, and it would also imply that atoms with a large number of electrons, or with a

large atomic number Z, ought to be very small, since the Bohr radius for an atom of

nuclear charge Ze is [if the nuclear charge is Ze instead of e, then in the denom-

inator of the formula (38.17) for the Bohr radius of hydrogen, we must replace one

of the factors of e by Ze]. These conclusions are in stark conflict with the observed

properties of atoms: the spectra of most atoms are quite different, and the sizes of

atoms of large Z—such as lead or bismuth—are considerably larger than hydrogen,

which is the smallest of all atoms.

The rule that governs the configuration of the electrons in an atom is the Pauli

Exclusion Principle, which was discovered by Wolfgang Pauli and is often known

simply as the Exclusion Principle. For atoms, the Exclusion Principle asserts

Each stationary state of quantum numbers n, l, m, and can be occupied by no

more than one electron.

Since for each stationary orbital state of quantum numbers n, l, and m there are

two possible spin states we can also rephrase the Exclusion Principle as

follows: each stationary orbital state of quantum numbers n, l, and m can be occupied

by no more than two electrons.

(ms � ; 12),

ms

a0

a0>Z

n � 1, l � 0, m � 0, ms � ; 12,

ms � ; 12;m � 0,l � 0,

n � 1,

ms.


periodic table

atomic number Z

Pauli Exclusion Principle

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39.2 The Exclusion Principle and the Structure of Atoms 1329

THE PERIODIC TABLE OF THE CHEMICAL ELEMENTSa

aIn each box, the upper number is the atomic number. The lower number is the atomic mass, that is, the mass (in grams) of one mole or, equivalently, the mass

(in atomic mass units) of one atom. Numbers in parentheses denote the atomic mass of the most stable or best-known isotope of the element; all other numbers

represent the average mass of a mixture of several isotopes as found in naturally occuring samples of the element.

TABLE 39.2

Pauli originally proposed this principle as an empirical rule, based on the observed

features of atomic spectra. It was later established that the Exclusion Principle is inti-

mately linked to the value of the spin of the electron; the Exclusion Principle can be

shown to be a necessary consequence of the quantum theory of particles of half-integer

spin. Thus, protons and neutrons also obey the Exclusion Principle, a fact of great impor-

tance for the configuration of these particles in the interior of the nucleus (see the next

chapter). In contrast, particles of integer spin, such as photons, do not obey the Exclusion

Principle. There is no limit to the number of such particles that can be packed into a

given stationary state, for instance, one of the standing-wave states in a cavity filled

with blackbody radiation.

For our investigation of the electron configuration of atoms, we will find it convenient

to start with a list of all the available states, in order of increasing energy (see Example

2). The states of lowest energy have there are two such states:n � 1;

Na Mg Al Si P S Cl Ar

Li Be B

H He

C N O F Ne

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

IA(1)

IIA(2)

IIIA(13)

VIIIA(18)

IVA(14)

VA(15)

VIA(16)

VIIA(17)

IIIB(3)

IVB(4)

VB(5)

VIB(6)

VIIB(7)

(8)

(9)

(10)

VIIIBIB

(11)IIB(12)

Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sb Te I Xe

Cs Ba *La Hf Ta W Re Os Ir Pt Au Hg

Ds Uuu Uub

Tl Pb

Sn

Uuq

Bi Po At Rn

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

Fr Ra †Ac Rf Db Sg Bh Hs Mt

11 12 13 14 15 16 17 18

3 4 5

1 2

6 7 8 9 10

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

58 59 60 61 62 63 64 65 66 67 68 69 70

90 91 92 93 94 95 96 97 98 99 100 101 102

87

3

2

1

Per

iod

s

Atomic number

Group designation

Atomic mass

Symbol for element

*Lanthanides

†Actinides

4

5

6

788 89 104 105 106 107 108 109 110 111 112 114

Lu

Lr

71

103

22.98977 24.3050 26.98154 28.0855 30.97376 32.065 35.453 39.948

6.941 9.012182 10.811

1.00794 4.002602

12.0107 14.0067 15.9994 18.99840 20.1797

39.0983 40.078 44.955910 47.867 50.9415 51.9961 54.938049 55.845 58.93320 58.6934 63.546 65.409 69.723 72.64 74.92160 78.96 79.904 83.798

85.4678 87.62 88.90585 91.224 92.90638 95.94 98.9072 101.07 102.90550 106.42 107.8682 112.411 114.818 118.710 121.760 127.60 126.90447 131.293

132.90545 137.327 138.9055 178.49 180.9479 183.84 186.207 190.23 192.217 195.078 196.96654 200.59 204.3833 207.2

(289)

208.98037 208.9824 209.9871 222.0176

140.116 140.90765 144.24 144.9127 150.36 151.964 157.25 158.92534 162.50 164.93032 167.26 168.93421 173.04

232.0381 231.0359 238.0289 237.0482 244.0642 243.0614 247.07003 247.0703 251.0796 252.083 257.0951 258.0984 259.1011

174.967

262.110

223.0197 226.0277 227.0277 261.1089 262.1144 263.118 262.12 265.1306 (268) (271) (272) (285)


Likewise, for there are 18 available states, and so on.

The groups of states of a given value of n are called shells, and they are conventionally

labeled with the letters K, L, M, etc., from the innermost to the outermost shell. Thus,

the two states with form the K shell, the eight states with form the L

shell, the eighteen states with form the M shell, etc. In the simple Bohr theory,

these groups of states were called shells because the sizes of the orbits in each group

are similar; thus, these groups of orbits form layers around the nucleus, like the layers

of an onion. But in wave mechanics there is no such simple way to visualize the shells.

According to the Exclusion Principle, each of the states listed above can accommodate

one, and only one, electron. Thus, if an atom with Z electrons is in its ground state, the

electrons will occupy the first Z of the states in the above list. We can therefore build

up the configurations for all the atoms in the periodic table of elements by beginning

with hydrogen and adding electrons one by one, sequentially filling the states in our list.

The second element in the periodic table is helium, which has two electrons. To

obtain its electron configuration, we must add one electron to the hydrogen configu-

ration; since the single electron of hydrogen occupies one of the states of the K shell,

we can place the second electron in the other available state in the K shell. Helium

therefore has a full K shell.

The third element is lithium, with three electrons. When we add one electron to

the helium configuration, we must place this third electron in the L shell, with

Detailed calculations show that in multielectron atoms, the states are slightly

lower in energy than the states, so this third electron is in one of the two states

with quantum numbers and

The next element is beryllium, with four electrons. Thus, we must add one more

electron in the L shell, in the other available state with quantum numbers

and m � 0.

l � 0,n � 2,

m � 0.l � 0,n � 2,

l � 1

l � 0

n � 2.

n � 3

n � 2n � 1

n � 3,


n � 1 l � 0 m � 0 ms � �12

n � 1 l � 0 m � 0 ms � �12

TABLE 39.3

ms � �12n � 2 l � 1 m � �1

ms � �12n � 2 l � 1 m � �1

ms � �12n � 2 l � 1 m � 0

ms � �12n � 2 l � 1 m � 0

ms � �12n � 2 l � 1 m � �1

ms � �12n � 2 l � 1 m � �1

ms � �12n � 2 l � 0 m � 0

ms � �12n � 2 l � 0 m � 0

TABLE 39.4

K shell (states with n � 1)

L shell (states with n � 2)

Next, consider there are eight available states:n � 2;


39.2 The Exclusion Principle and the Structure of Atoms 1331

We can continue in this way, filling up the states in our list one by one. With the

tenth element, neon, we will have filled the L shell. And with the eleventh element,

sodium, we must place one electron in the M shell, and so on.

This simple procedure for building up the electron configurations of the atoms

provides us with an immediate explanation of the similarities of the elements in a

column of the periodic table. For instance, the similarity in the chemical behavior and

the similarity in the spectra of helium and neon can be traced to a similarity of their

electron configurations: both these atoms have full shells of electrons.The full-shell elec-

tron configuration is quite stable, tending neither to capture nor to lose an electron; these

and other noble gases are chemically inert. Likewise, the chemical and spectroscopic

similarities of hydrogen, lithium, and sodium can be traced to a similarity of their elec-

tron configurations: they all have a single electron outside of a full shell of electrons.

This single, outer electron tends to come off the atom fairly easily, and in chemical

reactions these atoms all tend to lose an electron. In contrast, fluorine and chlorine

are one electron short of a full shell, and in chemical reactions they tend to capture an

electron to complete their shell.

Thus, the Exclusion Principle in conjunction with a simple counting procedure

for the available stationary states is sufficient to explain the broad, qualitative features

of the periodic table of elements. Detailed calculations, based on wave mechanics, pro-

vide quantitative theoretical results for ionization energies, spectral lines, atomic sizes,

etc., in agreement with the observed atomic properties.

For atoms other than hydrogen there is no simple formula for the energies of the

stationary states. However, in an atom of fairly large atomic number, say, the

dominant force on the innermost electrons is the attractive Coulomb force exerted by

the positive charge Ze of the nucleus, and the repulsive forces exerted by the other

electrons can mostly be neglected. Thus, these innermost electrons have hydrogenlike

orbitals. The energy of a single electron in such an orbital is given approximately by

Eq. (38.23), with one modification: the product of the electron and the proton

charge must be replaced by the product of the electron and the nuclear charge;

hence in Eq. (38.23) must be replaced by which leads to the following approx-

imate formula for the energy:

(39.6)

From this formula, we can estimate the frequency and the wavelength of light emit-

ted during a quantum jump from some initial state to a final state. However, such a

quantum jump between the innermost orbitals of the atom is not possible if the atom

has its full complement of electrons—all the orbitals are then occupied by electrons,

and the Exclusion Principle forbids quantum jumps into an already occupied orbital.Thus,

a jump is possible only if some external disturbance first removes one of the electrons

from the atom, leaving a gap into which some other electron can jump. Such a process

occurs when the target atoms in an X-ray tube are subjected to the impact of the elec-

tron beam. The target atoms are disturbed by this impact, and sometimes an electron

in one of the innermost orbitals is ejected, leaving a gap into which another electron

can jump.The photon emitted when an electron jumps into the vacant innermost orbital

has a very short wavelength; it is an X ray. Thus, the quantum jumps of the innermost

electrons of atoms give rise to the characteristic spectrum of X rays mentioned in Section

37.5. These characteristic X rays correspond to the two sharp peaks in Fig. 37.19.

When one of the electrons in the innermost orbital is removed, the one remain-

ing electron “shields” a part of the nuclear charge from the view of the electron

making the quantum jump. For a nearby electron in an state jumping into an � 2

n � 1

En � � meZ

2e4

2(4p�0)2U2

1

n2� �

Z2 � 13.6 eV

n2

e4Z2,e4e � Ze

e � e

Z 20,


single available state, the effect of the shielding electron is to reduce the effec-

tive nuclear charge by one elementary charge.Thus, we can obtain somewhat better esti-

mates of the energies of these states by replacing the nuclear charge Z in Eq. (39.6) by

the shielded charge

(39.7)

Henry Moseley used this predictable behavior of the transitions to the K shell to cor-

rect the positions of several heavy metals in the periodic table; these were previously

positioned by mass, not by the atomic number Z.

Suppose that in an atom of copper in the target of

an X-ray tube, one of the electrons in the state is ejected

during the impact of the electron beam on the target. Suppose that subsequently

one of the other electrons in the atom jumps from the state into this avail-

able empty state. What are the energy and the wavelength of the photon

emitted during this quantum jump?

SOLUTION: The initial energy of the electron is

and the final energy is

Hence the energy of the photon is

The frequency of the photon is

and the wavelength is

COMMENTS: This agrees well with the experimentally determined wavelength,

This characteristic X ray is known as copper radiation, where

the refers to the dominant transition, here to the K shell from the state. In

engineering and materials research, diffractometers often use X rays of this wave-

length, somewhat smaller than atomic spacings, to determine crystal structures.

Checkup 39.2

QUESTION 1: Consider the K, L, and M shells. Which of these has the largest number

of states? Which the least?

✔

n � 2�

K�l � 0.154 18 nm.

l �c

f�

3.00 � 108 m/s

1.93 � 1018 Hz� 1.55 � 10�10 m � 0.155 nm

f �1.28 � 10�15 J

h�

1.28 � 10�15 J

6.63 � 10�34 Js� 1.93 � 1018 Hz

E2 � E1 � 8.00 � 103 eV � 1.28 � 10�15 J

E1 � �

(Z � 1)2 � 13.6 eV

12� �

(29 � 1)2 � 13.6 eV

1� � 1.066 � 104 eV

E2 � �

(Z � 1)2 � 13.6 eV

22� �

(29 � 1)2 � 13.6 eV

4� � 2.67 � 103 eV

n � 1

n � 2

n � 1

(Z � 29)EXAMPLE 3

En � �

me(Z � 1)2e4

2(4p�0)2U2

1

n2� �

(Z � 1)2 � 13.6 eV

n2

Z � 1:

n � 1


HENRY MOSELEY (1887–1915) English

physicist, lecturer at Manchester, where he

worked under Rutherford. Moseley was a skill-

ful experimenter, and his brilliant investiga-

tions of the characteristics of spectral lines of X

rays led to firm assignments of atomic numbers

for chemical elements. He was killed in action

in the futile Gallipoli campaign, at age 28.


QUESTION 2: When a shell is full, does it always have equal numbers of electrons of

spin up and spin down?

QUESTION 3: The ionization energy for the hydrogen atom is 13.6 eV. What is the

energy required for the second ionization of a helium atom that is, the energy

for the removal of the second electron (if the first electron has already been removed

previously)?

(A) (B) (C)

(D) (E)

39.3 ENERGY LEVELS IN MOLECULES

The chemical bonds that bind two or more atoms together in a molecule, such as

or NaCl, arise from a rearrangement of the outer electrons, or valence electrons, of the

atoms. In some molecules (e.g., the outer electrons are shared between the atoms,

a sharing that produces an attractive force (covalent bond). In some other molecules

(e.g., NaCl), one atom loses an electron to the other atom, and the atom with the

missing electron is then electrically attracted by the atom with the extra electron

(ionic bond). Hydrogen atoms are especially susceptible to losing electrons to neigh-

boring atoms. When a hydrogen atom is located between two other atoms, it often

loses its electron to these neighbors, and the residual proton of the hydrogen atom

then electrically attracts the neighboring atoms, holding them together (hydrogen

bond).

The chemical bonds are elastic—they behave rather like springs tying the atoms

together. The spring holds the atoms at an average equilibrium distance, but permits

the atoms to oscillate back and forth about this average distance, with some kinetic

and potential energy. This means that the energy of the molecule is the sum of the

electronic energy of the atoms and the vibrational energy of the motion of the atoms

in relation to each other. We will first focus on the vibrational energy of a molecule

and examine its quantization.

The mass of the atom is concentrated in its center, in the nucleus, which is much

smaller than the interatomic distances in a molecule. We can therefore schematically

represent a molecule—for instance, a diatomic molecule—as a system of pointlike

masses connected by a massless spring, which represents the bond between the two

atoms (see Fig. 39.6). The atoms oscillate in unison relative to the center of mass,

which we can regard as fixed. Thus, the system is an oscillator, and the energy of this

oscillator is subject to Planck’s quantization condition, Eq. (37.3). If the frequency of

oscillation is f, the energy of the vibration is quantized according to 2

(39.8)

The corresponding energy-level diagram is shown in Fig. 39.7.The molecule will emit

a photon if it makes a transition from an upper level to a lower level.The vibrational tran-

sitions in a molecule are restricted by a selection rule: the transition must proceed from one

level to the next, that is, transitions spanning two or more levels in one jump are forbidden.

n � 0, 1, 2, pE � n hf

O2),

O2

9 � 13.6 eV4 � 13.6 eV

3 � 13.6 eV2 � 13.6 eV1 � 13.6 eV

(Z � 2),

39.3 Energy Levels in Molecules 1333

FIGURE 39.6 Representation of an oscil-

lating diatomic molecule.

In a diatomic molecule,we represent nuclei astwo pointlike masses…

…and the chemicalbond as a massless spring.

selection rule

vibrational energy of molecule

2More precisely, the energies are However, the additional the zero-point energy, while

important for the calculation of some quantities, does not affect any of the transitions or other properties

discussed here.

12hf,E � (n � 1

2)hf.


This selection rule can be deduced from wave mechanics.The arrows in Fig. 39.7 indi-

cate the permitted transitions. The frequencies of the spectral lines emitted during all

these transitions are therefore the same,

(39.9)

Thus, the frequency of the spectral lines equals the frequency of vibration of the mol-

ecule. Typically, the frequencies of vibration of molecules are of the order of

and the wavelengths of the emitted spectral lines lie in the infrared.

Besides the vibrational motion, the molecule can also perform rotational motion.

For the purposes of this rotational motion, we can regard the molecule as two point-

like masses linked by a massless rod, that is, a dumbbell (see Fig. 39.8). If the moment

of inertia of the dumbbell about a perpendicular axis through the center of mass is I,

then the kinetic energy of rotation is

(39.10)

where is the angular frequency of the rotation. Let us express this in terms of the

angular momentum.The magnitude of the angular momentum is and this is quan-

tized in the usual way [see Eq. (39.1)], so we can write

(39.11)

where J is called the rotational quantum number. Solving Eq. (39.11) for and sub-

stituting this into Eq. (39.10) implies that the energy of the rotational motion is quantized,

that is,

(39.12)

Figure 39.9 displays the energy-level diagram for the rotational states of a molecule.

The transitions are, again, subject to the selection rule that they must proceed from

one level to the next. Such transitions are indicated by the arrows in Fig. 39.9.

Note that here we considered only rotation about a perpendicular axis through the

center of mass. For rotation about the axis through the atoms, the moment of inertia

of a diatomic moelcule is many orders of magnitude smaller, since the mass is con-

J � 0, 1, 2, pE �J ( J � 1)U2

2I

�

J � 0, 1, 2, pIv � 1J ( J � 1)U

I�,

�

E � 12I�2

1013 Hz,

flight �¢E

h�

hf

h� f


E

n � 4

n � 3

n � 2

n � 1

4hf

3hf

2hf

hf

n � 0

For vibrational frequencyf, energy levels are evenlyspaced with �E � hf…

…and only transitions between adjacent levelsare allowed, so all emittedphotons have f light f.


the oscillating molecule. The arrows indicate

the possible transitions.

rotational energy of molecule

FIGURE 39.8 A rotating diatomic molecule

can be regarded as two pointlike masses linked

by a massless rigid rod.

Axis of rotationis through centerof mass…

…and is perpendicularto line joining two atoms.


centrated in the nuclei. Equation (39.12) tells us that for such small values of I, the

energy levels will be very high, too high for practical consequences.

The moment of inertia of the HCl molecule about its axis of

rotation is What is the energy of the first

excited rotational state of this molecule? The second excited rotational state?

SOLUTION: For the first excited state, and Eq. (39.12) gives

For the second excited state, and

The rotational transitions in a molecule involve much smaller energies than the elec-

tronic transitions in an atom, such as the electronic transitions in a hydrogen atom.This

means that the photons emitted in purely rotational molecular transitions are of rather

low energy, and their wavelengths lie in the far infrared region of the spectrum. However,

rotational molecular transitions are often observed in conjunction with a simultaneous electronic

transition in one of the atoms of the molecule or with a vibrational transition of the molecule.

This increases the energy of the transition and reduces the wavelength of the spectral

line. By inspection of Fig. 39.9 we see that successive rotational transitions have slightly

different energies and wavelengths; thus, they give rise to a group, or sequence, of adja-

cent spectral lines. This is called a spectral band. Figure 39.10 is a photograph of sev-

eral spectral bands in the spectrum of the N2 (nitrogen) molecule.

E �2 � 3 � U2

2I�

3 � (1.05 � 10�34 Js)2

2.66 � 10�47 kgm2� 1.24 � 10�21 J

J � 2

E �2U2

2I�

(1.05 � 10�34 Js)2

2.66 � 10�47 kgm2� 4.14 � 10�22 J

J � 1

2.66 � 10�47 kgm2.EXAMPLE 4

39.3 Energy Levels in Molecules 1335

E

J � 1

J � 2

J � 3

J � 410�2

I

6�2

I

3�2

I

�2

I

0J � 0

Spacing between adjacentrotational energy levelsvaries in proportion to Jof upper level.

Only transitions betweenadjacent levels are allowed.


the rotating molecule. The arrows indicate

the possible transitions.

FIGURE 39.10 Several bands of spectral lines emitted by the N2 molecule.

spectral band

Checkup 39.3

QUESTION 1: For a vibrational transition in a molecule, the frequency of the radia-

tion equals the frequency of vibration of the molecule. Is it likewise true that for a

rotational transition in a molecule, the frequency of the radiation equals the frequency

of the rotational motion?

QUESTION 2: Suppose that a molecule is initially in the state in Fig. 39.9. If it

makes the sequence of downward transitions indicated by the arrows in this figure,

how many photons does it emit? Which of these photons has the most energy? The

least energy? Which has the longest wavelength? The shortest?

J � 4

✔

Successive rotational transitions give rise tosequences of spectral lines, or spectral bands.


FIGURE 39.11 Potential energy of an

electron in a crystal. The plus signs mark the

positions of the atoms along a row in the

crystal.

QUESTION 3: The moment of inertia of the deuterium molecule is twice as large

as that of the hydrogen molecule Which has a larger energy difference between

the ground state and the first excited rotational state? By what factor?

QUESTION 4: The frequencies of the photons emitted during the four transitions

shown in Fig. 39.9 are (in units of )

(A) 1, 2, 3, 4 (B) 0, 1, 3, 6 (C) 1, 3, 6, 10 (D) 4, 7, 9, 10

39.4 ENERGY BANDS IN SOL IDS

As discussed in Section 22.5 in a metal the outermost, or valence, electrons of the

atoms are detached from their atoms, and they are free to wander all over the volume

of the metal. However, whenever such a “free” electron passes near one of the posi-

tively charged atoms of the metal, it experiences an attractive force. Figure 39.11 is a

rough sketch of the potential energy for an electron moving along a row of atoms in

the crystal lattice of a metal; the dips in the potential energy indicate the attractive

force. While the electron moves along this row of atoms, the attractive force acts repet-

itively, each time the electron passes near an atom. Under suitable conditions, such a

repetitive action of a force can lead to a large cumulative effect. According to wave

mechanics, we have to think of the electron as a wave, and an encounter with an atom

scatters the wave: some fraction of the wave proceeds in its original direction of motion

and some fraction is reflected. The repetitive scatterings at the atoms in the row will

build up a large reflected wave by constructive interference if the scattering by one

atom produces a reflected wave that is in phase with the reflected wave produced by

the next atom. This will happen if the extra distance for a round trip from one atom

to the next and back is equal to one de Broglie wavelength or an integer multiple of one

wavelength. Designating the distance between the atoms by a, we can express the con-

dition for constructive interference of all the reflected waves as

(39.13)

Since the de Broglie wavelength is related to the momentum of the electron by

[see Eq. (38.32)], we can express the condition for total reflection in terms of the

momentum of the electron:

(39.14)

or

(39.15)

If this condition is satisfied, the reflected wave will gain more and more strength at

each reflection at each atom, and finally match the strength of the incident wave. The

result is a standing wave, which travels neither right nor left.

The energy of such a standing wave is strongly affected by the interaction of the

electron with the atoms of the lattice. The energy of the standing wave will be low

if the peaks of intensity of the wave coincide with the locations of the atoms (see

Fig. 39.12a), because the electron then has a large probability for being found near

an atom, where the potential energy is low. The energy of the standing wave will be

high if the peaks of intensity of the wave fall between the locations of the atoms (see

Fig. 39.12b), because the electron then has a large probability of being found between

one atom and the next, where the potential energy is high. Thus, the wavefunction

p �h2a

, 2h2a

, 3h2a

, p

2a �hp

, 2hp

, 3hp

, p

l � h�p

2a � l, 2l, 3l, p

U �I

(H2).

(D2)


U

x+ + + +

Near each positive ion…

…potential energy ofelectron dips, indicatingan attractive force.

Conceptsin

Context


depicted in Fig. 39.12a corresponds to a low-energy state of the electron, and that

in Fig. 39.12b corresponds to a high-energy state. The transition from one of these

states to the other involves a quantum jump of the electron.

To appreciate the implications of such a quantum jump, suppose we take an elec-

tron of low initial momentum and we gradually increase its energy and momentum.

At first, the energy of the electron will increase smoothly with the momentum. But at

the critical value given by Eq. (39.15), the energy has to increase by a step that

represents the energy difference between the low-energy standing wave and the high-

energy standing wave. Beyond this critical value, the energy again increases smoothly

with the momentum, until we reach the second critical value p � 2h�2a, where there

is another step in the energy, etc. From this discussion we see that the permitted ener-

gies of a free electron in a crystal lattice occur in several permitted intervals, within

which the energy increases smoothly with momentum.These permitted intervals are sep-

arated by forbidden intervals, where the electron energy increases by a step. Figure 39.13

shows such permitted and forbidden energy intervals on an energy-level diagram. The

permitted intervals, shown with lines in the diagram, are called energy bands; the forbid-

den intervals are called energy gaps. The precise widths of the permitted energy bands

and the forbidden energy gaps depend on the details of the crystal lattice, but all crys-

tals with “free” electrons have some kind of band pattern in their energy-level diagram.

As in the case of the electron configuration of atoms, we can deduce the electron

configuration of crystals by means of the Exclusion Principle. In a crystal in its ground

state, the electrons settle in the available states of lowest energy. To discover the elec-

tron configuration, we proceed as before: we take all the “free” electrons and pack them,

one by one, into the available energy bands.The lowest energy bands will then be com-

pletely filled, but the upper energy band will be either filled or partially filled, depend-

ing on the number of “free” electrons and the number of available states.

The differences among the electric properties of conductors, semiconductors, and

insulators arise from the partial or complete filling of the energy bands. A conductor,

such as copper or silver, has a band partially filled with electrons (see Fig. 39.14). When

the electrons in this partially filled band are subjected to an electric field, they absorb

energy from the field, and they make transitions to some of the slightly higher, empty

states of the band.Thus, the electrons respond to the electric field, and they begin to carry

an electric current.

p � h�2a


FIGURE 39.12 Standing electron waves

in a crystal lattice. The diagrams show the

intensity of the wave, which is proportional

to the probability of finding the electron,

. (a) This standing wave has its peaks

of intensity at the locations of the atoms

(red dots). (b) This standing wave has its

peaks of intensity midway between the

locations of the atoms.

�c�2

(a)

0

(b)

intensity

x

0

intensity

x

A high probability of findingan electron near positive ionsimplies a low electron energy…

…and a high probability offinding an electron between atomsimplies a high electron energy.

E

gap

gap

gap

Permitted energies of anelectron in a crystal occurthroughout intervals calledenergy bands…

…that areseparated byforbiddenintervals calledenergy gaps.

FIGURE 39.13 Energy-level diagram

for an electron moving in a crystal.

FIGURE 39.14 In a conductor, an upper energy band

is only partially filled with electrons. The portion of the

band filled with electrons is shown in blue.

E

partiallyfilled band

In a conductor, emptystates are only very slightlyhigher in energy…

…than full states, so electrons can easily make transitions andcarry current.


In an insulator, such as diamond, the lower bands are completely filled with electrons

(see Fig. 39.15). When the electrons in a full band are subjected to an electric field, they

cannot make transitions into other states in the band, because all of these states are

already full, and the Exclusion Principle forbids transitions into already full states.

This implies that the electrons in the full band cannot respond to the electric field;

they therefore cannot accelerate in the direction of the electric force, and they cannot

begin to carry a current. The only way the electrons in the full band could respond to

the electric field is by making transitions to a higher, empty energy band; but to do

this without external help requires that an electron acquire a large amount of energy

from random thermal fluctuations, which is difficult, since the typical energy of random

thermal fluctuations is small compared with the energy gap at room temperature.

In a semiconductor, such as silicon, germanium, or gallium arsenide, the bands

are completely filled with electrons, as in an insulator. However, the energy gap between

the last full band and the next, empty band is much smaller than in an insulator (see Fig.

39.16); typically, the width of the energy gap separating the full band from the next,

empty band is less than or approximately equal to 1 eV, whereas in an insulator the

gap is often 5 eV or more. In a semiconductor at room temperature, the random ther-

mal fluctuations of the energy will permit many of the electrons at the top of the full

band to make a transition to the next, empty band. This means these electrons have

nearby empty states; they can respond to the electric field, and they can carry a current.

The uppermost full band in an insulator or a semiconductor is called the valence band,

and the empty band above it is called the conduction band (in a metal, the conduc-

tion band is the valence band, and this band is only partially filled).

The values of the resistivities of semiconductors are between those of conductors and

insulators. The resistivities of semiconductors vary over a wide range; the resistivities

may be to times as large as the resistivities of conductors. Semiconductors fall

into two categories: n type and p type. In an n-type semiconductor, the carriers of cur-

rent are free electrons that have reached the conduction band.Thus, the mechanism for con-

duction is the same as in a metallic conductor. However, the resistance of a semiconductor

is higher than that of a metallic conductor, because the semiconductor has fewer free elec-

trons in its conduction band than a metal in its partially filled upper band. Also, a (pure)

semiconductor differs from a metal in that the resistivity decreases as the temperature

increases (see also Fig. 27.8).This curious behavior is due to an increase in the number

of free electrons—as the temperature increases, more electrons are excited into the con-

duction band by random thermal fluctuations, and these extra free electrons more than

compensate for the extra friction experienced by each at the higher temperature.

In a p-type semiconductor, the carriers of current are “holes” of positive charge. This

type of semiconductor has a valence band that is almost, but not quite, filled with elec-

1015104


� 6 eV

E

conductionband

valenceband

In an insulator, emptystates are separated bya large energy gap…

…from full states, so electrons cannot readilymake transitions, and cannot carry current.

FIGURE 39.15 In an insulator, the lower

energy bands are completely filled with electrons.

The gap between the uppermost filled band and

the next, empty band is fairly large.

semiconductor

valence bandconduction band

n-type semiconductor

p-type semiconductor


trons. Thus, there are missing electrons, or “holes,” in the electron distribution, and if

these holes move, they will transport charge. To see how such a transport of charge

comes about, consider Fig. 39.17, showing an array of electrons and positive ions. In

Fig. 39.17a, these electrons and ions form neutral atoms. Suppose that the right end

of this array is connected to the positive pole of a battery (not shown) and the left end

to the negative pole. If the battery pulls an electron out of the right end, it will leave

the array with a hole, or missing electron, at the position of the last atom (Fig. 39.17b).

The electrons will then play a game of musical chairs: the electron from the next-to-

last atom will jump into this hole, leaving a hole at the position of the next-to-last

atom (Fig. 39.17c); and then the electron from the next atom will jump, etc. The col-

lective motion of the electrons from left to right can be conveniently described as the

motion of a hole from right to left. The hole virtually carries positive charge from the

right to the left. In essence, this is the mechanism for conduction in a p-type semi-

conductor. Instead of free electrons, this type of semiconductor has free holes. A flow

of current is then a flow of holes, and the direction of the current is the same as the direc-

tion of motion of the holes.

Semiconductors usually contain both free electrons and free holes. Whether a

semiconductor is n type or p type depends on which kind of charge carrier dominates.

The concentration of free electrons and of free holes is largely determined by the impu-

rities that are present in the material. Donor impurities consist of atoms that release a

valence electron when placed in the semiconductor, and they increase the number of

free electrons. Acceptor impurities consist of atoms that trap electrons when placed in the

semiconductor, and they thereby generate holes. Hence, a semiconductor with donor

impurities will be n type and one with acceptor impurities will be p type. For instance,

silicon with arsenic impurities, which have one more valence electron than silicon, is

an n-type semiconductor, and silicon with boron impurities, which have one less valence

electron than silicon, is a p-type semiconductor. Even though the deliberately added

impurity atoms may amount to only a few parts per million, they completely change

the conductivity because the semiconductor has so few current carriers to start with.

For typical devices at room temperature, such “doping” of silicon with impurities

increases the density of free charge carriers, and thus the conductivity, by a factor of

to compared with pure silicon.106103


� 1 eV

valenceband

conductionband

E

In a semiconductor,empty states ofconduction bandare separated byonly a small gap…

…so some electronscan be thermallyexcited, and canthen carry current.

…from full statesof valence band…

FIGURE 39.16 In a semiconductor, the lower

energy bands are almost completely filled with

electrons, but some electrons reach the conduction

band because the gap between the uppermost filled

band and the next, empty band is small.

FIGURE 39.17 A row of positive ions

(red balls marked and electrons (blue

dots marked �).

�)

(a)

(b)

(c)

(d)

– – – – – – –

– – – – – –

– – – – – –

– – – – – –

When electron is pulled out of right end…

…there is a missing electron, or “hole.”

Next electron jumps into hole…

…and each electron, in turn, moves to right.

Motion of many electrons to right is equivalent to one hole moving to left.

Conceptsin

Context

donor impurities

acceptor impurities


Checkup 39.4

QUESTION 1: Suppose that a crystalline solid has an energy band that is completely filled

and all bands above this filled band are empty. Is this solid a conductor, an insulator,

or a semiconductor? What extra information do you need to decide?

QUESTION 2: Must conductors always be n type? Can they be p type?

QUESTION 3: Suppose we cool a semiconductor to very low temperature, so all elec-

trons settle into the lowest possible states. Will the semiconductor conduct?

QUESTION 4: Equation (39.14) says that the forbidden values of the momentum are

evenly spaced. Are the forbidden values of the energy (the gaps) also evenly spaced?

QUESTION 5: How does p-type silicon differ from n-type silicon?

(A) p-type has acceptor atoms and free electrons; n-type has donor atoms and holes.

(B) p-type has acceptor atoms and holes; n-type has donor atoms and free electrons.

(C) p-type has donor atoms and free electrons; n-type has acceptor atoms and holes.

(D) p-type has donor atoms and holes; n-type has acceptor atoms and free electrons.

39.5 SEMICONDUCTOR DEVICES

The manipulation of the resistivity of semiconductor materials by intentional con-

tamination with carefully selected impurities plays a crucial role in the manufacture

of semiconductor devices, such as diodes, transistors, and integrated circuits. It is a

characteristic feature of semiconductor materials that the addition of impurities to the

material has a drastic effect on the resistivity. For instance, the silicon used in elec-

tronic devices is usually contaminated, or “doped,” with small amounts of arsenic or

boron; the addition of just one part per million of arsenic will decrease the resistivity

of silicon by a factor of more than Pure semiconductor materials are hardly ever

used in practical applications. In most cases, the presence of impurities is what gives

the semiconductor materials their interesting and useful electric properties.

Rec t i f ie r (D iode)A semiconductor rectifier consists of a piece of n-type and a piece of p-type semi-

conductor joined together. The n-type semiconductor has free electrons, and the p-

type semiconductor has free holes; when they are joined, some of the free electrons

will wander from the n region into the p region, and some of the holes will wander

from the p region into the n region. Wherever the electrons and the holes meet, they

annihilate each other—the electron falls into the hole and fills it, which means that

both the electron and the hole disappear. This recombination of some electrons and

holes leaves residual positive and negative ions near the interface of the two regions,

and the electric charges of these ions generate an electric field across the interface (see

Fig. 39.18). This electric field opposes any further wandering of holes or electrons

from one region into the other. Because of this lack of mobile charge carriers, the inter-

face is known as the depletion region.

When such a p–n junction is connected to a battery or some other source of emf,

it will permit the flow of current from the p region into the n region, but not in the oppo-

site direction. Figure 39.19 shows the p–n junction connected to the source of emf so

the p region is at high potential and the n region at low potential, a configuration

105.

✔


Conceptsin

Context


called “forward bias.” The source of emf pumps a steady flow of electrons into the n

region and it removes electrons from the p region, which is equivalent to pumping

holes into the p region. The electrons and the holes meet at the junction and they

recombine.This process can continue indefinitely, and therefore the source of emf can

continue to pump current around the circuit indefinitely. Figure 39.20 is a plot of the

current vs. the voltage applied to the p–n junction. The current increases steeply with

the voltage, because the electric field associated with the applied voltage tends to cancel

the internal electric field at the p–n junction, and this makes it easier for the electrons

and holes to meet at the center. Evidently, the current is not simply proportional to

the voltage, and the p–n junction does not obey Ohm’s Law.

Now, consider what happens if the p–n junction is connected to the source of emf

so the p region is at low potential and the n region is at high potential, as shown in

Fig. 39.21. This configuration is called “reverse bias.” Many of the free electrons in

the n region then flow away through the wire on the right, and many of the holes in

the p region flow away through the wire on the left. Consequently, the region depleted

of charge carriers near the interface becomes wider, and the flow of current stops almost

immediately—the p–n junction blocks the current.


FIGURE 39.19 A p–n junction connected

to a source of emf. The n region is at low

potential and the p region at high potential

(forward bias).

I

p n

� �

Under “forward bias,” source ofemf pumps holes into p region and electrons into n region…

…which meet and annihilateat junction, so a steady currentcan flow around circuit.

FIGURE 39.20 Plot of current vs. voltage

for the p-n junction.

I

V0.6 V

Current increasessteeply…

…with increasing forward bias voltage.

negativeion

positiveion

freeelectron

freehole

p type n type

Very near junction, free electrons in n region canfall into holes in p region…

…leaving a regiondepleted of free charges.

Without free charges, (fixed)ions produce an electric field.

FIGURE 39.18 Pieces of p-type and n-type

semiconductor in contact. The plus and minus

signs represent the ions of the lattice. The blue

dots represent electrons and the red dots holes.

The arrows indicate the electric field generated

by the ions at the interface.

p n

��

Under “reverse bias,” source ofemf pulls holes from p region and electrons from n region…

…which widens depletionregion at junction, and stopsthe current.

FIGURE 39.21 A p–n junction connected

to a source of emf. The n region is at high

potential and the p region at low potential

(reverse bias).


The p–n junction is called a rectifier because it can be used to convert an alter-

nating current into a direct current. If the junction is connected to a source of alter-

nating emf, it will pass current only during the “forward” part of the cycle. The alternating

positive and negative emf then yields a periodic sequence of positive current pulses

(see Fig. 39.22). Such solid-state rectifiers find many practical applications; for

instance, they are used in the “alternators” that generate DC power in the electrical

systems of automobiles, and they are found in AC-to-DC power adapters for portable

electronics.

Trans i s tor (B ipo lar Trans i s tor )A bipolar transistor consists of a thin piece of semiconductor of one type sandwiched

between two pieces of semiconductor of the other type. Figure 39.23 illustrates an n–p–n

junction transistor. The thin piece in the middle is called the base, and the pieces at the

ends are called the emitter and the collector, respectively.The transistor has three termi-

nals, which are connected to two sources of emf, and , so the emitter-base junc-

tion has a forward bias and the base-collector junction has a reverse bias. In this

configuration, the emitter-base junction acts as a diode with forward bias, and it permits

the flow of electrons from the emitter into the base. However, the electrons that enter the

p region fail to recombine with holes, because this base region is quite thin and contains only

a low density of holes, and the electrons pass through it before they have a chance to meet

with a hole. The electrons wander to the base-collector junction, and the electric field

across this junction (indicated by the longer red arrows in Fig. 39.23) pulls the electrons

into the collector.They leave the collector via the terminal connected to its end, and they

continue around the external circuit, forming the external collector current Of the

electrons that enter the base, a small fraction wander to the terminal connected to the

base, and they leave via the wire connected there, forming the external base current

The use of the transistor as an amplifier for currents and voltages hinges on the

relationship between the collector and the base currents: a small change in the base

current or the base voltage leads to a quite large change in the collector current

As in the case of the diode with forward bias, if we increase the base potential

we cause a drastic increase in the flow of electrons entering the base from the emitter.

But most of these electrons flow straight through to the collector, and only a small

fraction flow to the terminal connected to the base. Thus, the result of an increase of

base potential is a large increase of collector current, but only a small increase of base

current. For a typical transistor, the ratio of the collector current increment and the

base current increment is of the order of 100 or 200, and this ratio has a fixed value,

VB,IC.

VBIB

IB.

IC.

VCVB


FIGURE 39.22 Current passed by the

rectifier vs. time. The negative portions of

the alternating current are blocked by the

rectifier, and only the positive portions

remain.

t0

I

2��

4��

When alternating emf is appliedto diode, current flows during forward-biased part of cycle…

…and current is blockedduring reverse-biased part.

n p n

emitter base collector

IB IC

VB

VC

��

��

Electrons are injected into base at this forward-biased junction…

…but most wander intothis junction, whereelectric field pulls themto collector…

…so that IC ismuch larger than IB.

FIGURE 39.23 An n–p–n junction

transistor. Two sources of emf and

are connected to the base and the

collector, respectively.

VC

VB


over a wide range of currents. The ratio of these current increments is called the cur-

rent gain factor,3

(39.16)

This means that whenever we change the base current by some amount (by adjust-

ing the voltage we will change the collector current by an amount a hundred or so

times larger. The transistor amplifies the current—a small input current at the base results in

a much larger output current at the collector. For instance, in the “amplifier” in a radio

receiver, the weak current picked up by the radio antenna is amplified by sending it into

a transistor, as input current at the base. Further amplification can be achieved by con-

necting several transistors in tandem, so the output of each serves as input for the next.

Transistors are used in a wide variety of electronic circuits, to amplify and control

currents. Figure 39.24 shows some ordinary transistors. In an integrated circuit, such

as shown in Fig. 39.25, many transistors and other circuit elements of extremely small

size are built up on a single crystal of silicon. The small transistors are not manufac-

tured by sticking together separate pieces of n- and p-type material, but by diffusing

suitable concentrations of acceptor and of donor impurities into different adjoining

layers of the silicon crystal.

L igh t -Emi t t ing D iode ( LED)In principle, a light-emitting diode is simply a p–n junction operated with forward

bias. At such a junction, electrons arriving from the n region meet holes arriving from

the p region, and they recombine, that is, the electrons fall into the holes. But this

jump is a transition of the electron from a state of high energy in the conduction band

to a state of lower energy in the valence band, a transition that releases energy. In gal-

lium arsenide and some other semiconducting materials, the released energy takes the

form of a photon of visible light (see Fig. 39.26). Thus, the p–n junction emits light

when an electric current passes through it. Such light-emitting diodes have many

practical applications in luminous displays in the dials of measuring instruments,

watches, electronic calculators, clocks, automobile speedometers, and so on.

VB),

IB

[current gain factor] �¢IC

¢IB


3In electronics catalogs and transistor circuits, the symbol or is usually used for the current gain factor.hfe�

FIGURE 39.24 Some transistors.

FIGURE 39.25 Integrated circuit with many miniature circuit elements.

p n

� �

I

In forward-biased LED, electron andhole meet at junction and recombine, producing a photon.

FIGURE 39.26 A p–n junction used as a

light-emitting diode.

PhysicsPagesCh39_Ac.qxd 01/03/2007 7:12 PM Page 1343 PMAC-291 PMAC-291:Books:GK-Jobs:GK023/Ohanion: TechBooks [PPG -QUARK]

Solar Ce l l (Photodiode)A solar cell is simply a light-emitting diode operating in reverse. When sunlight is

absorbed at the p–n junction, it excites an electron from the valence band to the con-

duction band, creating a free electron and a free hole (see Fig. 39.27).The electric field

at the junction then pulls the electron toward the n region and the hole toward the p

region. This means negative charge flows into the n region and from there into the

external wire connected on the right; while positive charge flows into the p region and

from there into the external wire connected on the left. Thus, sunlight striking the

junction generates an electric current in the external circuit.

Solar cells and other photodiodes are commonly manufactured out of p-type sil-

icon and n-type silicon. The emf of such a silicon solar cell is only about 0.6 V, and

the current it delivers is fairly small; most photodiodes produce a few tenths of an

ampere of current per watt of incident light. For the solar cell device geometry shown

in Fig. 39.28, with an area of about the current delivered in full sunlight is about

0.1 A. Many calculators are powered by photodiodes, eliminating the need for batteries.

Solar cells are routinely used to generate electric power on communications satel-

lites and other satellites in orbit around the Earth (see Fig. 39.29). Solar cells have

been used to generate power to drive experimental vehicles (see Fig. 39.30). Arrays of

solar cells are commercially available (Fig. 39.31). Some attempts have also been made

to use them to generate fairly large amounts of electric power for residential and indus-

trial use (see Fig. 39.32). For instance, a solar power station at Pocking, Germany gen-

erates about 10 megawatts.

4 cm2,


FIGURE 39.27 A p–n junction used as a

solar cell.

p n

I

…generating current in external circuit.

Photons strike junction of solar cell, producing freeelectrons and holes, …

1 mm�

�

2 cm

3 �m

sunlight

p-typesilicon

n-type silicon

Top layer is made thin solight can reach junction…

…and area of junction ismade large so more lightcan be absorbed.

FIGURE 39.28 Device geometry for solar

cell (side view).

FIGURE 39.29 Solar panels on the International Space Station.


Checkup 39.5

QUESTION 1: If the n region of a p–n junction is at a potential of 3 V and the p region

at 5 V, is the bias forward or reverse?

QUESTION 2: In the rectifier, the light-emitting diode, and the solar cell, electrons

and holes are created or recombined at the p–n junction. Which devices involve cre-

ation, which recombination?

QUESTION 3: You place a piece of n-type silicon in contact with p-type silicon. What

is the direction of the electric field at the interface?

(A) Electric field points from n-type region to p-type region.

(B) Electric field points from p-type region to n-type region.

(C) Electric field points parallel to the plane of the interface.

(D) There is no electric field without an external bias.

✔

Summary 1345

FIGURE 39.30 A race between experimental vehicles powered by solar cells

covering their bodies.

FIGURE 39.31 Panel of solar cells.

FIGURE 39.32 Solar power station

at Carrisa Plains in California.

SUMMARY

QUANTUM NUMBERS OF ATOMIC STATESn, l, m, and ms (principal, orbital, magnetic, and

spin quantum numbers)

(39.1)L � 1l (l � 1)UMAGNITUDE OF ANGULAR MOMENTUM

n � 1, 2, 3, # # #

l � 0, 1, 2, # # # , n � 1

m � �l, � l � 1, # # # , 0, # # # , l � 1, l

ms � , � 12�

12



(39.3)Lz � mU

(39.8)n � 0, 1, 2, pE � n hf

(39.12)J � 0, 1, 2, pE �J ( J � 1)U2

2I

z COMPONENT OF ANGULAR MOMENTUM

ELECTRON SPIN (INTRINSIC ANGULAR MOMENTUM)

Magnitude:

z component:

Spin up:

Spin down:

PAULI EXCLUSION PRINCIPLE Each stationary state of

quantum numbers n, l, m, and ms can be occupied by no

more than one electron.

VIBRATIONAL ENERGIES OF MOLECULE

ROTATIONAL ENERGIES OF MOLECULE

CONDUCTOR Partially filled conduction band.

INSULATOR Full valence band, empty conduction

band, large gap between.

SEMICONDUCTOR Full valence band, empty

conduction band, small gap between.

ms � �12

ms � �12

msU21

2 (12 � 1)U

z

y

x

�

L

m � 2, Lz � 2�

m � 1, Lz � �

m � 0, Lz � 0

m � �1, Lz � ��

m � �2, Lz � �2�

E

n � 4

n � 3

n � 2

n � 1

4hf

3hf

2hf

hf

n � 0

E

J � 1

J � 2

J � 3

J � 410�2

I

6�2

I

3�2

I

�2

I

0J � 0

� 6 eV

E

conductionband

valenceband

� 1 eV

valenceband

conductionband

EE

partiallyfilled band

conductor insulator semiconductor


Problems 1347

9. A spring-loaded butterfly valve in a large water pipe is controlled

by a stream of water from a separate pipe (see Fig. 39.33). Is this

a reasonable hydraulic analog of a transistor? How could you use

such a device to amplify a water current?


1. According to classical mechanics, what kind of orbit would have

zero angular momentum?

2. If there were no Exclusion Principle, what would be the electron

configuration of lithium?

3. The bond in the NaCl molecule is similar to the bond in the

KBr molecule. Explain this similarity.

4. Rotational transitions in a molecule give a band spectrum, but

vibrational transitions do not. Explain.

5. How does n-type silicon differ from p-type?

6. If you dope silicon with phosphorus impurities, will the silicon be

n type or p type?

7. What kind of valve in a hydraulic circuit is analogous to a diode

rectifier? Draw a picture of a water pipe with such a valve.

8. Do the current and the voltage in a transistor obey Ohm’s

Law?

VCIC

ponent of the total magnetic moment of the electron? [Hint:

Eq. (30.23) relates the z component of the spin magnetic

moment, �B � �z,spin, to the z component of the spin angular

momentum, . The proportionality of the orbital magnetic

moment to the orbital angular momentum is half as large.]

6. What are the possible values of the orbital angular momentum

of the hydrogen atom in its first excited state? Taking into

account the spin, what is the maximum possible value of the z

component of the total angular momentum? What is the cor-

responding value of the z component of the magnetic moment

of the atom? (Hint: see Problem 5.)

7. Within a shell of an atom, the groups of states of the same l

are called a subshell. For instance, in the L shell, the two

states with form one subshell, and the six states with

form another subshell.

(a) How many subshells are there within the M shell? How

many states are there in each of these subshells?

(b) Neon has a complete subshell and a complete

subshell in the L shell. What atom has similar subshells in

the M shell?

8. The circular orbits of orbital angular momentum in Bohr’s

theory roughly correspond to the wave-mechanical states of

maximum orbital quantum number, that is, If

compare the magnitude of the angular momentum

given by Bohr’s theory with the magnitude given by wave

mechanics. Repeat for and n � 500.n � 10,n � 4,

n � 2,

l � n � 1.

nU

l � 1l � 0

l � 1

l � 0

U12

PROBLEMS†39.1 Pr inc ipa l , Orb i ta l , and Magnet i cQuantum Numbers ; Sp in

1. According to wave mechanics, what are the possible values of

the orbital quantum number l if the principal quantum

number is If If

2. Suppose that a state in the hydrogen atom has orbital quan-

tum number What are the possible values of the mag-

netic quantum number m?

3. Suppose that the magnitude of the orbital angular momentum

vector is What are the permitted values of the z com-

ponent of the angular momentum?

4. Pretend that the electron is a small sphere of uniform density

of radius

(a) What is the moment of inertia of the electron according

to this model?

(b) What angular velocity of rotation is required to give the

sphere an angular momentum of magnitude What

is the corresponding speed of rotation of a point on the

equator of the sphere? Does this model make any sense?

5. An electron in an atom is in a state of quantum numbers

and What is the z component of the

total angular momentum of the electron? What is the z com-

ms � �12.m � 2,l � 2,

234 U?

2.8 � 10�15 m.

120U .

l � 5.

n � 3?n � 2?n � 1?

IB

IC

FIGURE 39.33 The butterfly valve in the

large pipe is hinged at the center. The stream of

water from the small pipe strikes the upper por-

tion of the butterfly valve and pushes it open.




according to Bohr theory and the energy calculated

according to the improved formula.

(b) For find the difference, in eV, between the

energies of the states of spin parallel and antiparallel to

the orbital angular momentum calculated according to the

improved formula. Which of these states has the lower

energy?

*17. For an electron in a state of given orbital (l ) and magnetic (m)

quantum numbers, the magnitude of the angular momentum

and the z component of the angular momentum are well

defined, but the x and y components are completely uncertain.

Show that nevertheless the sum of the squares of the x and y

components is well defined according to the formula

With the additional assumption that, on the average, and

are equal, show that the rms values of the x and y compo-

nents of the angular momentum are

(This is similar to what happens in the case of the velocity

components of a molecule of gas in a container. The, say, x

component of the velocity is equally likely to be positive or

negative, and therefore is completely unpredictable; neverthe-

less, the rms value is well defined.)

†39.2 The Exc lus ion Pr inc ip le and theSt ruc ture o f Atoms

18. How many possible electron states are there in the M shell

of a hydrogen atom? Make a list of these states, like

the lists in Tables 39.3 and 39.4.

19. List the quantum numbers of all the electrons of a boron atom

in its ground state.

20. List the quantum numbers of all the electrons of a carbon

atom in its ground state.

21. List the quantum numbers of all the electrons of an Na� ion

in its ground state.

22. What are the quantum numbers n and l for the outermost elec-

tron of the Li atom? The Na atom? The K atom? In what ways

are these quantum numbers of these different atoms similar?

23. Suppose that the spin quantum number of the electron were

instead of How many possible spin directions would the

electrons have in this case? What would be the possible per-

mitted quantum numbers for the case (K shell)? For the

case (L shell)? How many electrons could be placed in

the K shell? The L shell? Compare the resulting periodic table

of elements with the familiar periodic table in Section 39.2.

*24. What energy is required to eject one of the electrons from the

state in molybdenum out of the atom?

Express your answer in eV.

(Z � 42)n � 1

n � 2

n � 1

12.

32

(n � 3)

2v2x

2L2x � 2L2

y � Bl (l � 1) � m2

2U

L2y

L2x

(L2x � L2

y) � l (l � 1)U2 � m2U2

l � 1,n � 2,

9. The meson is a particle of spin quantum number 1. What is

the magnitude of the spin angular momentum of the

meson? What are the possible values of for this particle?

10. In many atoms, the nucleus of the atom has a spin. For exam-

ple, the nucleus of one of the isotopes of magnesium

has spin, with a spin quantum number of . What is the

magnitude of the spin angular momentum of this nucleus?

What are the possible values of for this nucleus?

11. Consider the angular-momentum vector with quantum

number What is the smallest possible angle that this

angular momentum can make with the z axis?

12. The spin angular momentum of the electron has a magnitude

and a magnetic quantum number or

For each of these two values of ms, calculate the

angle between the direction of the spin angular momentum

vector and the z axis.

13. The electron orbital angular momentum gives rise to a mag-

netic moment with z component where

is the Bohr magneton and m is the magnetic

quantum number. In Chapter 30, we saw that the electron

spin angular momentum gives rise to a magnetic moment with

z component Write an expression that relates

to the spin quantum number Is this relation differ-

ent from that for

14. Consider the angular-momentum quantum numbers for a

macroscopic object, such as a toy top with moment of inertia

spinning at 25 revolutions per second. The

axis of the top makes an angle of with the vertical (with

the Earth’s gravitational field). What are approximate values

of the angular-momentum quantum numbers l and m?

15. Similar to the Bohr magneton [see Problem 13 or Eq.

(30.23)], a nuclear magneton is defined by ,

where is the mass of a proton. Unlike the electron, how-

ever, the z component of the nuclear spin magnetic moment

is not related to the nuclear magneton in a simple way (due

to the internal structure of nuclei). For example, the z

component of the magnetic moment of a proton is

and that of a nucleus (an isotope of

carbon) is Express each of these magnetic

moments as a mutiple of a nuclear magneton.

*16. As stated in Section 39.1, the energies of the stationary states

of the hydrogen atom depend slightly on the orbital angular-

momentum quantum number l. An improved formula for the

energy of the state of quantum numbers n and l for nonzero l is

where the term corresponds to the spin parallel and

antiparallel, respectively, to the orbital angular momentum.

(a) For the case of the first excited state, and

the spin antiparallel to the orbital angular momentum,

find the difference, in eV, between the energy calculated

l � 1,n � 2,

; 12

En,l �mee

4

2(4p�0)2U2n2

c1�e4

(4p�0)2U2c 2n

a 1

l � 12 ; 1

2

�3

4nb d

3.55 � 10�27 J/T.

13C1.41 � 10�26 J�T,

mp

mN � eU �2mp

25�

2.0 � 10�4 kgm2

mz,orbital?

ms.mz,spin

mz,spin � ; mB.

mB � eU �2me

mz,orbital � mBm,

ms � �12.

ms � �1221

2 (12 � 1)U

l � 3.

ms

5�2(25Mg)

ms

r

r



Problems 1349

*25. Calculate the energy and the wavelength of the X ray emitted

during a quantum jump of an electron from the to the

state in copper

*26. An X-ray tube has a tungsten target Calculate the

energies and the wavelengths of the X rays emitted in the

quantum jumps from the to the state and from

the to the state.

*27. For the molybdenum atom the wavelength of the

characteristic X ray emitted in the quantum jump from the

to the state is approximately 0.071 nm. For what

atom is the wavelength of the corresponding characteristic X

ray twice as large? Half as large?

*28. In addition to X rays from copper and molybdenum targets,

commercial X-ray diffractometers often provide X rays of

wavelengths 0.229 nm, 0.194 nm, or 0.179 nm. Each of these

is due to a transition from the to the state of a

different element. What are the three elements?

*29. For transitions of electrons to a vacant L-shell state

from the higher-energy states of a many-electron atom with

nuclear charge Z, the nuclear charge is shielded by both K-

shell and L-shell electrons. The effective charge seen by an

electron making a transition to the L shell is given by

What is the wavelength of the X ray emitted

when an electron in a gold atom makes a transition

from the state to the state?

*30. After H. G. J. Moseley measured the energy of characteristic

X-ray photons, he plotted the atomic number Z as a func-

tion of the inverse of the square root of the photon wave-

length, Such a plot is linear. Find (a) the slope and (b)

the y intercept (Z-axis intercept) of such a plot.

*31. A sample of an unknown element is being used as the target in

an X-ray tube. It is found that the characteristic X-ray spec-

trum displays a strong spectral line at

Assume that this spectral line results from the quantum jump

of an electron from the to the state. Can you

identify the unknown element?

39.3 Energy Leve ls in Molecu les

32. The frequency of vibration of the molecule is

What are the energies of the vibrational

states? What is the frequency of the emitted radiation? The

wavelength?

33. The atoms of deuterium (D) and of hydrogen (H) have the

same electron configuration (one electron), but the deuterium

is a heavier atom than hydrogen, because it has more mass

in its nucleus. The mass of the deuterium atom is 2.014 u,

whereas the mass of the hydrogen atom is 1.008 u. Given

that the frequency of vibration of the molecule is

deduce the frequency of vibration of the

molecule.

34. The photon emitted in a vibrational transition of the hydro-

gen bromide (HBr) molecule has frequency 7.7 � 1013 Hz.

D21.31 � 1014 Hz,

H2

1.31 � 1014 Hz.

H2

n � 1n � 2

l � 0.0228 nm.

1�1l.K�

n � 2n � 3

(Z � 79)

ZL � Z � 7.4.

(n � 2)

n � 1n � 2

n � 1n � 2

(Z � 42),

n � 1n � 3

n � 1n � 2

(Z � 74).

(Z � 29).n � 1

n � 3

What is the spring constant of the chemical bond holding this

molecule together? You may assume, for simplicity, that the

bromine atom remains essentially at rest, and only the hydro-

gen atom moves.

*35. In the molecule, the distance between the two oxygen

nuclei is 0.20 nm.

(a) What is the moment of inertia of the molecule for rotation

about the perpendicular axis through the center of mass?

(b) What are the energies of the first, second, and third

excited rotational states? Express these energies in eV.

*36. The distance between the K and the Br nuclei in the KBr

molecule is 0.282 nm, and the center of mass is at a distance of

0.093 nm from the Br nucleus.

(a) What is the moment of inertia of a KBr molecule rotating

about its center of mass?

(b) What are the energies of the first, second, and third

excited rotational states? Express these energies in eV.

*37. Consider the HD molecule, where one of the atoms of the

hydrogen molecule has been replaced by a deuterium

atom. Deuterium and hydrogen have the same electronic

structure, but the mass of a deuterium atom is 2.014 u, and

that of a hydrogen atom is 1.008 u. Given that the frequency

of vibration of the molecule is deduce the

frequency of vibration of the HD molecule. [Hint: You may

use the result of Problem 41 of Chapter 15, where such an

oscillation about the center of mass was shown to have an

angular frequency where m is the reduced mass

*38. In a vibrational transition, a nitric oxide (NO) molecule emits

a photon of frequency What is the spring con-

stant of the chemical bond holding this molecule together?

(Hint: See Problem 37.)

*39. In Section 19.4, we asserted that quantum mechanics enables

us to neglect rotations about an axis through the atoms of a

diatomic or other linear molecule. Show this explicitly by cal-

culating the energy of the first rotational excited state of

for (a) rotation about the axis through the atoms and (b) rota-

tion about a perpendicular axis. Compare these two energies.

For (a), assume the proton is a uniform sphere with radius

and the atomic electron is a uniform sphere

with radius for (b), use the interatomic dis-

tance of

*40. What are the ratios of the frequencies emitted in rotational

transitions in a molecule from the first excited state to the

ground state, from the second excited state to the first, from

the third to the second, from the nth to the

**41. According to spectroscopic measurements, the energy differ-

ence between the first and the second excited rotational states

of the molecule is Deduce the moment of

inertia of the molecule. Deduce the center-to-center distance

between the N atoms.

2.38 � 10�22 J.N2

(n � 1)th?

7.4 � 10�11 m.

5.0 � 10�11 m;

1.0 � 10�15 m

H2

5.6 � 1013 Hz.

m � m1m2�(m1 � m2).]

v � 1k�m,

1.31 � 1014 Hz,H2

H2

O2



39.4 Energy Bands in So l ids

42. Consider a crystal with a spacing of 0.10 nm between one

atom and the next. What are the de Broglie wavelengths and

the energies at which the electron wave cannot propagate

through this crystal? Express the energies in electron-volts.

43. In silicon, the energy gap between the valence and the con-

duction bands is 1.1 eV. If we want to excite an electron from

the top of the valence band to the conduction band by means

of a photon, what is the maximum permitted wavelength for

the photon?

*44. The spacing of atoms in the crystal lattice of a metal, and thus

the wavelength for maximum repetitive scattering of electron

waves by the atoms, varies with the direction of electron wave

propagation in the crystal. Consider a simple cubic arrange-

ment of atoms as in Fig. 39.34, where the length of the cube

edge is Find the two longest de Broglie

wavelengths and corresponding energies in eV at which the

electron wave cannot propagate (a) along a cube edge, (b)

along the diagonal of a square face of a cube, and (c) along the

body diagonal of a cube.

a � 2.0 � 10�10 m.

a

FIGURE 39.34 Structure of a simple cubic crystal.

n p n

9.0 V

IB IC � 1.0 mA

1.0 k 1.0 k

R1 R2��

FIGURE 39.36 A transistor circuit with a single

source of emf.

39.5 Semiconduc tor Dev ices

45. Figure 39.23 shows a circuit diagram for an n–p–n transistor.

Draw the analogous diagram for the p–n–p transistor, and

explain how a small current leads to a large current IC .

46. When a transistor is connected to a circuit as shown in Fig.

39.35, it serves as an amplifier of voltage. The voltage gain

factor is defined as the ratio of the output voltage (measured

across the resistor RC ) to the input voltage VB. Evaluate this

ratio if RB � 3000 and RC � 6000 . Assume that the cur-

rent gain factor for this transistor is 100, and that the internal

resistance of the emitter–base junction (a diode with forward

bias) is negligible.

IB

n p n

VB

VC

��

Vout

��

RB

RC

FIGURE 39.35 A transistor acting as a voltage amplifier.

The source of emf provides the input signal, and the voltage

across the two free terminals constitutes the output signal.

VB

*47. We want to connect two transistors in tandem, so that the net

current amplification of the combination is the product of the

individual current amplifications of the two transistors.

Design a circuit that will accomplish this.

*48. A solar cell delivers 0.10 A at 0.60 V. How many such solar

cells do you need, and how must you connect them, to obtain

2.0 A at 6.0 V for charging a battery?

*49. A solar cell of area facing the Sun delivers 0.10 A at

0.60 V. What is the power delivered by the solar cell?

Compare with the incident power of sunlight

and deduce the efficiency for the conversion of energy of light

into electric energy.

*50. In practical transistor circuits, a single source of emf is used to

provide the two emfs and of Fig. 39.23. A typical circuit

with a 9.0-V emf is shown in Fig. 39.36. The transistor has a

current gain factor of 100 and it is desired to operate with

with the two resistors in place. Assume

that the potential across the forward-biased base–emitter

junction is 0.60 V. The resistors and are known as bias

resistors. Use Kirchhoff ’s rules to find the required values of

and R1.

R2

R2R1

1.0-k IC � 1.0 mA

VCVB

(1.0 kW�m2)

5.0 cm2



REVIEW PROBLEMS

51. (a) How many of the stationary states (counting spin states)

of the hydrogen atom have energy Energy

(b) How many of the stationary states (counting spin states)

of the hydrogen atom have energy and

Energy and Energy and

52. Suppose that all you know about a state of the hydrogen atom

is that the magnetic quantum number is What conclu-

sions can you draw about the value of the orbital quantum

number? The principal quantum number? The energy of the

state? (Hint: Is compatible with the given value of m? Is

Is

*53. Consider the possible directions of the orbital angular-

momentum vector for an electron with orbital quantum

number What are the possible values of the magnetic

quantum number m? For each value of m, calculate the angle

between the angular-momentum vector and the z axis. Draw a

diagram showing the possible orientations of the angular-

momentum vector for all the different values of m.

*54. Suppose that we regard the proton as a sphere of uniform den-

sity with a radius of rotating rigidly about its

axis. According to classical mechanics, if the spin angular

momentum of this sphere is to have a magnitude of

what must be the angular velocity of rotation? What must be

the speed of a point on the equator?

55. List the quantum numbers of all the electrons of a magnesium

atom in its ground state.

56. What are the quantum numbers n and l for the two outer-

most electrons of a Be atom? The Mg atom? The Ca atom? In

what way are these quantum numbers of these different atoms

similar?

*57. The conventional range of wavelengths for X rays extends

from 10 nm to 0.01 nm. Suppose you want to generate char-

acteristic X rays within this range of wavelengths by means of

the transition of an internal electron in an atom, from the first

excited state to the ground state. What atomic numbers are

suitable? Which atoms do these correspond to?

*58. Calculate the wavelengths of the characteristic X rays emitted

by the inner electrons in molybdenum atoms in transitions

from the second and the third excited states to the ground

state.

*59. When a block of metal serving as target in an X-ray tube is

bombarded with a beam of fast electrons, it emits not only

Bremsstrahlung but also characteristic X rays of wavelengths

0.167 nm, 0.141 nm, and 0.133 nm. Assuming that these X

rays arise in transitions from excited states into the ground

state, identify the metal.

60. In the HF molecule, the chemical bonds holding the two

atoms together behave like a massless spring of a spring con-

stant k � 9.7 � 102 N/m.

13�4U ,

1.0 � 10�15 m

l � 1.

l � 2?)l � 1?

l � 0

m � 3.

l � 3?�3.4 eVl � 1?�3.4 eV

l � 0?�3.4 eV

�3.4 eV?

�13.6 eV?

(a) Calculate the frequency of vibration of the molecule. For

the sake of simplicity, assume that the fluorine atom

remains at rest and only the hydrogen atom moves.

(b) Calculate the energy of the first excited vibrational state of

the molecule.

61. Consider the HCl molecule described in Example 4. Suppose

that this molecule is initially in the rotational state. If

the molecule sequentially makes purely rotational transitions

to the state, then to the state, and finally to the

state, what are the energies and the wavelengths of the

photons emitted in each of these transitions?

62. The distance between the two nuclei in the molecule is

0.074 nm.

(a) What is the moment of inertia of an molecule rotating

about its center of mass? What are the energies of the

first, second, and third excited rotational states? Express

these energies in eV.

(b) Suppose we replace one of the hydrogen atoms by a deu-

terium atom (D, or whose mass is twice that of the

hydrogen atom. Where is the new center of mass? What is

the moment of inertia of the HD molecule about its

center of mass? What are the energies of the first, second,

and third excited rotational states? By what factor do these

energies differ from those of the molecule?

63. (a) The binding energy of the NaCl crystal (salt) is 765

kJ/mole; this is the energy required to dissociate the crys-

tal into separate ions. This crystal is cubic, as is illustrated

in Fig. 39.37. Pretend that each ion forms a bond with

only the six nearest ions. What is the energy per Na�–Cl�

bond? Express the answer in electron-volts.

(b) The distance between each ion and the nearest ion is

0.281 nm. What is the Coulomb energy of a pair of ions

separated by this distance? Express the answer in electron-

volts. Explain why the answers obtained in (a) and (b) are

of the same order of magnitude, although not exactly

equal.

H2

2H),

H2

H2

J � 0

J � 1J � 2

J � 3

ClNa

Na

FIGURE 39.37 Structure of NaCl crystal. The yellow balls

are sodium ions and the green balls are chlorine ions.


64. Figure 39.38 shows a full-wave rectifier consisting of four

diodes connected together. This rectifier not only blocks the

negative portion of an entering alternating current, but also

reverses this portion, so the current is positive at all times.

Describe the flow of current through the four diodes when the

entering alternating current is positive and when it is negative.


and For the L shell, we have two

similar states but with and six states with

for a total of eight. For the M shell, we have

the largest number: eight states similar to the L-shell states

but with plus ten states with

for a total of eighteen.

2. Yes, for each orbital state there are a spin up and a spin down

state, and if all states are full, for every spin up electron there is

a spin down electron.

3. (D) Neutral helium has two electrons.

After one is removed, the single remaining electron is in a

purely hydrogenic orbit, but with nuclear charge

Equation (39.6) gives the energies of such orbits as

To remove the electron, we take it

from the state to which requires an energy of

removal

Checkup 39.3

1. No. From Eq. (39.9), we recall that the energy of rotation is

proportional to the square of the frequency of rotation. We

know that the frequency of radiation is proportional to the

difference of the energies of two states, and so is proportional

to the difference of two frequencies squared, and thus is not

equal to any single rotational frequency.

¢E � Z2 � (13.6 eV) � 4 � 13.6 eV.

n � q,n � 1

En � Z2 � (13.6 eV)�n2.

Z � 2.

(Z � 2)4 � 13.6 eV.

[2 � (2l � 1)],l � 2n � 3,

[2 � (2l � 1)],

l � 1n � 2,

ms � ; 12.m � 0,l � 0,

Answers to Checkups

Checkup 39.1

1. According to wave mechanics, the ground-state orbital angular

momentum is according to Bohr theory, For

both theories, the quantum number n � 1 in the ground state.

2. The magnitude of the orbital angular momentum is given by

for this is

3. The orbital angular-momentum quantum number may take

values from 0 to so for the possible values of l

are and

4. The intensity drops to zero at the center of each plot, so the

probability of finding the electron at the nucleus is zero in

each of the electronic states corresponding to Figs. 39.3b, c,

and d.

5. (E) and For a given value of l, the magnetic quan-

tum number m may take on integer values from to

Checkup 39.2

1. For any principal quantum number n, the values of the orbital

angular momentum quantum number l range from 0 to

and for each l there are values of the magnetic quan-

tum number m. There are two values of spin for each state. For

the K shell, we have the least number of states, two: n � 1,

2l � 1

n � 1,

�l.�l

�1.�1, 0,

l � 2.l � 0, l � 1,

n � 3,n � 1,

� 213U .

L � 13 � (3 � 1)Ul � 3,L � 1l (l � 1)U ;

L � U .L � 0;

(b)

Iin

t

Iout

t

pp

pp

nn

nn

Iin

Iout

(a)

FIGURE 39.38 (a) Four diodes connected to

form a full-wave rectifier. (b) The current going

into the rectifier, and the current coming out.



2. The molecule emits four photons, corresponding to the four

transitions shown. From the spacings in Fig. 39.9, we see that

the photon emitted in the first (top) transition has the largest

energy, and the photon emitted in the last (bottom) transition

has the least energy. Since the first photon has

the shortest wavelength, and the last photon the longest.

3. The rotational energy of any state is inversely proportional to

the moment of inertia [see Eq. (39.12)], so the lighter

molecule has the larger energy difference, by a factor of 2.

4. (A) 1, 2, 3, 4. We can obtain the energy difference between

two adjacent states directly from the corresponding adjacent

values on the vertical axis in Fig. 39.9. The photon frequency is

, so starting with the bottom transition we have

and 4 times This sequence continues: for a

transition from a state with any J to the next lower state,

with J � 1, 2, 3, 4, etc.

Checkup 39.4

1. A crystalline solid with a full band and only empty bands

above it is an insulator or a semiconductor. To decide, we

would need to know the magnitude of the energy gap.

2. Most conductors are n type, but a conductor can be p type. A

p-type conductor must have holes that behave like moving

positive charges (as illustrated in Fig. 39.17); this happens

when a band is almost filled. Such behavior occurs in conduc-

tors when different bands overlap, so electrons that would oth-

erwise be near the top of one band spill over into the

lower-energy states of an overlapping band.

3. No. At sufficiently low temperature, there are, in essence, no

electrons with enough thermal energy to cross the energy gap.

4. No. When the lattice potential is not too strong, the evenly

spaced values of the momentum and the dominantly kinetic

¢E � (U2�2I ) � [ J ( J � 1) � ( J � 1) J ] � (U2�I ) � J.

(J � 1)

U �I.�photon � 1, 2, 3

�photon � ¢E�U

¢E

H2


energy imply that the forbidden values of the

energy will be spaced roughly in proportion to the square of

the corresponding momentum values. (When such behavior is

generalized to crystal structures in three dimensions, or when

the lattice potential is strong, then more complicated sequences

of forbidden values of energy occur.)

5. (B) p-type has acceptor atoms and holes; n-type has donor

atoms and free electrons. Acceptor atoms remove electrons

from the valence band and leave behind positively charged

(p-type) holes; donor atoms contribute electrons (negatively

charged, n-type) to the conduction band.

Checkup 39.5

1. Forward. A diode is forward-biased when the p-type region is

at a higher potential than the n-type region.

2. The forward-biased rectifier involves recombination, since the

external source of emf pumps many electrons into the n region

and many holes into the p region, and they recombine at the

junction. (A reverse-biased diode involves the creation of an

extremely small number of thermally generated electron–hole

pairs.) The light-emitting diode is a forward-biased rectifier and

so also involves recombination. The solar cell involves creation,

since the absorption of photons creates electron–hole pairs.

3. (A) Electric field points from n-type region to p-type region.

In the interface region, the mobile carriers are depleted, and

since these carriers came from neutral impurity atoms, the

impurity ions left behind are opposite in sign to the mobile

carriers of each region. The electric field thus points from the

positive donor ions of the n-type region to the negative accep-

tor ions in the p-type region (see Fig. 39.18).

(E � p2�2m),



This nuclear reactor, immersed in a pool of water, releases energy by the

fission, or splitting, of nuclei of uranium. Such nuclear reactions produce

“penetrating radiations,” that is, radiations capable of penetrating through

cloth, paper, and skin. For protection, nuclear reactors are shielded by thick

layers of concrete, or, as in the photo, by a thick layer of water.

With the concepts of this chapter, we can consider such questions as:

? Why does fission release energy, and what is the amount of energy

released per fission? (Section 40.5, pages 1377 and 1376)

? What are the penetrating radiations produced in nuclear reactions,

and how are they produced? (Section 40.3, page 1365; and Example

5, page 1371)

? What is the energy of the penetrating radiation released in the fis-

sion of uranium? (Example 3, page 1367)

Nuclei40

40.1 Isotopes

40.2 The Strong Force and theNuclear Binding Energy

40.3 Radioactivity

40.4 The Law of Radioactive Decay

40.5 Fission

40.6 Nuclear Bombs and NuclearReactors

40.7 Fusion

C H A P T E R

1354

Conceptsin

Context


40.1 Isotopes 1355

? A typical reactor releases a power of 1200 MW. How much uranium does this

consume per year? (Example 7, page 1382)

Rutherford’s first experiments on the bombardment of atoms with a beam of alpha

particles established that the nucleus of the atom is very small, but contains most

of the mass of the atom. The nucleus is therefore very dense, and it must be made of

massive particles packed very tightly together. In later experiments, Rutherford pro-

ceeded to explore the structure of the nucleus, again using a beam of alpha particles as

a probe. He found that if the projectiles were energetic enough to penetrate the nucleus,

they would often split it into two pieces, two smaller nuclei. The smallest such piece

that could be split off was a nucleus of hydrogen, or a proton, and Rutherford there-

fore conjectured that all other nuclei also contain protons. However, all these other

nuclei have more mass and less charge than expected if they contained nothing but

protons—there must be some neutral particles in the nucleus or, alternatively, some

electrically neutral combination of particles of opposite charges.The mystery of the neu-

tral constituent of the nucleus was not solved until 1932, when J. Chadwick discovered

the neutron, a particle of about the same mass as the proton but of zero electric charge.

This discovery led to the modern view of the nucleus as a tightly packed conglomer-

ate of protons and neutrons (see Fig. 40.1).

Since the average distance between the protons in the nucleus is quite short, the

repulsive electric force among the nuclear protons is very large.This force would burst

the nucleus apart if there were not an extra, even larger, attractive force holding the

protons and the neutrons together. This extra force is the nuclear force, or the “strong”

force. Acting on two adjacent protons in a nucleus, this attractive force is about 100

times as large as the repulsive electric force.Thus, the strong force completely overwhelms

the electric force. However, in heavy nuclei—such as uranium—with a large number

of protons and a large total electric charge, the electric repulsion becomes important.

The fission of uranium, as manifested in the explosion of a nuclear bomb, provides a

spectacular demonstration of the electric force overpowering the strong force.

40.1 ISOTOPES

Nuclei are made of protons and neutrons. Generically, these two kinds of constituents

of the nucleus are called nucleons. Table 40.1 lists the main properties of protons and

neutrons. The values of the masses listed in this table are expressed in atomic mass

units, where The value of listed for the spin is the spin

quantum number. According to the usual rule for the magnitude of an angular momen-

tum [see Eq. (39.1)], the magnitude of the spin is actually but physi-212(

12 � 1)U ;

121 u � 1.660 54 � 10�27 kg.

neutron

proton

8.2�10–15 m

Protons and neutrons aretightly packed in nucleus.

FIGURE 40.1 The nucleus of the argon

atom, consisting of 18 protons (red) and 22

neutrons (gray).

nucleon

atomic mass unit u

THE NUCLEONS

NUCLEON MASS SPIN RADIUS

proton

neutron 1 � 10�15 m12mn � 1.008 66 u

1 � 10�15 m12mp � 1.007 28 u

TABLE 40.1


cists often list the values of the spins of particles by giving the spin quantum number,

rather than the magnitude of the spin angular momentum.

It is instructive to compare these particles with the electron. Both the proton and

the neutron have masses about 1840 times as large as that of the electron.1 Their spin

quantum number is the same as that of the electron. In contrast to the electron,

which is a pointlike particle of no discernible size, both the proton and the neutron are

small spheres, of a radius of about

The number of protons in the nucleus of a (neutral) atom of a given element matches

the number of its electrons, that is, it matches the atomic number of the element. For

example, the carbon atom has six electrons and it has six protons in its nucleus.

All the atoms of a given chemical element, such as carbon, have exactly the same

chemical properties, because they all have exactly the same number of electrons and the

same electron configuration. However, the atoms of a chemical element can differ in

mass, because their nuclei can have different numbers of neutrons. Thus, all carbon

atoms have six protons in their nuclei, but some have six neutrons, some have seven,

some have eight, and so on. Atoms with the same number of protons in their nuclei but

different numbers of neutrons are called isotopes. Carbon has eleven known isotopes,

designated (see Fig. 40.2).

The superscript before the chemical symbol (for instance, the superscript “12” in “12C”)

indicates the sum of the number of protons and the number of neutrons; this sum is

called the mass number. If we designate the mass number by the symbol A, then

(40.1)

where N is the number of neutrons and Z is the number of protons, or the atomic

number. Since the mass of each proton and each neutron is approximately one atomic

mass unit, the mass number is approximately equal to the mass of the nucleus in

atomic mass units.

Natural samples of atoms of carbon or any other chemical element contain char-

acteristic percentages of different isotopes. Natural carbon, as found in coal, is a mix-

ture of 98.90% of the isotope 12C and 1.10% of the isotope 13C. Carbon dioxide, as

found in air, contains not only the isotopes 12C and 13C but also a very small amount

of the isotope The other isotopes of carbon do not occur14C.(about 2.4 � 10�10%)

A � N � Z

8C, 9C, 10C, 11C, 12C, 13C, 14C, 15C, 16C, 17C, and 18C,

10�15 m.

12,

1356 CHAPTER 40 Nuclei

1Expressed in atomic mass units, the electron mass is 5.49 � 10�4 u.

6 protons5 neutrons

6 protons6 neutrons

6 protons7 neutrons

6 protons8 neutrons

11C 12C

13C 14C

All nuclei of a givenelement have the samenumber of protons…

…but isotopes of an element have differentnumbers of neutrons.

FIGURE 40.2 Some of

the isotopes of carbon.

isotopes

mass number A

atomic number Z


40.1 Isotopes 1357

naturally; they can be produced only artificially in transmutation of elements in a

nuclear reactor or in a particle accelerator.

All chemical elements have several isotopes (see the excerpt from the chart of iso-

topes in Table 40.2). Hydrogen has three isotopes or ordinary hydrogen; or

deuterium; or tritium). Helium has five isotopes, lithium has six, and so on. Some

of these isotopes occur in nature, others can be produced only by artificial means.

Most of the isotopes listed in the chart of isotopes are unstable; they decay by a

spontaneous nuclear reaction and transmute themselves into another element. The

decay is accompanied by the emission of alpha rays, beta rays, or gamma rays. The

alpha rays are high-speed alpha particles (that is, helium-4 nuclei), the beta rays are high-

speed electrons or antielectrons, and the gamma rays are high-energy photons. We

will examine these decay processes in Section 40.3.

3H,

2H,(1H,

EXCERPT FROM THE CHART OF ISOTOPESa

a The number Z, increasing vertically along the chart, is the number of protons in the isotope; it coincides with the atomic number. The number N,

increasing horizontally, is the number of neutrons. In each box, the number directly below the symbol for the isotope gives the abundance in percent for

naturally occurring isotopes, or else the half-life for unstable, artificially produced isotopes (the half-life is the time required for one-half of a sample of

unstable isotope to decay). The Greek letters indicate the emissions that accompany the decay: � rays (helium nuclei), �� rays (electrons), �� rays (anti-

electrons), or � rays. The bottom number gives the mass of the neutral atom (nucleus plus Z electrons) in atomic mass units. The bottom number in the

shaded boxes gives the atomic mass averaged in proportion to the abundance of naturally occuring isotopes.

TABLE 40.2

10

Ne

20.179

17 Ne

17.0177

0.109 s18 Ne1.67 s

19 Ne1.74 s

20 Ne

19.992 439

90.5%21 Ne

20.993 847

0.27%22 Ne

21.991 384

9.22%23 Ne37.6 s

24 Ne3.38 min

25 Ne0.61 s

26 Ne

26.0005

27 Ne

27.0072

917F

18.9984

15 F

15.0180

16 F

16.011

~10–19 s17 F66.0 s

18 F109.8 min

19 F

18.998 403

100%20 F11.1 s

21 F4.36 s

22 F4.0 s

23 F2.2 s

24 F

24.0093

25 F

25.0138

816O

15.9994

13 O0.0089 s

14 O70.5 s

15 O

15.003 065

122 s16 O

15.994 915

99.756%17 O

16.999 131

0.037%18 O

17.999 159

0.204%19 O26.9 s

20 O13.6 s

21 O3.4 s

22 O

23.0101

23 O

23.0193

715N

14.0067

11 N

11.0267

12 N0.011 s

13 N

1.007 94

9.97 min14 N

14.003 074

99.63%15 N

15.000 109

0.37%16 N

16.006 100

7.11 s17 N4.16 s

18 N00.63 s

19 N

19.0176

0.42 s20 N

20.0238

21 N

21.0289

614C

12.011

9 C0.127 s

10 C19.4 s

11 C

11.011 433

20.4 min12 C

12.000 00

98.89%13 C

13.003 355

1.11%14 C

14.003 242

5730 yr15 C2.45 s

16 C0.74 s

17 C

17.0226

18 C

18.0267

19 C

19.0370

5

Z

N

13B

10.811

8 B0.774 s

9 B

9.013 33

~8 � 10–19 s10 B

10.012 938

19.8%11 B

11.009 305

80.2%12 B0.020 s

13 B0.017 s

14 B 17 B

17.0986

411 12Be

9.012 18

6 Be

6.019 73

�3 � 10–21 s7 Be

7.016 930

53.3 day8 Be

8.005 305

~1 � 10–16 s9 Be

9.012 183

100%10 Be

1.6 � 106 yr11 Be13.8 s

12 Be0.011 s

14 Be

14.0440

Li

39 10

6.941

5 Li

5.0125

~10–21 s6 Li

6.015 123

7.5%7 Li

7.016 005

92.5%8 Li0.85 s

9 Li0.17 s

11 Li0.009 s

He

27 8

4.002 60

3 He

3.016 029

0.00013%4 He

4.002 603

~100%5 He

5.0122

2 � 10–21 s6 He0.802 s

8 He0.122 s

H

15 6

1.0079

1 H

1.007 825

99.985%2 H

2.014 102

0.015%3 H

3.016 049

12.33 yr

0

0 1 2

3 41 n

1.008 665

10.6 min

� –�

� –�

� –�

� +� +

� –�

� –�

� –�

� –�

� +

EC� +no �

p

� –�

� –�

� –�

� +no �

� +�

� +

� –�

� –�

� –�

� +no �

� +�

� –� –�

� –no �

� +

EC� +�

� +

� –�

� –�

� –�

p� +

� –� –�

� –no �

2�EC�

p, �, Li5

� –� –� –p, �

� –�

� –no �

n, �

� –no �

� –

� +

~


Experiments on the bombardment of nuclei with alpha particles and other parti-

cles indicate that the size of the nucleus is proportional to the cube root of the mass

number. Specifically, these experiments indicate that the radius R of a nucleus of mass

number A is

(40.2)

For example, the carbon-12 nucleus has a radius

whereas the uranium-238 nucleus has a radius

We can gain some feeling for how small these radii are by comparing them with the

radius of an atom, typically about The comparison tells us that the radius of

the nucleus is less than 1/10000 of the radius of the atom, so the volume of the nucleus

is only or one-trillionth, of the volume of the atom.

The proportionality between R and A 1�3 implies that the number of nucleons per unit

volume is the same for all nuclei. Since the volume of a sphere is the number of

nucleons per unit volume is

(40.3)

The mass of each nucleon is about and the mass density of the

nuclear material is therefore

(40.4)

This means that one cubic centimeter, or of nuclear material would have a

mass of 230 million tons!

According to Eq. (40.3), the volume per nucleon is We can

think of this volume as a cube enclosing the nucleon; the edge of the cube or, equiva-

lently, the distance from one nucleon to its nearest neighbor is therefore the cube root

of the volume, By comparing this with the radius

of a proton or neutron, about we see that inside the nucleus the nucleons

are so tightly packed together that they touch or almost touch (see Figs. 40.1 and 40.2).

Checkup 40.1

QUESTION 1: How many protons, how many neutrons, and how many nucleons are

there in the nuclei of each of the following carbon isotopes:

QUESTION 2: How many protons, how many neutrons, and how many nucleons are

there in the nuclei of each of the following hydrogen isotopes:

QUESTION 3: Which of the isotopes in Table 40.2 has the largest number of neutrons?

The largest number of protons? The largest mass number?

1H, 2H, and 3H?

8C, 9C, and 10C?

✔

1 � 10�15 m,

1�(1.4 � 1044)1�3 m � 2 � 10�15 m.

1�(1.4 � 1044) m3.

10�6 m3,

1.67 � 10�27 kg �1.4 � 1044

m3� 2.3 � 1017 kg/m3

1.67 � 10�27 kg,

� 1.4 � 1044 nucleons/m3

A

(4p�3)R3�

A

(4p�3)(1.2 � 10�15A 1�3)3 m3

43pR3,

1�1012,

10�10 m.

R � 1.2 � 10�15 m � (238)1�3 � 7.4 � 10�15 m

R � 1.2 � 10�15 m � (12)1�3 � 2.7 � 10�15 m

R � (1.2 � 10�15 m) � A 1�3


radius of nucleus


40.2 The Strong Force and the Nuclear Binding Energy 1359

QUESTION 4: Which of the following isotopes has the largest ratio of the number of

neutrons to the number of protons?

(A) 2H (B) 3H (C) 4He (D) 6Li (E) 12C

QUESTION 5: One nucleus has twice the radius of a second nucleus. What is the ratio

of the number of nucleons in the first nucleus to the number in the second?

(A) 2 (B) 3 (C) 4 (D) 6 (E) 8

40.2 THE STRONG FORCE AND THENUCLEAR B INDING ENERGY

Since the protons within a nucleus are at such short distances from one another, they

exert very large repulsive electric forces on one another. For two neighboring protons,

separated by a center-to-center distance of Coulomb’s Law gives

an electric repulsive force of

(40.5)

This is, roughly, four times the weight of this book; acting on a mass of only 10�27 kg,

the magnitude of this force is colossal.

Obviously, some extra force must be present in the nucleus to prevent it from instan-

taneously bursting apart under the influence of the mutual electric repulsion of the pro-

tons. This extra force is the strong force, already mentioned in Section 6.4. This force

acts equally between any two nucleons, regardless of whether they are protons or neutrons (the

force is “charge-independent”).

Figure 40.3 is a plot of the potential energy associated with the strong nucleon–nucleon

force, calculated from experimental data on nuclear collisions. From Chapter 8, we know

that a potential energy that increases as the nucleons separate corresponds to an attrac-

tive force. A decreasing potential energy means that the force does positive work when

the nucleons separate; this corresponds to a repulsive force. We therefore see from the plot

of the potential energy that the strong force is attractive over a range of internucleon

distances from In this range of distances, the

strong force is much larger than the electric force, as much as 100 times larger.The strong

force is repulsive for internucleon distances less than this means that

the nucleons have a hard core that resists interpenetration. For distances larger than

the strong force decreases drastically and finally vanishes.Thus, in con-

trast to the electric force, which fades only gradually and reaches out to large distances,

the strong force cuts off sharply and has only a short range. In order to feel the strong force,

the nucleons must be touching or almost touching; that is, the force acts only between

nearest neighbors in the nucleus.

In consequence of the short-range character of the strong force, a nucleon deep

inside the nucleus does not experience any net force–the nucleon interacts only with

its nearest neighbors, and since these pull it with equal forces in almost all directions,

the net force on the nucleon is zero or nearly zero (see Fig. 40.4). However, a nucleon

at the nuclear surface has neighbors only on the side that is toward the interior, and hence

these will exert a net force pulling the nucleon inward (see Fig. 40.4). Altogether this

means that nucleons are more or less free to wander about the interior of the nucleus,

� 2 � 10�15 m,

� 0.7 � 10�15 m;

� 0.7 � 10�15 m to � 2 � 10�15 m.

� 58 N

F �1

4p�0

e2

r2�

1

4p�0

�(1.6 � 10�19 C)2

(2.0 � 10�15 m)2

r � 2.0 � 10�15 m,

strong force

U

r

200MeV

100

–100

1�10–151�10–15 2�10–15 m2�10–15 m1�10–15 2�10–15 m

Strong force is attractivefor internucleon distancesof ≈ 0.7–2 � 10–15 m, …

…resists nucleoninterpenetration at shorter distances, …

…and cuts off sharply, having only a short range.

FIGURE 40.3 Potential energy for the

strong force acting between two nucleons.


but whenever they approach the nuclear surface, strong forces pull them back and pre-

vent their escape.

This suggests that the nucleons in the nucleus behave somewhat like the mole-

cules in a drop of water; such molecules are free to wander throughout the volume of

the drop, but when they approach the water surface, intermolecular forces hold them

back. This similarity between nuclei and drops of water rests on a similarity of the

laws of force. The intermolecular force has general features rather similar to those dis-

played in Fig. 40.3; the force is attractive over a short range and then becomes strongly

repulsive when the molecules begin to interpenetrate. The hard repulsive core makes

the water nearly incompressible, whereas the short-range attraction provides a cohe-

sive force that prevents water droplets from falling apart. The balance of attraction

and repulsion encourages water molecules to stay at a particular distance from one

another, and this gives water a particular, uniform density.

Because of the similarities between a liquid and the nuclear material, the nucleus can

be crudely regarded as a droplet of incompressible “nuclear fluid” of uniform density. The

fluid is, of course, made of nucleons, but for some purposes we can ignore the indi-

vidual nucleons, and we can calculate the properties of nuclei in terms of the gross

properties of a liquid. For example, the spherical shape adopted by most nuclei can

easily be understood as follows: Any nucleon located on the surface of a globule of

nuclear fluid experiences an inward force pulling it back into the volume, and conse-

quently the fluid tends to shrink its exposed surface to the smallest value compatible

with its (fixed) volume. Since a sphere has the least surface area for a given volume,

the globule of fluid will take the shape of a spherical droplet.

In a stable nucleus, the repulsive electric forces among the protons are held in check

by the attractive strong forces.To achieve this balance of forces, the presence of neutrons

is an advantage: a nucleus with more neutrons will have a larger size and, therefore, a

large average distance between pairs of protons—the neutrons in the nucleus dilute the

repulsive effect of the electric force. Consequently, all stable nuclei, with

the exception of hydrogen and one isotope of helium, contain at least as

many neutrons as protons; heavy nuclei, such as uranium, contain sub-

stantially more neutrons than protons.

Figure 40.5 is a plot of the number of neutrons vs. the number of

protons (N vs. Z). On this plot, blue dots indicate the stable nuclei and

red dots indicate unstable nuclei, that is, radioactive isotopes. Note that

there is no stable nucleus beyond bismuth However, several

elements beyond bismuth have some isotopes with very long lifetimes;

these are therefore almost stable, and they occur naturally.

The energy stored in a nucleus is a sum of the potential energies

contributed by the electric and the strong forces and the kinetic energies

of the nucleons. The potential energy is negative, and its magnitude is

larger than that of the kinetic energy. Thus, the “stored” energy is neg-

ative, and energy is released when the nucleus is assembled out of its con-

stituent nucleons. Conversely, energy must be supplied to take the nucleus

apart into its constituent nucleons. The energy that must be supplied to

take the nucleus apart, or the energy released during the assembly of the

(Z � 83).


150

140

130

120

110

100

90

80

70

60

50

40

30

20

10

010 20 30 40 50 60 70 80 90 1000

N

Z

A � 220

A � 220

200200

180180

160160

140140

120120

100100

8080

6060

4040

A � 220

200

180

160

140

120

100

80

60

40

A � 20

N�

Z

Heavy nuclei containmany more neutrons than protons.

For Z � 2, stable nuclei(blue) contain at least asmany neutrons as protons.

FIGURE 40.5 Number of neutrons (N )

vs. number of protons (Z) for stable nuclei

(blue dots) and unstable nuclei (red dots).

Net strong force on nucleons nearsurface is directed inward…

…while net strong force on nucleonsinside nucleus is essentially zero.

FIGURE 40.4 Strong forces on a nucleon

at the nuclear surface, and strong forces on a

nucleon in the nuclear interior.



nucleus out of its constituent nucleons, is called the binding energy (B.E.). Figure

40.6 is a plot of B.E./A, the binding energy divided by the number of nucleons, or the

average binding energy per nucleon. The curve plotted in Fig. 40.6 is called the curve

of binding energy. The energy unit used in this plot is the MeV, where

(40.6)

This unit is widely used in nuclear physics.

The binding energy of a typical nucleus is a rather large amount of energy. As may

be seen from Fig. 40.6, the average binding energy per nucleon is in the vicinity of 8

MeV for almost all nuclei; thus a nucleus with a mass number A typically has a bind-

ing energy of about To put this number in perspective, let us compare it

with the rest-mass energy of the nucleons. Each nucleon has a mass of about one

atomic mass unit. The energy corresponding to one atomic mass unit is

or, in MeV units [see Eq. (40.6)],2

(40.7)

Thus, each nucleon has a rest-mass energy of about 930 MeV, and the A nucleons in

the nucleus have a rest-mass energy of about The ratio of binding

energy to rest-mass energy is then about which means the binding

energy is nearly 1% of the rest-mass energy!

By Einstein’s mass-energy formula, the mass associated with the binding energy is

B.E./c2. This mass is carried away by the energy released during the assembly of the

8�930 � 0.009,

A � 930 MeV.

1 u � c2 � 1.4924 � 10�10 J �1 MeV

1.6022 � 10�13 J� 931.5 MeV

1 u � c2 � 1.6605 � 10�27 kg � (2.9979 � 108 m/s)2 � 1.4924 � 10�10 J

A � 8 MeV.

1 MeV � 106 eV � 1.6022 � 10�13 J

B.E./AMeV

8

7

6

5

4

3

2

1

50 100 150 200

mass number, A

12C

11B

Ca Fe Zn Kr Mo TeSm

LuHg

Ra4He

16O

20Ne 24Mg

2H

3He

6LiBinding energy pernucleon increaseswith mass numberfor light nuclei,…

…and decreaseswith mass numberfor heavy nuclei.

Binding energy isabout 8 MeV pernucleon for almostall nuclei.

FIGURE 40.6 Average binding energy per nucleon vs. mass number.

curve of binding energy

energy equivalent of atomic mass unit

2We have retained four significant figures in this result because we will be needing a precise value of the

equivalence between energy and mass later on.


nucleus from its constituent protons and neutrons. The mass of a typical nucleus is

therefore about 1% less than the sum of the masses of these protons and neutrons.The

mass difference is called the mass defect,

� [mass defect] � [mass of Z protons and N neutrons]

� [mass of nucleus] (40.8)

Experimental values of the binding energy of a nucleus usually are obtained via the

mass defect. A precise measurement of the mass of the nucleus is compared with the

sums of the masses of the constituent protons and neutrons; the difference, or mass

defect, then gives the binding energy according to Eq. (40.8). This is how the exper-

imental points in the plot of the curve of binding energy were obtained.

What is the nuclear binding energy of the isotope Express

the energy in MeV. The mass of one atom of this isotope is

238.0508 u.

SOLUTION: The atomic number of uranium is and uranium-238 has

mass number so this isotope has 92 protons and

neutrons. According to Eq. (40.8),

(40.9)

We want to re-express this equation in terms of the mass of the atom, since we

know this mass.The mass of the uranium nucleus is the mass of the uranium atom

minus the mass of the 92 electrons of this atom. Hence

But 92 proton masses plus 92 electron masses equals 92 hydrogen masses, because

a hydrogen atom consists of one proton and one electron. We therefore obtain the

following convenient formula for the binding energy:

With (from Table 40.2, page 1357) and

this yields

The binding energy is therefore

Since [see Eq. (40.7)], the result is

B.E. � 1.934 � 931.5 MeV � 1802 MeV

u � c2 � 931.5 MeV

B.E. � 1.934 u � c2

� 1.934 u

B.E.

c 2� 92 � 1.007 825 u � 146 � 1.008 66 u � 238.0508 u

mn � 1.008 66 u,

[mass of 1H atom] � 1.007 825 u

B.E.

c 2� 92[mass of 1H atom] � 146mn � [mass of 238U atom]

B.E.

c 2� [92mp � 146mn] � ([mass of 238U atom] � 92me)

B.E.

c 2� [92mp � 146mn] � [mass of nucleus]

N � A � Z � 146A � 238,

Z � 92

238U?EXAMPLE 1

B.E.

c2


mass defect


Q value of reaction

Besides its role as nuclear binding force, the strong force also plays a crucial role

in nuclear reactions that occur when the nuclei in some target material are bombarded

by a beam of protons, neutrons, or alpha particles. When such a projectile strikes the

nuclear surface, the strong force pulls the projectile into the nucleus. The projectile

then either remains bound within the nucleus, forming a heavier nucleus; or else the

projectile disrupts the internal structure of the nucleus to such an extent that the

nucleus ejects one or several fragments, that is, one or several protons, neutrons, alpha

particles, or even other nuclei. Such nuclear reactions involve transmutation of

elements: the original nucleus is changed into a new nucleus of different mass number and

atomic number.

The first such transmutation of elements was discovered in 1919 by Rutherford

when he bombarded nitrogen with alpha particles. He found that this led to the trans-

mutation of nitrogen into oxygen, according to the reaction3

As we will see in Section 40.3, the alpha particle is actually the nucleus of a helium-

4 atom; hence the reaction can also be written3

(40.10)

In his early experiments, Rutherford used a naturally radioactive material as the

source of his beam of alpha particles. But in the 1930s, physicists began to build

machines for the artificial acceleration of beams of charged particles.The first of these

accelerators were electrostatic; they accelerated particles by means of strong static elec-

tric fields produced by a large amount of electric charge accumulated on a spherical

capacitor (see Fig. 40.7). Many of the later accelerators use a combination of electric

and magnetic fields. Thus, in the cyclotron (see Section 30.1), a uniform magnetic

field holds protons in a circular orbit, while an electric field acts on them periodically,

gradually increasing their energy step by step.

The investigation of nuclear reactions initiated by projectiles from accelerators led

to the discovery of a multitude of new isotopes, most of them short-lived and highly

radioactive. These investigations also led to a better understanding of the details of

the strong force.

The energy of the projectile required to initiate a nuclear reaction or the energy

released in a nuclear reaction can be calculated from the rest-mass energies of the iso-

topes that participate in the reaction. The energy released in the reaction is called its

Q value. For a reaction with initial masses and final masses

the Q value is the difference between the initial and the final sums of rest-mass energies:

(40.11)

If the rest-mass energy after the reaction is smaller than that before, the Q value

is positive, and energy is released in the reaction. Such a reaction can proceed even if

the energy of the projectile is very low (nearly zero).

If the rest-mass energy after the reaction is larger than that before, the Q value is

negative, and energy must be supplied to make the reaction proceed; that is, the reac-

tion absorbs energy, instead of releasing energy.The energy required to make the reac-

tion proceed must come from the kinetic energy of the incident projectile; thus, the

Q � (m1c2 � m2c2 � ) � (m1c2 � m2c 2 � )

m1, m2, p ,m1, m2, p

4He � 14N S 17O � 1H

� � 14N S 17O � p


transmutation of elements

FIGURE 40.7 The Cockcroft–Walton

accelerator built in 1932.

3These reactions as written include only nucleons, and do not account for the atomic electrons of the atoms

or ions involved.


projectile needs a minimum energy to initiate the reaction. The kinetic energy of the

projectile is then (partially) converted into rest-mass energy of the reaction products.

Calculate the Q value for the reaction (40.10). What can you

say about the minimum kinetic energy of the alpha particle

required to initiate this reaction?

SOLUTION: The Q value is the difference between the total rest-mass energy

before the reaction and the total rest-mass energy after the reaction.The total rest-

mass energy before the reaction (40.10) is the rest-mass energy of the alpha par-

ticle (or nucleus) plus the rest-mass energy of the 14N nucleus. The total

rest-mass energy after the reaction is the sum of the rest-mass energies of the

nucleus and the proton. Thus,

(40.12)

Since the chart of isotopes lists the masses of the atoms, rather than the masses of

nuclei, we want to express Eq. (40.12) in terms of atomic masses. For this purpose, we

add the rest-mass energy of 18 electrons to the first two terms on the right side of

Eq. (40.12), and we subtract the rest-mass energy of 18 electrons from the last two terms:

Each of the terms in brackets is now the mass of a complete atom:

Substituting the values of the masses listed on the chart of isotopes (Table 40.2, page

1357), we then find

Since we can express our final result as

This negative Q value indicates that the sum of rest-mass energies after the reac-

tion is larger than before, that is, the reaction absorbs energy.The absorbed energy

is 1.191 MeV, and this energy must be supplied by the alpha particle. The mini-

mum kinetic energy of the alpha particle must therefore be at least 1.191 MeV.

COMMENT: Although the minimum energy required for the reaction is 1.191

MeV, the reaction can proceed only with alpha particles of somewhat higher kinetic

energy, because momentum conservation dictates that some kinetic energy must

be given to the reaction products.

Q � �0.001 279 � 931.5 MeV � �1.191 MeV

u � c2 � 931.5 MeV,

� 1.007 825 u � c2 � �0.001 279 u � c2

Q � 4.002 603 u � c2 � 14.003 074 u � c2 � 16.999 131 u � c2

� [mass of 17O atom]c2 � [mass of 1H atom]c2

Q � [mass of 4He atom]c2 � [mass of 14N atom]c2

� [mass of 17O nucleus � 17me]c2 � [mp � me]c2

Q � [mass of 4He nucleus � 4me]c2 � [mass of 14N nucleus � 14me]c2

� [mass of 17O nucleus]c2 � mpc2

Q � [mass of 4He nucleus]c2 � [mass of 14N nucleus]c2

17O

4He

EXAMPLE 2



Checkup 40.2

QUESTION 1: By inspection of Fig. 40.3, determine where the strong force has its

largest attractive value.

QUESTION 2: In Example 1 we found that the binding energy of is 1802 MeV.

Accordingly, what is the value of B.E./A for this isotope?

QUESTION 3: Can a stable nucleus have a negative mass defect, that is, a mass excess?

QUESTION 4: Consider the reverse of the reaction (40.10), that is, bombardment of

by protons to form helium and nitrogen, What is the Q

value of this reaction? What can you say about the minimum kinetic energy of the

proton required to initiate this reaction? Ignore the recoil of the oxygen nucleus.

QUESTION 5: According to Fig. 40.6, which of the nuclei named in this figure has the

largest binding energy per nucleon (largest B.E./A)? Which has the largest binding

energy (largest B.E.)?

(A) 2H, Fe (B) 2H, 4He (C) 4He, Ra (D) Fe, Ra (E) Fe, 4He

40.3 RADIOACTIVITY

Radioactivity was discovered in 1896 by Henri Becquerel.Through an accident, he noticed

that samples of uranium minerals emitted invisible rays which could penetrate through sheets

of opaque materials and make an imprint on a photographic plate.Subsequent investigations

established that uranium, and many other radioactive substances, emit three kinds of rays:

alpha rays (�),beta rays (�), and gamma rays (�). Of these, the alpha rays are the least pen-

etrating; they can be stopped by a thick piece of paper. The beta rays are more penetrat-

ing; they can pass through a foil of lead or a plate of aluminum.The gamma rays are the

most penetrating; they can pass through a thick wall of concrete. When these three kinds

of rays are aimed into a magnetic field, the alpha and beta rays are deflected in opposite

directions, whereas the gamma rays proceed without deflection (see Fig. 40.8).This simple

experiment demonstrates that alpha rays and beta rays are electrically charged, with charges

of opposite signs, whereas gamma rays are neutral.

By more detailed experiments. Becquerel identified the beta rays as high-speed

electrons. Some years later, Rutherford investigated the nature of the alpha rays and

demonstrated that they are identical to nuclei of helium. Thus, electrons and helium

nuclei are, somehow, manufactured within the sample of uranium or other radioac-

tive substance, and they are ejected at high speed. Gamma rays are high-energy pho-

tons emitted by the radioactive substance; the energy of these gamma-ray photons is

typically a thousand times as large as that of X-ray photons.

The “manufacture” and ejection of these high-speed electrons and helium nuclei

occurs in the nuclei of the radioactive substance, by nuclear decay reactions. In the fol-

lowing, we will discuss the broad features of these nuclear reactions.

Alpha Decay

The alpha particle consists of two protons and two neutrons; it has the same structure as

the helium-4 nucleus. When a radioactive nucleus ejects an alpha particle consisting

of two protons and two neutrons, the radioactive nucleus does not create these protons

1H � 17O S 4He � 14N.

17O

238U

✔


alpha rays (�), beta rays (�), and gamma rays (�)

Conceptsin

Context

�

�

�

When rays enter magneticfield, � and � rays are deflectedin opposite directions,…

…and � raysare undeflected.

FIGURE 40.8 Alpha (�), beta (�), and

gamma (�) rays emitted by a radioactive

source. A magnetic field is perpendicular to

the plane of the page.

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and neutrons out of nothing—the nucleus merely takes two of its own protons and

neutrons and spits them out. Since the nucleus loses two protons and neutrons, both

the atomic number and the mass number of the nucleus decrease.The atomic number

decreases by 2, and the mass number by 4. Thus, the original isotope becomes a dif-

ferent isotope of a different chemical element.The following examples of alpha decays

illustrate such transmutations of chemical elements:

(40.13)

(40.14)

In the first of these alpha-decay reactions, uranium is transmuted into thorium, and in

the second, radium into radon. The original nucleus in a decay reaction is called the

parent, and the resulting nucleus is called the daughter.

The alpha-decay reaction can be regarded as a fission, or splitting, of the nucleus

into two smaller nuclei. The, say, uranium nucleus fissions into a thorium nucleus and

a helium nucleus. The total number of protons and of neutrons is unchanged in this

fission (see Fig. 40.9). The fission occurs spontaneously, because of an instability in

the original nucleus. Large nuclei, such as uranium or radium, contain many protons,

which exert repulsive electric forces on each other. Although the repulsive electric force

is balanced by the attractive strong force, the balance of these forces is rather precar-

ious because the electric force easily reaches from one end of the nucleus to the other,

whereas the strong force acts only between adjacent nucleons and therefore, in a large

nucleus, cannot reach directly from one end to the other. Thus, any accidental, spon-

taneous elongation of the nucleus can shift the balance in favor of the electric force—

the nucleus then elongates more and more, and ultimately bursts apart into two

fragments. In true fission, the two fragments are of approximately equal size. In alpha

decay, we are dealing with an extreme case of fission, with two fragments of very

unequal size. The ejection of an alpha particle is strongly favored over ejection of, say,

a hydrogen nucleus or a lithium nucleus, because the helium nucleus is an exception-

ally tightly bound nucleus. In consequence of this large binding energy, the formation

of an alpha particle, just before its ejection, makes more energy available for driving the

fission reaction.

All the large, heavy nuclei beyond bismuth are unstable; they all are subject to

alpha decay or other forms of spontaneous fission. Besides, many isotopes of some-

what smaller nuclei suffer from the same instability.

226Ra S 222Rn � �

238U S 234 Th � �


238U

92 protons146 neutrons

4He

2 protons2 neutrons

234Th


Alpha decay of anunstable nucleus isoften favored…

…since 4He nucleus istightly bound, availing more energy for fission.

Atomic number of daughter nucleus has decreased by two,and mass number by four.

FIGURE 40.9 Fission of uranium

into thorium and helium.

parent nucleus and daughter nucleus

fission

ANTOINE HENRI BECQUEREL (1852–1908) French physicist. He was awarded the

1903 Nobel Prize for his discovery of radio-

activity.


The kinetic energy of the alpha particles ejected in reactions such as (40.13) and

(40.14) can be calculated from the masses of the participating nuclei and the mass of the

alpha particle.

Calculate the energy released in the alpha decay of 238U, and

calculate the kinetic energy of the alpha particle ejected in this

reaction.The atomic masses of the isotopes of uranium, thorium, and helium in this

reaction are 238.0508 u, 234.0436 u, and 4.0026 u, respectively.

SOLUTION: The energy released in the reaction (40.13) is simply the difference

between the total rest-mass energy before the reaction and the (smaller) total rest-

mass energy after the reaction.The total rest-mass energy before the reaction is the

rest-mass energy of the uranium nucleus, The total rest-

mass energy after the reaction is the sum of the rest-mass energies of the thorium

nucleus and the alpha particle,

The energy released in the reaction is therefore

(40.15)

To express this in terms of the masses of the atoms rather than the masses of the

nuclei, we add the rest-mass energy of 92 electrons to the first term on the right

side of Eq. (40.15) and we subtract the same amount from the other two terms:

(40.16)

Now each of the terms in square brackets is the mass of a complete atom:

(40.17)

We know that these masses are 238.0508 u, 234.0436 u, and 4.0026 u, respec-

tively; hence,

or, with

Since the thorium nucleus is much heavier than the alpha particle, it suffers

almost no recoil, and the alpha particle carries away almost all of the energy released

in the reaction. Thus, the kinetic energy of the alpha particle will be 4.3 MeV.

In many cases of alpha decay, the daughter nucleus is also unstable, and decays

some time after its formation, either by alpha decay or by beta decay. The daughter of

the daughter then decays, and so on. The sequence of daughters descending from the orig-

inal parent is called a radioactive series. The series ends when it reaches a stable iso-

Q � 0.0046 � 931.5 MeV � 4.3 MeV

u � c2 � 931.5 MeV,

Q � 238.0508 u � c2 � 234.0436 u � c2 � 4.0026 u � c2 � 0.0046 u � c2

Q � [mass of 238U atom]c2 � [mass of 234Th atom]c2 � [mass of 4He atom]c2

� [mass of 4He nucleus � 2me]c2

Q � [mass of 238U nucleus � 92me]c2 � [mass of 234Th nucleus � 90me]c2

� [mass of 4He nucleus]c2

Q � [mass of 238U nucleus]c2 � [mass of 234Th nucleus]c2

[mass of Th nucleus]c2 � [mass of He nucleus]c2.

[mass of U nucleus]c2.

EXAMPLE 3


Conceptsin

Context

radioactive series


tope. For instance, the decay of uranium initiates a radioactive series, which ultimately

ends with a stable isotope of lead.

Beta Decay

The simplest beta-decay reaction is the decay of the neutron.The free neutron is unsta-

ble, and it decays into a proton, an electron, and an antineutrino:

n S p � e� � (40.18)

In this equation, the symbol (nu-bar) represents the antineutrino. Neutrinos (�) and

antineutrinos ( ) are particles of very small mass and spin quantum number which

travel at nearly the speed of light. Thus, they are somewhat similar to photons. How-

ever, in contrast to photons, which interact with electric charges, neutrinos and anti-

neutrinos do not interact directly with electric charges—in fact, they hardly interact with

anything at all, and they pass through the entire bulk of the Earth with little hindrance.

As their names indicate, neutrinos and antineutrinos are antiparticles of each other;

they can annihilate each other, producing a flash of light (gamma rays).

For a free neutron, the decay reaction (40.18), on the average, takes a time of 15

minutes.4 For a neutron in a nucleus, the reaction may proceed at a faster rate or at a slower

rate, depending on whether the nucleus promotes the reaction by supplying extra energy,

or inhibits the reaction by withdrawing energy. When the reaction occurs in a nucleus,

the electron and antineutrino are ejected, and the net effect is the conversion of a neu-

tron into a proton, which increases the atomic number of the nucleus by 1, while leav-

ing the mass number unchanged.Thus, the original isotope is transmuted into an isotope

of the next chemical element. Two examples of beta decays are the following:

60Co S 60Ni � e� � (40.19)

14C S 14N � e� � (40.20)

In each of these reactions, the number of neutrons (N ) decreases by 1, and the number

of protons (Z) increases by 1 (see Figs. 40.10 and 40.11).

The antineutrino emitted in these reactions is almost impossible to detect. Its existence,

however, can be inferred from the conservation of energy. In the beta decay of, say, radioac-

tive cobalt-60 [Eq. (40.19)], sometimes the electron emerges with one energy, and some-

times with another.This is because the energy of the decay is shared between the electron

and the neutrino, and sometimes one of these particles carries off most of the energy,

sometimes the other. Hence the energy of the electron is unpredictable—it can be any-

thing from zero up to a maximum amount that corresponds to the electron carrying away

all the energy of the decay. In fact, it was on the basis of this variability of the energy of

the ejected electron that Pauli first proposed the existence of the neutrino, since this was

the only way to preserve the Law of Conservation of Energy. When Pauli made this pro-

posal in 1931, there was no direct evidence for the existence of an extra particle; neutri-

nos and antineutrinos were detected only much later, in experiments with nuclear reactors.

Beta-decay reactions of the type (40.19) and (40.20) involve the conversion of a neu-

tron into a proton and the ejection of an electron and an antineutrino and occur for

many nuclei that have more neutrons than the stable isotopes of a given element. In

nuclei that have fewer neutrons than the stable isotopes, there are often beta-decay

�

�

12,�

�

�


4The average lifetime of free neutrons is 15.0 min, but the half-life (see next section) is 10.6 min.

neutrinos (�) and antineutrinos ( ) �

e–60Co


60Ni


In negative beta decay,electron and antineutrinoare ejected.

Atomic number of daughter nucleushas increased by one, transmutingcobalt into nickel. Mass numberis unchanged.

�—

FIGURE 40.10 Beta decay of 60Co into60Ni. The number of neutrons is 33 before

the decay and 32 after; the number of pro-

tons is 27 before the decay and 28 after.


reactions involving the conversion of a proton into a neutron and the ejection of an

antielectron, or positron, and a neutrino. The positron is the antiparticle of the elec-

tron; it has the same mass and spin as the electron, but the opposite electric charge. An

example of a beta-decay reaction with ejection of a positron is

22Na S 22Ne � e� � � (40.21)

Here the symbol e� represents the positron. After the positron is ejected in this decay,

it sooner or later collides with one of the abundant atomic electrons in the surround-

ing environment and annihilates with it, producing gamma rays. In the reaction (40.21),

the number of protons (Z) decreases by 1, and the number of neutrons (N ) increases

by 1 (see Fig. 40.12).

All the beta-decay reactions are instigated by a new kind of force, the “weak”

force. This is one of the four fundamental forces found in matter (see Sections 6.4 and

41.3). Whereas the electromagnetic and the strong forces play an important role in

determining the structure of atoms and of nuclei, the weak force plays no role in this.

Its only effect is to engender the beta-decay reactions (40.19)–(40.21) and other sim-

ilar reactions.

What is the maximum kinetic energy of the beta rays emitted in

the beta decay of The atomic masses of and of are

14.003 24 u and 14.003 07 u, respectively.

SOLUTION: The energy released in the reaction (40.20) is, again, the difference

between the total rest-mass energy before the reaction and the (smaller) total rest-

mass energy after the reaction. The total rest-mass energy before the reaction is

14N14C14C?EXAMPLE 4


positron (e�)

�

e–

—

14C

6 protons8 neutrons

14N

7 protons7 neutrons

In this negativebeta decay…

…carbon is transmutedinto nitrogen.

FIGURE 40.11 Beta decay of 14C into14N with ejection of an electron and an anti-

neutrino. The number of neutrons is 8

before the decay and 7 after; the number of

protons is 6 before the decay and 7 after.

�

e�22Na


22Ne


In positive beta decay,positron (antielectron)and neutrino are ejected.

Atomic number of daughternucleus has decreased by one,transmuting sodium into neon.Mass number is unchanged.

FIGURE 40.12 Beta decay of 22Na into 22Ne, with

ejection of an antielectron and a neutrino. The number

of neutrons is 11 before the decay and 12 after; the

number of protons is 11 before the decay and 10 after.

“weak” force


the rest-mass energy of the carbon nucleus, The total

rest-mass energy after the reaction is the sum of the rest-mass energies of the nitro-

gen nucleus and the electron, (the rest-mass energy

of the neutrino is negligible). The energy released in the reaction is therefore

If we add and subtract the rest-mass energy of 6 electrons on the right side of this

equation, we obtain

Here the two terms in brackets are the masses of the complete atoms, so

Substituting these masses and substituting MeV, we find

The nitrogen nucleus, being much heavier than the electron, remains at rest, or

nearly at rest. The decay energy is therefore shared between the electron and the

antineutrino. If the antineutrino carries away almost none of the energy, then the

electron will acquire nearly all the energy. Thus, the maximum possible electron

kinetic energy is 0.16 MeV.

Gamma Emiss ion

Gamma rays are high-energy photons emitted by nuclei when nucleons make transi-

tions from one stationary nuclear state to another. The emission of a gamma ray by a

transition of a nucleon is similar to the emission of visible photons or of X rays by a

transition of an atomic electron.

When a nucleus suffers alpha decay or beta decay, it is often left in an excited state,

and it then eliminates the excitation energy in the form of a gamma ray. Thus, gamma

emission is usually one step in a two-step process, with an alpha or beta decay preceding

the gamma emission. For example, the beta decay of leads to subsequent gamma

emission according to the sequence of reactions

�R

(40.22)

The isotope is commonly employed in high-intensity industrial irradiation cells,

called cobalt cells. But, as shown in Eq. (40.22), the emitter of the gamma rays is nickel,

not cobalt.

As mentioned above, gamma rays are also generated when an electron annihilates

with an antielectron. But this is a secondary process, which does not directly involve

the nucleus.

60Co

60Ni � g

�60Co S 60Ni � e�

60Co

� 0.000 17 � 931.5 MeV � 0.16 MeV

Q � 14.003 24 u � c2 � 14.003 07 u � c2 � 0.000 17 u � c2

u � c2 � 931.5

Q � [mass of 14C atom]c2 � [mass of 14N atom]c2

� [mass of 14C nucleus � 6me]c2 � [mass of 14N nucleus � 7me]c2

Q � [mass of 14C nucleus � 6me]c2 � [mass of 14N nucleus � 6me]c2 � mec2

Q � [mass of 14C nucleus]c2 � [mass of 14N nucleus]c2 � mec2

[mass of 14N nucleus]c2 � mec2

[mass of 14C nucleus]c2.



Figure 40.13 shows the energy-level diagram for the

nucleus. When this nucleus is formed in the beta decay of

it is initially in the second excited state (2.50 MeV above the ground state). It

makes a transition to the first excited state (1.33 MeV above the ground state),

and then a transition to the ground state. What are the energies of the gamma rays

emitted in these transitions?

SOLUTION: The first transition is from the energy level to the

energy level The energy of the gamma ray emitted in this tran-

sition is

The second transition is from the energy level E1 � 1.33 MeV to the ground state,

at The energy of the gamma ray emitted in this transition is

E � E1 � E0 � 1.33 MeV

E0 � 0.

E � E2 � E1 � 2.50 MeV � 1.33 MeV � 1.17 MeV

E1 � 1.33 MeV.

E2 � 2.50 MeV

60Co

60NiEXAMPLE 5


As illustrated in Examples 2, 3, and 4, the Q value of a nuclear

reaction is best calculated by adding electron masses to the

nuclear masses, so as to form masses of atoms. The atomic

masses can then be copied from charts or tables of isotopes

(which always list atomic masses, rather than nuclear masses).

However, it is necessary to take some extra care with this

technique in the case of beta decays with antielectrons, such

as the beta decay in Eq. (40.21). On the left side of Eq. (40.21),

the nucleus is sodium, which requires 11 electrons to be a

complete sodium atom. On the right side the nucleus is neon,

which requires only 10 electrons to become a complete neon

atom. Hence, if we add 11 electron masses to the nuclear

masses on each side, we will be left with one surplus electron

mass on the right side. Furthermore, the right side has an

antielectron mass (equal to the electron mass). Thus, the Q

value for the reaction (40.21) is

� [mass of 22Ne atom]c2 � 2mec2

Q � [mass of 22Na atom]c2

PROBLEM-SOLVING TECHNIQUES NUCLEAR REACTIONS

Conceptsin

Context

E

60Co 60Ni

� –

�

�

2.50 MeV

1.33 MeV

0 MeV

After beta decayof 60Co, …

… 60Ni is in itssecond excited state…

…and emits one gamma ray in transition to first excitedstate, and another in transitionto ground state.

FIGURE 40.13 Energy-level

diagram for the 60Ni nucleus

formed in the beta decay of 60Co.


Checkup 40.3

QUESTION 1: Why do alpha and beta emissions involve transmutation of elements

whereas gamma emission does not?

QUESTION 2: Figure 40.8 shows the deflections of alpha and beta rays in a magnetic

field. If instead of consisting of an electron, the beta ray consisted of an antielectron,

what would be its direction of deflection?

QUESTION 3: Can a nuclear decay reaction have a negative Q value?

QUESTION 4: An electron is emitted during the beta decay of a cesium-136 nucleus

(136Cs). Which of the following nuclei results?

(A) 135Xe (B) 136Xe (C) 137Cs (D) 136Ba (E) 137Ba

40.4 THE LAW OF RADIOACTIVE DECAY

In radioactive alpha or beta decay, the original isotope, or parent, becomes transmuted

into another isotope, or daughter. If we have some given initial amount of parent

material, say, 1 gram of radioactive strontium, some nuclei decay at one time, some at

another, and the parent material disappears only gradually. In the case of radioactive

strontium, one-half of the original amount disappears in 29 years, one-half of the

remainder in the next 29 years, one-half of the new remainder in the next 29 years,

and so on. Hence the amounts left at times 0, 29, 58, 87 years, etc., are gram,

etc. The time interval of 29 years is called the half-life, or of strontium.

Mathematically, we can represent the amount of strontium at different times by

the formula

(40.23)

Here n is the number of strontium nuclei at time t and is the number at the initial

time, With we can easily verify that at 29 years, 58 years,

etc., the formula (40.23) yields the expected results:

at

and so on. However, the formula (40.23) is valid not only at these times, but also at inter-

mediate times. For instance, at

Figure 40.14 is a plot of the number of remaining strontium nuclei vs. time.

� n0 � 0.34

n � n0 a 1

2b 45 years�29 years

� n0 a 1

2b 45�29

� n0 a 1

2b 1.55

t � 45 years,

at t � 58 yearsn � n0 a 1


� n0 a 1

2b 2

� n0 �1

4

t � 29 yearsn � n0 a 1


� n0 a 1

2b 1

� n0 �1

2

at t � 0n � n0 a 1

2b 0

� n0 � 1 � n0

t � 0,t1�2 � 29 years,t � 0.

n0

n � n0 a 1

2b t�t1�2

t1�2,

1, 12, 14,

18

✔


half-life t1/2

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We can conveniently express Eq. (40.23) in terms of the exponential function.

Since (see Math Help: The Exponential Function, page 910 in

Chapter 28), we can rewrite Eq. (40.23) as

If we adopt the notation

(40.24)

this becomes

(40.25)

Equation (40.25) is called the Law of Radioactive Decay, and is called the mean

lifetime, or the time constant. The inverse of is often called the decay constant

It can be shown that is indeed the average lifetime of all the radioactive

nuclei in the initial amount of parent material. Note that the mean lifetime is longer

than the half-life; for instance, for strontium, the half-life is 29 years, but the mean

lifetime is After one time constant, when

Eq. (40.25) tells us that the number of strontium nuclei has decreased to a factor of

times the original number (see Fig. 40.14).e�1 � 0.368

t � t,t � (29 years)�0.693 � 42 years.

tl � 1�t.t

t

n � n0e�t�t

t �t1�2

ln 2�

t1�2

0.693

n � n0(e�ln 2)t�t1�2 � n0e�t (ln 2)�t1�2

12 � eln 1�2 � e�ln 2


Law of Radioactive Decay

mean lifetime �

decay constant � � 1/�

1 g

1/2

0.368

1/4

1/8

1/161/32

29 �0 58 87 116 145 174 yrtime

amount ofradioactivestrontium

Amount of radioactive strontiumdecreases exponentially, to half ofinitial value after one half-life, …

…and to e–1 ≈ 0.368of initial value after one mean lifetime �.

FIGURE 40.14 Amount of remaining radioactive strontium

vs. time. In this plot, the amount of strontium is measured in

grams. The initial amount is 1 gram, which corresponds to

After 29 years the remaining amount is

gram, which corresponds to 3.35 � 1021 nuclei.

126.70 � 1021 nuclei.

MARIE SKLODOWSKA CURIE (1867–1934) and PIERRE CURIE (1859–1906)French physicists and chemists. For their dis-

covery of the radioactive elements radium and

polonium, they shared the 1903 Nobel Prize

with Henri Becquerel. On the death of Pierre,

Marie succeeded him in his professorship at the

Sorbonne. She received a second Nobel Prize,

in chemistry, for further work on radium. Her

daughter Irène Joliot-Curie, shared the 1935

Nobel Prize with Frédéric Joliot-Curie for

their work on production of radioactive sub-

stances by bombardment with alpha particles.


With appropriate values of the half-life—and thus, by Eq. (40.24), of the decay

time constant—the Law of Radioactive Decay (40.25) is, of course, applicable to the

decay of any radioactive isotope.The lifetimes of different radioisotopes vary over a wide

range. Some have half-lives of several billions of years; others have half-lives as short

as a fraction of a second.Table 40.3 lists the half-lives of some radioisotopes.The chart

of isotopes in Table 40.2 lists some more half-lives.

Since each decay produces one alpha or beta ray, the rate of emission of rays by

the parent material is equal to the decay rate, or the number of parent nuclei that decay

per second. To find the instantaneous decay rate we differentiate n with respect to

time, which gives us the instantaneous rate of change of the number of nuclei:

(40.26)

The negative sign of this rate of change indicates that the number of nuclei is decreasing.

The number of nuclei that decay per second is the negative of the rate of change dn�dt :

(40.27)

Thus, the decay rate is directly proportional to the (instantaneous) number of parent nuclei

and inversely proportional to the mean lifetime.This means that the decay rate is large

if we have a large sample of radioactive material, and it is large if the mean lifetime is

short. Equation (40.27) expresses a characteristic property of many random decay

processes: the negative of the rate of change of a quantity is proportional to the quan-

tity. It is straightforward to show by direct integration of (40.27) that for such a pro-

portionality, the exponential decay of Eq. (40.25) follows.

By substituting the definition (40.24) into Eq. (40.27), we can express the decay

rate in terms of the half-life:

(40.28)�

dn

dt� 0.693

n

t1�2

�

dn

dt�

n

t

dn

dt�

d

dt n0e�t�t � �

n0

t e�t�t � �

n

t


decay rate in terms of half-life

SOME RADIOISOTOPES

RADIOISOTOPE RADIOACTIVITY HALF-LIFE

5730 years

2.6 years

5.27 years

28.8 years

8.04 days

1600 years

4.5 � 109 years�, g238U

�, g226Ra

�, g131I

�, g90Sr

�, g60Co

�, g22Na

�14C

TABLE 40.3

decay rate in terms of mean lifetime

decay rate


Since tables of radioactive isotopes list half-lives, the form (40.28) is convenient for cal-

culations.

The radioactive decay rate of a sample, or the number of disintegrations per second,

is usually called the activity of the sample. The SI unit for the radioactive decay rate,

or the activity, is the becquerel (Bq).

(40.29)

In practice, a larger unit is traditionally preferred; this larger unit is the curie

(40.30)

What is the decay rate of 1.00 gram of radioactive strontium,

The atomic mass of this strontium isotope is 89.9 g.

SOLUTION: Since the atomic mass of this strontium isotope is 89.9 g, the number

of moles in 1.00 gram of strontium is mole, or (1�89.9) � 6.02 � 10 23

atoms � 6.70 � 1021 atoms. With t1�2 � 28.8 years from Table 40.3, Eq. (40.28)

then tells us that the decay rate is

Thus, one gram of strontium-90 emits about 5 trillion beta rays per second!

The effects of penetrating radiation on a material depend on the amount of energy

absorbed over time. The amount of absorbed radiation energy per kilogram of mate-

rial is called the absorbed dose. The SI unit of absorbed dose is the gray (Gy), where

A unit in common use is the rad,

The damage that radiation inflicts on biological tissue depends on the kind of radi-

ation; for example, an alpha particle damages many molecules in each cell in its path,

whereas gamma rays damage fewer molecules per cell.To obtain a measure of the bio-

logical damage, the absorbed dose is multiplied by a correction factor, the relative bio-

logical effectiveness (RBE). The RBE for alpha particles is in the range 10–20, for

beta particles it is 1–2, and for gamma rays and X rays it is 1. The measure of the bio-

logically equivalent absorbed dose is the rem5, where

[equivalent dose in rem] � [dose in rad] � RBE

1 rad � 0.01 J/kg � 0.01 Gy

1 Gy � 1 J/kg.

� 5.10 � 1012 s�1 � 5.10 � 1012 Bq

� 0.693 �6.70 � 1021

28.8 � 3.16 � 107 s

�

dn

dt� 0.693

n

t1�2� 0.693 �

6.70 � 1021

28.8 years

1�89.9

90Sr?EXAMPLE 6

1 curie � 1 Ci � 3.7 � 1010 disintegrations/s

1 becquerel � 1 Bq � 1 disintegration/s


activity

becquerel (Bq)

curie (Ci)

sievert (Sv)

absorbed dose

gray (Gy) and rad

relative biological effectiveness (RBE)

equivalent dose in rem

The SI unit of biological equivalent dose is the sievert (Sv), where 1 Sv � 100 rem5.

5An older measure of ionizing radiation is the roentgen, defined as the amount of radiation that would ionize

a charge of of air at standard temperature and pressure.The rem is an acronym for

roentgen equivalent in man.

3.34 � 10�10 C in 1 cm3


Checkup 40.4QUESTION 1: Table 40.3 lists some half-lives of radioactive isotopes. Suppose

you have of each of these isotopes. Which has the highest decay rate? Which

the lowest?

QUESTION 2: If you initially have of how long does it take for this to decay

to

QUESTION 3: According to the Law of Radioactive Decay, at time what frac-

tion of the initial amount of parent material remains? For radioactive strontium,

according to Fig. 40.14, how many grams of strontium remain at this

time?

QUEST ION 4 : The decay rate of a sample of (see Table 40.3) is currently

disintegrations per second. What will be the number of disintegrations

per second after 3200 years?

(A) (B) (C)

(D) (E)

QUESTION 5: Suppose you have each of 14C, 226Ra, and 238U. Which has the

larger decay rate, that is, which has the larger number of disintegrations per second?

(See Table 40.3).

(A) 14C (B) 226Ra (C) 238U

1.0 mg

2.5 � 1055.0 � 108

1.0 � 1092.0 � 1094.0 � 109

8.0 � 109

226Ra

t � 42 years;

t � t,

12 mg? To 14 mg? To 18 mg?

131I,1 mg

1.00 mg

✔


PHYSICS IN PRACTICE RADIOACTIVE DATING

Atmospheric carbon dioxide contains the stable isotopes 12C

and 13C, and also the radioactive isotope 14C, with a half-life

of 5730 years (the 14C isotope is produced by cosmic rays strik-

ing the upper atmosphere). While a plant or animal lives, it

absorbs carbon dioxide from the air and therefore maintains

the same relative abundance of the isotopes 12C and 14C as in

air. When the organism dies, it ceases to absorb carbon, and

then the 14C in its body decreases due to beta decay [Eq.

(40.25)] according to the Law of Radioactive Decay. Thus,

fresh wood or bones have the same abundance of 14C relative

to 12C as air, but ancient wood or bones have less 14C.

Archeologists take advantage of this decrease of 14C for

the radioactive dating of materials found in ancient tombs or

other sites. They determine the abundance of carbon (all

isotopes) in a sample of material by chemical analysis, and

they determine the abundance of 14C by measuring the beta

activity of the sample. From this, the age can be calculated. For

instance, samples of bone taken from the cadaver of the

“iceman” (see Fig. 1) discovered in 1991 in the Alps near the

Italian-Austrian border have about half the abundance of 14C

relative to 12C as air. We can then conclude that about one

half of the “normal” amount of 14C has decayed, and that the

age of the sample must be about one half-life, that is, about

5700 years (accurate analysis of the 14C abundance gives 5300

years). This means that the iceman died in the late Stone

Age. Radioactive dating by carbon can be employed to estab-

lish reliable ages for samples as old as about 60 000 years. In

older samples the residue of 14C is too small for accurate

measurements of beta activity.

FIGURE 1 This “iceman” was preserved for thousands of years by

the ice in which he was buried. An exceptionally warm summer

melted the ice and revealed his cadaver.


40.5 F ISS ION

In heavy, large nuclei—such as uranium—the balance between the attractive strong

force and the repulsive electric force is quite precarious. As already mentioned in our

discussion of alpha decay, in such a nucleus the electric force becomes dominant if the

nucleus suffers some extra elongation, either spontaneous or caused by an external dis-

turbance, such as the impact of a neutron on the nucleus. The repulsive electric force

then overpowers the attractive strong force, and the nucleus splits into two fragments.

Such a splitting of a nucleus is called fission. Figure 40.15 illustrates the fission of a

nucleus after it suffers a disturbance and elongation when hit by a neutron. Fission

was first discovered by Otto Hahn, Lise Meitner, and Fritz Strassmann, who bom-

barded a sample of uranium with a beam of neutrons and found that the uranium fis-

sions into barium and krypton, according to the reaction

More generally, it was found that uranium can fission in several different ways, with

the formation of a variety of fission fragments, each with a mass of about one-half of

the mass of the original uranium nucleus. The fission fragments are usually radioac-

tive, and they often emit one or several neutrons. Thus, the net reaction is actually

(40.31)

While the fission fragments fly apart, the repulsive electric force does positive work on

the fragments, giving them a large kinetic energy. This kinetic energy represents the

energy released in the reaction.We can calculate the amount of energy released in the reac-

tion in the usual way, by comparing the mass of the uranium nucleus with the (smaller)

sum of the masses of the fission fragments. But even without such a calculation we can

see from the curve of binding energy (Fig. 40.6) that the fission of a large nucleus leads

to a release of energy. By inspecting this figure we recognize that a large nucleus has a some-

what lower amount of binding energy per nucleon than a medium-sized nucleus (the

curve has a maximum at Hence the fission of a large

nucleus into two medium-sized nuclei results in an increase of the net binding energy,

that is, it results in a release of energy.

A � 56, corresponding to 56Fe).

neutron � 238U S fission fragments � 2 or 3 neutrons

neutron � 238U S 145Ba � 94Kr

40.5 Fission 1377

fission

238U


94Kr


145Ba


n

Impact of a neutron ona uranium nucleus…

…can result in fissionfragments, such asthese nuclei of bariumand krypton.

FIGURE 40.15 Fission of uranium triggered by the impact of a neutron.

LISE MEITNER (1878–1968) Austrian

physicist. She studied under Boltzmann, and

became the first woman to earn the Ph.D. in

physics at the University of Vienna. In Berlin

she and Otto Hahn collaborated for several

decades, and she became the first female full

profesor in Germany. She fled Nazi Germany

to Sweden in 1938, just before Hahn discov-

ered the splitting of uranium from chemical

evidence. With her nephew, Otto Frisch, she

coined the term nuclear fission and described

its physical mechanism, but only Hahn was

awarded the Nobel Prize in Chemistry for

nuclear fission in 1944. Hahn, Meitner, and

Hahn’s assistant Fritz Strassman shared the

Enrico Fermi Award for their discovery in

1966. Meitnerium (Mt), atomic number 109,

was named in her honor.

Conceptsin

Context


On the average, the energy released in the fission of a uranium nucleus is about

200 MeV. We can gain a better appreciation of the magnitude of this energy by look-

ing at the following numbers.The total energy released by the complete fission of 1 kg

of uranium (a lump slightly larger than a golf ball) is 200 MeV times the number of

atoms in 1 kg of uranium. Since a mole of uranium is 238 g, 1.00 kg of uranium

is and

the energy released by the fission of the nuclei of these atoms is

(40.32)

This is equivalent to the energy released in the burning of 2 million liters of gasoline. It

is also equivalent to the energy released in the explosion of about 20 000 tons of TNT.

To exploit the energy released by the fission of uranium we rely on the neutrons that

emerge in the reaction (40.31). If one initial fission occurs in a lump of uranium, the

2 or 3 neutrons that emerge from this reaction can strike other uranium nuclei and

trigger their fission, and the neutrons that emerge from these fissions can trigger fur-

ther fissions, and so on. Thus, one initial neutron can initiate an avalanche of neutrons

and an avalanche of fission reactions. Such an avalanche is called a chain reaction (see

Fig. 40.16). If no neutrons, or very few neutrons, are lost from the chain reaction, the

rate of fission and the rate of release of energy grow drastically with time. For instance,

if on the average two of the neutrons released in each fission succeed in generating

further fission reactions and further neutrons, then the numbers of fission reactions

in successive steps of the chain will be 2, 4, 8, 16, . . .. If this multiplicative growth

continues unchecked, the rate of release of energy will become explosive.

� 8.1 � 1013 J

200 MeV

nucleus� 2.53 � 1024 nuclei � 5.1 � 1026 MeV

1000�238 � 4.20 moles, or 6.02 � 1023 � 4.20 atoms � 2.53 � 1024 atoms,


140Xe

235U 235U

235U

235U

95Y

145Ba

141Ba

88Kr

neutron

92Kr

138I

92Sr

Three neutrons emergefrom one initial fission…

…and can strike otheruranium nuclei, trigg-ering their fission…

…and then more neutrons can trigger more fissions, resulting in a chain reaction.

FIGURE 40.16 A chain reaction. A 235U nucleus

absorbs a neutron and fissions, emitting several neutrons,

which are absorbed by other 235U nuclei.

chain reaction

Conceptsin

Context


In the case of conditions are not favorable for sustaining a chain reaction.

Uranium-238 is a fairly stable nucleus, which does not readily fission when a neutron

strikes it. Instead, the nucleus often merely absorbs the incident neutron, removing it

from the chain reaction.This loss of neutrons blocks the chain reaction. However, nat-

urally occurring uranium also contains a small amount (0.72%) of the isotope

which is unstable and much more susceptible to fission.This isotope is widely used as the

nuclear fuel in applications of chain reactions in nuclear bombs and nuclear reactors.

Another isotope used in such applications is an isotope of plutonium. But this

isotope does not occur naturally; it must be manufactured by nuclear transmutation reac-

tions involving uranium, in a nuclear reactor.

Checkup 40.5

QUESTION 1: Does the energy released in fission of a large, heavy nucleus originate from

the initial “strong” potential energy or the electric potential energy?

QUESTION 2: Why are small, light nuclei incapable of fission?

QUESTION 3: Is a chain reaction possible if, on the average, each fission of a nucleus

releases less than 1 neutron? Exactly 1 neutron? More than 1 neutron?

QUESTION 4: When a neutron strikes a nucleus resulting in fission, which of the

following is true?

(A) The rest mass of the products is greater than the sum of the masses of

and the neutron.

(B) The rest mass of the products is less than the sum of the masses of and the

neutron.

(C) The products always include Ba (barium) and Kr (krypton).

(D) The total number of nucleons in the products is less than 236.

(E) The total number of nucleons in the products is greater than 236.

40.6 NUCLEAR BOMBS ANDNUCLEAR REACTORS

In a given mass of or , neutrons produced by spontaneous fission or stray

neutrons coming from elsewhere can initiate the first step in the chain reaction. Whether

the reaction keeps going depends on how many neutrons are lost from the chain, by

absorption without fission (as in or by escape beyond the boundary of the mass.

If the mass is large, few neutrons will reach its boundary before they are intercepted by a

nucleus; thus a large mass inhibits escape of the neutrons and favors the chain reaction.

The mass is said to be critical if the number of neutrons lost from the chain reaction

(by escape or by absorption) equals the number of neutrons released by the fissions. In

this case the chain reaction merely proceeds at a constant rate—as in a nuclear reactor.

The mass is said to be supercritical if the number of neutrons lost from the chain is

smaller than the number of neutrons released in fission reactions. In this case the chain

reaction proceeds at an ever-increasing, runaway rate leading to an explosion—as in a

nuclear bomb. For pure arranged in a spherical configuration, the critical mass is

about 50 kg.

The simplest fission bomb, commonly known as an atomic bomb, or A-bomb,

consists of two pieces of whose separate masses are each less than the critical235U

235U

238U)

239Pu235U

235U

235U

235U

✔

239Pu,

235U,

238U,

40.6 Nuclear Bombs and Nuclear Reactors 1379

critical and supercritical mass

atomic bomb


FIGURE 40.18 Explosion of a fission bomb.

mass but whose combined mass is more than the critical mass. To detonate such a

bomb, the two pieces of uranium, initially at a safe distance from each other, must be

suddenly brought close together. In the first such bomb (see Fig. 40.17), the device

used for the assembly of the two pieces of uranium consisted of a gun which propelled

one piece of uranium toward the other at high speed. More sophisticated fission bombs

consist of a (barely) subcritical mass of 239Pu; if this mass is suddenly compressed to

higher than normal density, it will become supercritical. The sudden compression is

achieved by the preliminary explosion of a chemical high explosive arranged in a layer

around the mass of plutonium.

The energy released in the explosion of an A-bomb is typically equivalent to that

released in the explosion of about 20 kilotons of TNT (see Fig. 40.18).

A much higher explosive yield is achieved by a hydrogen bomb, or H-bomb, in

which an A-bomb is used to trigger fusion reactions similar to the fusion reactions that

power the Sun (see Section 40.7). The energy released in the explosion of an H-bomb

is typically equivalent to that in one or several megatons of TNT. Such an explosion would

level an entire city, with complete devastation and complete destruction of life by incin-

eration and crushing out to a radius of about 16 km from the center of the explosion.

For the peaceful exploitation of nuclear fission in a nuclear reactor, we must keep

the chain reaction under control, so it releases energy at a steady rate. This means that

the configuration of the uranium or other nuclear fuel must be critical rather than super-


propellantexplosive

uranium

neutronreflector

gun tubeneutron reflector

To detonate the first nuclear bomb, a gun propelled onepiece of uranium…

…toward another piece, exceeding critical mass andcausing a chain reaction.FIGURE 40.17 A fission bomb.

hydrogen bomb

nuclear reactor

ENRICO FERMI (1901–1954) Italian

and later American physicist. He worked on

experimental and theoretical investigations of

beta decay, the artificial production of isotopes

by neutron bombardment, for which he

received the 1938 Nobel Prize, and the fission

of uranium. Fermi was one of the leaders of

the Manhattan Project, and he provided the

first experimental demonstration of a chain

reaction.


critical. The most common type of reactor operates with “enriched” uranium consist-

ing of a few percent mixed with more than 95 percent of . Such a uranium mix-

ture cannot by itself maintain a chain reaction—the 238U soaks up too many of the

neutrons. However, if the uranium is enmeshed in a substance capable of slowing down

the neutrons released in fissions, a chain reaction becomes viable. The substance that

slows the neutrons is called the moderator. The role of the moderator in a fission reac-

tion is analogous to that of a catalyst in a chemical reaction. The moderator enhances

the chain reaction because slow neutrons are more efficient at producing fissions in

than fast neutrons, and they are also less likely to be absorbed by

Inside the reactor, the uranium is usually placed in long fuel rods, and these are

immersed in the bulk of the moderator (see Fig. 40.19). Fast neutrons released by fis-

sions travel from the fuel rods into the moderator; there they lose their kinetic energy

by collisions with the moderator nuclei; and then they wander back into one or another

of the fuel rods and trigger further fissions. The three best moderators are ordinary

water heavy water and graphite (pure carbon).

The configuration of the reactor—the size, number, and location of the fuel rods and

the shape of the moderator—must be designed so the reactor is nearly critical. Fine adjust-

ments in the number of neutrons and the reaction rate are made by means of control rods

of boron or cadmium. These substances greedily soak up neutrons, and by pushing the

control rods in or pulling them out, the reaction rate can be decreased or increased.

The main application of reactors is for the generation of electric power. In the

United States, most of the reactors used for this purpose have water-filled cores. The

water acts simultaneously as moderator and as coolant. The water circulates through

the core and removes the heat energy released by the fission reactions (see Fig. 40.20).

(D2O),(H2O),

238U.235U

238U235U


control rod

fuelrod

moderator(water) cool water

intake

reactor vessel

hotwateroutlet

To suppress the reaction, control rodsare inserted to absorb neutronsfrom uranium fission in fuel rods.

To enhance reaction, collisions in moderator slow neutrons,increasing fission in fuel rods.

FIGURE 40.19 Nuclear reactor.

steam generator

turbine

reactor

pump

pump

condenser

coolingwater

Water moderator is also coolant, transferringheat energy to steam, …

…and steam drives turbines thatgenerate electricity.

FIGURE 40.20 Schematic diagram of nuclear power plant.

moderator

fuel rods

control rods


The heat is transferred from the water to steam, which drives a steam turbine con-

nected to an electric generator. Thus, the nuclear reactor plays the role of the furnace

of a conventional steam engine and uranium replaces coal or oil as fuel.

A nuclear reactor in a power plant produces 1200 MW of heat.

How many kilograms of does this reactor consume per year?

SOLUTION: The energy that must be supplied by fission reactions in one year is

the product of the power and the time,

From Eq. (40.32) we know that the energy released in the fission of 1.0 kg of ura-

nium is Hence, the number of kilograms required per year is

Nuclear power plants could meet all our energy requirements for the next several

hundred years, or even the next several thousand years, if we exploit low-grade ores

containing nuclear fuels. Unfortunately, nuclear fission yields rather dirty energy—

the fission reactions generate dangerous radioactive residues. Nuclear power plants

must be carefully designed to hold these residues in confinement. The reactor core is

enclosed in a massive reactor vessel, and as an extra precaution this vessel and the

attached pumps and pipes are enclosed in a strong containment shell (see Fig. 40.21).

The elaborate safety features that are incorporated in the design of a nuclear power

plant make the construction and the maintenance extremely expensive. Furthermore,

when the load of fuel of the reactor has been spent, the residual radioactive wastes

must be removed to a safe place to be held in storage for thousands of years until their

radioactivity has died away.

If some of the radioactive material contained in a nuclear reactor is released in the

form of smoke or dust in an accidental explosion or fire, it can be carried away by

winds, and it can then descend to the ground as lethal radioactive fallout. After the

3.8 � 1016 J �1.0 kg

8.1 � 1013 J� 470 kg

8.1 � 1013 J.

� 1200 � 106 W � 3.16 � 107 s � 3.8 � 1016 J

E � P ¢t � 1200 MW � 1 yr

235UEXAMPLE 7


FIGURE 40.21 The containment

shells of the two nuclear reactors at

the San Onofre nuclear generating

station in California.

Conceptsin

Context


disastrous nuclear accident that happened at Chernobyl, near Kiev in the former

U.S.S.R., in 1986, a plume of fallout spread north and west into Europe and produced

hazardous contamination thousands of kilometers away. An area of 30-km radius

around the reactor was so heavily contaminated it had to be permanently evacuated.

The immediate cause of the Chernobyl disaster was operator error—through a series

of almost incredible blunders, the operators lost control of the reactor, and its power

level surged to 100 times normal values, which overheated the core, blew it apart, and

started a fire in the graphite moderator.

The water-moderated reactors used in the United States are thought to be much safer

than the Chernobyl reactor. But some experts question whether they are safe enough.

Scenarios for conceivable reactor accidents have been analyzed in great detail.The worst

that might happen is a loss-of-coolant accident, that is, the loss of the water that nor-

mally circulates through the core of the reactors serving both as moderator and as

coolant. Although such a loss of water would shut off the fission chain, the reactor con-

tains a large amount of radioactive residues, and the heat from the radioactivity can by

itself overheat the reactor vessel beyond safe limits. As a precaution, reactors are equipped

with emergency cooling systems, and they are also encased in a containment shell.

Nuclear power reactors in the United States all use 235U as fuel. Unfortunately, the

supply of this nuclear fuel is rather limited, and it is expected to become exhausted in

this century. There are, however, several other nuclear fuels in larger supply. One of

these is 238U. Although this isotope is incapable of supporting chain reactions, it can

be converted into 239Pu, which supports chain reactions. The manufacture of 239Pu is

an automatic side effect of the operation of nuclear reactors now in operation; in all of

these reactors, the fuel rods contain a mixture of 235U and 238U, and fission neutrons

striking the 238U gradually convert it into 239Pu. A reactor fueled with 239Pu not only

makes good use of a material that would otherwise go to waste, but, if the reactor is sur-

rounded by a blanket of 238U, it can also manufacture extra 239Pu.The number of neu-

trons released in the fission of 239Pu is so large that in an efficiently designed reactor

neutrons can be diverted to the 238U without hindering the chain reaction. Such a

reactor can produce more 239Pu (from 238U) than it consumes (from its original supply).

A reactor of this kind is called a breeder reactor. Because of worries over the safety of

these reactors, they have not been adopted in the United States, but dozens are oper-

ating in Europe.

Checkup 40.6

QUESTION 1: If each fission of a uranium nucleus produces three neutrons, how many

of these neutrons can you afford to lose, if you want to maintain a chain reaction?

QUESTION 2: Why is it impossible to maintain a chain reaction in a very small mass

of uranium, say, 100 g?

QUESTION 3: Why is the presence of 238U detrimental to the maintenance of a chain

reaction in a mass of 235U?

QUESTION 4: A breeder reactor produces more nuclear fuel than it consumes. Is this

in conflict with conservation of energy?

QUESTION 5: In a nuclear reactor, the moderator

(A) Coordinates the activities of the technicians operating the reactor.

(B) Controls the steam that drives the turbines of the electrical generator.

(C) Absorbs neutrons before they decay into protons and electrons.

(D) Slows neutrons so that they can be absorbed by 235U in fuel rods later.

(E) Slows the reaction by permitting more neutrons to be absorbed by 238U.

✔


breeder reactor


40.7 FUSION

Fusion is the merging of two small nuclei to form a larger nucleus. The fusion of two

small nuclei, such as the nuclei of hydrogen, deuterium, or tritium, releases energy.

For instance, the heat of the Sun is generated by a fusion reaction called hydrogen

burning, in which hydrogen nuclei fuse together to make helium nuclei. By inspec-

tion of the curve of binding energy, we can immediately recognize that the fusion of

small nuclei leads to a release of energy. Figure 40.6 shows us that the smallest nuclei

have exceptionally low binding energies per nucleon. Hence, the fusion of two such

nuclei into a larger nucleus of higher binding energy results in release of energy. Note

that the process of fusion is the reverse of fission: small nuclei (such as hydrogen) release energy

when they fuse; large nuclei (such as uranium) release energy when they split.

The hydrogen burning in the Sun proceeds in three steps; first hydrogen nuclei fuse

to make deuterium (2H), then deuterium fuses with hydrogen to make helium-3 (3He),

and finally helium-3 fuses with helium-3 to make helium-4 (4He):

1H � 1H S 2H � e� � � (40.33)

(40.34)

(40.35)

Each of the first two reactions must occur twice for the last reaction to occur once.

Each of these reactions releases energy; when the first two reactions occur twice and

the last occurs once (with the formation of one nucleus of 4He), the net energy released

is 24.7 MeV. Four nuclei of 1H are consumed in this process (6 are consumed when

the first two reactions occur twice, but 2 are regenerated when the last reaction occurs);

hence the amount of energy released per nucleon of “fuel” is 24.7 MeV per 4 nucle-

ons, or 6.2 MeV per nucleon.This number is to be compared with the energy released

in the fission of uranium, about 200 MeV per 235 nucleons of “fuel,” or 0.85 MeV

per nucleon. Thus, fusion of a given mass of 1H in the Sun releases about 7 times as

much energy as the fission of an equal mass of 235U.

The fusion reactions (40.33)–(40.35) are called thermonuclear, because they can

proceed only at extremely high temperatures and pressures, such as the temperature of

about in the core of the Sun. The high temperature is needed to over-

come the electric repulsion that the hydrogen nuclei experience whenever they come

close together. At high temperatures, the hydrogen nuclei have high speeds, and their

collisions are sufficiently violent to overcome the electric repulsion and bring the nuclei

into the intimate contact required for fusion.

The thermonuclear reaction based on hydrogen fuel cannot be duplicated on Earth,

because we cannot attain the pressure found in the core of the Sun. However, there

are some fusion reactions based on deuterium and tritium that proceed at

lower pressures, for example,

2H � 3H S 4He � neutron (40.36)

Although the reaction (40.36) can proceed at attainable pressures, it requires a tem-

perature of about even higher than in the center of the Sun. Such temperatures

have been attained in the explosions of H-bombs, where the preliminary explosion of

a fission bomb heats and compresses a mixture of deuterium and tritium and thereby

initiates fusion.

108 K,

(3H)(2H)

1.5 � 107 K

3He � 3He S 4He � 1H � 1H

1H � 2H S 3He � g


thermonuclear reaction

fusion

HANS ALBRECHT BETHE (1906–2005)German and later American Physicist. During

World War II he worked on the Manhattan

Project as director of the Theoretical Physics

Division. He received the Nobel Prize in

1967 for his investigations of nuclear reactions

in stars.


Experimental attempts at a peaceful exploitation of fusion power also rely on a deu-

terium–tritium fuel mixture, which is heated to the required temperature of about

either by an intense electric current, or by a particle beam, or by a laser beam. At such a tem-

perature, the deuterium and tritium will be in the form of a plasma, that is, a totally ion-

ized gas, consisting of a mixture of independent nuclei and electrons.The plasma cannot

be contained by the walls of a conventional reactor vessel, since any contact with the wall

of the vessel would cool the plasma, and would also melt the wall. Instead, the plasma

must be held suspended in the middle of the vessel, away from contact with the walls.

One scheme for controlled fusion attempts to suspend the plasma by means of

magnetic fields, that is, by magnetic confinement. Figure 40.22 shows the Princeton

Tokamak Test Reactor, in which the plasma is confined inside a large toroidal sole-

noid (a solenoid shaped like a donut). The plasma is heated by a combination of elec-

tric currents induced in the plasma and particle beams aimed into it. A larger test

reactor, a 500 MW fusion power generator known as ITER (International

Thermonuclear Experimental Reactor), is under construction at Cadarache, France.

Another scheme for fusion attempts to extract energy by exploding small pellets

of a deuterium–tritium mixture in a combustion chamber by hitting them with intense

laser beams. This scheme is called inertial confinement because the inertia of the

pellet holds it together long enough for the reaction to occur. The beams heat the

pellet so suddenly that it has no time to disperse before fusion begins. The pellet then

explodes like a miniature hydrogen bomb. After the thermal energy has been extracted

from the combustion chamber, the next pellet is placed in the chamber, and so on.

Figure 40.23 shows the combustion chamber of the National Ignition Facility (NIF)

at Lawrence Livermore National Laboratory, where 192 high-power laser beams con-

verge to heat a deuterium–tritium pellet (see also Chapter 26). Although both mag-

netic and inertial confinement schemes have been successful in initiating fusion, the

amounts of energy released in fusion have remained far below the amount of energy

that had to be fed into the reactor chambers to heat the plasma.

Nuclear fusion is an attractive source of energy because it bypasses many of the

safety problems associated with nuclear fission, especially the production of heavy,

long-lived radioactive nuclei. However, tritium is radioactive and thus raises safety

concerns, both when used as a fuel and when produced in the liquid lithium coolant

of proposed fusion reactors by absorption of fast neutrons. Since the half-life of tri-

tium is fairly short (12.3 years) and extraneous tritium could be recycled for fuel,

108 K

40.7 Fusion 1385

FIGURE 40.22 The Princeton Tokamak Test Reactor.

magnetic confinement

inertial confinement

FIGURE 40.23 Combustion chamber at the National

Ignition Facility in Livermore, California.



fusion is considered a potential source of much cleaner energy than fission. Besides,

we have available an enormous supply of deuterium. There is enough heavy water

mixed with the ordinary water in the oceans of the Earth to satisfy our energy needs

for millions of years.

Checkup 40.7

QUESTION 1: Why are large, heavy nuclei incapable of fusion?

QUESTION 2: What is the net result of the sequence of fusion reactions (40.33)–(40.35),

that is, what nuclei are consumed, and what nuclei are produced?

QUESTION 3: Fusion requires high temperature. Is this also true for fission?

QUESTION 4: Do the fusion reactions (40.33)–(40.35) produce any radioactive nuclei?

QUESTION 5: Tritium is produced in the reaction

What is the net result of this reaction followed by reaction (40.36)?

(A)

(B)

(C)

(D)

(E) 2H � 2H � 2H S 3He � 1H � 2 neutron.

2H � 2H � 2H S 4He � 1H � neutron.

2H � 2H � 3H S 4He � 1H � 2 neutron.



2H � 2H S 3H � 1H

✔

SUMMARY

PROBLEM-SOLVING TECHNIQUES Nuclear Reactions (page 1371)

(40.7)1 u � c2 � 931.5 MeV

RADIUS OF NUCLEUS (A is mass number.)

ENERGY EQUIVALENT OF ATOMIC MASS UNIT

PHYSICS IN PRACTICE Radioactive Dating (page 1376)

ATOMIC MASS UNIT u 1 u � 1.660 54 � 10�27 kg

ISOTOPES Atoms with the same number of

protons in their nuclei, but different numbers of

neutrons.

(40.2)R � (1.2 � 10�15 m) � A1�3

6 protons5 neutrons

6 protons6 neutrons

11C 12C

(40.8)[mass of N neutrons and Z protons] � [mass of nucleus] � B.E./c2MASS DEFECT AND BINDING ENERGY (B.E.)

(40.11)Q � (m1c2 � m2c2 � ) � (m1c2 � m2c2 � )Q VALUE OF REACTION Initial masses

m1, and final masses m1, m2, pm2, p


Questions For Discussion 1387

(40.28)�

dn

dt�

n

t� 0.693

n

t1�2

disintegration/s (40.29)1 Bq � 1

disintegrations/s (40.30)1 Ci � 3.7 � 1010

1 gray � 1 Gy � 1 J/kg � 100 rad

10–20 for alpha

1–2 for beta

1 for gamma

1 sievert � 1 Sv � 100 rem

DECAY RATE OR ACTIVITY

BECQUEREL

CURIE

SI UNIT OF ABSORBED DOSE

RELATIVE BIOLOGICAL EFFECTIVENESS (RBE)

SI UNIT OF EQUIVALENT ABSORBED DOSE

FISSION The splitting of a nucleus into two fragments,

usually with the emission of one or more neutrons.

FUSION The merging of two small nuclei to form a

larger nucleus.

or (40.23)

(40.25)

n � n0 a 1

2b t�t1�2

n � n0e�t�tLAW OF RADIOACTIVE DECAY

where is the half-life,

and is the mean lifetime, or time constant, of

the decay.

t

t1�2t �t1�2

ln 2�

t1�2

0.693,

3. According to Fig. 40.6, which isotope has the largest binding

energy?

4. Why do alpha and beta emissions involve transmutation of

elements whereas gamma emission does not?

5. If you irradiate a sample of material with alpha, beta, or

gamma rays, is the sample likely to become radioactive? What

if you irradiate it with neutrons?


1. What is the average overall density of a typical atom, such as

iron? Why is this much smaller than the density of a nucleus,

is given by Eq. (40.4)?

2. Naturally occurring magnesium has an atomic mass of 24.305.

On the basis of this information, can you conclude that natu-

ral magnesium contains a mixture of several isotopes? Can you

guess which isotopes?

�

�

�

KINDS OF RADIOACTIVITY alpha decay: � particle (4He nucleus)

beta decay: � particle (electron or antielectron)

gamma decay: � ray (high-energy photon)

EQUIVALENT ABSORBED DOSE [equivalent dose in rem] � [dose in rad] � RBE


6. Radiation is sometimes used to sterilize food or surgical sup-

plies. What are the advantages or disadvantages of this proce-

dure over sterilization by heating?

7. Can radioactive dating with be used to determine the age

of a book printed during the last century? The age of deposits

of mineral oil? The age of ancient Egyptian gold jewelry?

Ancient Egyptian furniture?

8. Why is an absorbed dose of alpha radiation more likely to kill

a biological cell than the same absorbed dose of gamma radia-

tion?

9. Tritium is a radioactive isotope of hydrogen which occurs

naturally in small concentrations in the ordinary water found

in the environment. The half-life of this isotope is 12 years.

Describe how you could take advantage of this isotope to

determine the age of a bottle full of wine that your wine mer-

chant claims is 25 years old.

10. Alpha particles are more massive than beta particles, yet for

equal kinetic energies, the alpha particles are stopped by a

thinner layer of material than the beta particles. Can you

explain why an alpha particle loses its energy more quickly?

(Hint: Think of the electric forces that act on the particles in a

collision with an atom.)

11. All the best moderators for reactors consist of fairly light-

weight nuclei. Why is a material with heavy nuclei, such as

lead, not a good moderator?

(3H)

14C

12. Why is it difficult to separate the isotopes of and

13. Some artificial satellites carry small nuclear reactors as power

supplies. What danger does this pose for people on Earth?

14. Nuclear fission bombs produce a long-lived radioactive iso-

tope of strontium an element that is chemically similar

to calcium. Explain why strontium poses a severe hazard to

humans and other vertebrates.

15. Hydrogen bombs operate with tritium, a radioactive isotope of

hydrogen with a half-life of 12.3 years. When a hydrogen

bomb is held in storage, its tritium gradually decays, and must

be replenished with fresh tritium every few years. According

to a recent proposal, arms control could be achieved by halting

production of tritium. Every few years, the superpowers would

then have to take some of their bombs out of service, extract

the remaining tritium, and use it to replenish their other

bombs. Suppose the United States and Russia each have 2000

hydrogen bombs now. Without fresh tritium, how many

bombs will each superpower have in 25 years? In 50 years?

16. Why is it more difficult to achieve controlled fusion in a reac-

tor than controlled fission?

17. A fusion reactor would produce a large amount of tritium

a radioactive isotope of hydrogen. If the tritium were

accidentally released into the environment, it would be likely

to contaminate the water. Explain.

(3H),

(90Sr),

235U?238U


PROBLEMS

40.1 I so topes

1. How many protons and neutrons are there in the nucleus of

the isotope

2. What isotope has 82 protons and 122 neutrons in its nucleus?

3. What are the number of protons and the number of neutrons

in each of the following isotopes:

?

4. What isotope has 17 protons and 18 neutrons in its nucleus?

What isotope has 18 protons and 17 neutrons? Such isotopes

in which the numbers of protons and neutrons are exchanged

are called mirror nuclei. Find another example of mirror

nuclei in the chart of isotopes in Table 40.2.

5. Use the chart of isotopes in Table 40.2 to make a list of all the

isotopes of oxygen. What are the number of protons and the

number of neutrons in the nucleus of each of these isotopes?

6. By inspection of Fig. 40.5, estimate the ratio of neutrons to

protons in stable light nuclei and in stable heavy

nuclei (Z � 80).

(Z � 20)

138La124Xe,63Zn,63Cu,

52Mn,52Cr,27Al,24Na,

238U?56Fe?16O?

7. In any nuclear reaction, the nuclear electric charge must be con-

served and the mass number must be conserved. Are the follow-

ing reactions in accord with these conservation laws? Explain.

8. Naturally occurring boron is a mixture of 80.2% 11B and

19.8% 10B. From the masses of these isotopes (listed in the

chart of isotopes in Table 40.2) calculate the atomic mass of

naturally occurring boron.

9. According to Eq. (40.2), what is the nuclear radius of the

smallest of the isotopes of carbon? The largest of these iso-

topes?

10. The largest known nucleus is that of the isotope 266 of ele-

ment 109, which was recently named meitnerium (Mt). What

is the radius of the 266Mt nucleus?

n � 238U S 121Ag � 118Pd

4He � 10B S 13C � n

2H � 12C S 4He � 9B


20. According to Fig. 40.6, what is the binding energy of 56Fe?

From this binding energy, calculate the mass of the isotope (in

atomic mass units).

21. What are the mass defect (in atomic mass units) and the

nuclear binding energy (in MeV) of the isotope 4He? The

mass of 4He is given in the chart of isotopes in Table 40.2.

22. What are the mass defect (in atomic mass units) and the

nuclear binding energy (in MeV) of the isotope 12C?

23. Nuclei with 2, 8, 14, 20, 28, 50, 82, or 126 protons or neutrons

are called magic nuclei because they have exceptionally large

binding energies and they are exceptionally stable (see also

Problem 12). Compare the binding energy of the magic

nucleus 16O with that of the nonmagic nucleus 16F. The

masses of these isotopes are given in Table 40.2.

24. How much energy (in MeV) is required to remove one neu-

tron from the nucleus of the isotope 14N? (Hint: Compare

the binding energies of 14N and 13N; the masses of these iso-

topes are given in the chart of isotopes in Table 40.2.

25. Nuclei with “magic numbers” of protons or neutrons are excep-

tionally stable (see also Problems 12 and 23), analogous to a

closed shell in atomic physics. When a nucleus has only one or a

few nucleons outside a closed shell, such nucleons are often rel-

atively weakly bound. Consider cerium-140, which has a magic

number of 82 neutrons. Compare 140Ce and 142Ce to find the

average binding energy for each of the two extra neutrons in142Ce. Compare your result with the average binding energy per

nucleon in 140Ce. The atomic masses of 140Ce, 142Ce, and the

neutron are 139.9054 u, 141.9092 u, and 1.0087 u.

26. Two nuclei with the same mass number but different atomic

numbers are called isobars. Calculate the binding energy per

nucleon for the stable isobars germanium-74 and selenium-

74. Which nucleus is more tightly bound? The atomic masses

of 74Ge and 74Se are 73.9218 u and 73.9225 u, respectively.

27. Verify that in the reaction (40.10), both sides have the same

number of protons and neutrons.

*28. When boron is bombarded with alpha particles, the following

reaction is observed:

How much energy is released in this reaction?

*29. When is bombarded with protons, the following reaction

occurs:

What is the minimum kinetic energy required for the proton

to initiate this reaction? Neglect the recoil energy of the

nucleus.

*30. Consider the reaction

(a) What is the energy absorbed in this reaction?

(b) We can initiate this reaction either by bombarding a tri-

tium target with protons, or by bombarding a hydrogen

1H � 3H S 3He � n

7Be

1H � 7Li S 7Be � n

7Li

4He � 10B S 13C � 1H

Problems 1389

11. In some reference books, isotopes are listed with their atomic

number as a left subscript; for example, deuterium (2H) would be

listed as 21H, and carbon-14 would be 146 C. Determine which

of the following are incorrectly written: 3115P, 44

22Sc, 11850Sn, 207

83 Pb.

Since the subscript is unique for each element, we could con-

sider using a notation without the atomic symbol of the ele-

ment. What isotope would be meant by 8939? By 197

79 ?

12. Many stable nuclei have an even mass number A, and all but a

few stable nuclei have an even number N of neutrons or an

even atomic number Z. Certain even numbers for N and Z

correspond to very stable nuclei; these “magic nuclei” have

These suggest a

nuclear shell structure analogous to atomic shells, with inde-

pendent shells for protons and neutrons. In the chart of iso-

topes in Table 40.2 for what value of N are there the most

stable isotopes (that is, for what value of N are there the most

naturally occurring isotopes, as indicated by any listed values

for the percent abundance)? For what value of Z? What do the

stable isotopes 36S, 37Cl, 38Ar, 39K, and 40Ca have in common?

The most abundant stable isotopes of barium, lanthanum,

cerium, praseodymium, and neodymium are 138Ba, 139La,140Ce, 141Pr and 142Nd. What do these have in common?

13. What stable isotope has one-half the radius of a 104Ru nucleus?

One-half the radius of a 128Xe nucleus? See the chart of iso-

topes in Table 40.2.

14. Chlorine has two stable isotopes, 35Cl and with atomic

masses 34.968 85 u and 36.965 90 u, respectively. In the peri-

odic table, the average atomic mass of chlorine is 35.453 u.

From these data, calculate the percent abundance of the two

stable chlorine isotopes.

15. Nuclei with the same mass number but different atomic num-

bers are called isobars. Find the number of isobars for each

mass number from 1 to 17. Which nuclei have no isobars? Use

the chart of isotopes in Table 40.2.

*16. Neutron stars are made (almost) entirely of neutrons, and they

have approximately the same density as that of a nucleus.

What is the radius of a neutron star of mass 0.50 times the

mass of the Sun?

*17. What fraction of the volume of your body is filled with nuclear

material? The average density of your body is about

*18. Suppose that an alpha particle of energy 4.4 MeV collides

head-on with a stationary gold nucleus. What is the distance of

closest approach? Does the alpha particle make contact with

the surface of the nucleus? Since the gold nucleus is so heavy,

assume for simplicity that it gains no energy in the collision.

40.2 The S t rong Force and theNuc lear B ind ing Energy

19. The binding energy of the electron in the ground state of the

hydrogen atom is 13.6 eV. Calculate the corresponding mass

defect of the hydrogen atom; express this in atomic mass units.

(Since the mass defects associated with atomic binding ener-

gies are small, they are usually ignored in nuclear physics.)

1000 kg/m3.

37Cl,

(Z or N ) � 2, 8, 14, 20, 28, 50, 82, or 126.


target with tritium ions. To initiate this reaction, the

kinetic energy of the bombarding particles must be some-

what larger than the energy absorbed in the reaction, since

the recoil of the reaction products retains some of the

energy supplied by the bombarding particle. If tritium

ions are used as bombarding particles, the required kinetic

energy is larger than if the protons are used as bombard-

ing particles. Explain the difference.

*31. Consider the spontaneous fission of uranium into barium and

krypton, according to the reaction

The masses of the isotopes are 238.050

79 u, 144.926 87 u, and 92.931 12 u, respectively.

(a) Find the Q value for this reaction.

(b) The uranium nucleus is initially at rest. Find the final

speeds of the barium and krypton nuclei after the reac-

tion, and find their individual energies.

*32. For large nuclei, the total electrostatic potential energy can be

approximated by that of a sphere of uniform positive charge

density This energy can be calculated by building up spheri-

cal shells of charge.

(a) Show that the potential at the surface of a small sphere of

charge q and radius r is

(b) Adding a shell of charge

requires an energy

Integrate such contributions from r � 0 to r � R to

obtain the total electrostatic energy of a sphere of radius R

and total charge Q.

(c) Use Eq. (40.2) and the result from (b) to estimate the

average electrostatic energy per proton for .

†40.3 Radioac t iv i ty

33. What isotope is formed by the alpha decay of

34. What isotope is formed by the negative beta decay of 85Kr? 63Ni?

35. What isotope is formed by the positive beta decay, or beta plus

decay, of

36. The alpha decay of uranium [see Eq. (40.13)] is the first step in

a radioactive series of decays. The next four steps are a negative

beta decay, another negative beta decay, another alpha decay,

and another alpha decay. What are the daughter nuclei pro-

duced in these four steps?

37. For some radioactive nuclei, the inverse of beta decay is favor-

able, a process known as electron capture; typically, the nucleus

22Na? 64Cu?

216Po? 239Pu?

(Z � 92)238U

dU � V dq

dq � r d (volume) � r 4pr 2dr

V �1

4p�0

q

r�rr 2

3�0

r.

145Ba, and 93Kr238U,

238U S 145Ba � 93Kr

absorbs one of its own atomic electrons. A neutrino is created in

this process. What is the energy of the emitted neutrino when

radioactive transmutes via electron capture into the stable

nucleus 37Cl? The atomic masses of 37Ar and 37Cl are 36.966

78 u and 36.965 90 u, respectively.

38. When beta-decays into what is the maximum pos-

sible kinetic energy of the emitted electron? The maximum

possible energy of the emitted neutrino? What are the mini-

mum energies? The atomic masses of and are

128.904 98 u and 128.904 78 u, respectively.

39. The nucleus is often used as a a source of gamma rays for

experiments in magnetism, since these gamma rays are readily

absorbed by iron nuclei. Suppose an atom of at rest

emits a 0.707-MeV gamma ray. What is the recoil velocity of

the cobalt atom?

*40. The alpha decay of results in Calculate the energy

of the emitted alpha particle. The masses of these two isotopes

are 209.9829 u and 205.9745 u, respectively.

*41. What is the maximum kinetic energy of the beta ray emitted

in the beta decay of a neutron, according to the reaction

(40.18)? Ignore the recoil of the proton.

*42. If neutrons had a somewhat smaller mass, then the (slow)

electrons in atoms could combine with protons in the nucleus

according to the reaction

e� � p S n � �

How much smaller would the mass of the neutron have to be to

make this reaction viable? What consequences would this reac-

tion have for the existence of atoms and the existence of life?

*43. What isotope is formed in the negative beta decay of

Calculate the maximum energy of the beta rays emitted in this

decay. The masses of the relevant isotopes are given in the

chart of isotopes in Table 40.2.

†40.4 The Law of Radioac t ive Decay

44. According to Fig. 40.14, at what time will the remaining

amount of radioactive strontium have fallen to of the ini-

tial amount?

45. According to the best available data, the half-life of 14C is

believed to be 5730 years. However, according to previous

data, the half-life was thought to be 5570 years, and age deter-

minations based on this value of the half-life were in error.

What percentage error in age determination does the error in

the half-life introduce? For a sample from the year 3000 B.C.,

what is the error in years?

46. The isotope found on the Earth was originally synthe-

sized in nuclear reactions in supernovas that exploded in our

Galaxy about years ago and scattered debris

through the Galaxy, some of which became trapped in the

Earth during its formation, about years ago. What

fraction of the original amount of uranium synthesized in the

4.6 � 109

6.8 � 109

238U

1�10

14C?

206Pb.210Po

57Co

57Co

129Xe129I

129Xe,129I

37Ar




Galaxy is still in existence? What fraction of the amount of

uranium trapped in the Earth during its formation is still in

existence? The half-life of

47. The activity of an industrial irradiation cell is

when the cell is new. What will be the activity

after the cell has been in use for 2.0 years? The half-life of

is 5.3 years.

48. What is the activity of 1.00 gram of pure radium, The

half-life of this isotope is listed in Table 40.3.

49. What is the activity of 1.0 microgram of The half-life of

this isotope is listed in Table 40.3.

50. Some modern theories of elementary particles propose that

the proton might be unstable with a very long half-life.

Suppose the half-life for proton decay is

What would be the decay rate for the protons in an experi-

ment using of water? What is the inverse of

this decay rate, that is, how long on average would you have to

wait for one disintegration?

51. The activity of a sample of is initially 12.2 mCi (milli-

curies). After 2.00 years, the activity has dropped to 1.87 mCi.

From these data, what is the half-life of If the initial

sample was pure (atomic mass 56.9), what was the initial

mass of this sample?

52. 1.00 mole of naturally occurring rubidium emits beta rays at a

decay rate of due to the radioactive isotope

which has a natural isotopic abundance of 27.8%. From

these data, calculate the half-life of

53. A 75-kg worker receives an accidental whole-body dose of 15

rads of beta radiation with an RBE of 1.7. What energy is

absorbed? What is the absorbed dose in grays? What is the

equivalent absorbed dose in rems? In sieverts?

54. What absorbed dose in rads of alpha radiation with an RBE

of 20 causes the same amount of biological damage as 10 rads

of gamma radiation?

55. During a radiation treatment of a 75-g tumor, a patient

absorbs 5.0 J of gamma radiation. The tumor absorbs 30% of

the radiation; by rotating the gamma-ray beam, the remaining

energy is distributed throughout 1.5 kg of surrounding tissue.

What is the radiation dose absorbed by the tumor? By the sur-

rounding tissue?

40.5 F i ss ion40.6 Nuc lear Bombs andNuc lear Reac tors

56. According to the curve of binding energy in Fig. 40.6, what is

the binding energy of a nucleus with mass number

What is the binding energy of a nucleus with mass number

What is the energy released in the fission of the

nucleus into two nuclei?

57. The bomb exploded at Hiroshima had an explosive yield of

about 20 kilotons of TNT, or How many kilo-8.4 � 1013 J.

A � 100A � 200

A � 100?

A � 200?

87Rb.

87Rb,

7.75 � 104 Bq

57Co

57Co?

57Co

5.0 � 104 liters

1.0 � 1033 years.

22Na?

226Ra?

60Co

1.0 � 1016 Bq

60Co

238U is 4.5 � 109 years.

grams of uranium actually underwent fission to release this

amount of energy?

58. Use the curve of binding energy to find the nucleus of smallest

mass number A for which fission into two nuclei of mass

number A�2 is energetically viable. (Hint: For fission to be

viable, the height of the curve at A�2 must be above the height

at A. Why?)

59. Naturally occurring uranium contains 99.28% of the isotope238U and 0.72% of the isotope How many kilograms of

natural uranium must we process to extract the 50 kg of

required for a bomb? Pretend that all of the can be

extracted.

60. Some artificial space probes that travel far from the Sun are

powered by nuclear energy, typically the alpha decay of pluto-

nium-238. Suppose such a satellite uses 2.0 kg of At

what rate is energy generated from alpha decay? The atomic

masses of and are 238.0496 u and 234.0409 u,

respectively, and the half-life of is 88 years.

61. How much energy is released in the following fission reaction?

The atomic masses of rubidium-87 and cesium-137 are

86.9092 u and 136.9071 u, respectively.

*62. Consider the fission reaction

Given that the masses of the isotopes are

235.043 94 u, 143.922 85 u, and 91.926 27 u, respectively, cal-

culate the energy released in this reaction.

40.7 Fus ion

63. Some designs for fusion reactors include a liquid lithium blan-

ket for the absorption of fast neutrons and energy. The lithium

also generates more tritium for fuel according to the reaction

How much energy is generated in this reaction? Use the chart

of isotopes in Table 40.2.

64. To achieve fusion, a sufficient number density n of nuclei must

be confined for a long enough time to provide a sustained

reaction. For deuterium–tritium fusion the condition is

known as Lawson’s criterion.

(a) In magnetic confinement test reactors, confinement times

of about 0.50 s have been achieved. What is the corre-

sponding required density of nuclei? How does this com-

pare with the density of an ideal gas at standard

temperature and pressure?

(b) For inertial confinement, suppose the fuel is in a liquid pellet

with the same number density as that of protons in water.

The laser pulse lasts only take this to be the

confinement time. By what factor must the pulse compress

the volume of the liquid to achieve Lawson’s criterion?

1.0 � 10�10 s;

nt � 1.0 � 1020 s/m3,

t

n � 6Li S 3H � 4He

235U, 144Ba, and 92Kr

n � 235U S 144Ba � 92Kr

n � 235U S 87Rb � 137Cs � 12 n

238Pu

234U238Pu

238Pu.

235U

235U

235U.

Problems 1391


*65. For deuterium and tritium to fuse, they must get close enough

to touch.

(a) Use Eq. (40.2) to estimate the radii of deuterium and tri-

tium and thus obtain their center-to-center distance when

barely touching.

(b) Calculate the electrostatic potential energy of the two

nuclei when touching.

(c) If the nuclei approach head-on, each with kinetic energy

equal to about half the amount obtained in (b), they can

fuse. At a temperature T, there will be a broad distribution

of thermal velocities. Suppose that at a temperature T,

enough nuclei with kinetic energy equal to 10 times the

average thermal kinetic energy are present for fusion.

From these considerations, estimate the temperature

needed for fusion.

*66. Consider the three fusion reactions (40.33)–(40.35) that occur

in the core of the Sun. From the masses of the nuclei that par-

ticipate in these reactions (see the chart of isotopes in Table

40.2), calculate the energy released in each reaction. What is

the net energy if the first two reactions occur twice and the

last reaction occurs once?

*67. (a) The rate at which the Sun radiates heat and light is

3.9 � 1026 W. If the energy for all of this radiation comes

from the fusion reactions (40.33)–(40.35), what is the rate

at which the Sun consumes hydrogen?

32 kT

(b) The amount of hydrogen in the Sun is about

How long will this fuel last if the Sun goes on radiating at

a steady rate?

*68. Whenever the first two of the fusion reactions (40.33)–(40.35)

proceed twice and the last proceeds once, 24.7 MeV is released,

and so are two neutrinos. Hence, the number of neutrinos

released is 1 per 12.3 MeV. Assume that the fusion reactions

(40.33)–(40.35) account for all of the of heat

and light radiated by the Sun. Calculate the rate at which the

Sun releases neutrinos.

**69. A sequence of reactions that can be used in controlled fusion

is the burning of deuterium “fuel,” as follows:

The net result of this sequence of reactions is the transmuta-

tion of six nuclei of deuterium (2H) into two nuclei of 4He,

two protons (1H), and two neutrons. Calculate the energy

released in each of these reactions. What is the net energy

released per nucleon of fuel?

2H � 3He S 4He � 1H

2H � 3H S 4He � n

2H � 2H S 3H � 1H

2H � 2H S 3He � n

3.9 � 1026 W

1.5 � 1030 kg.


deuterons (nuclei of deuterium) of 1.2 MeV? Neglect the

recoil of the carbon nucleus.

75. Bombardment of with protons produces an isotope of

nitrogen, according to the reaction

What is the minimum energy that the proton must have to

initiate this reaction? Neglect the recoil of the carbon nucleus.

*76. Nuclear physicists sometimes deduce the mass of an isotope

from the energy of the beta rays emitted by the isotope. For

instance, the maximum energy of the beta rays emitted in the

decay of into 27Al is 2.610 MeV. Given that the mass of

is 26.9815 41 u, what mass can you deduce for

*77. Find the maximum kinetic energy of the beta rays emitted in

the decay

16N S 16O � e� �

78. One of the dangerous radioactive isotopes in the radioactive

waste produced by nuclear reactors is with a half-life of

10.8 yr. How long must we hold the radioactive waste in stor-

85Kr,

�

27Mg?27Al

27Mg

1H � 13C S 13N � n

13C

REVIEW PROBLEMS

70. How many protons and neutrons are there in the nucleus of

the isotope 18Na? 39K? 231Pu?

*71. Suppose you bombard a target of magnesium with alpha parti-

cles. If, in a head-on collision with a stationary magnesium

nucleus, an alpha particle is to reach the nuclear surface just

barely before being halted by the repulsive electric force, what

must be the energy of the alpha particle? Express your answer

in MeV.

72. A nucleus of radius fissions into two equal

spherical pieces. What is the radius of each piece?

73. Chadwick discovered the neutron when he bombarded boron

with alpha particles, which resulted in the reaction

Calculate the energy released in this reaction.

74. Consider the following nuclear reaction:

What is the energy absorbed in this reaction? Can this reac-

tion be initiated by bombarding carbon with a beam of

2H � 12C S 4He � 10B

4He � 11B S 14N � n

7.4 � 10�15 m


age to reduce the activity of the material to 0.10% of its

initial value?

79. Bananas contain potassium, in the form of (93.258%),

(0.0117%), and (6.7302%). The isotope is radioactive,

with a half-life of The total amount of potas-

sium in a banana is 0.50 g. Calculate the activity of a banana.

80. For diagnostic medical imaging, the isotope iodine-123 is

absorbed by healthy thyroid tissue and emits 1.4-MeV gamma

rays. Similarly, technetium-99 is absorbed by abnormal brain

tissue and emits 0.14-MeV gamma rays. Thus, the absence or

presence of gamma emission from a region of the thyroid or

brain, respectively, is indicative of abnormal tissue. Assume

that a compound containing 0.10 g of each isotope is

injected into the bloodstream of a 25-kg child, and that 10%

of the radiation is absorbed throughout the body. What is the

absorbed dose in each case?

*81. The partial eradication of the thyroid in patients suffering

from hyperthyroidism can be accomplished by injecting a

compound containing radioactive iodine into the body;

the iodine then concentrates in the thyroid and kills its cells. If

the thyroid is to be subjected to an activity of 0.10 Ci, how

many grams of should be injected? The half-life of is

8.04 days.

82. In a mass spectrometer, ions are accelerated through a poten-

tial and enter a region of uniform magnetic field, where they

travel in a semicircular arc before reaching a detector (see Fig.

30.26). Atoms from an unknown substance are singly ionized

and analyzed using a magnetic field of 0.425 T. The detector is

fixed for an arc radius of 6.50 cm. When the accelerating volt-

age is varied, a large number of ions are detected at 5249 V,

and a small number at 6123 V. From the chart of isotopes in

Table 40.2, what is the unknown substance?

131I131I

131I

1.28 � 109 yr.

40K41K

40K39K

85Kr 83. When a neutron combines with a proton to form a deuteron

a gamma ray with energy 2.223 MeV is released. The

atomic masses of and are 1.007 825 u and 2.014 102 u,

respectively. From these data, calculate the mass of the neutron.

84. Consider the fission of a nucleus into two nuclei.

In the first instant after this fission, the two 119Pd nuclei are

just barely in contact, and they then accelerate away from each

other, impelled by their mutual electrostatic repulsion. If the

center-to-center distance between the nuclei is twice

their radius (see Fig. 40.24). what is the electric potential

energy? What is the force? What is the acceleration of each

fragment?

119Pd

119Pd238U

2H1H

(2H),


2r

FIGURE 40.24 Two palladium nuclei.

Answers to Checkups

Checkup 40.1

1. The number of protons is 6 for all three isotopes; all isotopes

of a given chemical element have the same atomic number.

The number of nucleons, or mass number, is given by the

superscript before the element symbol, and so is 8, 9, and 10

for the respective isotopes. The number of neutrons is the

number of nucleons minus the number of protons, and so is 2,

3, and 4, respectively.

2. Each hydrogen isotope has one proton (all isotopes of a given

element have the same atomic number), and the three isotopes

have 0, 1, and 2 neutrons, and thus 1, 2, and 3 nucleons,

respectively.

3. The rightmost column,containing only corresponds to the

largest number of neutrons The top row contains

isotopes of Ne; these have the most protons The

isotope has the largest mass number in the chart.

4. (B) Tritium, or has two neutrons and one proton;

each of the other isotope choices has an equal number of neu-

trons and protons.

5. (E) 8. Since Eq. (40.2) asserts that the radius is proportional

to the cube root of the atomic number, the atomic number is

thus proportional to the cube of the radius, so the ratio is

23 � 8.

3H,3H.

27Ne

(Z � 10).

(N � 17).

27Ne,

85. The total worldwide supply of in high-grade ores is esti-

mated at The total worldwide demand for

energy is about Suppose that the entire

energy demand were supplied by 1000-MW nuclear reactors

fueled with

(a) How many reactors would we need?

(b) How long would the high-grade fuel last?

235U.

5.0 � 1020 J/yr.

2.6 � 108 kg.

235U


Checkup 40.2

1. Since the force is the negative of the gradient of the potential,

the largest attractive (negative) force occurs where the slope of

the potential vs. distance is most positive. From the plot of

Fig. 40.3, this is at about

2. The binding energy per nucleon for is 1802 MeV/(238

at the heavy-nuclei (low) end

of the values common to all but the lightest nuclei.

3. No; this would mean the nucleus has more energy than the

separate protons and neutrons, and the nucleus could and

would fly apart.

4. The Q value of the reverse reaction is the negative of that of

the forward reaction, and so is The energy

required would be nearly zero, although for the reaction to

occur, some energy is required to overcome the electric repul-

sion of the nucleus.

5. (D) Fe; Ra. The largest binding energy per nucleon occurs at

the peak of the curve in Fig. 40.6, or, of the nuclei named, for

Fe (iron). Since the curve per nucleon is almost constant,

decreasing very slowly for large mass numbers, the largest

binding energy occurs at the far right end of the curve, or for

the named nucleus with the largest mass number, Ra (radium).

Checkup 40.3

1. Because alpha and beta decays involve the emission of charged

particles, they result in a change of the atomic number of the

nucleus; gamma decay involves the emission of a photon, which

carries no charge, and so does not result in transmutation.

2. The direction of deflection for a positron would be to the left,

in the same direction as the � ray, since the antielectron carries

an elementary positive charge.

3. No. To decay, that is, to spontaneously overcome the binding

energy and emit particles, the initial energy of the nucleus

must be greater than the final energy of the system, and so the

Q value (initial minus final rest-mass energy) must be positive.

4. (D) The negative beta decay results in a unit increase in

atomic number Z, transmuting cesium into barium. Beta

decay does not change the mass number so the

decay product is

Checkup 40.4

1. The decay rate is proportional to the number of nuclei and

inversely proportional to the half-life [Eq. (40.28)]. Since the

atomic masses of the isotopes in Table 40.3 differ by less than

a factor of 20, so do the number of nuclei in The

extreme values of half-life in the table differ from the nearest

values by much more than a factor of 20, so the smallest and

largest half-life determine the highest and lowest decay rate,

respectively. Thus, has the highest decay rate and

the lowest.

238U,131I

1.0 mg.

136Ba.

A � 136,

136Ba.

�1.191 MeV.

� 8 MeV

nucleons) � 7.6 MeV/nucleon,

238U

0.9�1.0 � 10�15 m.

2. As discussed before and after Eq. (40.23), the number of

radioactive nuclei is the initial number after one half-life, or

8.04 days for (see Table 40.3); the number is the initial

number after two half-lives, or 16.08 days, and is the initial

number after three half-lives, or 24.12 days.

3. The fraction of radioactive nuclei remaining at is,

according to Eq. (40.25), According to Fig. 40.14,

the initial amount is 1.00 g, and so the amount remaining

after is 0.37 g.

4. (B) From Table 40.3, 3200 years is two half-lives,

so the number of radioactive nuclei and the decay rate will

decrease by a factor of to the value

5. (A) has a half-life times that of

However, of has more nuclei by a factor of

of and thus has a higher

decay rate by a factor of about

Checkup 40.5

1. It is the stored electric potential energy of protons held close

together that is released when the nuclear fragments separate.

2. From the curve of binding energy, Fig. 40.6, we can see that

the fission of a light nucleus would result in a smaller binding

energy per nucleon; that is, it would require a net increase of

energy.

3. A release of less than one neutron per fission would result in

less than one additional fission, and so would not provide a

chain reaction; similarly, exactly one neutron could in principle

provide a steady reaction, but since some neutrons would be

lost, it would not provide a true chain reaction. A release of

even slightly more than one neutron per fission could indeed

result in a chain reaction, since one or more additional fissions

could be caused by each fission.

4. (B) The rest mass of the products is less than the sum of the

rest masses of and the neutron. Since energy is released in

the reaction, the rest mass of the system decreases. The other

choices are not valid, because a variety of products are possible,

and because the number of nucleons, 236, remains the

same.

Checkup 40.6

1. To ensure a chain reaction, you could lose up to two; that is, to

continue a steady reaction, each fission must produce one neu-

tron that is later absorbed.

2. A mass of 100 g is too small in size to ensure that enough

neutrons are reabsorbed to support a chain reaction; that is,

too many neutrons escape from the mass.

3. The nuclei absorb too many neutrons in nonfission events,

preventing those neutrons from causing the fission of

4. No, the new nuclear fuel is produced from extra uranium

placed around the reactor, not from fuel being consumed

inside the reactor.

235U.

238U

235U

0.3 � 16 � 5.

226Ra,226�14 � 16 than 1.0 mg

14C1.0 mg14C.

1600�5730 � 0.3226Ra14C.

2.0 � 109.(12)

2 � 14

2.0 � 109.

t � 42 years

e�1 � 0.368.

t � t

18

14

131I

12



5. (D) Slows neutrons so that they are more likely to be absorbed

by in fuel rods later. Thus, the moderator enhances the

reaction.

Checkup 40.7

1. Heavy nuclei contain so many protons that electrostatic repul-

sive forces push the nuclei apart, so that fusion is not energeti-

cally favorable.

2. The reaction (40.35) requires that each of the reactions

(40.33) and (40.34) occur twice. The net result is that four

nuclei are consumed and one nuclei is produced (along

with two positrons, two neutrinos, and two gamma rays).

4He

1H

235U

3. No; fission can proceed spontaneously or with the absorption

of slow (low-temperature) neutrons.

4. No; all the nuclear products of reactions (40.33)–(40.35) are

stable nuclei. However, penetrating radiation is produced,

including the gamma ray produced in reaction (40.34) and the

gamma rays produced when the positron of reaction (40.33)

annihilates an electron.

5. (D) While all the

net reactions listed are in principle possible, only (D) repre-

sents the sum of the given reaction (production of tritium

from two deuterium nuclei) and the reaction (40.36) (produc-

tion of from the tritium product of the given reaction and

a third deuterium nucleus).

4He





Trails of small bubbles reveal the passage of high-energy particles in a

bubble chamber filled with liquid hydrogen. A high-energy proton entered

the chamber and struck a proton in a hydrogen nucleus, and the collision

produced a spray of several kinds of new particles.

While learning about elementary particles in this chapter, we will con-

sider such questions as:

? How do charged particles make trails of bubbles in a bubble chamber?


? The several tracks emerging from the collision shown in Fig. 41.5

reveal the creation of new particles in the collision. What reaction

created these particles? (Section 41.1, page 1400)

? How can we calculate the momentum of a particle from the radius of

curvature of the track? (Example 2, page 1401)

Elementary Particlesand Cosmology41

41.1 The Tools of High-EnergyPhysics

41.2 The Multitude of Particles

41.3 Interactions andConservation Laws

41.4 Fields and Quanta

41.5 Quarks

41.6 Cosmology

C H A P T E R

1396

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? Besides energy and momentum, what quantities are conserved in a reaction that

creates particles, such as the reaction that created the new particles in the photos?


In this final chapter we briefly explore two fascinating questions: What is the struc-

ture of the physical world on the smallest scale? And what is the structure of the

physical world on the largest scale? To answer the first of these questions we must turn

to particle physics and the search for the elementary, indivisible building blocks of

matter. And to answer the second question we must turn to cosmology and the inves-

tigation of the dynamics of the Universe on a large scale. We will see that there are

profound links between the physics of elementary particles and the evolution of the

Universe. The study of these links relies on a wide range of tools—from high-energy

accelerators used to re-create the kinds of particles that populated the Universe imme-

diately after the Big Bang, to large optical and radio telescopes needed to observe dis-

tant regions of the Universe.

Early in the last century physicists discovered that the atom is not an elementary,

indivisible unit—each atom consists of electrons orbiting around a nucleus. The elec-

tron is a truly elementary particle, the first elementary particle to be discovered. But

the nucleus is not an elementary, indivisible unit—each nucleus consists of protons

and neutrons packed tightly together. In the 1930s, physicists began to build acceler-

ating machines producing beams of energetic protons or electrons that could serve as

projectiles; with these atom smashers physicists could split the nucleus. In the 1950s,

they built much larger and more powerful accelerating machines; with these new

machines they attempted to split the proton and neutron. But the result of these

attempts was chaos: when struck by very energetic projectiles, the proton and neutron

do not split into any simple subprotonic pieces. Instead, such violent collisions gener-

ate a multitude of new, exotic particles by the conversion of kinetic energy into mass.

For want of a better name, these new particles were called “elementary particles.”

However, most of these new particles are more massive and more complicated than

protons and neutrons—they do not appear to be truly elementary, indivisible units.

After much effort, theoretical physicists imposed some order on this chaos.They found

convincing evidence that protons, neutrons, and many other “elementary particles” are

made of very small, compact subunits. The subprotonic building blocks are called

quarks.

41.1 THE TOOLS OF HIGH-ENERGY PHYSICS

Protons and neutrons are much “harder” than atoms or nuclei. A projectile of a bom-

barding energy of a few eV can shatter an atom, and a projectile of a bombarding

energy of a few MeV can shatter a nucleus. But to make a dent in a proton, we need

a bombarding energy of a few hundred or thousand MeV. Elementary-particle physi-

cists like to measure the energies of their projectiles in billion electron-volts (GeV) or

in trillion electron-volts (TeV). Expressed in joules, these energy units are

1 TeV � 1012 eV � 1.60 � 10�7 J

1 GeV � 109 eV � 1.60 � 10�10 J


The acceleration of a projectile to such a high energy requires large, sophisticated

accelerator machines. High energy—one GeV to thousands of GeV—characterizes

the realm of particle physics. Collisions between particles at these high energies are

capable of creating a large variety of new particles by the conversion of kinetic energy

into mass.

The two largest accelerators are the proton synchrotrons at the Fermi National

Accelerator Laboratory (Fermilab, near Chicago) and at the Conseil Européen pour la

Recherche Nucléaire (CERN, on the Swiss–French border near Geneva). Figure 41.1

shows an overall view of Fermilab; the accelerator is buried underground in a circular

tunnel 6 km in circumference. The CERN accelerator tunnel is even larger, about

27 km in circumference.The large hadron collider (LHC) under construction at CERN

will provide proton collisions with an energy of 14 TeV. The Fermilab Tevatron accel-

erator provides the largest energy now available, producing a beam of protons with an

energy of 1 TeV and a speed of 99.9995% of the speed of light. The protons travel in

an evacuated circular beam pipe (Fig. 41.2). Large magnets placed along this pipe exert

forces on the protons, preventing their escape—the protons move as if on a circular

racetrack. At regular intervals the pipe is joined to cavities connected to high-voltage

oscillators; in each of these cavities, oscillating electric fields impel the protons to

higher energy. After several hundred thousand circuits around the racetrack, the pro-

tons reach their final energy of 1 TeV.

Before the protons are allowed to enter the main accelerator ring, they must pass

through several smaller preliminary accelerators.The protons generated by a proton gun

first pass through an electrostatic generator, then through a linear accelerator, and then

through a “small” circular accelerator. Only after the protons have passed through these

preliminary stages do they enter the main ring. Once the particles have been given

their maximum energy, they are guided out of the accelerator by steering magnets and

made to crash against a target consisting of a block of metal or a tankful of liquid.

Within the material of the target, the high-energy particles collide violently with the

protons or neutrons of the nuclei. The reactions that take place in the collisions create

new particles by conversion of energy into mass. Unfortunately, not all of the kinetic

energy of the incident particles can participate in these reactions. As we saw in Section

11.3, the velocity of the center of mass remains constant during the collision, and

therefore the particles must retain some kinetic energy. A relativistic calculation indi-

cates that, in the collision of a high-energy proton and a stationary proton, the parti-

1398 CHAPTER 41 Elementary Particles and Cosmology

FIGURE 41.1 Panoramic view of Fermilab.

FIGURE 41.2 View of the underground tunnel housing the

Tevatron accelerator at Fermilab. The long row of magnets

encases the beam pipe.

accelerator



cles retain most of the kinetic energy, and only a small fraction becomes

available for reactions. Such collisions are quite inefficient.

The efficiency improves drastically if two high-energy protons are

made to collide head-on.The available energy is then the sum of the ener-

gies of the two protons. At Fermilab, a second beam pipe has been installed

in the tunnel housing the original beam pipe. A beam of antiprotons trav-

els in this beam pipe, in a direction opposite to that of the protons in the

original beam pipe. These two opposite beams cross at two intersection

points (Fig. 41.3), where the violent head-on collisions create a wide vari-

ety of new particles.

In order to observe the particles that emerge from the scene of these

collisions, physicists have used several kinds of particle detectors. Some

of these—scintillation counters and Cerenkov counters—signal the pas-

sage of each electrically charged particle by giving off brief (and weak)

flashes of light; these flashes of light are detected by sensitive photomultiplier tubes (see

Chapter 37, Physics in Practice: Photomultiplier , page 1268). Other detectors—bubble

chambers and multiwire chambers—render the tracks of electrically charged particles

visible, either on a photographic record or on a computer-generated picture.

A bubble chamber is a tank filled with superheated liquid, usually liquid hydro-

gen, whose temperature is slightly above the boiling point. Such a liquid is unstable—

it is about to start boiling but it will usually not start until some disturbance triggers

the formation of the first few bubbles. A charged particle zipping through the cham-

ber provides just the kind of disturbance the liquid is waiting for—a fine trail of small

bubbles forms in the wake of the particle’s passage. High-speed cameras can take a

picture of these bubble tracks before they disperse and disappear in the turmoil of sub-

sequent widespread bubbling and boiling of the liquid.

Figure 41.4 shows the BEBC bubble chamber that was used at CERN for many years

in a wide variety of experiments. The bubble chamber is surrounded by a large electro-

magnet, which aids in the identification of the particles passing through the bubble cham-

ber. The magnetic field generated by this magnet pushes the particles into curved orbits

as they pass through the chamber.The direction of the curvature depends on the sign of

the electric charge, and the radius of the curved orbit is proportional to the momentum

FIGURE 41.4 (a) The Big European Bubble Chamber (BEBC) at

CERN, surrounded by the magnet, which almost completely hides it

from view. (b) The BEBC before its installation into the magnet. (b)

(a)

FIGURE 41.3 The Collider Detector at

Fermilab (CDF).

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bubble chamber


of the particles (Section 30.1).Thus, measurement of the radius of a bubble track tells us

the sign of the electric charge of the particle and the momentum of the particle.

The photograph of Fig. 41.5 was taken with a bubble chamber. It shows the track

of a high-energy proton that entered the bubble chamber and collided with a proton

at rest in the core of one of the hydrogen atoms in the liquid filling the chamber. The

collision destroyed one of the two protons, but produced a handful of new particles: seven

positive pions, seven negative pions, a positive kaon, and a neutral lambda particle.

Using the respective symbols for these particles (see Tables 41.2 and 41.3), the reac-

tion can be summarized as follows:

(41.1)

The lambda particle subsequently decayed into a proton and a negative pion:

(41.2)

The sum of the rest masses of the particles after the reaction (41.1) is much larger

than the sum of the rest masses of two protons. The excess mass comes from the con-

version of energy into mass—some of the kinetic energy of the incident proton has

been converted into mass.

This conversion of kinetic energy into mass plays a crucial role in the discovery of

new particles. Almost all the new particles discovered during the past 30 years are con-

siderably heavier than protons and neutrons. Physicists need powerful accelerators to

produce the large kinetic energies that must be supplied for the manufacture of these

new heavy particles.

Elementary-particle physicists prefer to measure the masses of

particles in units of MeV/c2, using the conversion 1 u � 931.5

MeV/c 2. Thus a proton has a mass of 938 MeV/c 2, a pion has 139.6 MeV/c2, a

kaon has 494 MeV/c2, and lambda has 1115 MeV/c2 (see Tables 41.2 and 41.3).

EXAMPLE 1

¶ S p � p�

p � p S p � 7p� � 7p� � K� � ¶


This collision of a high-energyproton with a proton at rest…

…creates positive particles:one proton, one kaon, andseven positive pions; …

…and a neutral lambda thathere decays into a proton anda negative pion.

…and negative particles:seven negative pions; …

FIGURE 41.5 Tracks of particles in a bubble chamber

(the tracks have been colored to distinguish different

kinds of particles; in the original bubble-chamber photo-

graph the tracks are all white). A proton enters the bubble

chamber (yellow track from top of photograph) and col-

lides with one of the protons at rest in the nucleus of one

of the hydrogen atoms in the liquid filling the chamber.

The tracks of positively charged particles (red) and nega-

tively charged particles (blue) created in this collision can

be seen emerging from the scene of the accident. Besides

the charged particles, a neutral lambda was created in the

collision. Such a neutral particle leaves no track in the

bubble chamber, but it reveals itself when it decays into a

proton (yellow track at bottom) and a negative pion

(purple track)

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Given these masses, calculate the minimum kinetic energy required for each proton

if we want to initiate the reaction (41.1) in a head-on collison between two pro-

tons of equal kinetic energies.

SOLUTION: In a head-on collison between two protons of equal energies, all of

the kinetic energy can be converted into rest-mass energy (this would not be true

for the collision of a high-energy proton with a stationary proton; that is, it would

not be true for the collision as photographed in Fig. 41.5). If the kinetic energy of

each proton is K, the kinetic energy of both is 2K, and this must equal the differ-

ence of rest-mass energy before and after the reaction:

(41.3)

Thus, the kinetic energy of each proton must be

In a detector region with a uniform magnetic field of 4.38 T, a

pion leaves a track with a radius of curvature of 1.46 m. What

is the momentum of the pion? The magnetic field is applied in the downward

direction, and the motion of the pion is horizontal, clockwise when looking from

above. What is the sign of the elementary charge on the pion?

SOLUTION: From Eq. (30.6), the magnitude of the momentum of a pion in cir-

cular motion in a uniform magnetic field is

As mentioned in Section 30.1, this expression for the momentum is relativistically

correct. In relativistic calculations, the momentum is often expressed in MeV/c,

or GeV/c :

From this form it is apparent that p2c2 is much greater than the square of the pion

rest-mass energy, 140 MeV (Example 1). Recalling that we

see that which means that this pion is extremely relativistic.

With the magnetic field downward, the magnetic force implies

counterclockwise motion for a positive particle.The clockwise motion implies that

the motion is that of a negative pion.

Although bubble chambers yield excellent pictures of particle tracks, they are very

complex, very large, and very expensive machines, and they cannot take picures fast

enough to satisfy the needs of experimenters. Bubble chambers have now been replaced

by various types of electronic tracking chambers, for example, multiwire chambers.

These chambers are strung with thousands of fine wires, in an array that resembles a

F � qv � B

E2 W (mc2)2,

E2 � p2c2 � (mc2)2,

� 1.92 � 103 MeV/c � 1.92 GeV/c

p � 1.02 � 10�18 kg�m/s �1.00 MeV

1.60 � 10�13 J�

3.00 � 108 m/s

c

p � qBr � 1.60 � 10�19 C � 4.38 T � 1.46 m � 1.02 � 10�18 kg�m/s

EXAMPLE 2

2625 MeV�2 � 1313 MeV.

� 2625 MeV

� 14 � 139.6 MeV � 494 MeV � 1115 MeV � 938 MeV

�[mass of lambda]c2 � 2mpc2

2K � mpc2 � 14[mass of pion]c2 � [mass of kaon]c2


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tracking chamber

multiwire chamber


coordinate grid (Fig. 41.6). The wires are connected to voltage supplies and to cur-

rent sensors. When a charged particle passes through the gas in the chamber, it ion-

izes the gas along its track, and the electrons released in this ionization are attracted

to the wires near the track, where they form miniature electric discharges, which are

detected as current pulses by the sensors individually connected to the wires. These

current pulses reveal which wires are near the track, and a computer can then imme-

diately reconstruct the location of the track and draw a picture of the track. An impor-

tant advantage of multiwire chambers over bubble chambers is that the raw data are in

the form of current pulses, which can be directly processed and stored in digital form

by a computer. In contrast, the raw data from bubble chambers were in the form of

photographs, and the tracks recorded on these photographs had to be measured and

converted into numbers before they could be processed and analyzed by a computer;

the conversion of photographic data into digital data is very time-consuming.

The UA1 detector at CERN incorporates several multiwire chambers. Figure 41.7

shows a computer-generated picture of tracks of particles in the multiwire chamber

of this detector.


Each wire in array is con-nected to a voltage sourceand a current detector…

…so that when a passing particleionizes gas atoms, a sequence ofcurrent pulses is detected.

FIGURE 41.6 Wires strung in a cylindri-

cal array for a multiwire chamber. The array

is shown during construction, before it was

placed inside the vacuum chamber.

FIGURE 41.7 Computer-generated picture of particle tracks in the

multiwire chamber of the UA1 detector at CERN. These are the tracks

of particles produced in the collision of a proton and an antiproton that

entered the chamber horizontally from opposite directions.

Checkup 41.1

QUESTION 1: Do neutrons make tracks in bubble chambers? In multiwire chambers?

QUESTION 2: For what purpose are particle detectors placed in magnetic fields?

QUESTION 3: Figure 41.5 shows the tracks of positive and of negative particles in a

bubble chamber immersed in a magnetic field. The magnetic field is perpendicular to

the plane of the page. Is it directed into the page or out of the page?

QUESTION 4: Is the Q value for the reaction (41.1) positive or negative? For the reac-

tion (41.2)?

QUESTION 5: What is the advantage of head-on collisions between protons of oppo-

site motion over collisions with a stationary target?

(A) The head-on collision is easier to achieve than hitting the stationary target.

(B) In the head-on collision, the full kinetic energy is available for reaction.

(C) In the head-on collision, the backward recoil makes detection easier.

(D) In the head-on collision, both protons and antiprotons can be used.

(E) In the head-on collision, exactly twice as much energy is available.

✔


41.2 THE MULT ITUDE OF PART ICLES

As physicists built accelerators of higher and higher energy, they discovered more and

more new particles. By now, several hundred particles are known. Whenever a new,

more powerful accelerator makes available more energy for collisions, more particles

and heavier particles are produced—there seems to be no end in sight.

The known particles fall into three groups: leptons, baryons, and mesons. The leptons are

distinguished from baryons and mesons by the interactions in which they participate.

Leptons interact via the electromagnetic force and the weak force (we already became

acquainted with the weak force in Chapter 40, where we found that the weak force pro-

duces some radioactive decay reactions involving neutrinos). In contrast, baryons and

mesons interact via the electromagnetic force, the weak force, and the strong force. Because

the strong force is stronger than the other forces, it predominates in reactions among

baryons and mesons; accordingly, baryons and mesons are called strongly interacting

particles, or hadrons (from the Greek hadros, thick). The distinction between baryons

and mesons is based on their spin: baryons have half-integer spin ( U, U, U, . . .) and

mesons have integer spin (0, U 2U, . . .). Particles with half-integer spin are called

fermions, and particles with integer spin are called bosons. Thus, baryons are fermi-

ons, and mesons are bosons.

There are six different leptons (see Table 41.1). Among these, the electron (e) is

the most familiar. The muon is very similar to the electron; it has the same electric

charge as an electron but its mass is about 200 times as large. The tau is also simi-

lar to the electron, but its mass is even larger than that of the muon. The neutrinos

(�e, ��, and ��) are particles of zero electric charge and very small mass. Until recently,

neutrinos were believed to have zero mass, like photons. But now there is clear evidence

that neutrino masses are not zero. In Table 41.1, the masses of the particles are expressed

in units of MeV/c2 and the electric charge and the spin are expressed in multiples of

the elementary charge e and in multiples of Planck’s constant .The choice of MeV/c2

for the unit of mass is very convenient in calculations involving the rest-mass energies;

for instance, from the electron mass of 0.511 MeV/c2 listed in the table, we immediately

see that the electron rest-mass energy is 0.511 MeV. All the leptons have spin .

Besides the six leptons of Table 41.1, there are six antileptons: the antielectron (or

positron), the antimuon, the antitau, and the three antineutrinos. These antiparticles

have the opposite electric charge and some other opposite properties, but they have

12

U

(t)

(m)

52

32

12

41.2 The Multitude of Particles 1403

THE LEPTONS

PARTICLE SYMBOL MASSa SPIN CHARGE COMMENTS

electron e constituent of atoms

muon � produced in decay of pion; abundant in cosmic rays

tau �

electron neutrino �e 0 produced in beta decay; abundantly emitted by Sun

muon neutrino �� 0

tau neutrino �� 0

a Neutrinos are now known to undergo quantum-mechanical oscillations between different mass states. Many experiments are underway to clarify such processes.

12� 18.2 MeV�c 2

12� 0.19 MeV�c 2

12� 3 eV�c 2

�1121784 MeV�c 2

�112105.6 MeV�c 2

�1120.511 MeV�c 2

TABLE 41.1

leptons

hadrons

bosons and fermions



SOME BARYONS

PARTICLE SYMBOL MASS SPIN CHARGE QUARK CONTENT COMMENTS

proton p uud constituent of nucleus

neutron n 939.6 0 ddu constituent of nucleus

lambda 1115 0 uds has strangeness

sigma-plus 1189 uus has strangeness

sigma-minus 1197 dds has strangeness

sigma-zero 1192 0 uds has strangeness

xi-minus 1321 dss has strangeness

xi-zero 1315 0 uss has strangeness

omega-minus 1672 sss has strangeness

charmed lambda 2280 udc contains charmed quark�112¶c

�3�132��

�212�0

�2�112��

�112©0

�1�112©�

�1�112©�

�112¶

12

�112938.3 MeV�c2

TABLE 41.2

exactly the same mass and spin as the corresponding particles. (The notation for an

antiparticle is a bar over the letter or, alternatively, a superscript indicating the elec-

tric charge; thus, the notation for the antielectron is or

The baryons are the most numerous group of particles.The most familiar baryons

are the proton and the neutron; these are the baryons of the least mass.Table 41.2 lists

some of the other baryons. For every baryon in Table 41.2, there exists an antibaryon.

As in the case of leptons, these antiparticles have an opposite charge but the same

mass and spin as the corresponding particles.

Finally, the mesons are another numerous group of particles. Some of them are listed

in Table 41.3.The most familiar mesons are the three pions (with electric charges 0, �e,

�e); these were the first mesons to be discovered.

For every meson there exists an antimeson; however, these antiparticles have already

been included in Table 41.3. For example, the antiparticle to the �� is the and vice

versa.The antiparticle to the �0 is the �0; this means that when two �0 mesons meet, they

can annihilate each other [see Section 40.3 and the discussion before Eq. (36.47)]. Also,

the �0 usually decays into two photons, in essence annihilating itself.

Most of the particles are unstable; they decay, spontaneously falling apart into sev-

eral other particles.The only absolutely stable particles are the electron, the proton, and

the neutrinos. The lifetimes of the other particles range from a fairly long 15 minutes

for the neutron to about for many of the exotic particles.

A particle that lives only is incapable of making a measurable track in a

tracking chamber or other detector. Thus, such a short-lived particle cannot be

detected directly, but its existence can be inferred from a careful study of the rates of

reactions of longer-lived particles engaged in collisions. The short-lived particle par-

ticipates in these reactions as an intermediate, ephemeral state, which causes a char-

acteristic increase of the reaction rate whenever the energy of the colliding particles

coincides with the energy required for the production of the short-lived particle.

Because of their enhancing effects on reaction rates, the short-lived particles are often

called resonances.

10�23 s

10�23 s

p�,

e�.)e

resonances

baryons

mesons


Checkup 41.2

QUESTION 1: Do all leptons have spin All baryons?

QUESTION 2: What are the mass, spin, and electric charge of an antitau? An electron

antineutrino?

QUESTION 3: Which is the heaviest particle listed in Tables 41.1–41.3?

QUESTION 4: Particles not listed in Tables 41.1–41.3 include the neutral eta particle

, which has spin and the neutral B particle (B0), a massive particle with spin

0. To what group of particles does the belong? The

(A) Lepton, baryon (B) Lepton, meson (C) Baryon, lepton

(D) Baryon, meson (E) Meson, baryon

41.3 INTERACTIONS ANDCONSERVATION LAWS

The reactions that occur among the particles in a high-energy collision are governed

by the four fundamental forces: the “strong” force, the electromagnetic force, the “weak”

force, and the gravitational force. However, at the microscopic level, particles are sub-

ject to quantum uncertainties in position and momentum, and the force acting on

them is not well defined. Hence, physicists usually prefer to speak of four fundamen-

tal types of interactions, instead of forces.

The strong force acts on baryons and mesons, but not on leptons. As we will see

in Section 41.5, the strong force is a special case of an even stronger force, called the

“color” force.

B0?h0

12;(h0)

12?

✔


SOME MESONS

PARTICLE SYMBOL MASS SPIN CHARGE QUARK CONTENT COMMENTS

pion-zero 0 0 and abundant in cosmic rays;carrier of nuclear force

pion-plus 139.6 0 abundant in cosmic rays;carrier of nuclear force

pion-minus 139.6 0 abundant in cosmic rays;carrier of nuclear force

kaon-zero 498 0 0 has strangeness

kaon-plus 494 0 has strangeness

kaon-minus 494 0 has strangeness

J/psi 3097 1 0 contains charmed quarks

D-zero 1865 0 0 contains charmed quark

D-plus 1869 0 contains charmed quark

upsilon 9460 1 0 contains bottom quarksbb�

cd�1D�

cuD0

ccJ�c

�1su�1K�

�1us�1K�

�1dsK0

du�1p�

ud�1p�

dduu135.0 MeV�c 2p0

TABLE 41.3

strong force


The electromagnetic force acts primarily on charged particles, but it also acts on

neutral particles—such as the neutron—that contain an internal distribution of elec-

tric charge and are endowed with magnetic moments.

The weak force acts on leptons, baryons, and mesons. However, its effect on baryons

and mesons is often hidden behind the much larger effects produced by the strong or

the electromagnetic force. To see the purest manifestation of the weak force, we have

to examine reactions involving leptons. The weak force is deeply involved in many

reactions that bring about the decay of unstable particles. For example, the weak force

is responsible for the beta decay of the neutron. The weak force is also responsible for

other beta decays.

The gravitational force is of no direct interest in particle physics. Although all

particles and all forms of energy interact gravitationally, the gravitational effects pro-

duced by individual particles are too feeble to be of any significance at the energies

available in laboratories on Earth. However the gravitational force is of great interest

to theorists who study the ultrahigh energies of the early Universe.

Table 41.4 lists the strength of each of the fundamental forces and also the range,

or the maximum distance over which this force can reach from one particle to another.

In the table, the strengths of the forces are expressed relative to that of the strong force,

to which a strength of 1 has been assigned arbitrarily.1

All the forces, and all the reactions that they produce, obey the usual conservation laws for

energy, momentum, angular momentum, and electric charge. Besides these conservation laws

for familiar quantities, experiments with high-energy particles have led to the discov-

ery of new conservation laws involving several esoteric quantities, such as baryon

number, lepton number, isospin, strangeness, and parity. All reactions obey a conser-

vation law for baryon number, which is a generalization of the conservation law for

mass number from nuclear physics. Each baryon has a baryon number of each

antibaryon and all other particles have baryon number 0.The conservation law for

baryon number then simply states that the net baryon number remains unchanged in

any reaction.This is a mathematical expression of the requirement that any baryons that

disappear in a reaction must be replaced by an equal number of other baryons, with

antibaryons counting as negative baryons. For example, consider the reaction (41.1).

The baryon numbers for p, , and � are 1, 0, 0, and 1, respectively; hence, in

the reaction (41.1), the net baryon number is before and

after; that is, the net baryon number remains unchanged.

1 � 0 � 0 � 0 � 11 � 1

p , K�

�1,

�1,


1The strengths of the forces depend on the energies of the particles. The values in the table are appropri-

ate for low energies.

baryon number

THE FOUR FUNDAMENTAL FORCES

RELATIVE FORCE ACTS ON STRENGTH RANGE

strong/color baryons and mesons (hadrons) � 10�15 m

electromagnetic particles with charge or magnetic moment infinite

weak leptons, baryons, and mesons � 10�18 m

gravitational all forms of matter infinite10�38

10�6

10�2

1

TABLE 41.4

Conceptsin

Context

electromagnetic force

weak force

gravitational force


Lepton number is for leptons what baryon number is for baryons. Each lepton

has lepton number each antilepton and all other particles have lepton number

0. The net lepton number remains unchanged in any reaction. See also Problem 18.

Isospin is somewhat more complicated because it is a quantity with several com-

ponents; that is, isospin is like a vector quantity.The conservation law for isospin states

that the net sum of the isospin “vectors” of all the particles involved in a reaction

remains unchanged.

Strangeness is similar to the baryon and lepton numbers. Each hadron has a strange-

ness number: the proton has strangeness 0, the kaon-plus has the lambda has

the pion has 0, etc. (values of the strangeness are given in the “Comments” column in

Tables 41.2 and 41.3).The strangeness of an antiparticle is the negative of the strange-

ness of the corresponding particle.The conservation law for strangeness states that the

net strangeness number remains unchanged in a reaction. For example, in the reaction

(41.1), the net strangeness is before and after; that is, there

is no change.

Parity characterizes the behavior of a quantum-mechanical wave under a reversal

of the x, y, and z coordinates. Such a reversal is physically equivalent to forming a

mirror image of the wave. It can be shown that the mirror image of the quantum-

mechanical wave for a stationary state is either equal to the original wave

or else equal to the negative of the original wave Conservation of parity

means that the net parity (the product of all the individual parities) of all the parti-

cles participating in a reaction is unchanged.

The conservation laws for energy, momentum, angular momentum, electric charge,

baryon number, and lepton number are absolute—no violation of any of them has ever

been discovered. In contrast, the conservation laws for isospin, strangeness, and parity

are approximate—they are valid for some reactions, but fail in some others. This, of

course, raises the question of what possible meaning can be attached to a “law” that

works sometimes and fails sometimes. The answer is that whether a conservation law

is obeyed or not depends on the kind of interaction, or the kind of force, that drives the

reaction. The reactions caused by the strong force obey all the conservation laws, but

reactions caused by the electromagnetic or the weak force do not. It is usually easy to

tell what force is involved in a reaction: reactions involving the strong force tend to be

fast; reactions involving the other forces tend to be (relatively) slow. For example, reac-

tion (41.1) is a fast reaction brought about by the strong force, whereas reaction (41.2)

is a slow reaction brought about by the weak force.The difference is evident in Fig. 41.5,

where reaction (41.1) appears to occur at a single point whereas reaction (41.2) occurs

only after the lambda particle has traveled a measurable distance.

Table 41.5 lists the conservation laws obeyed by the strong (color), electromag-

netic, and weak forces.

(parity � �1).

(parity � �1)

0 � 0 � 0 � 1 � 10 � 0

�1,�1,

�1,�1,


lepton number

isospin

strangeness

parity

FORCES AND CONSERVED QUANTITIES

ENERGY, MOMENTUM,AND BARYON LEPTON

FORCE ANGULAR MOMENTUM CHARGE NUMBER NUMBER STRANGENESS PARITY ISOSPIN

strong/color ✓ ✓ ✓ ✓ ✓ ✓ ✓

electromagnetic ✓ ✓ ✓ ✓ ✓ ✓

weak ✓ ✓ ✓ ✓

TABLE 41.5


Do the following reactions conserve electric charge? Do they

conserve baryon number? Do they conserve strangeness? Are the

reactions possible?

(41.4)

(41.5)

SOLUTION: In the first reaction, all particles are electrically neutral, and in the

second, there is one negative elementary charge before and after, so both reactions

conserve electric charge. Reaction (41.4) conserves baryon number, since the baryon

number is 1 for the and for But Eq. (41.5) does not conserve

baryon number, since the baryon number is 1 for the and for

This violation of an absolute conservation law implies that the reaction (41.5) is not

a possible reaction.

Reaction (41.4) does not conserve strangeness, since the strangeness is �1 for

the and for n � �0. Reaction (41.4) is possible, but since strangeness is

not conserved, it is not mediated by the strong force. Reaction (41.5) does not

conserve strangeness, since the strangeness is for the and for

Checkup 41.3

QUESTION 1: What is a hadron? Is the proton a hadron? The neutron? The electron?

The photon? The tau neutrino?

QUESTION 2: What are the mass, spin, electric charge, and strangeness of an anti-

xi-zero An anti-xi-minus, or xi-plus

QUESTION 3: What are the quantities conserved in all interactions? Conserved in

electromagnetic interactions, but not in weak interactions? Conserved in strong inter-

actions, but not in electromagnetic?

QUESTION 4: Consider the wavefunction What is the parity of this wave-

function, that is, what happens to it when you change x to

QUESTION 5: The reaction does not conserve strangeness. Which of the

strong, electromagnetic, or weak interactions can cause this reaction?

(A) Strong only

(B) Weak only

(C) Electromagnetic only

(D) Electromagnetic or weak only

(E) Strong, electromagnetic, or weak

QUESTION 6: In which of the four interactions does the electron participate? In which

does the neutron participate?

(A) All, all

(B) All, all except electromagnetic

(C) All except weak, all except strong

(D) All except strong, all

(E) All except weak, all except electromagnetic

¶ S n � p0

�x?

c � sin kx.

(�, or �)?(0)?

✔

K� � p0.

�1 � 0��2

0 � 0¶

K� � p0.0 � 0�n � p0.1 � 0¶

� S K� � p0

¶ S n � p0

EXAMPLE 3



41.4 F IELDS AND QUANTA

We know from our study of electricity and magnetism that forces are mediated by

fields. Distant particles do not act on one another directly; rather, each particle gen-

erates a field of force, and this field acts on the other particles. As we saw in Section

23.1, the existence of fields is required by conservation of energy and momentum.

Fields play the role of storehouses of energy and momentum; the energy and momen-

tum stored in the fields balance any excess or deficit in the energy and momentum of

the interacting particles engaged in (nonuniform) motion.The most spectacular exam-

ple of the conversion of particle energy into field energy occurs in the annihilation of

matter with antimatter: if an electron collides with an antielectron, the two particles

annihilate one another, giving off a burst of very energetic light, or gamma rays.

According to classical physics, such a burst of light consists of electric and magnetic

fields; thus, the annihilation converts all of the energy of the particles—including their

rest-mass energy—into field energy. The reverse reaction is also possible: if a gamma

ray collides with a charged particle, it can create an electron–antielectron pair. In such

pair creation, the field energy of the gamma ray is converted into the energy of the pair

of particles (Fig. 41.8)

Each of the four fundamental forces is mediated by fields of its own. Hence there are

gravitational fields, electromagnetic fields, strong fields, and weak fields. According

to quantum physics, the energy stored in fields is not smoothly distributed; rather the

energy is found in quanta, that is, small packets or lumps of energy that can be regarded

as particles.

In Chapter 37, we became familiar with the quanta of the electromagnetic field;

these quanta are the photons. Each of the other fundamental fields has quanta of its

own.Table 41.6 lists the quanta of all four fundamental fields. Like photons, the quanta

of the gravitational field, or gravitons, and the quanta of the strong field, or gluons,

are both massless. In contrast, the quanta of the weak field, or W�, W�, and Z0 par-

ticles, are endowed with mass.

At the quantum level, we can picture the field of force generated by a particle as a

swarm of quanta buzzing around the particle. For example, we can picture the electric field

surrounding an electron, or any other charged particle, as a swarm of photons.The swarm

is in a state of everlasting activity—the charged particle continually emits and reabsorbs

the photons of the swarm. Emission is creation of a photon; absorption is annihilation

of a photon. Hence we can say that the electric field arises from the continual interplay

of three fundamental processes: a photon is emitted by one particle, propagates through


A gamma ray strikes an electron, creating an electronand an antielectron.

This track was madeby recoiling electron.

FIGURE 41.8 The two spiraling tracks in

this bubble-chamber photograph were made

by an electron (green) and an antielectron (red).

These particles were created by a high-energy

gamma ray in a collision with the electron of a

hydrogen atom in the bubble chamber.

gravitonsgluonsW� particlesW� particlesZ0 particles

FIELD QUANTA MASS

gravitational gravitons

weak W particles 80.4

Z particles 91.2

electromagnetic photons 0

strong/color gluons 0

0 MeV�c 2

TABLE 41.6 FIELDS AND THEIR QUANTA


the intervening distance, and is absorbed by the other particle. This exchange process

can be represented graphically by a Feynman diagram (Fig. 41.9), invented by Richard

Feynman.The photon exchanged between the two electrons is called a virtual photon,

because it lasts only a very short time and, being reabsorbed by an electron, is unde-

tectable by any direct experiment. The steady attractive or repulsive force between two charged

particles is generated by continual repetition of the photon exchange process. The photon is the

carrier of the electromagnetic forces. This is action-by-contact with a vengeance—at a

fundamental level, all the electromagnetic forces reduce to local acts of creation and

destruction involving particles and photons in direct contact.

In terms of a simple analogy, we can easily understand how the exchange of par-

ticles brings about forces. Imagine two boys tossing a ball back and forth between

them (Fig. 41.10); it is obvious from momentum conservation that this produces a net

repulsive force between the boys due to the recoil they suffer when throwing or catch-

ing the ball. Our intuition suggests that no such exchange process can ever produce

attraction. However, imagine two Australian boys tossing a boomerang back and forth

between them in the manner shown in Fig. 41.11; it is then obvious that this produces

an attractive force between the boys. Whether a photon exchanged between two charges

behaves like a ball or like a boomerang depends on the signs of the charges. Quantum

calculations, which take into account the wave nature of all the particles involved, show

that the net force is attractive for unlike charges and repulsive for like charges, as it

should be.


t

x

�

e

e

e

e

Time increasesvertically.

…and later, thiselectron absorbsphoton.

This electron emits photon…

FIGURE 41.9 Feynman diagram representing

the exchange of a virtual photon between two

electrons. The solid blue lines indicate the motion

of the two electrons, with the t axis plotted verti-

cally and the x axis plotted horizontally. The wavy

colored line indicates the motion of the photon.

For this ball throwand catch…

…recoils push boys apart.

FIGURE 41.10 Two boys throwing

a ball back and forth.

Feynman diagram

virtual photon

carrier

RICHARD PHILLIPS FEYNMAN (1918–1988) American physicist. His invention of

the Feynman diagram revolutionized the

computations of relativistic quantum processes.

He shared the 1965 Nobel Prize with Julian

Schwinger (1918–1994), American physicist,

and Sin-Itiro Tomonaga (1906–1979),

Japanese physicist, for work on the quantum

theory of electrons.

For this boomerangthrow and catch…

…recoils push boys together.

FIGURE 41.11 Two boys throwing a

boomerang back and forth.


The gravitational, weak, and strong forces are also generated by the exchange of vir-

tual particles: gravitons, W and Z particles, and gluons. It is a general rule of quan-

tum theory that the range of the force is inversely related to the mass of the particle that

serves as the carrier of the force. Thus, the photons and the gravitons that are the car-

riers of the electromagnetic and the gravitational force have zero mass—the ranges of

these forces are infinite. The W and Z particles that are the carriers of the weak force

have a very large mass—the range of this force is very short. The gluons that are car-

riers of the strong force have zero mass, and accordingly the range of the force ought

to be infinite. But because this force is very strong, its action over any appreciable dis-

tance triggers the spontaneous creation of particle-antiparticle pairs, and this effec-

tively obstructs the force and limits it to short distances.

Although the four fundamental forces seem drastically different, theoretical physi-

cists have sought to formulate a unified theory that treats several or all of these forces

as aspects of a single, more fundamental force. Electromagnetism is the most familiar

example of a unified field theory; that is, it is a theory that treats electric and magnetic

forces as two aspects of a single, underlying force. To appreciate fully the unification

of electricity and magnetism we would have to examine what relativity says about elec-

tric and magnetic fields; we could then see that electric and magnetic forces are merely

two aspects of a single force called the electromagnetic force. But even without adopt-

ing a relativistic point of view, we can see from Maxwell’s equations that electricity

and magnetism are intimately connected.

If we seek to unify the electromagnetic and weak forces, we must regard the carri-

ers of these forces—the quanta whose exchange generates the force—as closely related.

This would seem to contradict the large mass difference between these particles: the

photon is massless, but the W and Z particles are the heaviest particles now known.

The unified theory of the electroweak force formulated by S. Weinberg, A. Salam, and

S. Glashow attributes this mass difference to an imperfect symmetry (a “broken


unified field theory

electroweak force

Big Bang

10–40 s

1027 K

1018 K

109 K

100 K

10–30 s

10–20 s

10–10 s

100 s

1010 s

1020 s

temperature

age ofUniverse

gravity

strong

electromagnetic

weak

the present

A grand unified force…

…separated intogravity and the combinedstrong/electroweak force; …

…“later,” the strong andelectroweak forces separated…

…as finally didthe electromagneticand weak forces.

FIGURE 41.12 The evolution of the

Universe after the Big Bang.


symmetry”) between the photon and the W and Z particles.The theory asserts that per-

fect symmetry between these particles can be restored by giving them very high energies,

in excess of 100 GeV; at such high energies, the photon and the W and Z particles should

become essentially identical. Such high energies are difficult to achieve in our laboratories,

but they were readily available during the early stages of the Big Bang (see Section 41.6),

when the Universe was younger than 10�10 s and had a temperature in excess of

1015 K. It is believed that at these early times, there was no difference between the

photon and the W and Z particles, and there was no difference between the electro-

magnetic and weak forces. Furthermore, many theoretical physicists are now striving

to formulate a grand unified theory (GUT) which presumes that the gravitational,

electroweak, and strong forces were all originally a single, unified force. It is believed

that at about 10�34 s after the beginning of the Big Bang, gravity separated from the still-

unified strong and electroweak forces, and the strong force separated from the elec-

troweak force slightly later (see Fig. 41.12).

The most impressive success of the unified electroweak theory was its prediction of

the masses of the W and Z particles. The W and Z particles were detected in 1982 in

experiments at the proton–antiproton collider at CERN (and at the electron–position

collider at Stanford). The experiments at CERN involved the observation of about a

billion head-on collisions between protons and antiprotons of the same energy,

270 GeV. A few dozen W and Z particles were produced in these collisions (Fig.

41.13). The measured masses of the W and Z particles are, respectively,

and These measured values are within a fraction of a of the the-

oretically predicted values. This excellent agreement constitutes a brilliant confirma-

tion of the unified theory of weak and electromagnetic interactions.

Checkup 41.4

QUESTION 1: Compare the masses of the W and Z particles with the masses listed in

Tables 41.1–41.3. Are the W and Z the heaviest of all known particles?

QUESTION 2: Order the electromagnetic, strong, and weak forces in order of increas-

ing range.

(A) Strong, electromagnetic, weak.

(B) Weak, strong, electromagnetic.

(C) Electromagnetic, strong, weak.

(D) Strong, weak, electromagnetic.

(E) Weak, electromagnetic, strong.

41.5 QUARKS

Let us now return to our initial question. What are the ultimate, indivisible building

blocks of matter? We know of more than 300 particles. It is implausable that all these

hundreds of particles are truly elementary particles. It is likely that most of them, or

maybe all of them, are composite particles made of just a few elementary building

blocks.

To discover these building blocks, physicists have tried to break protons into pieces by

bombarding them with projectiles of very high energy. Unfortunately, if the energy of the

projectile is large enough to make a dent in a proton, then it is also large enough to create

new particles during the collision.This abundant creation of particles confuses the issue—

✔

GeV/c291 GeV/c2.

80 GeV/c2


FIGURE 41.13 Tracks of particles pro-

duced in a very energetic head-on collision

between a proton of energy 270 GeV that

entered from the right and an antiproton of

energy 270 GeV that entered from the left.

In this collision a particle was created.

It immediately decayed into an electron and

a neutrino. The track of the neutrino is not

visible.

W�

grand unified theory

Antiproton from left and protonfrom right collide head-on, creatingmany particles, including one W�.

Red arrow points to track of an electron, a decayproduct of W� particle.


we can never be quite sure which of the pieces that come flying out of the scene of the col-

lision are newly created particles and which are fragments of the original proton. In fact,

none of the pieces ever found in such collision experiments seems a likely candidate for

an elementary building block.Typically, the particles that emerge from a collision between

a high-energy projectile and a proton are pions, kaons, lambdas, deltas, and so on, all of

which seem to be less elementary than the proton.

Although brute-force collision experiments have failed to fragment protons into

elementary building blocks, somewhat more subtle experiments have provided us with

some evidence that distinct building blocks do indeed exist inside protons. At the

Stanford Linear Accelerator Center (SLAC), very high-energy electrons were shot at

protons; these electrons served as probes to “feel” the interior of the protons.The exper-

iments showed that occasionally the bombarding electrons were deflected through

large angles, bouncing off sharply from the interior of a proton.These deflections indi-

cate the presence of some lumps or hard kernels in the interior of the proton, just as,

in Rutherford’s experiments, the large deflections of alpha particles by atoms indi-

cated the presence of a hard kernel (nucleus) in the interior of the atom. Protons and

all the other baryons and mesons seem to be composite bodies made of several dis-

tinct pieces. In contrast, electrons and the other leptons seem to be indivisible bodies

with no internal structure. In some experiments the electron was probed with beams

of extremely energetic particles to within of its center. Even at these extremely

short distances, no substructures of any kind were found. Thus, the electron seems to

be a pointlike particle with no size at all, a truly elementary particle.

Even before the experimental evidence for lumps inside protons became available, the-

oretical physicists had noticed that particles could be classified into groups or families of

similar particles on the basis of their quantum numbers and their behavior in reactions.

To explain these similarities, they had proposed theories in which all the known baryons

and mesons are regarded as constructed out of a few fundamental building blocks.

According to these theories, the similarity between particles within a given family reflects

the similarity of their internal construction, just as the similarities between atoms in a group

of the periodic table reflect the similarity of their internal construction.

The most successful of these theories was the quark model proposed by M. Gell-

Mann and by G. Zweig. In this model, all particles were constructed of three kinds of fun-

damental building blocks called quarks. (Gell-Mann took the word quark from Finnegan’s

Wake, a book by James Joyce.) The three quarks are labeled up, down, and strange, or

simply u, d, and s. They all have spin and they have electric charges of

and elementary charge, respectively (see Table 41.7). Each quark—like any

other particle—has an antiparticle, of opposite electric charge.

�13

23, �

13,

12

10�18 m

41.5 Quarks 1413

QUARK MASSa SPIN ELECTRIC CHARGE STRANGENESS

u 5 MeV�c2 0

d 10 0

s 200 �1

aBased on theoretical estimates.

�13

12

�13

12

23

12

TABLE 41.7 THE FIRST THREE QUARKS

quarksup quark (u)down quark (d)strange quark (s)


To make the ordinary particles out of the quarks, the latter must be glued together

in diverse ways. For example, a proton is made of two u quarks and one d quark (Fig.

41.14). A neutron is made of two d quarks and one u quark (Fig. 41.15). A positive

pion is made of one u quark and one d antiquark (Fig. 41.16), and so on (Tables 41.2

and 41.3 list the quark composition of baryons and mesons). By gluing quarks together,

we can build up all the known baryons and mesons and explain their quantum numbers

and their similarities. Besides, we can predict some properties, such as magnetic moments

and reaction rates, of the composite particles from the assumed properties of the quarks.

There is only one snag: all experimental searches for free quarks have been unsuc-

cessful. Physicists now believe that quarks are permanently confined inside the ordi-

nary particles so there is no way to break the quark out of, say, a proton. It seems that

the quarks are held in place by an exceptionally strong force, which prevents their

escape. This new force is the “color” force.

The concept of “color” was first introduced to remedy an unacceptable violation

of the Exclusion Principle (see Chapter 39). Quarks, like leptons and other particles

of half-integer spin, ought to obey the Exclusion Principle. But investigations of the

quantum states of quarks within protons, neutrons, sigmas, etc., disclosed that several

apparently identical quarks were often found in the same quantum state. To avoid this

apparent violation of the Exclusion Principle, physicists postulated that each of the

quarks exists in three varieties, and that whenever two apparently identical quarks are

found in the same quantum state, they actually are of different varieties. The varieties

of quarks are characterized by a new property called color. Of course this “color” has

nothing to do with real color; it is merely a (somewhat unimaginative) name for a new

property of matter. The different quark colors are red, green, and blue. Thus there is

a red u quark, a green u quark, and a blue u quark, and so on. The antiquarks have anti-

colors; the different antiquark colors are antired, antigreen, and antiblue. Color is a very

subtle property of matter; it usually remains hidden inside the ordinary particles. All

the normal particles are “colorless”—they consist of several quarks with an equal mix-

ture of all three colors. For instance, one of the three quarks inside the proton is red,

one is green, and one is blue. Nevertheless, color plays a crucial role in the theory of

the forces that confine the quarks inside the ordinary particles.

The quarks are confined by extremely strong mutually attractive forces. These

forces between quarks are called color forces because the source of these forces is the

color just as the source of the electric force is the electric charge. Each of the three

varieties of color is analogous to a kind of electric charge. A body is color-neutral, or

“colorless,” if it contains equal amounts of all three colors or if it contains equal amounts

of color and anticolor, just as a body is electrically neutral if it contains equal amounts

of positive and negative electric charge.

The color force is a fundamental force that is included in our table of fundamen-

tal forces together with the strong force (Table 41.4). The color force is closely related

to the strong force—the latter is actually a special instance of the former. The rela-

tionship between the color force and the strong force is analogous to the relationship

between the electric force and the intermolecular force. As we saw in Section 22.1 the

force between two electrically neutral atoms or molecules is a residual electric force

resulting from an imperfect cancellation among the attractions and repulsions of the

charges in the two atoms or molecules. Likewise, the strong force between, say, two

“colorless” protons is a residual color force resulting from an imperfect cancellation

among the attractions and repulsions of the quarks in the two protons.Thus the “strong”

force between two protons is no more than a pale reflection of the much stronger color

forces acting within each proton.

At a fundamental level, the color force between two quarks is due to an exchange

of virtual particles between the quarks. The particle that acts as the carrier of the color


d

proton

u u

Proton is made of one d quark andtwo u quarks.

FIGURE 41.14 Structure of the proton.

The sizes of the quarks are not drawn to

scale. Their sizes are probably much smaller

than the size of the proton.

Neutron is madeof one u quarkand two d quarks.

u

neutron

d d

FIGURE 41.15 Structure of the neutron.

FIGURE 41.16 Structure of a positive

pion (p�).

color force

ud

positive pion

Positive pion is madeof one d antiquarkand one u quark.


force is the gluon, the same as the carrier of the strong force. Figure 41.17 shows a

Feynman diagram representing the exchange of a gluon between two quarks. Such an

exchange of a gluon between two quarks is analogous to the exchange of a photon

between two charged particles (see Fig. 41.9).

Another modification of the simple quark model with three quarks arose from the

theory of the unification of the electromagnetic and weak forces. In order to make this

theory fit some experimental data, physicists had to postulate the existence of a fourth

quark, different from the u, d, and s quarks. This new hypothetical quark was labeled

charmed, or simply c.

The hypothesis of the charmed quark soon received firm experimental support.

In 1974, teams of experimenters at the Brookhaven accelerator and at the Stanford

accelerator discovered the meson and some other related mesons (see Table 41.3),

all of which have exceptionally long lifetimes.These particles contain charmed quarks

with an electric charge of and a mass of 60 MeV/c2.

The proliferation of quarks did not stop with four quarks. In 1977, a team of exper-

imenters at Fermilab discovered the (Greek Upsilon) mesons. These are by far the

most massive mesons known (see Table 41.3). Each of these contains a new quark;

this fifth kind of quark has been labeled bottom, or b.Theoretical considerations then

suggested that there should exist a sixth quark, labeled top, or t. In 1995, another team

of experimenters at Fermilab confirmed the existence of the top quark.

Figure 41.18 summarizes all the constituents of particles and the carriers of forces;

these form the foundation of the Standard Model of particle physics. This model

describes all the known particles in terms of the elementary particles listed in Fig. 41.18,

and it describes all interactions in terms of four kinds of carriers (however, the graviton

is not fully incorporated into the model, because there is as yet no complete and coher-

ent theory of quantum gravity). The four kinds of carriers are the gluons (which come

in various color combinations), the photon, the W�, W�, Z, and the graviton. The six

leptons fall into three families: the electron family, the muon family, and the tau family.

Correspondingly, the six quarks also fall into three families: the up–down family, the

strange–charm family, and the bottom–top family. Each of the six leptons has an antipar-

ticle. Each of the six quarks (up, down, strange, charmed, top, bottom) comes in three

colors (red, green, blue); furthermore, each quark has an antiquark, which comes in three

varieties of anticolor (antired, antigreen, antiblue). Altogether, this amounts to 36 quarks,

12 leptons, and 24 carriers (counting all the color combinations of gluons and their

�

23

J�c

41.5 Quarks 1415

charmed quark (c)bottom quark (b)top quark ( t)

Standard Model

t

x

gluon

green quark

blue quark

blue quark

green quark

This quark emits a gluon, changescolor from green to blue; …

…and this quark absorbs a gluon,changes color from blue to green.

FIGURE 41.17 Exchange of a gluon

between two quarks.

FIGURE 41.18 The forces, carriers, and elementary particles of the Standard

Model.

Leptons

electronfamily

muonfamily

taufamily

Quarks

Forces

up-downfamily

strange-charmfamily

bottom-topfamily

du

cs

tb

color forceelectromagnetic

force weak forcegravitational

force

Carriers gluon photon W and Z graviton


antiparticles). This proliferation of constituents raises the question whether matter

really has such a large number of elementary building blocks. Pushing forward our

search for the ultimate building blocks, we have uncovered layers of structures within

layers of structures—electrons and nuclei within atoms, protons and neutrons within

nuclei, quarks within protons and neutrons. Is there another, more elementary layer

within quarks?

Seeking to answer this question, some theorists have been exploring string theories,

according to which all particles consist of tiny, vibrating strings, only about 10�33 cm

in length.The different masses, spins, electric charges, and other properties of the par-

ticles are supposed to arise from different vibrational patterns of the strings. For this

scheme to work, the strings may have to exist in a 9-dimensional space, that is, a space

with 6 more dimensions than the 3-dimensional space of our everyday experience.

The extra 6 dimensions are thought to be imperceptible because they are tightly curled

up, so they extend over distances of only about It is the ambition of string

theorists to deduce the masses and all the other properties of particles from the char-

acteristics of the strings. But in spite of prodigious effort, they have not yet succeeded

in this, and they have not demonstrated any clear experimental connection between string

theories and our physical world.

Checkup 41.5

QUESTION 1: Do the charges of two u quarks and one d quark add up to the charge

of the proton? Do the charges of two d quarks and one u quark add up to the charge

of a neutron?

QUESTION 2: Each quark has spin How can the net spin of three quarks add to

form the spin of the proton?

QUESTION 3: What is a gluon?

QUESTION 4: What is the electric charge of the u antiquark? The d antiquark? The s

antiquark?

(A)

(B)

(C)

(D)

(E)

41.6 COSMOLOGY

Cosmology is the study of the Universe at large, its size, its shape, and its evolution.

Seeking to grasp the Universe, our minds must wander over distances as great as 10 bil-

lion light-years and over times as long as 10 billion years or longer.

Until the early part of the last century, astronomers thought the Universe to be

much smaller. They thought that the farthest stars at the edge of our Galaxy were

about 30 000 light-years away and that there was nothing but dark, empty space beyond

that distance.2 But in the 1920s, Edwin Hubble used the 100-inch telescope on Mount

�23e, �1

3e, �13e

�23e, �2

3e, �13e

�23e, �1

3e, �13e

�23e, �1

3e, �13e

�23e, �1

3e, �13e

12

12.

✔

10�33 cm.


2A light-year is the distance traveled by light in 1 year, 9.5 � 1015 m.

string theories


Wilson to establish that the faint, wispy “nebulae” found in all parts of the sky were actu-

ally gigantic conglomerations of stars, similar to our own Galaxy but at very great dis-

tance from us.

Our Milky Way Galaxy contains about 1011 stars arranged in an irregular, disklike

region some 105 light-years in diameter. The disk has a central bulge, and it has spiral

arms along which stars are concentrated (Fig. 41.19). There are many external galax-

ies beyond our Galaxy. With our large telescopes we can see altogether about 1011

galaxies! There are supergiant galaxies with 1013 stars each, and there are dwarf galax-

ies with “only” 106 stars. There are spherical galaxies and elliptical galaxies, like lumi-

nous globes and eggs; there are spiral galaxies and barrel-shaped galaxies, like whirling

pinwheels; and there are irregular galaxies with the weirdest shapes (see Figs. 41.20a–c).

41.6 Cosmology 1417

FIGURE 41.19 These pictures taken with the Hubble Space Telescope give an idea what our

Milky Way Galaxy looks like when viewed (a) face on and (b) edge on. The pictures are actually

photographs of two distant galaxies (NGC 5457 and NGC 4631) similar to our Galaxy.

FIGURE 41.20 (a) Spiral galaxy (NGC 3031) in Ursa Major. The plane of this galaxy is

inclined to our line of sight; face on, this galaxy would look circular. (b) Unusual galaxy (NGC

5128) in Centaurus. Note the thick lane of dust surrounding this galaxy. (c) Spiral galaxy (NGC

4565) in Coma Berenices. Note how thin this galaxy is.


All the galaxies are in motion. Many of them congregate in clusters, orbiting about

each other, sometimes accidentally colliding. But let us ignore the fine details of the

motion of galaxies and concentrate on the large-scale features of the motion. We then

find that on a large scale, the galaxies have a motion of recession—all the distant galax-

ies are moving away from us. The recessional velocities of galaxies are determined by

the red-shift method (see Fig. 41.21). The light from the receding stars is Doppler-

shifted to lower frequencies, or longer (redder) wavelengths [see Eq. (36.13) in Section

36.3]. To find the recessional velocity, astronomers need only measure how much the

wavelength of the light received from the atoms in some distant galaxy is shifted rel-

ative to the wavelength emitted by similar atoms in our laboratories on Earth.

By measuring the velocities and distances of galaxies, Hubble discovered that the

motion of recession obeys a very simple rule: the velocity of each galaxy is directly pro-

portional to its distance. This means that from the point of view of our Galaxy, nearby

galaxies move slowly and distant galaxies move fast (see Fig. 41.22). This proportion-

ality between velocity v and distance r is called Hubble’s Law and can be expressed as

v � H0r (41.6)

where is the Hubble constant. If r is expressed in billions of light-years, the numer-

ical value of the Hubble constant is

(41.7)

Astronomers determine the enormous cosmic distances from us to other galaxies

by the brightness or “headlight” method. In essence, this method relies on the fol-

lowing. Stars in faraway galaxies look faint to us, and stars in nearby galaxies look

bright—just as, on a dark road, the headlights of a faraway automobile look faint and

the headlights of a nearby automobile look bright. If all stars generated precisely the

same quantity of light, then differences in their apparent brightness as observed by

our telescopes would be entirely due to differences in their distances—there would

then be a simple mathematical relationship between apparent brightness and distance.

H0 � 2.1 � 107 (m/s)�(billion light-years)

H0


FIGURE 41.21 (a) A distant galaxy seen together with a star in our galaxy. (b) The spectrum of each,

shown on the same scale. Note that lines in the galaxy spectrum are red-shifted to longer wavelengths. The

H�, H�, H� and H absorption lines are the first four lines of Bolmer Series of hydrogen (see Fig. 38.7).

Hubble’s Law

Hubble constant

(a) (b)

A galaxy withv = 7,600 km/s

A star in the Milky Way

Wavelength (nm)

450 500 550400 600 650 700 750

750

H H� H�

rest

λobserved

H�

Absorption lines in anearby star appear attheir rest wavelengths…

…but absorption lines in thespectrum of a distant galaxyare redshifted by the galaxy’smotion away from us.

Recessional velocity vis calculated from themeasured redshift.

red-shift method


However, in practice, there can be some complications; two stars can be at the same dis-

tance yet differ in apparent brightness because one generates more light than the other.

Such intrinsic differences between stars must be taken into account when using the

brightness method. Astronomers have developed clever techniques for selecting stars

of standard brightness, but some observational errors remain in the distance determi-

nations. Numerous analyses support the direct proportionality expressed by Eq. (41.6),

but there are substantial uncertainties in the numerical value of Eq. (41.7)—the Hubble

constant is uncertain by at least — 10%.

Although Fig. 41.22 gives the misleading impression that our Galaxy is at the

center of the Universe and that all other galaxies are fleeing away from us, our Galaxy

does not occupy any special spot in the Universe. The other galaxies are not just flee-

ing away from us; they are fleeing away from each other. The Universe is expanding.

An extraterrestrial astronomer sitting on a distant galaxy would see our Galaxy and

all other galaxies fleeing away from her. Hence our spot in the Universe is pretty much

the same as every other spot. Cosmologists believe that this overall uniformity holds

not only in regard to the expansion, but also in regard to all other general features of

the Universe. For instance, the numbers and types of galaxies that the extraterrestrial

astronomer finds in her neighborhood will, on the average, be the same as we find in

our neighborhood—the Universe is pretty much the same everywhere. This assertion

of the large-scale uniformity of the Universe is called the Cosmological Principle.

The motion of recession of galaxies can be described by a very crude analogy. When

a grenade explodes in midair, the fragments of shrapnel spurt out in all directions.

Different fragments may have different velocities, and in a given amount of time, they

will reach different distances. After a time t, the position of a fragment having veloc-

ity v will be

(41.8)

If we rewrite this as

(41.9)

we see that at any given time the fragments that are at the greatest distances are those

with the highest velocities. This proportionality of velocity and distance has the same

v � r�t

r � vt

41.6 Cosmology 1419

our Galaxy

Relative to our Galaxy, velocitiesof other galaxies are radial…

…and directly pro-portional to distance.

FIGURE 41.22 The motion of

recession of distant galaxies.

Cosmological Principle


form as Hubble’s Law [see Eq. (41.6)]. Thus, Hubble’s Law suggests that the galax-

ies were set in motion by a primordial cosmic explosion billions of years ago and have

been more or less coasting along ever since. Incidentally: In the expansion of the

Universe, only the distances between the galaxies increase; the galaxies themselves do

not expand. This is also in agreement with the grenade analogy, where, of course, only

the distances between the shrapnel fragments increase while the fragments themselves

remain of constant size. (But the grenade analogy has serious defects: It posits a center

for the explosion, whereas the Universe has no special center. Also, in the modern view,

the expansion of the Universe involves the expansion of space itself, not just the motion

of matter in ordinary space.)

The explosion that started the expansion of the Universe is called the Big Bang.

We can calculate how long ago this happened by comparing Eqs. (41.6) and (41.9). We

thus see that the inverse of the Hubble constant must coincide with the expansion time:

(41.10)

or

(41.11)

However, in this calculation of the age of the Universe, we have ignored the possibil-

ity that the velocity of galaxies may change with time. For instance, since gravity pulls

the galaxies toward each other, we might expect that gravity tends to inhibit the motion

of recession and tends to slow down the expansion of the Universe. This would imply

that the velocities of all galaxies were somewhat larger in the past and, consequently,

the true age of the Universe ought to be somewhat smaller than the 14 billion years indi-

cated by our naive calculation. But in any case, this number gives us a rough estimate

of the age.

Since the Universe started some finite time ago, only light from those parts of it that

are sufficiently near can have reached us. The speed of light is c � 3.00 � 108 m/s � 1

light-year/year, and in the time t light travels a distance

This distance is the radius of the observable universe3. Everything within this radius

we can see (given sufficiently powerful telescopes); anything beyond we cannot see

because the light has not yet had time to reach us. Note that as time increases, the radius

ct increases, that is, the observable universe includes more and more of the total Universe.

The idea of the Big Bang can be put to several direct tests. First, nothing we observe

can be older than about 14 billion years. Our Sun has an age of only 5 billion years.The

oldest objects that we can reliably date are the globular clusters of stars found near our

Galaxy (see Fig. 41.23). Old stars enter a red-giant stage, turning a reddish color and

swelling to several hundred times normal size. A young cluster contains few red giants,

and an old cluster contains many. Careful calculations of stellar evolution indicate that

the oldest globular clusters have ages of about 10 billion to 13 billion years, in good agree-

ment with the expansion time of Eq. (41.11).

� 1.4 � 1010 light-years

ct � (1 light-year/year) � (1.4 � 1010 year)

� 4.5 � 1017 s � 1.4 � 1010 years

t �1

2.1 � 107�

billion light-years

m/s�

1

2.1 � 107�

9.5 � 1024 m

m/s

t � 1�H0


Big Bang

observable universe

3 This value of the radius of the observable universe is only an approximation. For an exact calculation, we

need to use the theory of General Relativity, taking into account that space and time are curved.


Another method by which we can date the birth of stars is by the age of chemical

elements. All of the elements, with the exception of hydrogen and helium, were syn-

thesized in very massive stars soon after our Galaxy came into being. These stars sur-

vived only a short time and then exploded as supernovas, scattering these elements

into the cloud of gas and dust that was destined to become our Solar System. Radioactive

dating tells us that these elements are somewhere between 7 and 15 billion years old,

in agreement with the expansion time to within experimental uncertainties.

More evidence for the Big Bang emerges from studies of the abundances of hydro-

gen and helium. These elements were formed in the hot, primordial fireball of the Big

Bang. When the Universe was about 3 minutes old, the temperature was 109 K, sub-

jecting hydrogen to nuclear fusion. Theoretical calculations show such fusion led to

abundances of about 75% hydrogen and 25% helium, in remarkable agreement with

the observed abundances of 74% and 24% (plus traces of other elements).4 This con-

firms our picture of a hot Big Bang.

But the most decisive item of evidence for the Big Bang is that some of the radi-

ant heat given off by the primordial explosion can still be found in the sky today.

Originally, the radiant heat emitted by the dense, primordial fireball was in the form

of very penetrating gamma rays and X rays. But as the Universe expanded, the elec-

tromagnetic radiation expanded with it, and at present the wavelengths of the fireball

radiation are much longer.Theoretical calculations indicate that the radiant heat must

have a blackbody spectrum (see Chapter 37) with a peak around a wavelength of

1 mm, the wavelength of microwaves.

This kind of radiation was discovered in 1964 by Arno A. Penzias and Robert W.

Wilson, two scientists working at Bell Laboratories with very sensitive microwave

communication equipment (Fig. 41.24). They found that the entire sky is noisy; there

is radiation coming at the Earth from all directions. Physicists at Princeton immedi-

ately recognized the cosmological significance of this discovery, and identified the

radiation as cosmic background radiation, a relic of the Big Bang. This radiation is

that of a blackbody with temperature of about 3 K (more precise data give 2.73 K).

What has happened here is that the extremely hot radiant heat from the primordial fire-

ball has gradually cooled down as the Universe expanded, and by now its temperature

has come pretty close to absolute zero. The cosmic background radiation is residual

radiant heat left in the sky after the Big Bang. It is direct material evidence for the

Big Bang.

41.6 Cosmology 1421

4 Fusion within stars continues to produce helium, but the total amount of helium produced by all the stars

since the Big Bang is only a few percent.

cosmic background radiation

FIGURE 41.23 The globular cluster

(NGC 5272) in Canes Venatici.

FIGURE 41.24 Horn antenna at Bell

Laboratories, Holmdel, New Jersey. This

antenna was designed for microwave com-

munication experiments with the Echo and

Telstar satellites.


The big question in cosmology is this: Will the Universe expand forever? Or will

the gravitational attraction between galaxies eventually stop the expansion and cause

the Universe to contract, ultimately leading to to a terminal cosmic implosion? To

help find the answer, we must calculate the deceleration of the motion of recession

caused by gravity.

Consider a spherical region of our Universe, and assume that galaxies are uniformly

distributed throughout this spherical region and the entire Universe (see Fig. 41.25).

Gauss’ Law applies to the inverse-square gravitational force as it does to the inverse-

square electric force. Accordingly, we know that the motion of the galaxies in the spher-

ical region is unaffected by the rest of the Universe. Consider now one of the galaxies

at the surface of this region, presently at a radial distance r0 from us. That galaxy will

continue to recede from us forever provided its present velocity v0 is greater than the

escape velocity, Eq. (9.27),

(41.12)

where M is the mass inside the spherical region. This mass can be expressed in terms

of the average density of mass in the Universe,

(41.13)

and the velocity is given by Eq. (41.6), Thus the condition for a perma-

nently expanding Universe becomes

(41.14)

or

(41.15)

If we insert the numerical value [see Eq. (41.11)] and the

numerical value for the gravitational constant, we obtain the conditions

(permanent expansion)

(ultimate contraction) (41.16)

In principle, this makes it very simple to predict the future evolution of the Universe;

but in practice, we are severely handicapped by the uncertainty in the mass density.

When reckoning the mass of the Universe, we must take into account the mass

belonging to galaxies and also whatever mass is to be found in the intergalactic space

between galaxies. Table 41.8 lists all the known contributions to the mas density of

the Universe, as a percentage of the critical mass density The visi-

ble, luminous matter contained in stars makes only a small contribution, less than

of the critical mass. Thus, the bulk of the mass of the Universe is in nonluminous,

invisible forms, which can be detected only by indirected means.

The mass of baryonic matter (protons and neutrons) listed in the table is inferred

from studies of the abundance of deuterium formed in the fireball of the Big Bang.

Apparently, most of this baryonic matter did not get captured into luminous stars

during the later evolution of the Universe, and it must still be lurking in and around

galaxies, in the form of small, nonluminous stars (“brown dwarfs” and cold stellar

12%

9 � 10�27 kg/m3.

r0 � 9 � 10�27 kg/m3

r0 � 9 � 10�27 kg/m3

H0 � 1�(4.5 � 1017 s)

r0 �3

8pG H2

0

H0r0 � A8p

3 Gr2

0 r0

v0 � H0 r0.

M � r0 4p

3 r3

0

r0

v0 � 12GM�r0


r

A galaxy near the edgeof a spherical region…

…experiences a net gravitationalforce only from galaxies within the spherical region.

FIGURE 41.25 A spherical region

of our Universe.


remnants), perhaps interstellar planets, and large clouds of low-density gas surrounding

the galaxies.

The mass of the nonbaryonic dark matter is inferred from studies of galactic

dynamics. Astronomers can detect the presence of dark matter by its gravitational

effects on the rotational motion of galaxies and on the orbital motion of galaxies around

each other. From such studies, we know that galaxies are surrounded by large,

extragalactic clouds, or halos, of some kind of dark matter. This is sometimes called

the “missing mass” because we cannot see it, and we do not know what it consists of.

In the absence of any information about this dark matter, astronomers and physicists

have felt free to speculate about various possibilities. One such speculation is that this

dark matter might consist of clouds of neutrinos. We know that neutrinos have a small,

nonzero mass, but the magnitude of this mass has so far proved too small to be meas-

ured. If it is about 4 eV�c2, then the clouds of neutrinos left over from the Big Bang

in the form of thermal neutrino blackbody radiation might be sufficient to account

for the dark matter. Another speculation is that the dark mass might consist of some

new, exotic particles invented by theorists working on unified theories of interactions.

But no such particles have so far been confirmed experimentally.

The final contribution to the mass of the Universe listed in Table 41.8 is the most

mysterious. Beginning in 1998, astronomers combined luminosity and red-shift data

for a class of very bright supernovas to determine their recessional velocities and dis-

tances.These data indicate that instead of the presumed state of slowing expansion, the

Universe is actually experiencing an accelerating expansion. To account for an acceler-

ating Universe, cosmologists now assume there exists some sort of dark energy that

permeates the space between galaxies. This dark energy apparently makes up about

70% of the critical density and provides an effective repulsive gravitational force that

accelerates the expansion. We do not yet have any firm grasp on exactly what this dark

energy might be.

Will the Universe continue to expand or will it ultimately contract? A forever-

expanding and accelerating Universe gives the best fit to all the facts as we currently

know them. But there are enough uncertainties in our measurements and enough loop-

holes in our arguments that the possibility of an ultimately contracting Universe cannot

be dismissed. It will be a while before we know the ultimate fate of the Universe.

Checkup 41.6

QUESTION 1: Why do astronomers not rely on triangulation to measure the distances

of galaxies?

QUESTION 2: If the expansion of the Universe is accelerating or is slowing down, can

the Hubble constant be truly a constant?

QUESTION 3: Imagine a galaxy very close to the edge of the observable Universe.This

galaxy is moving away from us at a speed very close to the speed of light. Does this

mean that this galaxy will reach the edge of the observable universe and disappear

from sight?

QUESTION 4: Where were the atoms of most of the elements inside your body made?

(A) In the core of the Earth soon after the formation of the Sun

(B) In supernovas soon after our Galaxy was born

(C) In the Sun, later transmitted to Earth by cosmic radiation

(D) In nuclear fission of heavier elements such as uranium

(E) In the first of the Big Bang10�34 s

✔

41.6 Cosmology 1423

dark energy

CONTRIBUTIONSTO THE MASS

DENSITY OF THE UNIVERSE

FORM OF MATTER PERCENTAGE OFCRITICAL DENSITY

luminous matter 0.4%(stars)

baryonic matter 4%

dark matter 25%(nonbaryonic)

dark energy 71%

TABLE 41.8

dark matter



SUMMARY

leptons

baryons

mesons

carriers

Strength Carrier

strong/color 1 gluon

electromagnetic photon

weak W and Z

gravitational graviton10�38

10�6

10�2

Absolute Approximate

energy isospin

momentum strangeness

angular momentum parity

electric charge

baryon number

lepton number

PARTICLES

FORCES

CONSERVED QUANTITIES

LEPTONS electron (e), muon tau

and neutrinos (�e, ��, and ��).

(t),(m),

QUARKS up (u), down (d), strange (s),

charmed (c), bottom (b), and top (t).

COSMOLOGICAL PRINCIPLE On a large scale, the Universe is uniform.

HUBBLE’S LAW (v is the recessional velocity of a

galaxy and r is the distance to the galaxy.)

fermions

bosons

hadrons ef

f

3. The names baryon, meson, and lepton come from the Greek

barys (heavy), mesos (middle), and leptos (thin, slender). These

names were originally intended to indicate the masses of the

particles. According to the lists of particles and masses given

in this chapter, is it true that the baryons have the largest

masses and the leptons the smallest?


1. Why are high-energy accelerators necessary for the produc-

tion and discovery of new, massive particles?

2. Physicists were planning to construct the Superconducting

Super Collider (SSC), a 20-TeV accelerator, which would have

cost some $4 billion. Can such an expenditure be justified?

v � H0r (41.6)

H0 � 2.1 � 107 (m/s)�(billion light-years) (41.7)

d

proton

u u

u

neutron

d d

Leptons

electronfamily

muonfamily

taufamily

e �e

� ��

� ��

Quarks

Forces

up-downfamily

strange-charmfamily

bottom-topfamily

du

cs

tb

color forceelectromagnetic

force weak force gravitationalforce

Carriers gluon photon W and Z graviton


Problems 1425

4. How does the antineutron differ from the neutron?

5. In experiments at CERN, an antielectron and an antiproton

have been put together to form an antihydrogen atom. In prin-

ciple, two of these atoms can form an antihydrogen molecule.

Could we confine a sample of antihydrogen gas in an ordinary

steel bottle? Can you think of any way of confining it?

6. How would you refute the proposition that the Sun is made of

antimatter?

7. Why does a particle that lives only 10�23 s not make a track in

a bubble chamber? (Hint: Suppose the particle moves at the

maximum conceivable speed; how far will it travel in 10�23 s?)

8. The strengths of the fundamental forces depend on the ener-

gies of the particles. In the case of the gravitational force, the

strength increases with the energy. Why would you expect this

to be true?

9. The boomerang analogy described in Fig. 41.11 is defective in

that the boomerang requires the presence of air. What would

be the motion of a boomerang in vacuum?

10. In Fig. 41.7, a large number of particles emerge in the longitu-

dinal direction (toward the right and the left). Why is this

expected, whereas the emergence of particles in the transverse

direction (upward and downward) is surprising? (Hint:

Consider a head-on collision between two rockets; which way

do you expect most fragments to spurt out?)

11. Neglecting dark energy, describe the difference between the

final states of the Universe for and

12. How will life ultimately end if the Universe continues to

expand forever? If it contracts?

r0 � 3H 20 �(8pG ).

r0 � 3H 20 �(8pG )

PROBLEMS

41.1 The Too ls o f H igh-Energy Phys ics41.2 The Mul t i tude of Par t i c les

1. According to Eq. (30.6), the radius of the orbit of a charged

particle of momentum p in a magnetic field is

Expressed in this way, in terms of the momentum, this for-

mula remains valid even if the particle is relativistic (although

for a relativistic particle).

(a) Show that for an ultra-relativistic particle the formula

becomes

(b) At a CERN accelerator, protons of energy 450 GeV travel

in a circular orbit of radius 1.1 km. Calculate the strength of

the magnetic field required to achieve this orbital radius.

2. The Large Hadron Collider (LHC) accelerator at CERN will

have a radius of 4.2 km and produce protons of momentum

7.0 TeV/c, or

(a) What magnetic field is required to hold the protons in a

circular orbit of this radius? (Hint: Use the formula given

in Problem 1.)

(b) What is the period of the orbital motion? (Hint: The

speed of the proton is nearly equal to the speed of light.)

3. The relativistic expression gives the

energy available for inelastic reactions when a particle of mass

m and kinetic energy K is incident on a stationary particle of

the same mass. Use this formula to calculate the available

energy (in GeV) for an antiproton incident on a stationary

proton in the following cases:

22mc 2(2mc 2 � K)

3.7 � 10�13 kg�m/s.

r �E

qcB

p � mv

r �p

qB

(a) The kinetic energy of the incident antiproton is 10 MeV.

(b) The kinetic energy of the incident antiproton is 1.0 TeV

(as in the Tevatron).

(c) The kinetic energy of the incident antiproton is 7.0 TeV

(as in the Large Hadron Collider).

4. Which is the most massive particle listed in the tables of

Section 41.2? Express the mass of this particle in atomic mass

units and compare the mass with that of the helium atom and

that of the lithium atom.

5. Count the number of particles (including antiparticles) that

are leptons.

*6. A particle detector is in a uniform magnetic field of 1.20 T.

For motion perpendicular to the field, what are the radii of

curvature for the following particles and kinetic energies:

(a) a 5.0-GeV proton; (b) a 1.0-MeV electron; and

(c) a 300-MeV positive pion.

*7. Show that for a particle of (total) energy E, the time-dilation

factor can be expressed as E�mc2. What is the time-dilation

factor for a muon of energy 950 MeV? The lifetime of this

muon is in its own reference frame. What is its

lifetime in the laboratory reference frame?

*8. Suppose we want to produce a particle by the head-on

collision of two protons of equal energies according to the

reaction

What is the minimum kinetic energy required for each proton

to initiate this reaction? The masses of the particles are given

in Tables 41.2 and 41.3.

p � p S p � K0 � ©�

©�

2.2 � 10�6 s



17. Is strangeness conserved in the following reactions?

18. Conservation of lepton number is actually three separate con-

servation laws; experiments indicate that electron lepton

number, muon lepton number, and tau lepton number are

each individually conserved. For example, the electron lepton

number is �1 for e� and �e, it is �1 for and e, and it is

zero for all other particles. For the following reactions, deter-

mine what missing particles are needed to conserve all lepton

numbers:

19. Show that the emission of a photon by a free electron is

impossible because it conflicts with energy conservation.

(Hint: Consider the emission process in the reference frame in

which the electron is initially at rest.)

20. Show that the annihilation of an electron and an antielectron

into a single photon is impossible, because it

conflicts with conservation of momentum. (Hint: Consider

the reaction from the reference frame in which the two elec-

trons have opposite velocities of equal magnitudes.)

21. Consider the following reactions produced by a beam of

particles in a bubble chamber filled with liquid hydrogen:

Verify that all of these reactions conserve baryon number and

strangeness.

22. The particle decays according to the reaction

Does this reaction conserve baryon number? Strangeness?

41.4 F ie lds and Quanta

23. The W particle can have either a positive charge or a

negative charge Figure 41.26 shows the Feynman dia-

gram for the decay of the neutron (n) via exchange of a the

end products are a proton (p), an electron (e), and an electron

antineutrino ( e). Can you guess the Feynman diagram for the

decay of the antineutron?

�

W�;

(W�).

(W�)

� S ¶ � p�

�

K� � p S ¶ � p� � p�

K� � p S ¶ � p0

K� � p S ©0 � p0

K� � p S ©� � p�

K�

(e� � e� S g)

t� S �m � ____ � ____

p S n � e� � ____

m� S e� � ____ � ____

�e�

K0 � n Sp� � p�

¶ S p � p�

p� � n S K� � ¶

*9. In 1908, a meteoroid struck near Tunguska, in Siberia. The

destructive effects of this impact have been estimated as

equivalent to the explosion of 12.5 megatons of TNT, or

One possible explanation for this explosion is

that the meteoroid was made of antimatter, which annihilated

with an equal amount of matter when it came in contact with

the Earth’s surface. If so, how much antimatter would account

for the explosion?

*10. The particle decays in two alternative ways:

and

Which of these reactions releases the most energy and gives

the decay products the largest kinetic energy?

*11. A particle at rest decays into a pion and a neutron:

What is the net kinetic energy of the decay products?

*12. Consider the annihilation of an electron and an antielectron,

resulting in two gamma rays,

Suppose that the electron and antielectron are initially at rest.

What are the energies of the resulting gamma rays? What are

their wavelengths?

*13. The meson decays into two gamma rays:

If the pion is initially at rest, what are the energies of the two

gamma rays? What are their wavelengths?

*14. A particle at rest decays into two pions:

What is the kinetic energy of each of these pions? What is the

momentum of each? (Hint: Use the relativistic relation

between energy and momentum.)

41.3 Interact ions and Conser vat ion Laws

15. Verify that the reaction (41.2) conserves baryon number. Does

the reaction conserve strangeness?

16. Which of the following reactions are forbidden by an absolute

conservation law?

e� � ne Sp� � p0

K� � n S ©� � p0

p� � n Sp� � p0 � p0

K� � p S K� � p � p0

p� � p S ¶ � K0

K0 Sp0 � p0

K0

p0 S g � g

p0

e� � e� S g � g

©� S p� � n

©�

� S 0 � p�

� S ¶ � K�

�

5.3 � 1016 J.


Problems 1427

*26. According to a speculative theory of the strong interactions,

the proton should be unstable and decay with a lifetime of

about 1033 years. Consider the protons in a mass of 1.0 � 106

kg of water. How many of these protons would decay in one

year?

*27. In the hot, early Universe, at a temperature of 1015 K, what

was the average kinetic energy of the random thermal motion

of gas particles? Compare this energy with the energy of about

100 GeV required to achieve symmetry between W particles

and photons.

41.5 Quarks

28. Table 41.2 lists the quark composition of baryons. For all the

and particles, verify that the listed quark composition

gives the correct values of electric charge, baryon number, and

strangeness.

29. The antiproton is made of three quarks. What kind of

quarks are these?

30. How many quarks are there in a hydrogen atom? In a water

molecule? (The oxygen nucleus contains eight protons and

eight neutrons.)

31. How many quarks are created in the reaction (41.1)?

32. A particle is made of one d quark and one u antiquark. What

is the electric charge of this particle? What is this particle?

33. Antiparticles contain antiquarks; for example, the proton

quark content is uud and the antiproton quark content is

. Are any of the baryons in Table 41.2 their own antipar-

ticle? Are any of the mesons in Table 41.3 their own antiparti-

cle? Which ones? What can you say about the quark content

of a particle that is its own antiparticle?

34. According to a theoretical prediction based on the quark

model, the masses of the nucleon

the the �, and the should be related by

Since this is intended as an approximate relation, the mass dif-

ferences between the positive, negative, and neutral kinds of

nucleon, or are to be neglected. Check this relation

against the experimental values of the masses.

*35. As described in Section 41.5, high-energy electrons produced

by the Stanford Linear Accelerator were used to probe the

internal structure of protons. To detect small lumps in the

proton, the wavelength of the electrons must be smaller than

the lumps. The electrons had an energy of 20 GeV (which is

an ultra relativistic energy). What is the de Broglie wave-

length, of these electrons?l � h�p,

©,

(mN � m) � 12(3m¶ � m©)

©,

(N � proton or neutron),

u u d

p

©

t

x

p

n

W–

e �e–

FIGURE 41.26 Feynman diagram for

the decay of a neutron.

24. The strong force can also be crudely interpreted as pion

exchange. Figure 41.27 shows the Feynman diagram for the

exchange of a between a proton and a neutron; note that

the proton changes into a neutron, and vice versa, so electric

charge is conserved at each vertex. Draw corresponding dia-

grams for the exchange of a and of a p0.p�

p�

t

x

n

p

��

p

n

FIGURE 41.27 Feynman diagram for

the exchange of a between a proton

and a neutron.

p�

25. A virtual particle may exist for a time provided that its

energy is less than the uncertainty in energy The two are

related by an uncertainty principle, To mediate

the strong nuclear force, how long may a virtual pion of rest-

mass energy 140 MeV exist? To mediate the weak nuclear

force, how long may a virtual Z0 of rest-mass energy 91 GeV

exist? How far does each particle travel in this time? (For sim-

plicity, use to estimate the range of the force.)v � c

¢E ¢t � U .

¢E.

¢t



41.6 Cosmology

36. Figure 41.28 shows the positions of four galaxies, with our

own Galaxy at the center. Draw a figure showing the positions

of these five galaxies at a later time, when the Universe is twice

as old.

Galaxy) in the reference frame of the galaxy P ; do this graphi-

cally, by subtracting the velocity vectors.

38. If the velocity of recession of the galaxies were not propor-

tional to distance but, rather, proportional to the

distance squared or proportional to some other

power of the distance, then our Galaxy would occupy a pre-

ferred, central spot in the Universe. Explain.

39. The value of the Hubble constant that Hubble had deduced

from the available data in 1936 was 1.0 � 108 (m/s)�(billion

light-years). The corresponding expansion time [see Eq.

(41.11)] is 1.9 � 109 years. How does this compare with the

age of the Earth and the age of globular clusters? With what

problem was Hubble faced?

(v � H�0r2)

(v � H0r)

44. Two of the baryons in Table 41.2 have identical quark content.

Which two are these? Consider the various conserved quanti-

ties discussed in Section 41.3. Which of these quantities do

you think might be different for the two baryons?

45. A particle is made of two d quarks and one s quark. What is

the electric charge of this particle? What is this particle?

46. The particle has quark content uuu and decays into two

particles in a reaction where only one quark–antiquark pair is

created. If one particle is a proton, what is the other particle in

the decay?

47. Sketch a Feynman diagram in which two masses are initially

moving apart along the x axis, then come to rest after

exchanging a graviton, and then approach after exchanging

another graviton.

48. The recessional velocity of a galaxy in the constellation Ursa

Major is and the distance to this galaxy is

approximately 1.0 � 109 light-years. Deduce a value of the

Hubble constant from these data alone. How does your value

compare with the average accepted value in Eq. (41.7)?

Obtain an estimate of the age of the Universe in years from

your Hubble constant.

1.5 � 107 m/s,

¢��

REVIEW PROBLEMS

*40. A particle at rest decays into two pions, one positive and one

negative. In a magnetic field of 1.25 T, each pion leaves a curved

track perpendicular to the field with a radius of 54.9 cm. What

is the mass of the particle that decayed? What is this particle?

*41. The neutron decays to a proton, an electron, and an electron anti-

neutrino. Assume that for one particular decay, the neutron is at

rest and the neutrino carries away negligible energy and momen-

tum. What is the net kinetic energy of the proton and electron?

*42. In Dan Brown’s novel Angels and Demons, 0.25 gram of anti-

matter is released about 3.0 km above ground level. Assuming

that the subsequent annihilation results in isotropic radiation,

what is the total energy incident on a square meter of ground

directly below the annihilation? If the annihilation occurs over

10 seconds, compare the average energy flux at ground level

during the annihilation with that of direct sunlight, approxi-

mately

43. In each of the following forbidden reactions, what conserva-

tion law is violated?

g � m� � n Sp0 � p

p � n S p � n

p � p S p � p�

1.0 kW/m2.

FIGURE 41.28 Four galaxies ( ) at

various distances from our Galaxy ( ).

37. Consider the galaxies shown in Fig. 41.29; the arrows are the

velocity vectors of those galaxies in the reference frame of the

Earth. According to the Galilean addition law for velocities,

find the velocity vectors of these galaxies (including our

P

FIGURE 41.29 Motion of a galaxy P and

other galaxies relative to our Galaxy ( ).



Answers to Checkups

Checkup 41.1

1. No to both. Charged particles are required to boil the liquid in

a bubble chamber to create a bubble track. Similarly, charged

particles are required to ionize the gas in a multiwire chamber

and produce the electrons that create current pulses.

2. The magnetic field in a particle detector diverts a charged par-

ticle into a circular orbit that reveals both the sign of the

charge and the momentum of the particle.

3. The positively charged particles (red tracks) are diverted to the

left as they travel downward. With the initial velocity down-

ward, the magnetic field B must be directed out of the page in

order to produce a magnetic force to the left.

4. In reaction (41.1), the masses of the products greatly exceed

the masses of the two initial protons, so the Q value is nega-

tive. We saw that reaction (41.2) proceeds spontaneously, so its

Q value must be positive.

5. (B) In the head-on collision between protons of opposite

motion, the full kinetic energy is available for reaction. For a

stationary target, the nonzero momentum of the projectile

must be conserved, requiring much of the final energy to

remain kinetic.

Checkup 41.2

1. All leptons have spin , but some baryons have other half-

integer spins; for example, the has spin

2. An antiparticle has the same mass and spin but the opposite

charge as the corresponding particle; thus, from Table 41.1,

the antitau has mass spin and electric charge

and the electron antineutrino has a mass less than 3 eV�c2,

spin and zero electric charge.

3. The heaviest particle listed is the upsilon particle, one of

the mesons of Table 41.3.

4. (D) Baryon; meson. All known leptons are listed in Table

41.1; there are no others. Other particles with half-integer

spin, such as the are baryons; particles with mass and with

integer spin, such as the B0, are mesons.

Checkup 41.3

1. A hadron is a particle that participates in the strong interac-

tion, that is, a baryon or a meson. Thus the proton and neu-

tron are hadrons, but the electron and the tau neutrino, both

leptons, are not. Similarly, the photon, the quantum of electro-

magnetism, is not a hadron.

2. An antiparticle has the same mass and spin as the particle, but

the opposite charge and strangeness. From the xi-zero data of

h0,

(�)

12,

�e,

12,1784 MeV�c2,

32.�

12

F � q v � B

Table 41.4, the anti-xi-zero then has mass 1315 MeV/c2,

spin zero charge, and strangeness similarly, the ant-xi-

minus, or xi-plus, has mass 1321 MeV/c2, spin charge �1, and

strangeness �2.

3. All interactions conserve energy, momentum, angular

momentum, charge, baryon number, and lepton number.

Strangeness and parity are conserved in electromagnetic (and

strong) interactions, but not in weak interactions. Isospin is

conserved in strong interactions, but not in electromagnetic

(nor in weak) interactions.

4. The parity of sin kx is �1, since the function becomes the

negative of itself when you reverse the x coordinate, sin

(�ky) � sin kx.

5. (B) Weak only. Only the weak interaction does not conserve

strangeness.

6. (D) All except strong; all. As a lepton, the electron does not

participate in strong interactions. The neutron participates in

all four—even though the neutron is net charge-neutral, it has

an internal charge distribution and a magnetic moment, and

participates in the electromagnetic interaction.

Checkup 41.4

1. Yes. Although heavier particles continue to be sought, the W

and Z are the heaviest particles now known.

2. (B) Weak, strong, electromagnetic. The range of a force is

inversely related to the mass of the particle that mediates the

force. The W and Z particles (weak force) have the largest

masses, photons (electromagnetic force) and gluons (strong

force) are massless, but the range of the strong force is limited

by pair production.

Checkup 41.5

1. Yes to both. Since the charge of the u quark is and the

charge of the d quark is the charge of is and

the charge of is zero.

2. Two of the spins are up, one is down, resulting in a net spin

of

3. The gluon is the particle that acts as carrier of the color force

and the strong force.

4. (E) The charge of an antiquark is the negative

of the charge of the corresponding quark.

�23e, �1

3e, �13e.

12.

2d � u

�e2u � d�13e,

�23e

12,

�2;12,

(0)



Checkup 41.6

1. Other galaxies are so distant that the angles required for trian-

gulation cannot be measured accurately. Even with the diame-

ter of the Earth’s orbit as a baseline, no difference in the angle

to a distant galaxy from either end of such a baseline can be

detected.

2. No. The Hubble “constant” reflects the present rate of expan-

sion, and its value would increase if the expansion accelerates

or decrease if the expansion slows; it would even change sign if

contraction ultimately occurs.

3. No. The edge of the observable universe is expanding at the

speed of light, and so a galaxy near the edge, moving slower

than the speed of light, becomes further from the edge, and

will not disappear.

4. (B) In supernovas soon after our Galaxy was born. These early

supernovas spewed out dust and gas with the heavy elements

that later coalesced to become our Solar System.


A-1

A alpha N nu

B beta � xi

gamma omicron

delta pi

epsilon P rho

� zeta sigma

eta T tau

theta upsilon

iota phi

kappa chi

� lambda psi

M mu � omega�m

c°l

x�k�

f£i�

yu™th

s©�

r��

pßd¢o g�

��

n�

Appendix 1: Greek Alphabet

Appendix 2: Mathematics Review

A 2.1 Symbols

means a equals b

means a is not equal to b

means a is greater than b

means a is less than b

means a is not less than b

means a is not greater than b

means a is proportional to b

means a is approximately equal to b

means a is much greater than b

means a is much less than b

e � 2.718 28 . . .

A 2.2 Powers and Roots

For any number a, the nth power of the number is the number multiplied by itself n times.

This is written as and n is called the exponent. Thus,

For instance,

A negative exponent indicates that the number is to be divided n times into 1; thus

a�1 �1

a a�2 �

1

a2 a�3 �

1

a3 etc.

32 � 3 � 3 � 9 33 � 3 � 3 � 3 � 27 34 � 3 � 3 � 3 � 3 � 81 etc.

a4 � a # a # a # a etc.a3 � a # a # aa2 � a # aa1 � a

an,

p � 3.141 59 . . .

a V b

a W b

a � b

a r b

a � b

a � b

a � b

a � b

a � b

a � b

GK023-Expand-AP[01-50].qxd 02/03/2007 6:39 PM Page 1 PMAC-291 PMAC-291:Books:GK-Jobs:GK023/Ohanion: TechBooks [PPG -QUARK]

A zero exponent yields 1, regardless of the value of a:

The rules for the combination of exponents in products, in ratios, and in powers

of powers are

For instance, it is easy to verify that

Note that for any two numbers a and b

For instance,

The nth root of a is a number such that its nth power equals a. The nth root is writ-

ten . The second root is usually called the square root, and designated by

As suggested by the notation roots are fractional powers, and they obey the usual

rules for the combination of exponents:

A 2.3 Ar i thmet i c in Sc ien t i f i c Nota t ion

The scientific notation for numbers (see the first page of the Prelude) is quite handy

for the multiplication and the division of very large or very small numbers, because

we can deal with the decimal parts and the power-of-10 parts in the numbers separately.

For example, to multiply by we multiply 4 by 5 and by

as follows:

To divide these numbers, we proceed likewise:

4 � 1010

5 � 1012�

4

5�

1010

1012� 0.8 � 1010�12 � 0.8 � 10�2 � 8 � 10�3

� 20 � 1010�12 � 20 � 1022 � 2 � 1023

(4 � 1010) � (5 � 1012) � (4 � 5) � (1010 � 1012)

1012,10105 � 1012,4 � 1010

(a1>n)m � am>n

(a1>n)n � an>n � a

a1>n,

a1>2 � 1a

1a:a1>2a1>n

(2 � 3)3 � 23 � 33

(a # b)n � an # bn

(32)3 � 32�3 � 36

32

33� 3�1 �

1

3

32 � 33 � 35

(an)m � anm

an

am � an�m

an # am � am�n

a0 � 1

A-2 APPENDIX 2 Mathematics Review


APPENDIX 2 Mathematics Review A-3

When performing additions or subtractions of numbers in scientific notation, we must

be careful to begin by expressing the numbers with the same power of 10. For exam-

ple, the sum of and is

A 2.4 Algebra

An equation is a mathematical statement that tells us that one quantity or a combination

of quantities is equal to another quantity or combination. We often have to solve for

one of the quantities in the equation in terms of the other quantities. For instance, we

may have to solve the equation

for x in terms of a and b. Here a and b are numerical constants or mathematical expres-

sions which are regarded as known, and x is regarded as unknown.

The rules of algebra instruct us how to manipulate equations and accomplish their

solution. The three most important rules are:

1. Any equation remains valid if equal terms are added or subtracted from its left

side and its right side.

This rule is useful for solving the equation We simply subtract a from both

sides of this equation and find

that is,

To see how this works in a concrete numerical example, consider the equation

Subtracting 7 from both sides, we obtain

or

Note that given an equation of the form we may want to solve for a

in terms of x and b, if x is already known from some other information but a is a math-

ematical quantity that is not yet known. If so, we must subtract x from both sides of

the equation, and we obtain

Most equations in physics contain several mathematical quantities which sometimes

play the role of known quantities, sometimes the role of unknown quantities, depend-

ing on circumstances. Correspondingly, we will sometimes want to solve the equation

for one quantity (such as x), sometimes for another (such as a).

2. Any equation remains valid if the left and the right sides are multiplied or divided

by the same factor.

This rule is useful for solving

ax � b

a � b � x

x � a � b,

x � �2

x � 5 � 7

x � 7 � 5

x � b � a

x � a � a � b � a

x � a � b.

x � a � b

1.5 � 109 � 3 � 108 � 1.5 � 109 � 0.3 � 109 � 1.8 � 109

3 � 1081.5 � 109



We simply divide both sides by a, which yields

or

Often it will be necessary to combine both of the above rules. For instance, to solve

the equation

we begin by subtracting 10 from both sides, obtaining

or

and then we divide both sides by 2, with the result

or

3. Any equation remains valid if both sides are raised to the same power.

This rule permits us to solve the equation

Raising both sides to the power we find

or

As a final example, let us consider the equation

(as established in Chapter 2, this equation describes the vertical position of a particle

that starts at a height and falls for a time t ; but the meaning of the equation need not

concern us here). Suppose that we want to solve for t in terms of the other quantities in

the equation. This will require the use of all our rules of algebra. First, subtract x from

both sides and then add to both sides. This leads to

and then to

Next, multiply both sides by 2 and divide both sides by g; this yields

t

2 �2

g(x0 � x)

12 gt

2 � x0 � x

0 � �12 gt

2 � x0 � x

12 gt

2

x0

x � �12 gt

2 � x0

x � b1>3

(x3)1>3 � b1>313,

x3 � b

x � 3

x �6

2

2x � 6

2x � 16 � 10

2x � 10 � 16

x �b

a

ax

a�

b

a


Finally, raise both sides to the power or, equivalently, extract the square root of both

sides. This gives us the final result

A 2.5 Equat ions wi th Two Unknowns

If we seek to solve for two unknowns simultaneously, then we need two independ-

ent equations containing these two unknowns. The solution of such simultaneous

equations can be carried out by the method of elimination: begin by using one equa-

tion to solve for the first unknown in terms of the second, then use this result to

eliminate the first unknown from the other equation. An example will help to make

this clear. Consider the following two simultaneous equations with two unknowns

x and y:

To solve the first equation for x in terms of y, subtract 2y from both sides and then

divide both sides by 4:

Next, substitute this expression for x into the second equation:

To simplify this equation, multiply both sides by 4:

and combine the two terms containing y:

This is an ordinary equation for the single unknown y, and it can be solved by the

methods we discussed in the preceding section, with the result

It then follows from the above expression for x that

A 2.6 The Quadrat i c Formula

The quadratic equation has two solutions:

A 2.7 Logar i thms and the Exponent ia l Func t ion

The base-10 logarithm of a (positive) number is the power to which 10 must be raised

to obtain this number. Thus, from and and and

we immediately deduce that10 000 � 1041000 � 103100 � 10210 � 101

x ��b ;2b2 � 4ac

2a

ax2 � bx � c � 0

x �8 � 2y

4�

8 � 2 � 3

4�

2

4�

1

2

y � 3

16 � 8y � �8

2 � (8 � 2y) � 4y � �8

2 �8 � 2y

4� y � �2

x �8 � 2y

4

2x � y � �2

4x � 2y � 8

t � B2

g(x0 � x)

12,

APPENDIX 2 Mathematics Review A-5



etc.

Likewise

etc.

Thus, the logarithm of a number between 1 and 10 is somewhere between 0 and 1,

but to find the logarithm of such a number, we need the help of a computer program

(many calculators have built-in computer programs that yield the value of the loga-

rithm at the touch of a button). For some calculations, it is convenient to remember that

log2 � 0.301 ≈ 0.3 and log5 � 0.699 ≈ 0.7.

The logarithm of the product of two numbers is the sum of the individual loga-

rithms, and the logarithm of the ratio of two numbers is the difference of the indi-

vidual logarithms. This rule makes it easy to find the logarithm of a number expressed

in scientific notation. For example, the logarithm of is

Note that the logarithm of any (positive) number smaller than 1 is negative. For exam-

ple,

The exponential function exp(x) is defined by the following infinite series:

This function is equivalent to raising the constant e � 2.718 28 … to the power x:

The natural logarithm ln x is the inverse of the exponential function, so

and

Natural logarithms obey the usual rules for logarithms,

ln(xa) � a ln x

ln a x

yb � ln x � ln y

ln(x # y) � ln x � ln y

x � ln(ex)

x � eln x

exp(x) � ex

exp(x) � 1 � x �x2

2�

x3

3�2�

x4

4�3�2� p

log (5 � 10�3) � log 5 � log 10�3 � 0.699 � 3 � �2.301

log (2 � 106) � log 2 � log 106 � 0.301 � 6 � 6.301

2 � 106

log 0.001 � �3,

log 0.01 � �2

log 0.1 � �1

log 1 � 0

log 10 000 � 4,

log 1000 � 3

log 100 � 2

log 10 � 1


APPENDIX 3 Geometry and Trigonometry Review A-7

Note that

and

If we designate the base-10 logarithm, or common logarithm, by log x, then the rela-

tionship between the two kinds of logarithm is as follows:

ln x � ln(10log x) � (log x)(ln 10) � 2.3026 log x

ln 10 � 2.3026

ln e � 1

Appendix 3: Geometry and TrigonometryReview

A3.1 Per imeters , Areas , and Volumes

[perimeter of a circle of radius r] � 2�r

[area of a circle of radius r] � �r2

[area of a triangle of base b, altitude h] � hb�2

[surface area of a sphere of radius r] � 4�r2

[volume of a sphere of radius r] � 4�r3�3

[area of curved surface of a cylinder of radius r, height h] � 2�rh[volume of a cylinder of radius r, height h] � �r2h

A3.2 Angles

The angle between two intersecting straight lines is defined as the fraction of a com-

plete circle included between these lines (Fig. A3.1). To express the angle in degrees,

we assign an angular magnitude of 360� to the complete circle; any arbitrary angle is

then an appropriate fraction of 360�. To express the angle in radians, we assign an

angular magnitude of radians to the complete circle; any arbitrary angle is then an

appropriate fraction of For example, the angle shown in Fig. A3.1 is of a com-

plete circle, that is, 30�, or radian. In view of the definition of angle, the length of

arc included between the two intersecting straight lines is proportional to the angle

� between these lines; if the angle is expressed in radians, then the constant of pro-

portionality is simply the radius:

(1)

Since it follows that

(2)

Each degree is divided into 60 minutes of arc (arcminutes), and each of these into 60

seconds of arc (arcseconds). In degrees, minutes of arc, and seconds of arc, the radian

is

(3)1 radian � 57� 17� 44.8–

1 radian �360�

2p�

360�

2 � 3.141 59� 57.2958�

radians � 360�,2p

s � r u

p>61122p.

2p

θ

r s

O

FIGURE A3.1 The angle � in this

diagram is � � 30�, or �/6 radian.


A3.3 The Tr igonometr i c Func t ions

The trigonometric functions of an angle are defined as ratios of the lengths of the

sides of a right triangle erected on this angle. Figure A3.2 shows an acute angle and

a right triangle, one of whose angles coincides with The adjacent side OQ has a

length x, the opposite side QP a length y, and the hypotenuse OP a length r. The

sine, cosine, tangent, cotangent, secant, and cosecant of the angle are then defined

as follows:

sine (4)

cosine (5)

tangent (6)

cotangent (7)

secant (8)

cosecant (9)

Find the sine, cosine, and tangent for angles of 0�, 90�, and

45�.

SOLUTION: For an angle of 0�, the opposite side is zero and the adja-

cent side coincides with the hypotenuse Hence

(10)

For an angle of 90�, the adjacent side is zero and the opposite side coin-

cides with the hypotenuse Hence

(11)

Finally, for an angle of 45� (Fig. A3.3), the adjacent and the opposite sides have the

same length and the hypotenuse has a length of times the length of

either side Hence

(12)

The definitions (4)–(9) are also valid for angles greater than 90�, such as the angle

shown in Fig. A3.4. In the general case, the quantities x and y must be interpreted as

the rectangular coordinates of the point P. For any angle larger than 90�, one or both

of the coordinates x and y are negative. Hence some of the trigonometric functions

will also be negative. For instance,

(13)

Figure A3.5 shows plots of the sine, cosine, and tangent vs. u.

sin 135� �1

12 cos 135� � �

1

12 tan 135� � �1

sin 45� �1

12 cos 45� �

1

12 tan 45� � 1

(r � 12x � 12y).

12(x � y)

sin 90� � 1 cos 90� � 0 tan 90� � q

(y � r).

(x � 0),

tan 0� � 0cos 0� � 1sin 0� � 0

(x � r).

(y � 0),

EXAMPLE 1

csc u � r>ysec u � r>x

cot u � x>ytan u � y>xcos u � x>r sin u � y>r

u

u.

u

A-8 APPENDIX 3 Geometry and Trigonometry Review

FIGURE A3.4 The angle � in

this diagram is larger than 90�.

yr

x

45°

FIGURE A3.3 A right

triangle with an angle of 45�.

xx

y

y

P

θ

O

θ

ry

P

QO x

FIGURE A3.2 A right triangle.


A3.4 Tr igonometr i c Ident i t ies

From the definitions (4)–(9) we immediately find the following identities:

(14)

(15)

(16)

(17)

Figure A3.6 shows a right triangle with angles and Since the adjacent

side for the angle is the opposite side for the angle and vice versa, we see

that the trigonometric functions also obey the following identities:

(18)sin (90� � u) � cos u

90� � uu

90� � u.u

csc u � 1>sin u

sec u � 1>cos u

cot u � 1>tan u

tan u � sin u>cos u

APPENDIX 3 Geometry and Trigonometry Review A-9

FIGURE A3.5 Plots of the sine, cosine, and tangent functions.

θsin

θ,

θ,

degrees

radians/22π 5π/23π/2π π

90°

1

–1

180° 270°

(a)

360° 450°

θcos

π

θ,

θ,

degrees

radians23 π/2π/2π

90°

1

0

–1

180° 270°

(b)

360°

90°

2

1

0

–1

–2

180° 270°

(c)

360°

θtan

π

θ,

θ,

degrees

radians2π/23π/2π0°

FIGURE A3.6 A right triangle with

angles � and 90� � �.

θ

θ90° –


(19)

(20)

According to the Pythagorean theorem, With and

this becomes or

(21)

The following are a few other trigonometric identities, which we state without proof:

(22)

(23)

(24)

(25)

(26)

(27)

A3.5 The Laws of Cos ines and S ines

In an arbitrary triangle the lengths of the sides and the angles obey the laws of cosines

and of sines. The law of cosines states that if the lengths of two sides are A and B and

the angle between them is � (Figure A3.7), then the length of the third side is given by

(28)

The law of sines states that the sines of the angles of the triangle are in the same

ratio as the lengths of the opposite sides (Figure A3.7):

(29)

Both of these laws are very useful in the calculation of unknown lengths or angles of

a triangle.

sin �

A�

sin �

B�

sin g

C

C

2 � A

2 � B

2 � 2AB cos g

cos(� � �) � cos � cos � � sin � sin �

sin(� � �) � sin � cos � � cos � sin �

cos 2u � 2 cos2 u � 1

sin 2u � 2 sin u cos u

csc2 u � 1 � cot2

u

sec2 u � 1 � tan2

u

cos2 u � sin2

u � 1

r 2 cos

2 u � r

2 sin 2 u � r

2,y � r sin u,

x � r cos ux2 � y2 � r 2.

tan (90� � u) � cot u � 1>tan u

cos (90� � u) � sin u

A-10 APPENDIX 4 Calculus Review

FIGURE A3.7 An arbitrary triangle.

B

C

A γ

β

α

Appendix 4: Calculus Review

A4.1 Der iva t ives

We saw in Section 2.3 that if the position of a particle is some function of time, say,

then the instantaneous velocity of the particle is the derivative of x with

respect to t:

(1)

This derivative is defined by first looking at a small increment that results from a

small increment and then evaluating the ratio in the limit when both

and tend toward zero. Thus

(2)dx

dt� lim

¢tS0

¢x

¢t

¢t

¢x¢x>¢t,¢t,

¢x

v �dx

dt

x � x(t),


Graphically, in a plot of position vs. time, the derivative is the slope of the straight

line tangent to the curved line at the time t (see Figure A4.1).

In general, if is some given function of a variable u, the derivative of f

with respect to u is defined by

(3)

In a plot of f vs. u, this derivative is the slope of the straight line tangent to the curve

representing f (u).

Starting with the definition (3) we can find the derivative of any function (provided

the function is sufficiently smooth so the derivative exists!). For example, consider the

function If we increase u to the function f (u) increases to

(4)

and therefore

(5)

The derivative is then

(6)

(7)

The second term on the right side vanishes in the limit the first term is simply

2u. Hence

(8)

or

(9)

This is one instance of the general rule for the differentiation of

(10)

This general rule is valid for any positive or negative number n, including zero.The proof

of this rule can be constructed by an argument similar to that above. Table A4.1 lists

the derivatives of the most common functions.

A4.2 Impor tant Ru les for D i f feren t ia t ion

1. Derivative of a constant times a function:

(11)d

du (cf ) � c

df

du

d

du (un) � nun�1

un:

d

du (u2) � 2u

df

du� 2u

¢u S 0;

� lim¢uS0

(2u) � lim¢uS0

(¢u)

df

du� lim

¢uS0

¢f

¢u� lim

¢uS0

2u ¢u � (¢u)2

¢u

df>du

� 2u ¢u � (¢u)2

¢f � (u � ¢u)2 � f � (u � ¢u)2 � u2

f � ¢f � (u � ¢u)2

u � ¢u,f (u) � u2.

df

du� lim

¢uS0

¢f

¢u

f � f (u)

dx�dt

APPENDIX 4 Calculus Review A-11

x (t)

tt

FIGURE A4.1 The derivative of

x (t) at t is the slope of the straight

line tangent to the curve at t.

SOME DERIVATIVES

(In all the following formulas, u is in radian:)

d

du tan�1

u �1

1 � u2

d

du cos�1 u � �1>21 � u2

d

du sin�1

u � 1>21 � u2

d

du csc u � �cot u csc u

d

du sec u � tan u sec u

d

du cot u � �csc

2 u

d

du tan u � sec2

u

d

du cos u � �sin u

d

du sin u � cos u

d

du eu � eu

d

du ln u �

1

u

d

du un � nun�1

TABLE A4.1


For instance,

2. Derivative of the sum of two functions:

(12)

For instance,

3. Derivative of the product of two functions:

(13)

For instance,

4. Chain rule for derivatives: If f is a function of g and g is a function of u, then

(14)

For instance, if and then

5. Partial derivatives: If f is a function of more than one variable, then the partial

derivative of f with respect to one of the variables, say x , is denoted 0f�0x, and is

obtained by treating all the other variables as constants when differentiating.

For instance, if f � x2y � y2z, then

A4.3 In tegra ls

We have learned that if the position of a particle is known as a function of time, then

we can find the instantaneous velocity by differentiation. What about the converse

problem: if the instantaneous velocity is known as a function of time, how can we find

the position? In Section 2.5 we learned how to deal with this problem in the special

case of motion with constant acceleration. The velocity is then a fairly simple func-

tion of time [see Eq. (2.17)]

(15)v � v0 � at

0f

0x� 2xy,

0f

0y� x2 � 2yz, and

0f

0z� y2

� cos(2u) � 2

d

du sin(2u) �

d sin(2u)

d(2u) d (2u)

du

f (g) � sin g,g � 2u

d

du f (g) �

df

dg dg

du

� sin u � 2u � u2 � cos u

d

du (u2 sin u) � sin u

d

du u2 � u2

d

du sin u

d

du ( f � g) � g

df

du� f

dg

du

d

du (6u2 � u) �

d

du (6u2) �

d

du (u) � 12u � 1

d

du ( f � g) �

df

du�

dg

du

d

du (6u2) � 6

d

du (u2) � 6 � 2u � 12u



and the position deduced from this velocity is [see Eq. (2.22)]

(16)

where and are the initial position and velocity at the initial time Now we

want to deal with the general case of a velocity that is an arbitrary function of time,

(17)

Figure A4.2 shows what a plot of v vs. t might look like. At the initial time the

particle has an initial position (for the sake of generality we now assume that

We want to find the position at some later time t. For this purpose, let us divide the

time interval into a large number of small time intervals, each of the duration

The total number of intervals is N, so The first of these intervals

lasts from to the second from to etc.

In Figure A4.3 the beginnings and the ends of these intervals have been marked

etc., with etc. If is sufficiently small, then during

the first time interval the velocity is approximately during the second, etc.This

amounts to replacing the smooth function by a series of steps (see Fig. A4.3).

Thus, during the first time interval, the displacement of the particle is approximately

during the second interval, etc. The net displacement of the parti-

cle during the entire interval is the sum of all these small displacements:

(18)

Using the standard mathematical notation for summation, we can write this as

(19)

We can give this sum the following graphical interpretation: since is the area

of the rectangle of height and width the sum is the net area of all the rectan-

gles shown in Figure A4.3, i.e., it is approximately the area under the velocity curve.

Note that if the velocity is negative, the area must be reckoned as negative!

Of course, Eq. (19) is only an approximation. To find the exact displacement of

the particle we must let the step size tend to zero (while the number of steps N

tends to infinity). In this limit, the steplike horizontal and vertical line segments in

Fig. A4.3 approach the smooth curve. Thus,

(20) x(t) � x0 �lim

¢tS0NSq

aN�1

i�0

v(ti) ¢t

¢t

¢t,v(ti)

¢tv(ti)

x(t) � x0 � aN�1

i�0

v(ti) ¢t

x(t) � x0 � v(t0) ¢t � v(t1) ¢t � v(t2) ¢t � � � �

t � t0

¢t ;v(t1)v(t0) ¢t ;

v(t)

v(t1);v(t0);

¢tt2 � t0 � 2¢t,t1 � t0 � ¢t,t2,t1,

t0,

t0 � 2¢t ;t0 � ¢tt0 � ¢t ;t0

t � t0 � N ¢t.¢t.

t � t0

t0 � 0).x0

t0,

v � v(t)

t0 � 0.v0x0

x � x0 � v0t � 12 at2


FIGURE A4.2 Plot of a function v(t).

t0 tt

v (t)

FIGURE A4.3 The interval has

been divided into N equal intervals of

duration so etc.t1 � t0 � ¢t,¢t,

t � t0

t1 t2 t3 tN – 2tN – 1

tt0

v (t1)

v (t2)

v (t0)

v (t)



FIGURE A4.4 The area under the velocity

curve.

t0 tt

v (t)

In the notation of calculus, the right side of Eq. (20) is usually written in the follow-

ing fashion:

(21)

The right side is called the integral of the function The subscript and the super-

script on the integration symbol are called, respectively, the lower and the upper

limit of integration; and is called the variable of integration (the prime on the vari-

able of integration merely serves to distinguish that variable from the limit of inte-

gration t). Graphically, the integral is the exact area under the velocity curve between

the limits and t in a plot of v vs. t (see Fig. A4.4). Areas below the t axis must be reck-

oned as negative.

In general, if f (u) is a function of u, then the integral of this function is defined

by a limiting procedure similar to that described above for the special case of the func-

tion v(t). The integral over an interval from to is

(22)

where As in the case of the integral of this integral can again be

interpreted as an area: it is the area under the curve between the limits a and b in a

plot of f vs. u.

For the explicit evaluation of integrals we can take advantage of the connection

between integrals and antiderivatives. An antiderivative of a function is simply

a function such that For example, if and then an

antiderivative of is The fundamental theorem of calculus

states that the integral of any function can be expressed in terms of antideriva-

tives:

(23)

In essence, this means that integration is the inverse of differentiation. We will not

prove this theorem here, but we remark that such an inverse relationship between inte-

gration and differentiation should not come as a surprise. We have already run across

an obvious instance of such a relationship: we know that velocity is the derivative of the

position, and we have seen above that the position is the integral of the velocity.

We will sometimes write Eq. (23) as

(24)

where the notation means that the function is to be evaluated at a and at

b, and these values are to be subtracted. For example, if

(25)

Table A4.2 lists some frequently used integrals. In this table, the limits of inte-

gration belonging with Eq. (24) have been omitted for the sake of brevity.

�b

a

un du �un�1

n � 1` ba

�bn�1

n � 1�

an�1

n � 1

n � �1,

F (u)F (u)�ba

�b

a

f (u) du � F (u) ` ba

�b

a

f (u) du � F (b) � F (a)

f (u)

F(u) � un�1>(n � 1).f (u)

n � �1,f (u) � undF>du � f.F(u)

f (u)

v(t),ui � a � i ¢u.

�b

a

f (u) du �lim

¢uS0NSq

aN�1

i�0

f (ui) ¢u

u � bu � a

t0

t�

t��

v(t).

x(t) � x0 � �t

t0

v(t�) dt�


A4.4 Impor tant Ru les for In tegrat ion

1. Integral of a constant times a function:

(26)

For instance,

(27)

2. Integral of a sum of two functions:

(28)�b

a

[ f (u) � g(u)] du � �b

a

f (u) du � �b

a

g(u) du

�b

a

5u2 du � 5�b

a

u2 du � 5 a b3

3�

a3

3b

�b

a

cf (u) du � c�b

a

f (u) du


SOME INTEGRALS

for

for

(where ku is in radians)

(where ku is in radians)

� du

(u2 � k2)3/2�

u

k22u2 � k2

� du

u2k2 ; u2� �

1

k ln a k � 2k2 ; u2

ub

� du

k2 � u2�

1

k tan�1

a u

kb

�2k2 � u2 du �1

2 c u2k2 � u2 � k2 sin

�1 a u

kb d

� du

2u2 ; k2� ln a u � 2u2 ; k2 b

� du

2k2 � u2� sin�1

a u

kb

� du

1 � ku�

1

k ln(1 � ku)

�cos(ku) du �1

k sin(ku)

�sin(ku) du � �1

k cos(ku)

� ln u du � u ln u � u

�eku du �eku

k

u � 0�1

udu � ln u

n � �1�undu �un�1

n � 1

TABLE A4.2


For instance,

(29)

3. Change of limits of integration:

(30)

(31)

4. Change of variable of integration: If u is a function of v, then

(32)

For instance, with

(33)

Finally, let us apply these general results to some specific examples of integration

of the velocity.

A particle with constant acceleration has the following veloc-

ity as a function of time [compare Eq. (15)]:

where is the velocity at

By integration, find the position as a function of time.

SOLUTION: According to Eq. (21), with

Using rule 2 and rule 1, we find that this equals

(34)

The first entry listed in Table A4.2 gives � (for and �(for Thus,

(35)

This, of course, agrees with Eq. (16).

� v0t � 12 at2

x(t) � x0 � v0 t� `0

t

� 12 a t�2 `

0

t

n � 1).

t�dt� � t�2>2n � 0)dt� � t�

x(t) � x0 � �t

0

v0 dt� � �t

0

at� dt� � v0�t

0

dt� � a�t

0

t� dt�

x(t) � x0 � �t

0

v(t�) dt� � �t

0

(v0 � at�) dt�

t0 � 0,

t � 0.v0

v(t) � v0 � at

EXAMPLE 1

�b

a

u3 du � �

b

a

v6 du � �2b

2a

v6(2v) dv

u � v2,

�b

a

f (u) du � �v(b)

v(a)

f (u)

du

dv dv

�b

a

f (u) du � ��a

b

f (u) du

�b

a

f (u) du � �c

a

f (u) du � �b

c

f (u) du

�b

a

(5u2 � u) du � �b

a

5u2 du � �b

a

u du � 5 a b3

3�

a3

3b � a b2

2�

a2

2b



The instantaneous velocity of a projectile traveling through air

is the following function of time:

where is measured in meters per second and t is measured in seconds. Assuming

that at what is the position as a function of time? What is the posi-

tion at

SOLUTION: With and Eq. (21) becomes

When evaluated at this yields

The acceleration of a mass pushed back and forth by an elastic

spring is

(36)

where B and are constants. Find the position as a function of time. Assume

and at

SOLUTION: The calculation involves two steps: first we must integrate the accel-

eration to find the velocity, then we must integrate the velocity to find the position.

For the first step we use an equation analogous to Eq. (21),

(37)

This equation becomes obvious if we remember that the relationship between

acceleration and velocity is analogous to that between velocity and position. With

and we obtain from Eq. (29)

(38)

Next,

(39) � �B

v2 cos vt �

B

v2

x(t) � �t

0

v(t�) dt� � �t

0

B

v sin vt�dt� �

B

v a�

1

v cos vt� b ` t

0

�B

v sin vt

v(t) � �t

0

B cos vt� dt� � B

1

v sin vt� `

0

t

t0 � 0,v0 � 0

v(t) � v0 � �t

t0

a(t�) dt�

t � 0.x � 0v � 0

�

a(t) � B cos vt

EXAMPLE 3

� 1722 m

x(3.0) � 655.9 � 3.0 � 61.14 � (3.0)2>2 � 3.26 � (3.0)3>3t � 3.0 s,

� 655.9t � 61.14t2>2 � 3.26t3>3 � 655.9(t�) `

0

t

� 61.14(t�2>2) `0

t

� 3.26(t�3>3) `0

t

� 655.9�t

0

dt� � 61.14�t

0

t�dt� � 3.26�t

0

t�2 dt�

x(t) � �t

0

(655.9 � 61.14t� � 3.26t�2) dt�

t0 � 0,x0 � 0

t � 3.0 s?

t � 0,x � 0

v(t)

v(t) � 655.9 � 61.14t � 3.26t2

EXAMPLE 2



A4.5 The Tay lor Ser ies

Suppose that is a smooth function of u in some neighborhood of a given point

so the function has continuous derivatives of all orders. Then the value of the

function at an arbitrary point near a can be expressed in terms of the following infi-

nite series, where all the derivatives are evaluated at the point a:

(40)

This is called the Taylor series for the function about the point a. The series

converges, and is valid, provided u is sufficiently close to a. How close is “sufficiently

close” depends on the function f and on the point a. Some functions, such as sinu,

, and are extremely well behaved, and their Taylor series converge for any choice

of u and of a. The Taylor series gives us a convenient method for the approximate eval-

uation of a function.

Find the Taylor series for sin u about the point

SOLUTION: The derivatives of evaluated at are

etc.

Hence Eq. (32) gives

Note that for very small values of u, we can neglect all higher powers of u, so

which is an approximation often used in this book.

A4.6 Some Approximat ions

By constructing Taylor series, we can obtain the following useful approximations, all

of which are valid for small values of u. It is often sufficient to keep just the first one

or two terms on the right side.

(41)21 � u � 1 �1

2 u �

1

8 u2 �

1

16 u3 � ��

sin u � u,

� u �1

6u3 � � � �

�1

4�3�2� 0 � (u � 0)4 � � � �

sin u � 0 � 1 � (u � 0) �1

2� 0 � (u � 0)2 �

1

3�2� (�1) � (u � 0)3

d 4

du4 sin u �

d

du (�cos u) � sin u � 0,

d 3

du3 sin u �

d

du (�sin u) � �cos u � �1

d2

du2 sin u �

d

du cos u � �sin u � 0

d

du sin u � cos u � 1

u � 0sin u

u � 0.EXAMPLE 4

eu,cos u

f (u)

f (u) � f (a) �df

du (u � a) �

1

2 d2f

du2 (u � a)2 �

1

3�2 d 3f

du3 (u � a)3 � � � �

u � a,

f (u)



(42)

(43)

(44)

(45)

(46)

In all the following formulas, u is in radians:

(47)

(48)

(49)

(50)

(51)tan�1 u � u �

1

3 u3 �

1

5 u5 � ��

sin�1 u � u �

1

6 u3 �

3

40 u5 � ��

tan u � u �1

3 u3 �

2

15 u5 � ��

cos u � 1 �1

2 u2 �

1

24 u4 �

1

720 u6 � ��

sin u � u � 1

6 u3 �

1

120 u5 � ��

ln(1 � u) � u �1

2 u2 �

1

3 u3 � ��

eu � 1 � u �1

2 u2 �

1

2�3 u3 � ��

1

(1 � u)n � 1 � nu �n(n � 1)

2 u2 �

n(n � 1)(n � 2)

2�3 u3 � ��

1

21 � u� 1 �

1

2 u �

3

8 u2 �

5

16 u3 � ��

1

1 � u� 1 � u � u2 � u3 � ��

APPENDIX 5 Propagating Uncertainties A-19

Appendix 5: Propagating Uncertainties

Experimentalists carefully work to measure physical quantities and to determine the uncer-

tainty in each quantity. We must often calculate a new result from a measured quantity or

from several quantities; we must therefore understand the propagation of uncertainties

through functions and formulas.

To keep things simple, we will make the assumption that the uncertainties in each

quantity are symmetrically distributed about its measured value and that the various meas-

ured quantities are independent of each other. This is not always true. But by ignoring

correlations and assuming symmetry, we can reduce all the necessary propagation of uncer-

tainties to some simple formulas.

Suppose we have a measured quantity and its uncertainty, x ± �x, where �x is a posi-

tive quantity and has the same units as x, and is also known as the absolute uncertainty in

x. What, then, is the uncertainty of some function, f (x), of this data? Under the assump-

tion that the uncertainty is small, we can obtain the uncertainty from the first terms of the

Taylor series expansion of f : f (x � �x) � f (x) � (df (x)�dx)�x� … From this we find the

uncertainty �f � | f (x � �x) � f (x)| in the function value f (x) is

(1)

with the derivative evaluated at the point x. We can generalize this result to functions

of several variables as follows: given the data x ± �x, y ± �y, . . ., the function

f (x, y, . . .) has the associated uncertainty

¢f � ` df

dx ¢x `


(2)

where all the partial derivatives (see App. 4.2) are evaluated at the point x, y, . . .. If

we recall that we defined absolute uncertainties to be positive, we can write this as

(3)

From this relationship, we can derive several simple results for uncertainty propagation.

Addition and Subtraction.

Given f (x, y) � 3x � y � z � 5, find �f:

Thus in addition or subtraction, the uncertainties add, and in multiplication by

a constant, the uncertainty is multiplied by the same constant.

Multiplication, Division, and Exponentiation.

Given f (x, y) � x2y�(5z), find � f :

Equivalently, for multiplication and division, we add relative uncertainties (e.g.,

�x/x), and for exponentiation, we multiply the relative uncertainty by the mag-

nitude of the exponent, to get the relative uncertainty of the product, quotient,

or power.

Numerical Application to Ohm's Law, V � I R.

Given V � 1.5 � 0.1 Volt and I � 0.50 � 0.02 A, find R and �R:

Rearranging we find R � V�I � (1.5 Volt)�(0.50 A) � 3.0 �, and

EXAMPLE 3

� � 2xy >(5z)� ¢x � �x2>(5z)� ¢y � ��x2y>(5z2)� ¢z

¢f � ` 0f

0x ` ¢x � ` 0f

0y ` ¢y � ` 0f

0z ` ¢z

EXAMPLE 2

� 3¢x � ¢y � ¢z

� �3� ¢x � �1� ¢y � ��1� ¢z

¢f � ` 0f

0x ` ¢x � ` 0f

0y ` ¢y � ` 0f

0z ` ¢z

EXAMPLE 1

¢f � ` 0f

0x ` ¢x � ` 0f

0y ` ¢y � ��

¢f � ` 0f

0x ¢x ` � ` 0f

0y ¢y ` � ��

A-20 APPENDIX 5 Propagating Uncertainties


� 0.2 � � 0.12 � � 0.4 �

Note in the last step that unlike an ordinary calculation, we have rounded this

final result up; uncertainties should always be rounded up, never down.

� ` 1

0.50 A ` (0.1 Volt) � ` �1.5 Volt

(0.50 A)2 ` (0.02A)

� ` 1I

` ¢V � ` �V

I 2

` ¢I

¢R � ` 0R

0V ` ¢V � ` 0R

0I ` ¢I

APPENDIX 6 The International System of Units (SI) A-21

A6.1 Base Uni t s

The SI system of units is the modern version of the metric system. The SI system rec-

ognizes seven fundamental, or base, units for length, mass, time, electric current, ther-

modynamic temperature, amount of substance, and luminous intensity.b The following

definitions of the base units were adopted by the Conférence Générale des Poids et

Mesures in the years indicated:

meter (m) “The metre is the length of the path travelled by light in vacuum during

a time interval of 1/299 792 458 of a second.” (Adopted in 1983.)

kilogram (kg) “The kilogram is . . . the mass of the international prototype of the

kilogram.” (Adopted in 1889 and in 1901.)

second (s) “The second is the duration of 9 192 631 770 periods of the radiation cor-

responding to the transition between the two hyperfine levels of the ground state of the

cesium-133 atom.” (Adopted in 1967.)

ampere (A) “The ampere is that constant current which, if maintained in two straight

parallel conductors of infinite length, of negligible circular cross section, and placed

one meter apart in vacuum, would produce between these conductors a force equal to

newton per meter of length.” (Adopted in 1948.)

kelvin (K) “The kelvin . . . is the fraction 1/273.16 of the thermodynamic tempera-

ture of the triple point of water.” (Adopted in 1967.)

2 � 10�7

Appendix 6: The International System of Units (SI)

b At least two of the seven base units of the SI system are redundant. The mole is merely a certain number

of atoms or molecules, in the same sense that a dozen is a number; there is no need to designate this number

as a unit. The candela is equivalent to watt per steradian; it serves no purpose that is not served equally

well by watt per steradian. Two other base units could be made redundant by adopting new definitions of

the unit of temperature and of the unit of electric charge. Temperature could be measured in energy units

because, according to the equipartition theorem, temperature is proportional to the energy per degree of

freedom. Hence the kelvin could be defined as a derived unit, with joule per

degree of freedom. Electric charge could also be defined as a derived unit, to be measured with a suitable

combination of the units of force and distance, as is done in the cgs system.

Furthermore, the definitions of the supplementary units—radian and steradian—are gratuitous.These

definitions properly belong in the province of mathematics and there is no need to include them in a system

of physical units.

1 K � 12 � 1.38 � 10�23

1683


mole “The mole is the amount of substance of a system which contains as many ele-

mentary entities as there are atoms in 0.012 kilogram of carbon-12.” (Adopted in 1967.)

candela (cd) “The candela is the luminous intensity, in a given direction, of a source

that emits monochromatic radiation of frequency and that has a

radiant intensity in that direction of watt per steradian.” (Adopted in 1979.)

Besides these seven base units, the SI system also recognizes two supplementary

units of angle and solid angle:

radian (rad) “The radian is the plane angle between two radii of a circle which cut off

on the circumference an arc equal in length to the radius.”

steradian (sr) “The steradian is the solid angle which, having its vertex in the center

of a sphere, cuts off an area equal to that of a [flat] square with sides of length equal

to the radius of the sphere.”

1683

540 � 1012 Hz

A-22 APPENDIX 6 The International System of Units (SI)

NAMES OF DERIVED UNITS

QUANTITY DERIVED UNIT NAME SYMBOL

frequency 1/s hertz Hz

force newton N

pressure pascal Pa

energy joule J

power J/s watt W

electric charge coulomb C

electric potential J/C volt V

electric capacitance C/V farad F

electric resistance V/A ohm

conductance A/V siemen S

magnetic flux weber Wb

magnetic field tesla T

inductance henry H

temperature K degree Celsius

luminous flux lumen lm

illuminance lux lx

radioactivity 1/s becquerel Bq

absorbed dose J/kg gray Gy

dose equivalent J/kg sievert Sv

cd�sr>m2

cd�sr

�C

V�s>AV�s>m2

V�s

�

A�s

N�m

N�m2

kg�m�s2

TABLE A6.1 PREFIXES FOR UNITS

FACTOR PREFIX SYMBOL

yotta Y

zetta Z

exa E

peta P

tera T

giga G

mega M

kilo k

hecto h

10 deka da

deci d

centi c

milli m

micro

nano n

pico p

femto f

atto a

zepto z

yocto y10�24

10�21

10�18

10�15

10�12

10�9

m10�6

10�3

10�2

10�1

102

103

106

109

1012

1015

1018

10 21

10 24

TABLE A6.2


A6.2 Der ived Uni t s

The derived units are formed out of products and ratios of the base units. Table A6.1

lists those derived units that have been glorified with special names. (Other derived units

are listed in the tables of conversion factors in Appendix 8.)

A6.3 Pref ixes

Multiples and submultiples of SI units are indicated by prefixes, such as the familiar

kilo, centi, and milli used in kilometer, centimeter, and millimeter, etc. Table A6.2 lists all

the accepted prefixes. Some enjoy more popularity than others; it is best to avoid the use

of uncommon prefixes, such as atto and exa, since hardly anybody will recognize those.

APPENDIX 7 Best Values of Fundamental Constants A-23

Appendix 7: Best Values of FundamentalConstants

The values in the following table are the “2002 CODATA Recommended Values” by

P. J. Mohr and B. N. Taylor Listed at the website physics.nist.gov/constants of the

National Institute of Standards and Technology. The digits in parentheses are the

one–standard deviation uncertainty in the last digits of the given value.

RELATIVE UNCERTAINTYQUANTITY SYMBOL VALUE UNITS (PARTS PER MILLION)

UNIVERSAL CONSTANTS

speed of light in vacuum c 299 792 458 m.s�1 (exact)

magnetic constant 0 4� � 10�7 N·A�2

� 12.566 370 614 ... � 10�7 N·A�2 (exact)

electric constant 1�0c2 �0 8.854 187 817 ... � 10�12 F·m�1 (exact)

gravitational constant G 6.6742(10) � 10�11 m3·kg�1·s�2 1.5 � 10�4

Planck constant h 6.626 0693(11) � 10�34 J·s 1.7 � 10�7

in eV.s 4.135 667 43(35) � 10�15 eV·s 8.5 � 10�8

h>2� 1.054 571 68(18) � 10�34 J·s 1.7 � 10�7

in eV.s 6.582 119 15(56) � 10�16 eV·s 8.5 � 10�8

ELECTROMAGNETIC CONSTANTS

elementary charge e 1.602 176 53(14) � 10�19 C 8.5 � 10�8

magnetic flux quantum h�2e 0 2.067 833 72(18) � 10�15 Wb 8.5 � 10�8

quantum 2e 2�h 7.748 091 733(26) � 10�5 S 3.3 � 10�9

Josephson constant 2e�h 483 597.879(41) � 109 Hz·V�1 8.5 � 10�8

Bohr magneton �2me B 927.400 949(80) � 10�26 J·T�1 8.6 � 10�8

in eV.T�1 5.788 381 804(39) � 10�5 eV·T�1 6.7 � 10�9

nuclear magneton �2mp N 5.050 783 43(43) � 10�27 J·T�1 8.6 � 10�8

in eV.T�1 3.152 451 259(21) � 10�8 eV·T�1 6.7 � 10�9e U

e U

U

BEST VALUES OF FUNDAMENTAL CONSTANTSTABLE A7.1

(continued )



ATOMIC AND NUCLEAR CONSTANTS

General

fine-structure constant � 7.297 352 568(24) � 10�3 3.3 � 10�9

inverse fine-structure constant ��1 137.035 999 11(46) 3.3 � 10�9

Rydberg constant �2mec/2h 10 973 731.568 525(73) m�1 6.6 � 10�12

Bohr radius a0 0.529 177 2108(18) � 10�10 m 3.3 � 10�9

Electron

electron mass me 9.109 3826(16) � 10�31 kg 1.7 � 10�7

in u 5.485 799 0945(24) � 10�4 u 4.4 � 10�10

energy equivalent in Me V mec 2 0.510 998 918(44) MeV 8.6 � 10�8

electron-proton mass ratio me�mp 5.446 170 2173(25) � 10�4 4.6 � 10�10

electron charge to mass quotient �e�me �1.758 820 12(15) � 1011 C·kg�1 8.6 � 10�8

Compton wavelength h>mec lC 2.426 310 238(16) � 10�12 m 6.7 � 10�9

classical electron radius �2a0 re 2.817 940 325(28) � 10�15 m 1.0 � 10�8

Thomson cross section �e 0.665 245 837(13) � 10�28 m2 2.0 � 10�8

electron magnetic moment e �928.476 412(80) � 10�26 J·T�1 8.6 � 10�8

to Bohr magneton ratio e�B �1.001 159 652 1859(38) 3.8 � 10�12

to nuclear magneton ratio e�N �1838.281 971 07(85) 4.6 � 10�10

electron magnetic moment

anomaly |e|�B � 1 ae 1.159 652 1859(38) � 10�3 3.2 � 10�9

electron g-factor �2(1 � ae) ge �2.002 319 304 3718(75) 3.8 � 10�12

Muon

muon mass m 1.883 531 40(33) � 10�28 kg 1.7 � 10�7

in u

energy equivalent in MeV mc2 0.113 428 9264(30) u 2.6 � 10�8

105.658 3692(94) MeV 8.9 � 10�8

muon-electron mass ratio m�me 206.768 2838(54) 2.6 � 10�8

muon Compton wavelength h/mc �C, 11.734 441 05(30) � 10�15 m 2.5 � 10�8

muon magnetic moment �4.490 447 99(40) � 10�26 J·T�1 8.9 � 10�8

to Bohr magneton ratio �B �4.841 970 45(13) � 10�3 2.6 � 10�8

muon magnetic moment anomaly

||�(e /2m) � 1 a 1.165 919 81(62) � 10�3 5.3 � 10�7

muon g-factor �2(1 � a) g �2.002 331 8396(12) 6.2 � 10�10

Proton

proton mass mp 1.672 621 71(29) � 10�27 kg 1.7 � 10�7

in u 1.007 276 466 88(13) u 1.3 � 10�10

energy equivalent in MeV mpc2 938.272 029(80) MeV 8.6 � 10�8

proton-electron mass ratio mp�me 1836.152 672 61(85) 4.6 � 10�10

proton-neutron mass ratio mp�mn 0.998 623 478 72(58) 5.8 � 10�10

proton charge to mass quotient e�mp 9.578 833 76(82) � 107 C·kg�1 8.6 � 10�8

proton Compton wavelength �C, p 1.321 409 8555(88) � 10�15 m 6.7 � 10�9

h/mpc

proton magnetic moment p 1.410 606 71(12) � 10�26 J·T�1 8.7 � 10�8

to Bohr magneton ratio p�B 1.521 032 206(15) � 10�3 1.0 � 10�8

to nuclear magneton ratio p�N 2.792 847 351(28) 1.0 � 10�8

Neutron

neutron mass mn 1.674 927 28(29) � 10�27 kg 1.7 � 10�7

in u 1.008 664 915 60(55) u 5.5 � 10�10

energy equivalent in MeV mnc2 939.565 360(81) MeV 8.6 � 10�8

neutron-electron mass ratio mn�me 1838.683 6598(13) 7.0 � 10�10

neutron-proton mass ratio mn�mp 1.001 378 418 70(58) 5.8 � 10�10

U

(8p�3)r 2e

4p�0 U2>mee

2

Rq

e2�4p�0Uc

A-24 APPENDIX 7 Best Values of Fundamental Constants

(continued )


APPENDIX 7 Best Values of Fundamental Constants A-25


neutron Compton wavelength h>mnc �C, n 1.319 590 9067(88) � 10�15 m 6.7 � 10�9

neutron magnetic moment n �0.966 236 45(24) � 10�26 J·T�1 2.5 � 10�7

to Bohr magneton ratio n�B �1.041 875 63(25) � 10�3 2.4 � 10�7

to nuclear magneton ratio n�N �1.913 042 73(45) 2.4 � 10�7

Deuteron

deuteron mass md 3.343 583 35(57) � 10�27 kg 1.7 � 10�7

in u 2.013 553 212 70(35) u 1.7 � 10�10

energy equivalent in MeV mdc 2 1875.612 82(16) MeV 8.6 � 10�8

deuteron-electron mass ratio md�me 3670.482 9652(18) 4.8 � 10�10

deuteron-proton mass ratio md�mp 1.999 007 500 82(41) 2.0 � 10�10

deuteron magnetic moment d 0.433 073 482(38) � 10�26 J·T�1 8.7 � 10�8

to Bohr magneton ratio d�B 0.466 975 4567(50) � 10�3 1.1 � 10�8

to nuclear magneton ratio d�N 0.857 438 2329(92) 1.1 � 10�8

Alpha Particle

alpha particle mass m� 6.644 6565(11) � 10�27 kg 1.7 � 10�7

in u 4.001 506 179 149(56) u 1.4 � 10�11

energy equivalent in MeV m�c 2 3727.379 17(32) MeV 8.6 � 10�8

alpha particle to electron mass ratio m��me 7294.299 5363(32) 4.4 � 10�10

alpha particle to proton mass ratio m��mp 3.972 599 689 07(52) 1.3 � 10�10

PHYSICO-CHEMICAL CONSTANTS

Avogadro constant NA 6.022 1415(10) � 1023 mole�1 1.7 � 10�7

atomic mass constant

� 1 u mu 1.660 538 86(28) � 10�27 kg 1.7 � 10�7

energy equivalent in MeV muc2 931.494 043(80) MeV 8.6 � 10�8

Faraday constant NAe F 96 485.3383(83) C·mole�1 8.6 � 10�8

molar gas constant R 8.314 472 (15) J ·mole�1·K�1 1.7 � 10�6

Boltzmann constant R�NA k 1.380 6505(24) � 10�23 J·K�1 1.8 � 10�6

in eV.K–1 8.617 343(15) � 10�5 eV·K�1 1.8 � 10�6

molar volume of ideal gas RT>pT � 273.15 K, p � 101.325 kPa Vm 22.413 996(39) � 10�3 m3.mole–1 1.7 � 10�6

Loschmidt constant NA>Vm n0 2.686 7773(47) � 1025 m�3 1.8 � 10�6

Stefan-Boltzmann constant

(�2�60)k4� � 5.670 400(40) � 10�8 W·m�2·K�4 7.0 � 10�6

Wien displacement law constant

b � �maxT b 2.897 7685(51) � 10�3 m·K 1.7 � 10�6

U3c 2

mu � 112 m(12C)


Appendix 8: Conversion Factors

The units for each quantity are listed alphabetically, except that the SI unit is always

listed first. The numbers are based on “American National Standard; Metric Practice”

published by the Institute of Electrical and Electronics Engineers, 1982.

Angle

Length

T ime

1 hour (h) � 3600 s � 124 day � 60 min � 1.141 � 10�4 yr

2.738 � 10�3 yr

1 day � 8.640 � 104 s � 24 h � 1440 min � 1.003 sidereal days �

1.161 � 10�5 sidereal day � 3.169 � 10�8 yr

1 second (s) � 1.157 � 10�5 day � 13600 h � 1

60 min �

1 yard (yd) � 0.9144 m � 91.44 cm � 3 ft � 36 in. � 11760 mi

3.086 � 1013 km � 3.262 light-years

1 parsec (pc) � 3.086 � 1016 m � 2.063 � 105 AU � 3.086 � 1018 cm �

1.609 km � 0.8690 nmi � 1760 yd

1 statute mile (mi) � 1.609 � 103 m � 1.609 � 105 cm � 5280 ft �

� 1.852 km � 1.151 mi

1 nautical mile (nmi) � 1.852 � 103 m � 1.852 � 105 cm � 6.076 � 103 ft

� 3.281 � 10�6 ft � 3.937 � 10�5 in.

1 micron, or micrometer (mm) � 1 � 10�6 m � 1 � 104 A� � 1 � 10�4 cm

9.461 � 1012 km � 5.879 � 1012 mi � 0.3066 pc

1 light-year � 9.461 � 1015 m � 6.324 � 104 AU � 9.461 � 1017 cm �

� 0.6214 mi � 1.094 � 103 yd

1 kilometer (km) � 1 � 103 m � 1 � 105 cm � 3.281 � 103 ft � 0.5400 nmi

1 inch (in.) � 2.540 � 10�2 m � 2.54 cm � 112 ft � 2.54 � 104 mm � 1

36 yd

1.894 � 10�4 mi � 13 yd

1 foot (ft) � 0.3048 m � 30.48 cm � 12 in. � 3.048 � 105 mm �

1 fermi, or femtometer (fm) � 1 � 10�15 m � 1 � 10�13 cm � 1 � 105 A�

� 0.3937 in. � 1 � 10�5 km � 1.057 � 10�18 light-year � 1 � 104 mm

1 centimeter (cm) � 0.01 m � 1 � 108 A�

� 1 � 1013 fm � 3.281 � 10�2 ft

1.496 � 108 km � 1.581 � 10�5 light-year � 4.848 � 10�6 pc

1 astronomical unit (AU) � 1.496 � 1011 m � 1.496 � 1013 cm �

1 � 10�5 fm � 3.281 � 10�10 ft � 1 � 10�4 mm

1 angstrom (A� ) � 1 � 10�10 m � 1 � 10�8 cm � 1 � 10 nm �

3.241 � 10�17 pc � 1.094 yd

light-year � 1 � 106 mm � 5.400 � 10�4 nmi � 6.214 � 10�4 mi �

1 � 1015 fm � 3.281 ft � 39.37 in. � 1 � 10�3 km � 1.057 � 10�16

1 meter (m) � 1 � 10�9 nm � 1 � 1010 A�

� 6.685 � 10�12 AU � 100 cm �

1 second of arc (–) � 4.848 � 10�6 radian � 13600� � 1

60 � � 7.716 � 10�7 rev

1 revolution (rev) � 2p radians � 360� � 2.160 � 104� � 1.296 � 106 ��

1 minute of arc (�) � 2.909 � 10�4 radian � 160� � 4.630 � 10�5 rev � 60–

1 degree (�) � 1.745 � 10�2 radian � 60� � 3600– � 1360 rev

1 radian � 57.30� � 3.438 � 103 � � (1/2p) rev � 2.063 � 105 ��

A-26 APPENDIX 8 Conversion Factors


Mass

Area

Volume

1728 in.3 � 28.32 liters � 127 yd3

1 cubic foot (ft3) � 2.832 � 10�2 m3 � 2.832 � 104 cm3 � 7.481 gal �

2.642 � 10�4 gal � 6.102 � 10�2 in.3 � 1 � 10�3 liter

1 cubic centimeter (cm3) � 1 � 10�6 m3 � 3.531 � 10�5 ft3 �

6.102 � 104 in.3 � 1 � 103 liters � 1.308 yd31 cubic meter (m3) � 1 � 106 cm3 � 35.31 ft3 � 264.2 gal �

1 square yard (yd2) � 0.8361 m2 � 8.361 � 103 cm2 � 9 ft2 � 1296 in.22.788 � 107 ft2 � 2.590 km2

1 square statute mile (mi2) � 2.590 � 106 m2 � 2.590 � 1010 cm2 �

� 1.076 � 107 ft2 � 0.3861 mi21 square kilometer (km2) � 1 � 106 m2 � 1 � 1010 cm21 square inch (in.2) � 6.452 � 10�4 m2 � 6.452 cm2 � 1

144 ft2

3.587 � 10�8 mi2 � 19 yd2

1 square foot (ft2) � 9.290 � 10�2 m2 � 929.0 cm2 � 144 in.2 �

� 1 � 10�10 km2 � 3.861 � 10�11 mi21 square centimeter (cm2) � 1 � 10�4 m2 � 1.076 � 10�3 ft2 � 0.1550 in.21 barn � 1 � 10�28 m2 � 1 � 10�24 cm2

1 � 10�6 km2 � 3.861 � 10�7 mi2 � 1.196 yd21 square meter (m2) � 1 � 104 cm2 � 10.76 ft2 � 1.550 � 103 in.2 �

1 slug � 14.59 kg � 1.459 � 104 g � 32.17 lb

1 short ton � 907.2 kg � 9.072 � 105 g � 0.9072 t � 2000 lb

12000 short ton � 3.108 � 10�2 slug

1 pound (lb)c � 0.4536 kg � 453.6 g � 4.536 � 10�4 t � 16 oz �

1 ounce (oz) � 2.835 � 10�2 kg � 141.7 carats � 437.5 grains � 28.35 g � 116 lb

1.102 short tons � 68.52 slugs

1 metric ton, or tonne (t) � 1 � 103 kg � 1 � 106 g � 2.205 � 103 lb �

� 6.852 � 10�5 slug

1 � 10�6 t � 3.527 � 10�2 oz � 2.205 � 10�3 lb � 1.102 � 10�6 short ton

1 gram (g) � 1 � 10�3 kg � 6.024 � 1023 u � 5 carats � 15.43 grains �

1 grain � 6.480 � 10�5 kg � 6.480 � 10�2 g � 2.286 � 10�3 oz � 17000 lb

1 carat � 2 � 10�4 kg � 0.2 g � 7.055 � 10�3 oz � 4.409 � 10�4 lb

1 atomic mass unit (u) � 1.6605 � 10�27 kg � 1.6605 � 10�24 g

ton � 6.852 � 10�2 slug

1000 g � 1 � 10�3 t � 35.27 oz � 2.205 lb � 1.102 � 10�3 short

1 kilogram (kg) � 6.024 � 1026 u � 5000 carats � 1.543 � 104 grains �

5.259 � 105 min � 366.24 sidereal days

1 year (yr) � 3.156 � 107 s � 365.24 days � 8.766 � 103 h �

� 2.730 � 10�3 yr

1 sidereal day � 8.616 � 104 s � 0.9973 day � 23.93 h � 1.436 � 103 min

1 minute (min) � 60 s � 6.944 � 10�4 day � 160 h � 1.901 � 10�6 yr

APPENDIX 8 Conversion Factors A-27

c This is the “avoirdupois” pound. The “troy” or “apothecary” pound is 0.3732 kg, or 0.8229 lb avoirdupois.


Dens i ty1 kilogram per cubic meter

1 gram per cubic centimeter

1 lb per cubic foot

1 pound-per gallon

1 short ton per cubic yard

1 slug per cubic foot

Speed1 meter per second

1 centimeter per second

1 foot per second

1 kilometer per hour

1 knot, or nautical mile per

1 mile per hour

Acce lera t ion1 meter per second squared

1 centimeter per second squared

1 foot per second squared

Force1 newton

1 pound-force

1 short ton-force � 8.896 � 103 N � 8.896 � 108 dynes � 2000 lb-f

(lb-f ) � 4.448 N � 4.448 � 105 dynes � 12000 � short ton-force

1.124 � 10�9 short ton-force1 dyne � 1 � 10�5 N � 2.248 � 10�6 lb-f �

1.124 � 10�4 short ton-force(N) � 1 � 105 dynes � 0.2248 lb-f �

1 g � 9.807 m>s2 � 980.7 cm>s2 � 32.17 ft>s2(ft>s2) � 0.3048 m>s2 � 30.48 cm>s2 � 3.108 � 10�2 g

3.281 � 10�2 ft>s2 � 1.020 � 10�3 g

(cm>s2) � 0.01 m>s2 �

(m>s2) � 100 cm>s2 � 3.281 ft>s2 � 0.1020 g

1.609 km�h � 0.8690 knot

(mi>h) � 0.4470 m�s � 44.70 cm�s � 1.467 ft�s �

1.688 ft>s � 1.852 km>h � 1.151 mi>hhour � 0.5144 m�s � 51.44 cm�s �

� 0.5400 knot � 0.6214 mi>h(km>h) � 0.2778 m>s � 27.78 cm>s � 0.9113 ft>s

0.5925 knot � 0.6818 mi>h(ft>s) � 0.3048 m>s � 30.48 cm>s � 1.097 km>h �

3.600 � 10�2 km>h � 1.944 � 10�2 knot � 2.237 � 10�2 mi>h(cm>s) � 0.01 m>s � 3.281 � 10�2 ft>s �

1.944 knots � 2.237 mi>h(m>s) � 100 cm>s � 3.281 ft>s � 3.600 km>h �

32.17 lb>ft3 � 4.301 lb>gal

(slug>ft3) � 515.4 kg>m3 � 0.5154 g>cm3 �

(short ton>yd3) � 1.187 � 103 kg>m3 � 74.07 lb>ft3(1 lb>gal) � 119.8 kg>m3 � 7.481 lb>ft3 � 0.2325 slug>ft3

0.1337 lb>gal � 1.350 � 10�2 short ton>yd3 � 3.108 � 10�2 slug>ft3(lb>ft3) � 16.02 kg>m3 � 1.602 � 10�2 g>cm3 �

8.345 lb>gal � 3.613 � 10�2 lb>in.3 � 0.8428 short ton>yd3 � 1.940 slugs>ft3(g>cm3) � 1 � 103 kg>m3 � 62.43 lb>ft3 �

8.428 � 10�4 short ton>yd3 � 1.940 � 10�3 slug>ft36.243 � 10�2 lb>ft3 � 8.345 � 10�3 lb>gal � 3.613 � 10�5 lb>in.3 �

(kg>m3) � 1 � 10�3 g>cm3 �

1 cubic yard (yd3) � 0.7646 m3 � 7.646 � 105 cm3 � 27 ft3 � 202.0 gal

1 liter (l) � 1 � 10�3 m3 � 1000 cm3 � 3.531 � 10�2 ft3 � 0.2642 gal

1 cubic inch (in.3) � 1.639 � 10�5 m3 � 16.39 cm3 � 5.787 � 10�4 ft31 gallon (gal)d � 3.785 � 10�3 m3 � 0.1337 ft3


d This is the U.S. gallon; the U.K. and the Canadian gallon are 4.546 � 10�3 m3, or 1.201 U.S. gallons.


Energy

1 British thermal unit

1 calorie

1 electron-volt

1 foot-pound-force

1 kilocalorie (kcal), or large calorie

1 kilowatt-hour

Power1 watt

1 British thermal unit per hour

1 calorie per second

1 erg per second

1 foot-pound-force per second

1 horsepower

1 kilowatt

Pressure1 newton per square meter

1 atmosphere

1 dyne per square centimeter

1.450 � 10�5 lb-f>in.29.869 � 10�7 atm � 7.501 � 10�4 mm-Hg � 2.089 � 10�3 lb-f>ft2 �

(dyne>cm2) � 0.1 N>m2 �

1 bar � 1 � 105 N>m2 � 0.9869 atm � 750.1 mm-Hg

� 14.70 lb-f>in.21.013 � 106 dynes>cm2 � 29.92 in.-Hg � 2.116 � 103 lb-f>ft2

(atm) � 1.013 � 105 N>m2 � 760.0 mm-Hg �

� 2.089 � 10�2 lb-f>ft2 � 1.450 � 10�4 lb-f>in.2 � 7.501 � 10�3 torr

1 � 10�5 bar � 7.501 � 10�3 mm-Hg � 10 dynes>cm2 � 2.953 � 10�4 in.-Hg

(N>m2), or pascal (Pa) � 9.869 � 10�6 atm �

737.6 ft� lb-f>s � 1.341 hp

(kW) � 1 � 103 W � 3.412 � 103 Btu>h � 238.8 cal>s �

� 550 ft� lb-f>s(hp) g � 745.7 W � 2.544 � 103 Btu>h � 178.1 cal>s

4.626 Btu>h � 1.356 � 107 ergs>s � 1.818 � 10�3 hp

(ft� lb-f>s) � 1.356 W � 0.3238 cal>s �

7.376 � 10�8 ft� lb-f>s � 1.341 � 10�10 hp

(erg>s) � 1 � 10�7 W � 2.388 � 10�8 cal>s �

4.187 � 107 ergs>s � 3.088 ft�lb-f>s � 5.615 � 10�3 hp

(cal>s) � 4.187 W � 14.29 Btu>h �

7.000 � 10�2 cal>s � 0.2162 ft� lb-f>s � 3.930 � 10�4 hp

(Btu>h) � 0.2931 W �

0.7376 ft� lb-f>s � 1.341 � 10�3 hp

(W) � 3.412 Btu>h � 0.2388 cal>s � 1 � 107 ergs>s �

3.6 � 1013 ergs � 2.655 � 106 ft� lb-f

(kW� h) � 3.600 � 106 J � 3412 Btu � 8.598 � 105 cal �

(Cal) � 4.187 � 103 J � 1 � 103 cal

1.356 � 107 ergs � 8.464 � 1018 eV � 3.766 � 10�7 kW� h

(ft� lb-f ) � 1.356 J � 1.285 � 10�3 Btu � 0.3239 cal �

1.182 � 10�19 ft� lb-f

(eV) � 1.602 � 10�19 J � 1.602 � 10�12 erg �

6.242 � 1011 eV � 7.376 � 10�8 ft� lb-f � 2.778 � 10�14 kW�h

1 erg � 1 � 10�7 J � 9.478 � 10�7 Btu � 2.388 � 10�8 cal �

3.088 ft� lb-f � 1 � 10�3 kcal � 1.163 � 10�6 kW� h

(cal) f � 4.187 J � 3.968 � 10�3 Btu � 4.187 � 107 ergs �

1.055 � 1010 ergs � 778.2 ft� lb-f � 2.931 � 10�4 kW� h

(Btu)e � 1.055 � 103 J � 252.0 cal �

6.242 � 1018 eV � 0.7376 ft� lb-f � 2.778 � 10�7 kW� h

1 joule (J) � 9.478 � 10�4 Btu � 0.2388 cal � 1 � 107 ergs �

APPENDIX 8 Conversion Factors A-29

e This is the “International Table” Btu; there are several other Btus.f This is the “International Table” calorie, which equals exactly 4.1868 J. There are several other calories;

for instance, the thermochemical calorie, which equals 4.184 J.g There are several other horsepowers; for instance, the metric horsepower, which equals 735.5 W.


1 inch of mercury

1 pound-force per square inch

1 torr, or millimeter of mercury

1.333 � 10�3 bar � 1.333 � 103 dynes/cm2 � 0.03937 in.-Hg � 0.01934 lb-f/in.2

E lec t r i c Charge h

1 coulomb (C) statcoulombs, or esu of charge 0.1 abcoulomb,

or emu of charge

E lec t r i c Curren t1 ampere (A) statamperes, or esu of current 0.1 abampere, or

emu of current

E lec t r i c Po ten t ia l1 volt (V) statvolt, or esu of potential abvolts, or emu

of potential

E lec t r i c F ie ld1 volt per meter statvolt/cm abvolts/cm

Magnet i c F ie ld1 tesla (T), or weber per square meter gauss

E lec t r i c Res i s tance1 ohm (Ω) statohm, or esu of resistance abohms, or

emu of resistance

E lec t r i c Res i s t iv i ty1 ohm-meter statohm-cm abohm-cm

Capac i tance

1 farad (F) statfarads, or esu of capacitance abfarad,

or emu of capacitance

Induc tance

1 henry (H) stathenry, or esu of inductance abhen-

rys, or emu of inductance

3 1 � 1093 1.113 � 10�12

3 1 � 10�93 8.988 � 1011

3 1 � 1011(��m) 3 1.113 � 10�10

3 1 � 1093 1.113 � 10�12

(Wb>m2) � 1 � 104

3 1 � 106(V>m) 3 3.336 � 10�5

3 1 � 1083 3.336 � 10�3

33 2.998 � 109

33 2.998 � 109

(mm-Hg) � 1.333 � 102 N>m2 � 1>760 atm �

7.031 � 10�2 kp>cm26.805 � 10�2 atm � 6.895 � 104 dynes>cm2 � 2.036 in.-Hg �

(lb-f>in.2, or psi) � 6.895 � 103 N>m2 �

25.40 mm-Hg � 0.4912 lb-f>in.2(in.-Hg) � 3.386 � 103 N>m2 � 3.342 � 10�2 atm �


h The dimensions of the electric quantities in SI units, electrostatic units (esu), and electromagnetic units

(emu) are usually different; hence the relationships among most of these units are correspondences

rather than equalities (�).

( 3 )


Appendix 9: The Periodic Table and ChemicalElements

APPENDIX 9 The Periodic Table and the Chemical Elements A-31

Na Mg Al Si P S Cl Ar

Li Be B

H He

C N O F Ne

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

IA(1)

IIA(2)

IIIA(13)

VIIIA(18)

IVA(14)

VA(15)

VIA(16)

VIIA(17)

IIIB(3)

IVB(4)

VB(5)

VIB(6)

VIIB(7)

(8)

(9)

(10)

VIIIBIB

(11)IIB(12)

Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sb Te I Xe

Cs Ba *La Hf Ta W Re Os Ir Pt Au Hg

Ds Uuu Uub

Tl Pb

Sn

Uuq

Bi Po At Rn

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

Fr Ra †Ac Rf Db Sg Bh Hs Mt

11 12 13 14 15 16 17 18

3 4 5

1 2

6 7 8 9 10

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

58 59 60 61 62 63 64 65 66 67 68 69 70

90 91 92 93 94 95 96 97 98 99 100 101 102

87

3

2

1

Per

iod

s

Atomic number

Group designation

Atomic mass

Symbol for element

*Lanthanides

†Actinides

4

5

6

788 89 104 105 106 107 108 109 110 111 112 114

Lu

Lr

71

103

22.98977 24.3050 26.98154 28.0855 30.97376 32.065 35.453 39.948

6.941 9.012182 10.811

1.00794 4.002602

12.0107 14.0067 15.9994 18.99840 20.1797

39.0983 40.078 44.955910 47.867 50.9415 51.9961 54.938049 55.845 58.93320 58.6934 63.546 65.409 69.723 72.64 74.92160 78.96 79.904 83.798

85.4678 87.62 88.90585 91.224 92.90638 95.94 98.9072 101.07 102.90550 106.42 107.8682 112.411 114.818 118.710 121.760 127.60 126.90447 131.293

132.90545 137.327 138.9055 178.49 180.9479 183.84 186.207 190.23 192.217 195.078 196.96654 200.59 204.3833 207.2

(289)

208.98037 208.9824 209.9871 222.0176

140.116 140.90765 144.24 144.9127 150.36 151.964 157.25 158.92534 162.50 164.93032 167.26 168.93421 173.04

232.0381 231.0359 238.0289 237.0482 244.0642 243.0614 247.07003 247.0703 251.0796 252.083 257.0951 258.0984 259.1011

174.967

262.110

223.0197 226.0277 227.0277 261.1089 262.1144 263.118 262.12 265.1306 (268) (271) (272) (285)

THE PERIODIC TABLETABLE A9.1


A-32 APPENDIX 9 The Periodic Table and the Chemical Elements

Data were obtained from the National Institute for Standards and Technology; values are for the elements as they exist naturally on Earth or for the most stable isotope, with

carbon-12 (the reference standard) having a mass of exactly 12 u. The estimated uncertainties in values between ± and ± 9 units in the last digit of an atomic mass are in

parentheses after the atomic mass.

(Source: http://physics.nist.gov/PhysRefData/Compositions/index.html)

ATOMIC ATOMICELEMENT SYMBOL NUMBER ATOMIC MASS (u) ELEMENT SYMBOL NUMBER ATOMIC MASS (u)

ATOMIC MASSES AND ATOMIC NUMBERS OF CHEMICAL ELEMENTS

Actinium Ac 89 227.027 7

Aluminum Al 13 26.981 538 (2)

Americium Am 95 243.061 4

Antimony Sb 51 121.760 (1)

Argon Ar 18 39.948 (1)

Arsenic As 33 74.921 60 (2)

Astatine At 85 209.987 1

Barium Ba 56 137.327 (7)

Berkelium Bk 97 247.070 3

Beryllium Be 4 9.012 182 (3)

Bismuth Bi 83 208.980 38 (2)

Bohrium Bh 107 264.12

Boron B 5 10.811 (7)

Bromine Br 35 79.904 (1)

Cadmium Cd 48 112.411 (8)

Calcium Ca 20 40.078 (4)

Californium Cf 98 251.079 6

Carbon C 6 12.010 7 (8)

Cerium Ce 58 140.116 (1)

Cesium Cs 55 132.905 45 (2)

Chlorine Cl 17 35.453 (9)

Chromium Cr 24 51.996 1 (6)

Cobalt Co 27 58.933 200 (9)

Copper Cu 29 63.546 (3)

Curium Cm 96 247.070 3

Darmstadtium Ds 110 271

Dubnium Db 105 262.114 4

Dysprosium Dy 66 162.500 (1)

Einsteinium Es 99 252.083

Erbium Er 68 167.259 (3)

Europium Eu 63 151.964 (1)

Fermium Fm 100 257.095 1

Fluorine F 9 18.998 403 2 (5)

Francium Fr 87 223.019 7

Gadolinium Gd 64 157.25 (3)

Gallium Ga 31 69.723 (1)

Germanium Ge 32 72.64 (1)

Gold Au 79 196.966 55 (2)

Hafnium Hf 72 178.49 (2)

Hassium Hs 108 265.130 6

Helium He 2 4.002 602 (2)

Holmium Ho 67 164.930 32 (2)

Hydrogen H 1 1.007 94 (7)

Indium In 49 114.818 (3)

Iodine I 53 126.904 47 (3)

Iridium Ir 77 192.217 (3)

Iron Fe 26 55.845 (2)

Krypton Kr 36 83.798 (2)

Lanthanum La 57 138.905 5 (2)

Lawrencium Lr 103 262.110

Lead Pb 82 207.2 (1)

Lithium Li 3 6.941 (2)

Lutetium Lu 71 174.967 (1)

Magnesium Mg 12 24.305 0 (6)

Manganese Mn 25 54.938 049 (9)

Meitnerium Mt 109 268

Mendelevium Md 101 258.098 4

Mercury Hg 80 200.59 (2)

Molybdenum Mo 42 95.94 (1)

Neodymium Nd 60 144.24 (3)

Neon Ne 10 20.179 7 (6)

Neptunium Np 93 237.048 2

Nickel Ni 28 58.693 4 (2)

Niobium Nb 41 92.906 38 (2)

Nitrogen N 7 14.006 7 (2)

Nobelium No 102 259.101 1

Osmium Os 76 190.23 (3)

Oxygen O 8 15.999 4 (3)

Palladium Pd 46 106.42 (1)

Phosphorus P 15 30.973 761 (2)

Platinum Pt 78 195.078 (2)

Plutonium Pu 94 244.064 2

Polonium Po 84 208.982 4

Potassium K 19 39.098 3 (1)

Praseodymium Pr 59 140.907 65 (2)

Promethium Pm 61 144.912 7

Protactinium Pa 91 231.035 88 (2)

Radium Ra 88 226.025 4

Radon Rn 86 222.017 6

Rhenium Re 75 186.207 (1)

Rhodium Rh 45 102.905 50 (2)

Rubidium Rb 37 85.467 8 (3)

Ruthenium Ru 44 101.07 (2)

Rutherfordium Rf 104 261.108 9

Samarium Sm 62 150.36 (3)

Scandium Sc 21 44.955 910 (8)

Seaborgium Sg 106 263.118 6

Selenium Se 34 78.96 (3)

Silicon Si 14 28.085 5 (3)

Silver Ag 47 107.868 2 (2)

Sodium Na 11 22.989 770 (2)

Strontium Sr 38 87.62 (1)

Sulfur S 16 32.065 (6)

Tantalum Ta 73 180.947 9 (1)

Technetium Tc 43 98.907 2

Tellurium Te 52 127.60 (3)

Terbium Tb 65 158.925 34 (2)

Thallium Tl 81 204.383 3 (2)

Thorium Th 90 232.038 1 (1)

Thulium Tm 69 168.934 21 (2)

Tin Sn 50 118.710 (7)

Titanium Ti 22 47.867 (1)

Tungsten W 74 183.84 (1)

Ununbium Uub 112 285

Unununium Uuu 111 272

Ununquadium Uuq 114 289

Uranium U 92 238.028 9 (1)

Vanadium V 23 50.941 5 (1)

Xenon Xe 54 131.293 (2)

Ytterbium Yb 70 173.04 (3)

Yttrium Y 39 88.905 85 (2)

Zinc Zn 30 65.409 (4)

Zirconium Zr 40 91.224 (2)

TABLE A9.2


APPENDIX 10 Formula Sheets A-33

Chapters 1–21

U � 12kx

2

Fx � �dU

dx

U(x) � ��x

x0

Fx

(x�) dx�

E � K � U � �constant

U � mgy

K � 12mv2

W � �F � d s

W � F � s

W � Fx ¢x

F � �kx

fs � ms N

fk � mk N

w � mg

m a � Fnet

v� � v � VO

a � v2>r�A � B� � AB sin f

� Ax Bx � Ay By � Az Bz

A � B � AB cos f

A � 2A2x � A2

y � A2z

Ax � A cos u

a(x � x0) � 12(v

2 � v20)

x � x0 � v0t � 12at

2

a � dv>dt � d 2x>dt 2

v � dx>dt

.

L � r � p

L � Iv

P � tv

I� � t

t � FR sin u

I � ICM � Md 2

25MR2 (sphere); 112ML2 (rod)

ICM � MR2 (hoop); 12MR2 (disk);

I � �R 2 dm

K � 12I�2

v � R�

� � dv>dt � d 2f>dt2

� � df>dt

v�1 �m1 � m2

m1 � m2

v1; v�2 �2m1

m1 � m2

v1

I � �¢t

0

F dt

rCM �1

M �r dm

p � m v

U � �GMm>rg � GME

>R2E

v2 � GMS >r

F � GMm>r2

P � F � v

P � dW>dt

E � mc 2

¢S � �B

A

dQ>T

e � 1 � T2>T1

¢E � Q � W

pV g � [constant]; g � Cp>CV

TV g�1 � [constant];

vrms � 23kT>mTC � T � 273.15

pV � NkT

12rv

2 � rg y � p � �constant

p � p0 � �rgy

sin u � v>VE

f � � f >(1 < VE >v)

f � � f (1 ; VR>v)

f beat � f 1 � f 2

v � 2F>(M>L)

l � 2p>k; f � v>l; v � 2pf

y � A cos k(x ; vt) � A cos(kx ; vt)

v � 2mgd>Iv � 2g>l ; T � 2p2l>g� � 2k>mm d

2x>dt 2 � �kx

T � 2p>v; f � 1>T � v>2px � A cos(vt � d)

d L

dt� r � F

Appendix 10: Formula Sheets

RE � 6.37 � 106 m

ME � 5.98 � 1024 kg

G � 6.67 � 10�11 N.m2>kg2

g � 9.81 m>s2

c � 3.00 � 108 m>smp � 1.67 � 10�27 kg

me � 9.11 � 10�31 kg

1 cal � 4.19 J

k � 1.38 � 10�23 J>KNA � 6.02 � 1023>mole


;

r �p

qB

B � m0nI

HB � d s � HB‘ ds � m0I

d B �m0

4p I ds � r

r 3

F � qv � B

F �m0

2p qvI

r

P � I ¢VP � IE

R � rl>AI � ¢V>Ru � 1

2k�0E2

¢Y � ��PP0

F � ds

HkE� d A �Qfree, inside

�0

E � Efree>kC � �0A>dC � Q>¢V

u � 12�0E2

U � 12Q1V1 � 1

2Q2V2 � 12Q3V3 � �

�

�

Ez � �0V

0zEy � �

0V

0y ,Ex � �

0V

0x ,

V �1

4p�0

q�

r

HE � d A � HE� d A �Qinside

�0

!E � �E � dA

U � �p � E

t � p � E

p � lQ

E � s�2�0

E �1

4p�0

q�

r2

F �1

4p�0

qq�

r2

1

s�

1

s��

1

f

f � ; 12R

n1 sin u1 � n2 sin u2

v � c/n

c � 1�2m0�0

02E

0x2� m0�0

02E

0t2

[pressure] � S�c

S �1

m0

E � B

B � E>cHB � ds �HB ds � m0I � m0�0

d£E

dt

HB�dA � HB1dA � 0

E2 � E1 N2

N1

Z � BR2 � avL �1

vCb 2

�0 � 1>2LC

u �1

2m0

B2

U � 12 LI2

E � �LdI

dt

£B � LI

HE � ds � HE‘ ds � �d£B

dt

£B � �B � dA

E � �d£B

dt

E � vBl

U � �� B

� � � � B

m � I � �area of loop

d F � I d l � B Interference minima:

Interference maxima:

Diffraction minima:

n � n0e�t�t; t � t1�2�0.693

R � (1.2 � 10�15 m) � A1�3

E �J ( J � 1) U2

2I

mspin �e U

2me

l � h>pEn � �

mee4

2(4p�0)2 U2

1

n2� �

13.6 eV

n2

L � n U

¢y ¢py � h�4p

p � hf >cE � hf

E2 � p2c2 � m2c4

p �mv

21 � v2>c2; E �

mc2

21 � v2>c2

v�x �vx � V

1 � vxV>c2

L � 21 � V 2>c

2 L�

¢t �¢t�

21 � V 2>c

2

t� �t � Vx>c

2

21 � V 2>c

2

y� � y

x� �x � Vt

21 � V 2>c

2

f � � B1<v>c

1 ; v>c f

a sin u � 1.22l

a sin u � l, 2l, 3l, . . .

d sin u � 0, l, 2l, . . .

d sin u � 12l,

32l,

52l, ��

Chapters 22–41

�0 � 8.85 � 10�12 F�m

e � 1.60 � 10�19 C

h � 2p U � 6.63 � 10�34 J�s

c � 3.00 � 108 m/s

m0 � 1.26 � 10�6 H�mmp � 1.67 � 10�27 kg

me � 9.11 � 10�31 kg


ANSWERS A-35

Appendix 11: Answers to Odd-Numbered Problems and Review Problems

Chapter 1

1. 5.87 ft; 1.78 m (Assuming a height of 5 ft 10 in)

3. 48.7 m

5. 66 picas long and 51 picas wide

7. 12.7 mm; 6.35 mm; 3.18 mm; 1.59 mm; 0.794 mm; 0.397 mm

9. a) 1 mm; 3 � 106 m (Assuming grapefruit diameter � 0.1

m); b) 7 mm; 0.5 km (Assuming head diameter � 0.2 m)

11. 1 mm

13. 4.41 m; 6.94 m

15. 6.3 � 106 m

17. 1.4 � 1017 s

19. 7761 s

21. 23 h 56 min

23. 12 days

25. 3.7 � 107 beats/year

27. 0.25 min of arc; 0.463 km

29. 0.134 % in planets; 99.9% in sun

31. 0.021 % electrons; 99.98 % nucleus

33. 373.24 g

35. a) 8.4 � 1024 molecules; b) 4.3 � 1046 molecules;

c) 1680 molecules

37. 28.95 g/mol

39. 6.9 � 108 m

41. 2.1 � 1022 m

43. a) 1 pc � 2.06 � 105 AU; b) 1 pc � 3.08 � 1016 m;

c) 1 pc � 3.25 ly

45. 35.31 ft3

47. 2.72 m

49. 8.9 � 103 kg/m3; 5.6 � 102 lb/ft3; 0.32 lb/in3

51. 8.0 m3/day

53. 108; 1013

55. a) 7.4 � 102; b) 1.855 � 102; c) 8.47 � 10�3

57. 6.0 � 107 metric tons/cm3

59. 5.00 � 10�3 m3/s; 5.00 kg/s

61. 7.1 � 10�15 m; 3.0 � 10�15 m

63. 354 m2

65. 11�; 5.7�; 570 atoms

67. 359.76�; 1440.0�

69. 8.9 m; 9295 tons

71. 3.902 � 10�25 kg; 235.0 u

73. 2.8 � 1019 molecules

75. 0.125 mm

77. 88.5 km/h; 80.7 ft/s; 24.6 m/s

79. 3.81 � 109 s

81. yes, because the distance traveled while gliding �

18.7 km

83. a) 3840 km; b) 296 km; c) 0.315 or 1:3.2

Chapter 2

1. 0.3 s

3. 6.3 � 10�7 m/s; 5.4 cm/day

5. 32.5 km/h

7. 600 km/h

9. 14 km/h

11. 2.5 � 104 yr; 2.5 � 107 yr

13. 12.8 m/s; 46 km/h

15. 5.87 h; 150 h

17. 0.06 m

19. a) 14 s; 380 m; b) 72 m

21. 4.83 m/s

23. 2.0 m/s

25. a)

ORBIT LOG LOGPLANET CIRC (km) PERIOD (s) SPEED SPEED CIRC

Mercury 3.64 � 108 7.61 � 106 47.8 1.68 8.56

Venus 6.79 � 108 1.94 � 107 35.0 1.54 8.83

Earth 9.42 � 108 3.16 � 107 29.8 1.47 8.97

Mars 1.43 � 109 5.93 � 107 24.1 1.38 9.16

Jupiter 4.89 � 109 3.76 � 108 13.0 1.11 9.69

Saturn 8.98 � 109 9.31 � 108 9.65 0.985 9.95

Uranus 1.80 � 1010 2.65 � 109 6.79 0.832 10.26

Neptune 2.83 � 1010 5.21 � 109 5.43 0.735 10.45

Pluto 3.71 � 1010 7.83 � 109 4.74 0.676 10.57

b)

slope � �2.01

27. 20 m/s; 16.3 m/s

10 9.5

log (circumference)

log

(sp

eed

)

8.5 9

1.5

2

0.5

0

1

10.5 11


A-36 ANSWERS

29. 1.2 m/s; 0.5 m/s

31. 12 m/s; 0 m/s

33. 0.67 m/s; 0.53 m/s

35. 32.4 m/s

37. 3.4 � 103 m/s2

39. a) t(s) (m/s2) (in g)0 6.1 0.62 Method:

10 1.4 0.14 i) draw tangent to curve20 0.83 0.085 ii) get slope of line by 30 0.56 0.057 counting squares 40 0.49 0.050 iii) to find Δv and Δt

convert from km/h to m/sb) t(s) (m/s2) (in g)

0 �0.74 �0.07510 �0.44 �0.04520 �0.44 �0.04530 �0.31 �0.03240 �0.22 �0.022

41. 0 s; 1 s; x(0) � 0 m; x(1) � 1.2 m

43. 1 m/s2; 0.9 m/s2; 1.3 m/s2

45. at t � 0, a � 0; at t � 2 s, a � �2.5 m/s2; at t S ", a S 0

47. a)

aa

aa

49. a)

b) 1.6 s; 4.7 s; c) 0 s; 3.1 s; 6.3 s; v(0) � v(3.1) � v(6.3) �

0 m/s; a(0) � a(6.3) � �2 m/s2; a(3.1) � 2 m/s2

51. 2.4 m/s2

53. 6.36 � 107 s; 6.2 � 108 m/s

55. �350 m/s2; will probably survive

57. �7.1 m/s2; 3.8 s

59. 30 m/s; 300 m

61. 16 s

63.

t (s) 1 2 3 4 5 6

1.0

–1.0

–2.0

0 x (m)

2.0

7

Time, t (s)

Sp

eed

, v

(m

/s)

0 0.6 1.2 1.8 2.4

660

640

620

600

560

580

540

520

500 3.0

b)

TIME INTERVAL (s) AVG SPEED (m/s) DISTANCE TRAVELED (m)

0–0.3 647.5 194

0.3–0.6 628.5 189

0.6–0.9 611.5 183

0.9–1.2 596.0 179

1.2–1.5 579.5 174

1.5–1.8 564.0 169

1.8–2.1 549.5 165

2.1–2.4 535.0 161

2.4–2.7 521.0 156

2.7–3.0 508.0 152

c) 1722 � 2 m

t (s)

v (

m/s

)

4 10 5 6 7 8 9 31 2

18

t (s)

4 10 5 6 7 8 9 31 2

60 54

(m)

x

65. 32.9 m/s; 40.4 m/s

67. 0.875 m/s2; 4.4 m/s

69.

TOTAL STOPPING

v0(km/h) v0(m/s) v0 �t (m) DISTANCE (m)

15 4.17 8.3 1.1 9.4

30 8.33 16.7 4.3 21.0

45 12.5 25.0 10 35.0

60 26.7 33.3 18 51.3

75 20.8 41.7 27 68.7

90 25.0 50.0 39 89.0

�

v20

2a (m)


ANSWERS A-37

45. (�3.9 � 106 m)k

47. Because the vectors are nonzero, a zero result for the dot

product means they must be perpendicular.

49. Bx � �6.83, Bz � �4.5, Cz � 1.34

51. 0.44i � 0.22j � 0.87k

53. 0.49i � 0.81j � 0.32k

55. 24

59. �12i � 14j � 9k

61. 0.45i � 0.59j � 0.67k

65. Coordinate system rotated at �26.6�

67. 415 m, 29.8� W of N

69. x � 1.0, y � 1.7

71. A � B � 5.4i � 12.7j; A � B � 5.4i � 6.5j

73. 4.58

75. �304 m2

77. 4.0, 5.0

Chapter 4

1. a) 7 km, 5 degrees E of N; b) 5.6 km/h, 5 degrees E of N;

c) 8.24 km/h

3. 3.93 m

5. a) 2 i � (5 � 8t) j � (2 � 6t) k; b) 8 j � 6 k, magni-

tude � 10 m/s2, direction � 37� below the y-axis in the

y-z plane

7. 19.6 m at 90� below the direction of travel of the airplane

at 2 s; 24.7 m at 83� below the direction of travel of the

plane at 3 s

9. �13.3 km/h i � 123 km/h j

11. velocity � (90 i � 15 j) m/s, speed � 91 m/s; direction �

9.5� below the x-axis

13. a) v � (3t i � 2t j) m/s; b) r � [(3t2/2)] i � t2 j m

15. 2.4 m/s

17. 38 m/s

19. 65.8 m/s, 93.4 m/s

21. a) 7.25� b) 13 m

23. 1.74 sec, 14.9 m, 59.5 m

25. 3.13 � 103 m/s, 2.5 � 105 m/s, 452 sec

27. 64.8 m, 3.04 sec

29. 76�

31. 12 m/s, r � �21 m i � 55 m j

33. 21 m/s

35. The lake surface is 34.3 m below the release point and the

horizontal distance from release point is 68.8 m

37. Yes, puck passes 2.2 m above the goal, 0.391 sec

41. 63.4�

43. 9.29� and 80.5�

45. 5.19�, when angle off 0.03� in vertical direction arrow still

hits bull’s-eye (arrow hits 4.6 cm off center, which is still

within 12 cm diameter), when angle off 0.03� in horizontal

71. 15.5 m/s

73. a � � ge�t/

75. 66 m

77. 6.1 m/s

79. 44 m

81. 7.96 m or 3 floors; 22.1 m or 8 floors; 43.5 m or 15 floors

83. 2.8 s; 14 m/s up

85. 3.7 m above launching point; 8.6 m/s

87. 1.1 m/s; 5.5 m/s

89. 0.22%

91. 1.6 � 104 m/s2

93. 1.9 � 103 m/s; 2.6 � 102 s

95. 802 m/s; 1.9 s

97. 14.9 m/s; 5.1 m/s

99. a) b) (3/4)h above the ground; c) (2/3)h101. 18.3 m/s; 26.7 m/s; 33.3 m

105. 13.7 m

107. average speed � 1.3 m/s; average velocity � 0 m/s

109. 0.95 s; 28.8 m

111. 2.9 s

113. a) 4.3 m; b) 3.0 m/s; c) �6.0 m/s2

115. 21.1 m/s

117. 33.1 m/s; 2.21 � 103 m/s2

119. a) 8.10 m above ground; 11.1 m above ground; b) 9.8 m/s

down; c) 0 m/s2

Chapter 3

1. 11.8 km, 30� N of E

3. 11.2 km, 27.7� S of E

5. 612 m, 11.3� W of S

7. 436 km, 7.4� W of N

9. B � (�1.26 m)i � (3.2 m)j

11. 13.6 nmi, 88� E of N

13. 1.88 � 104 km, 1.98 � 104 km

15. 6.07 mi, 78.3� W of S

17. 9.19 km N, 7.71 km W

19. 1.7 m

21. (2i � 5j) cm

23. Az � � 4.2 units

25. a) �3i � 2j �2k b) � 7i �4j � 4k c) � 16i � 9j � 11k

27. x � � 9.9 m, y � 9.9 m

29. (1/3) i � (2/3) j � (2/3) k

31. c1 � �8/7, c2 � 9/7

33. 4940 km

35. (6/7) i � (12/7) j � (4/7) k

37. 9

39. �8, 112�

41. 56.1�

43. 45�

n12h/g ;


A-38 ANSWERS

direction arrow still hits bull’s-eye (arrow hits 4.7 cm off

center, which is still within 12 cm diameter)

47. 7.49�

49. No, projectiles will never collide

51. 45.8 km

53.

55. the maximum possible

height is 4.4 m

57. � � 9.94�, 17 km

59. 2.1 rev/min

61. 9.4 m/s2

63. 3.95 m/s2, 4 � 105 g

65. 8.99 � 1013 m/s2, 9.16 � 1012 g

67. 5.9 � 10�13 m/s2

69. aeq � 3.39 � 10�2 m/s2, a45 � 2.4 � 10�2 m/s2

71. aM � 0.0395 m/s2, aV � 0.0113 m/s2, aE � 0.00595 m/s2

73. 5.5 m/s, 2.5 m/s

75. 633 m/s i � 226 m/s j

77. V � 12 m/s, � � 83�

79. 4.60 m/s

81. 60 cm/s at 34� above the horizontal

83. speed � 27 km/h, � � 33�

85. a) 50 � 103 m; b) 33 � 103 m, 67 � 103 m

87. 15.1 km/h at 15� E of N

89.

91. 528 km/h at 8.5� N of W

93. a) vertical component � 62.1 km/h down, horizontal com-

ponent � 232 km/h in direction of plane’s travel; (b) 1.9 min

95. vE � 3.9 km/h, vN � �1.4 km/h (1.4 km/h south)

97. a) 13 m/s; b) 56.3�

99. a) 25 km; b) 50 km; c) No

101. 26 m/s

103. 8.9 m/s2

105. 10.8 sec

Chapter 5

1. 442 kg

3. 2.69 � 10�26 kg; 3.7 � 1025 atoms

5. 3.8 m/s2; 6.2 � 103 N

7. 6.6 � 103 N; 12 times the weight

9. �1.2 � 103 N

Fnet

Directionof motion

vrel �2v0

(h2�4v20t2 � 1)1>2

h � R(1 � sin u) �u2cos2u

2g ;

u �1

2 (d � p/2)

11. �1.8 � 103 m/s2; �1.3 � 105 N

13. 35 N

15. �4.2 m/s2; �2.4 m/s2; �1.0 � 103 N; �5.7 � 102 N

17. 0.063 m/s2

19. �2.90 � 103 N; �1.82 � 103 N

21. v � bx0 sin(bt); a � b2x0 cos(bt); F � mb2x0 cos(bt);

F � �mb2 (x � x0)

23. No, since the tension in the rope � 150 000 N � breaking

tension

25. 36o south of east; 260 N

27. 3.7 m/s2; 23.4o east of north

29. 4.7 � 1020 N; 25o clockwise from the Moon-Sun direction

31. 770 N in the positive x-direction

33. 2.6 kg; 34 N

35. 285 N on Mars; 1900 N on Jupiter

37. a) 9.9904 � 10�4; b) no

39. 128 N in the upper cord; 29.4 N in the lower cord

41.

43. Mg; Mg/2

45. at the upper end; at the midpoint

47. 1.2 m/s2; 36 N to the right; 36 N to the left

49. 600 N

51. 5.2 � 103 N

53. 165 N; 19.5o clockwise from positive x

57. 1.8 � 103 N upward

59. F in the first cable; in the second

cable; in the third cable;

61. 1.14 � 103 N or 265 lb; 820 N or 184 lb

63. 0.51 N

65. 1.9 m/s2; 14 m/s

67. 64 m; 5.1 s

69.mgR

2l (l � 2R)

a �F � (m1 � m2 � m3)g

m1 � m2 � m3

F a m3

m1 � m2 � m3

bF a m2 � m3

m1 � m2 � m3

b

T �pd

2lrg

8T �

pd 2lrg

4

T3 � m3g

F � T1 � (m1 � m2 � m3)g ; T2 � (m2 � m3)g ;

y

x

mg

y

N

R

R

T

θθ θ


ANSWERS A-39

87. a) 71. N � 680 N, F � 340 N

30°

F cos 30°

F sin 30°

N

F

588 cos 30°

60 × 9.8 = 588 N

588 sin 30°30°

73. a) Incline forward; b) 22.3 m/s2

75. π/4;

77.

79. 7.9 � 10�5 m/s2; 0.14 m

81. a) Fnet � �2i � 3j � 4k N; b) a � �0.33i � 0.50j � 0.67k

83. a)

m/s2, 0.9 m/s2

a �2g

l x ; x(t) � x0e22g/l

t

2B¢x

g

mg

x7000 N

T

B

y

θT sin θ

T cos θ

b) 7.1 � 103 N; c) 2.6 � 104 N

85. a)

Fboy on rope

Frope on boy

T

Fground on boy

boy

Fgirl on rope

Frope on girl

T

Fground on girl

girl

b) 250 N; c) 250 N; d) 250 N

mg

N

P

θ

θmg cos

θmg sin

b) 1.8 � 104 N; c) same as b)

89. a) 590 N; b) 700 N; c) 590 N; d) 0 N

91. a) �0.98 m/s2; b) 99 m; c) 50 km/h, same as speed when

first decoupled

93. a) b)

Chapter 6

1. 5.7 � 103 people

3. 0.83

5. 1.6 � 102 m

7. 53 m

9. 3.4 m, so he will reach the plate; 1.9 m/s

11. 0.48

13. 2.1 � 103 N

15. 0.27

17. 2.0 � 102 N

19. 2.8 m/s

21. 1.9�

23. 0.78

25. 1.4 � 10�3 m/s (1.4 mm/s)

27. 39.5 m

29. a)

T �m1m2

m1 � m2

ga �m2

m1 � m2

g;

mg

F sin

μsNFf =

N θ

F cos θ F

θ

mg

μ θscosFf =

N

F sin θ

F cos θ

F

θ


A-40 ANSWERS

b) c)

d) but now � is the

angle between the force F and the positive x direction

31. The quantity sin � must be positive in order to

find a solution for P

This gives the condition

33. 3.6 m/s2

35.

37. 23 N; 11 N

39. No, since k is not constant

41. 50 N/m

43. 8.8 � 102 N/m

45. 1.1 � 10�3 m

49. 4.4 � 102 N

51. 7.8 � 102 N at top; 8.1 � 102 N at bottom

53. 6.3�

55. 0.13 m

57. 0.224 N

59. 6.9 m/s

61. 22 m/s

63. 68�

65.

67. 1.40 � 103 m

69. The equilibrium conditions when the balls are at maximum

angular displacement is and the

condition when they are both vertical is (m1 � m2)g

These conditions cannot both be satisfied,

so the motion described is impossible

71.

73. at � � 45�, � � 0.099�

75. 0.89 m/s

77. 40 m; 3.2 s

79. a)

f � tan�1 a tan u

1 � 4p2 R�t 2gb�u;

T �mv2

2pr

�m2v2

2 � m1v21

l.

cos u2�cos u1 � m1�m2,

2gl sin u tan u

T �mkm1m2

g cos u

(m1 � m2) cos f � (m1 � 2m2)mk sin f

tan u �1

mk

cos u � mk

u � tan�1 ms,F �mg

2(1 � m2s )

;

u � tan�1 ms;F �mg

cos u � ms sin u; b) 3.9 � 102 j N; c) �3.1 � 102 i N.; d) �3.9 � 102 j N;

�3.1 � 102 i N; e) a � �7.85i m/s2; 31 m

81. 0.15 m

83.

85. 4.1 cm; 2.5 cm, 1.6 cm

87. a) 1.2 � 10�2m; b) 1.7 m/s2 up the incline; c) 1.0 � 10�2 m

89. Yes, since the centripetal force exceeds the maximum fric-

tional force

Chapter 7

1. 1.5 � 103 J

3. 252 J

5. 3.7 � 103 J

7. 2.35 � 105 J; 357 J/s

9. 2.2 � 107 J by first tugboat; 1.0 � 107 J by second tugboat;

3.2 � 107 J total

11. 2.6 � 103 J

13. 7 � 104 J by gravity; �7 � 104 J by friction

15. 54�

17. a) 1.3 � 104 J; b) 290 N; 1.3 � 104 J

19. a) 1.4 � 104 N; 8.3 � 103 N; b) 4.3 � 103 J; c) 2.2 � 104 J

21. a) 7.1 � 103 N; b) 2.2 � 105 J; 8.1 � 103 N

23. 6 J

25. �26 J

27. 3W0; (2N � 1)W0

29. a) ;

b)

31. 17 J

33. a) ; b)

35. 2.7 � 1033 J

37. 1.3 � 105 J; 5.8 � 103 J; 22

39. a) 4.0 � 105 J; b) 2.5 � 104 J; c) 1.2 � 106 J

41. 4.1 � 106 J

43. Kball � 46 J; Kperson � 38 J; they are of the same order of

magnitude

45. 1.9 J; 0.44 m

47. 6.2 � 109 J

49. 196 m/s

51. 3.4 � 10�18 J

53. a) 80 J; b) �1.295 � 103 J; c) 1.375 � 103 J

55. 7.4 � 103 J; 1.6 % of the energy acquired by eating an apple

57. 8.2 � 106 m3

�B2

12Axeq � B6

A

B

P � 2ky c1 � (l/2)

2(l/2)2 � y2d

k c y2 �l2

2�

l

22(l/2)2 � y2 d

a1 �m2 � mkm1

m1 � m2

g

mg

fk

N


ANSWERS A-41

29.

31. x � � 1.0 m; unbound for E � 0 J

33. a) 0.382 nm; b) �1.67 � 10�19 J; c) �0.34 nm, �0.89 nm

35. 6.3 eV/molecule

37.ENERGY PER PASSENGER

VEHICLE ENERGY PER MILE (J/mi) PER mi (J/ passenger-mi)

Motorcycle 1/60 gal/mi � 1.3 � 108 J/gal

� 2.2 � 106 J/mi 2.2 � 106

Snowmobile 1/12 � 1.3 � 108 � 1.1 � 107 1.1 � 107

Automobile 1/12 � 1.3 � 108 � 1.1 � 107 1.1 � 107/4 � 2.7 � 106

Bus 1/5 � 1.3 � 108 � 2.6 � 107 2.6 � 107/ 45 � 5.8 � 105

Jetliner 1/0.12 � 1.3 � 108 � 1.1 � 109 1.1 � 109/ 110 � 9.8 � 106

Concorde 1/0.1 � 1.3 � 108 � 1.3 � 109 1.3 � 109/ 360 � 3.6 � 106

Most efficient is the bus, least efficient is the snowmobile.

39. 183 m, assuming a mass of 70 kg for the climber

41. 1.05 � 104 kJ

43. Walking 1.7 kcal/kg; Slow running plus standing 2.8

kcal/kg; Fast running plus standing 2.8 kcal/kg

45. 1.88 � 109 eV

47. Thermal energy is 0.0001 % of mass energy

49. 511 keV; 939 MeV

51. 1.4 � 10�9 kg; .00000005% of the mass of the gasoline

53. 9.40 � 108 eV

55. 542 kcal

57. 18 kWh

59. 2.88 � 108 J

61. 1.5 � 10�3 hp; 23 kcal

63. a) 769 gal; b) 5.8 kW

65. 526 kWh/year; $79

67. 1100 W; 0 W

69. 0.61 hp

71. 746 J

73. 50�

75. �2.0 W

77. 4.24 � 105 W

79. 2500 km2

81. a) 1.7 � 1010 J; b) 17 min; c) 3.73 km

83. 1.2 � 10�4 W

85. 195 m diameter

87. a) 487 hp; b) 2593 hp

89. 52 hp

91. 37%

93. a) 3.2 � 104 W; b) �784 W; c) �3.1 � 104 W

95. 2.3 W

97. 2.1 � 107 kW

99. 3.4 kW

x � ; mv2

2 A

59. 5.1 m

61. 99 m/s; 9.8 � 1010 J; 23 tons

63. 0.16

65. 79%

67. 7.7 m/s

69. 1.1 � 104 J

71. 53 N

73. 100 m/s

77. a) 24 m/s; b) 7.1 m; c) 26 m

79. 48.2�

81. 2.1 � 103 J

83. 1.69 � 105 J; 2.06 � 105 J

85. a) �8.8 m/s2; 35.4 m; b) 1.06 � 104 N; 3.75 � 105 J

87. a) U � 2.35 � 107 J; K � 2.89 � 106 J; b) U � 17.7 �

106 J; K � 8.7 � 106 J; 120 m/s (or 430 km/h)

89. a) 150 J; b) 150 J; c) 122 m/s; d) 1520 m; e) 122 m/s

91. 4.1 � 104 J

93. a) 25.8 m/s; b) 10.3 m/s

Chapter 8

1. 0.076 m

5. assuming that x0 � 0; 5.6 m/s

7. assuming that x0 � 0; 1.3 J; 8 J; 29.3 J

9. F � �4x � 4x3

11. 64.5 m/s

13. a) bxy; b) �bxy

15. 2.61 � 106 J

17.

19. a) 13 kN; b) 13 kN. The force is independent of the

rope length

21.

23. 1.89 � 105 N

25. a) �4.58 m; 14.9 m/s; b) �14.7 m; 21.7 m/s2

27. a)

F � a x i � y j

(x2 � y2)3/2

U(x) �A

12x12�

B

6x6

U (x) � x2 �x4

4,

U (x) �Ax4

4,

0.5 1.0x

U (x)

–1.0

b) c) x � (26 � 2)(b/c)x1,2 � (2 ;22) (b/c);


A-42 ANSWERS

101. a) 4.3 � 10�12 J; b) 6.39 � 1014 J; c) 6.1 � 1011 kg/s;

d) 7.8 � 1010 years

103. a) 5 J; b) �4 J; no

105. a) 7.2 � 106 N; b) 0.414 m/s

107. 4.2 � 109 W

109. a) P � �1.82 � 106 � 3.63 � 105 t � 2.71 � 104 t2 �

964 t3; b) 9.757 � 106 J; 5.714 � 106 J; c) �1.35 �

106 W

111. 14 min

113. a) 1.6 � 107 kWh; b) 3.8 � 102 m3/s

Chapter 9

1. 8.2 N

3. 3.46 � 108 m

5. FSun � 0.41 N, FMoon � 2.3 � 10�3 N

7. FAlpha � 1.5 � 1017 N, FEarth � 3.5 � 1022 N

9. aJ � 24.9 m/s2, aS � 10.5 m/s2, aU � 8.99 m/s2

11. 1 � 109 N

13. 2.54 � 10�10 N at 52�

15. 2.76 � 10�4 g

17. �aEarth-Moon � 2.21 � 10�6 m/s2, �aEarth-Moon /g � 2.25 �

10�7, �aJupiter -Io � 0.0123 m/s2, �aJupiter - Io /g � 0.00687

19. 101 m/s

21. 3.08 � 103 m/s

23. 5.8 � 1015 sec � 1.8 � 108 years, 3.1 � 105 m/s

25. TIo � 1.77 days, TEuropa � 3.55 days, TGanymede � 7.15 days

27. 0.927 days

29. About 10 times

31. Same latitude 22.6� West, around Lincoln, Nebraska

33. m1/m2 � 1.6

35. 3.0 � 1010 m

37. a) 7.50 � 103 m/s, 8.32 � 103 m/s; b) 3.94 � 108 J,

4.85 � 108 J

39. 8.2 � 103 m/s

41. TS � 96.5 min, TE � 115 min

43. 5.33 � 1010 km, about 10 times Pluto’s mean orbital

distance from sun

45. 7.8 � 103 m/s, �1.4 � 1011 J

47.

49. U � �1.04 � 106 J, K � 5.2 � 105 J, E � �5.2 � 105 J

51. 8.86 mm

53. 0.253

55. a) 1.11 � 104 m/s; b) 1.23 � 1011 J � 29 tons of TNT;

c) 1.23 � 105 m/s2

57. 2270 m/s, 1.11 � 104 m/s

59. elliptical

61. a) speed � 1680 m/s, time � 6510 sec � 109 min; b) this

will give an elliptical orbit; c) this orbit will not be closed

because the launch speed is greater than the escape speed

22

63. a) No, speed is less than that needed for circular orbit;

b) 1.22 � 104 m/s

65. ES � �4.4 � 109 J, EE � �6.06 � 108 J

67. vperigee � 6.96 � 104 m/s, vapogee � 5.75 � 103 m/s

71. h � 4.26 � 106 m, v � 817 m/s

73. a) 2.6 � 103 m/s; b) 2.8 � 103 m/s; c) 0.401 years;

d) Venus moves 234.7�, Earth moves 144�

75. 1 rev/min

77. 1.6 � 10�9 N

79. 4.9 years

81. 1.90 � 1027 kg

83. a) vrel � 1.53 � 104 m/s; b) 3.91 g

85. height above earth � 9.89 � 106 m

87. a) –3.30 � 109 J; b) 3.30 � 109 J

Chapter 10

1. 9.0 kg·m/s; 3.2 kg·m/s.

3. 1.8 � 1029 kg·m/s; 4.3 � 107 kg·m/s; 3.8 � 104 kg·m/s;

95 kg·m/s; 2.0 � 10�24 kg·m/s

5. (1.6 � 10�25 i � 7.7 � 10�26 j) kg·m/s

7. a) (9.7i � 5.6j) kg·m/s; b) (9.7i � 0j) kg·m/s; c) (9.7i � 5.6j)

kg·m/s

9. 9.81 kg·m/s down; 98.1 kg·m/s down

11. �9.0 i m/s

13. �2.2 � 105 j

15. �2.0�2 � 10�5 i m/s

17. 8.26 � 103 m/s; 1.29 � 10�17 J

19. �(1.3 m/s)i � 0j

21. 66 N; 1.3 � 106 J

23. 150 N�s

25. �(4.10 � 103i � 929j) m/s

27. 5.2 N

29. 5 � 10�11 kg/s; 5 � 10�5 N

31.

33. 1.9 m from woman

35. 7.42 � 105 m; 0.107% of the sun’s radius

37. h/3 along the height, away from the unequal side

39. 0.027 cm directly away from the 40 g piece

41. 0.23 nm from the hydrogen atom

43. (950 � 103, 180 � 103, 820 � 103) light-years

45. (L/3, L/3, L/3)

47. (�0.061L, 0, 0)

49. 950 m from the base

51. 31.5 J; 63 J

53. 9.0 � 107 J

55. CM is on the axis of symmetry, a distance R/2 away from

either the base or the top of the hemisphere

57. 6.9 � 106 m/s in the direction of motion of the proton

59. 953 kg

u m

Man

k�1

1

1 � km/M


ANSWERS A-43

43. 9.3 m/s

45. a) 9.8 m/s; b) 4.8 � 105 J; c) �130 m/s2, 850 m/s2

47. 4.0 � 10�13 J

49. a) 3.9 � 105 J, 3.9 � 105 J; b) 7.8 � 105 J, 3.9 � 105 J

51. 210 m/s

53. a) 440 m/s; b) �1200 J; c) 9.6 J; d) missing kinetic energy

is energy that shows up as heat in bullet and block,

compression/deformation, and noise

55. 620 m/s

57. 860 m/s

59. (a) 3 m/s (b) 79 m/s2

61. 21 m/s

63. a) ; b)

65. (1 � cos�) i � sin� j; (1 � cos�) i

� sin � j (1 � (sin2 � � cos2 �)) � 0

so

67. 3.2 kg�m/s, 3200 N

69. v�1 � 2.6 km/h, v�2 � 13 km/h

71. 1.2 � 10�12 J

73. a) v�1 � 0 and v�2 � 20 m/s; b) The ball that had an initial

velocity lands on ground next to fence and the ball with no

initial velocity lands 11 meters away from the fence

75. 0.964, 0.036

77. a) 3.3 � 10�12 m/s; b) 2.4 � 107 tons of TNT;

c) 8.1 � 1014 N

79. a) Ball height h � �(1 – cos�) where � is the angle with the

vertical; b) height h � L�4 (1 – cos�) where � �

cos�1

80. a) 21 m/s and 11� west of north; b) 1.4 � 105 J

Chapter 12

1. 1.7 � 10�3 rad/s; 3.5 � 10�4 m/s

3. 81.5 rad/s; 13 rev/s

5. a) 0.52 m/s; 0.17 m/s; b) 1.8 m/s2; 0.61 m/s2

7. 22.0 rad/s; 1.28 m/s; 55.4 rad/s; 8.83 rev/s

9. 9.4 rad/s; 0.94 m/s; 1.9 rad/s2

11. 88 rev/s for aluminum; 2.1 rev/s for steel

13. At t � 0, � � 0; � � 0; � � 40 rad/s2; at t � 1.0 s, � � 15

rad; � � 25 rad/s; � � 10 rad/s2; at t � 2.0 s, � � 40 rad;

� � 20 rad/s; � � �20 rad/s2

15. 5.5 � 10�3 cm/s; 51 cm/s; 22 cm/s

17. 611 rad/s2; 70 revolutions

19. 1.5 � 102 rad/s2

21. �4.36 � 10�2 rad/s2; 0.89 revolutions

Q34�14 cos uR

v�1 � v�2

v�1� v�2 �v2

4

v

2

v�2 �v

2

v

9v�1 �

v

2

2mv sin�1 a1 �b2

4R2bp � 2 sin�1 a b

2Rb

i

61. 1.05 m

63. 1.6 m/s in the direction of motion of the bullet

65. 4D from the launch point

67. (�17.4 km/s, �17.4 km/s)

69. KCM � 3.8 � 104 J; KTOT � 6.38 � 105 J

71. 4.76 � 104 J; 3.6 � 105 J; 4.76 � 104 J

73. 3.9 � 103 J; 4.0 � 103 J

75. 0.955 � 10�3

77. 1.1 m/s

79. aboy � 5.0 m/s2 toward girl; agirl � 6.7 m/s2 toward boy;

1.7 m from boy

81. 8 km/h

83. 0.0927 nm

85. 59 cm

87. If the stack of two books is at the top of the triangle, the

CM is at a point halfway between the other two books and

0.43 m above the line connecting them

89. Halfway along the line joining the centers of the plates

Chapter 11

1. 1.13 � 103 kg�m/s, 2.3 � 104 N

3. a) 12 m/s, 7m/s, 3 m/s, 1 m/s, �1 m/s, 250 m/s2, 200 m/s2,

100 m/s2, 100 m/s2; b) 4.2 � 105 N, 3.4 � 105 N, 1.7 �

105 N, 1.7 � 105 N; c) 1.1 � 105 N�s

5. 12.6 kg�m/s, 4200 N

7. –1400 N

9. –1.8 kg�m/s, �1.35 � 104 N

11. 8.1 kg�m/s, 0.045 s

13. 18 kg�m/s

15. 7.5 � 10�2 s, 2.8 � 104 N

17. a) vproj� � �0.27 m/s, vtarg� � 0.53 m/s; b) Kproj � 1.9 �10�2

J, Ktarg � 0 J Kproj� � 2 � 10�3 J, Ktarg� � 1.7 � 10�2 J

19. 39 m/s

21. 0.57 J

23. 0.22v

25.

27. v1� � 15 m/s, v2� � 17 m/s

29. Last ball has velocity � v and other two balls have

velocity � 0

31. a) mass m rises to , mass 2m rises to ; b) mass m rises

to h, mass 2m stops and does not rise

33. 13.5 m/s

35. a) The 1400 kg mass has a velocity � 1.3 m/s and the

800 kg mass has a velocity � 6.1 m/s; b) t � 0.98 s x � 1.7 m

37. Yes

39. a) �7.5 m/s; b) �15 m/s; c) �15 m/s

41. �0.17 m/s, 0.18 m/s, 0.41 m/s, �0.34 m/s

4h9

h9

M

7


A-44 ANSWERS

23. �23 rad/s2; 1.7 s

25. �9.5 � 10�22 rad/s2

27. 10 rad/s; 8 rad

31. 3.8 � 104 J

33. 1.21 � 10�10 m

35. 6.50 � 10�46 kg·m2

37. 0.44 kg·m2

39. 0.46 kg·m2

41. 2.13 � 1029 J

43. 0.96 kg·m2; 1.5 � 107 J

45. �1.10 � 1022 kg·m2

47. 0.379MER2E

49.

51.

53. 0.426 MR2

55.

57.

59.

61. a) 225 kg·m2; b) 4.4 � 103 J

63. 3.49 � 105 J; 3.9%

65. 2 � 1025 J/s; 1.05 � 1015 s or 3.3 � 107 yr

67.

69.

71.

73.

75. (160 km/h) i at top; 0 km/h at bottom; (80 km/h) i � (80

km/h) j at front

77. 12 rad/s2; 74.1 rad/s; 35 turns

79. 0.012 J; 0.37 m/s

81.

83.

85. 4.4 m/s

MR2 a 5

8�

1

27pb

I2 � 2 a 2

5 MR2 � MR2 b �

14

5 MR2

I1 � 2 �2

5 MR2 �

4

5 MR2;

3

10 mR2

MR2

4

Ml2

6

MR2

4

I �3

5 MI

2

2

5 M(R5

2 � R51)

R32 � R3

1

a 1

2�

1

10 � 30h�4R bMR2

I �Ml2

3

ML2 sin2 u

12

Chapter 13

1. (4610 N·m)R, 613 kg, 940 kg

3. 310 N

5. 59 N·m

7. 130 hp, 176 N·m

9. 2900 N·m

11. 5.4 m/s, 7.7 m/s

13. 230 W

15. 1.9 � 106 J

17. 19 J, 75 J

19. 5.6 � 104 W, 1.1 � 104 N·m

21. 4.6 rad/s

23. a) 1140 N·m; b) 2.1 m/s2

25. 820 N

27. 2.7 � 104 N·m

29. 9.7 m/s2

31. Proof required.

33. 9.6 m/s2

35. rolls to the right, f � 2F/3

37. 17 rev

39. 0.024 N·m, 0.16 N

41. Proof required.

43. 2.83 m/s2, 2.94 m/s2, 3.89 %

45. 1.6 m

47. 2.8 � 1034 kg·m2/s

49. 1.6 kg·m2/s

51. 1.05 � 10�34 kg·m2/s, 2.11 � 10�34 kg·m2/s, 3.15 �

10�34 kg·m2/s

53. a) 1.8 � 109 kg·m2/s, upward; b) zero kg·m2/s

55. 7.9 � 1022 rad/s, 7.9 � 107 m/s, 2.1 � 10�12 J, 1.5 � 10�10 J

57. 0.051 kg·m2/s2

59. 5.6 � 1041 kg·m2/s, 3.14 � 1043 kg·m2/s, 1.8 %

61. 0.57 rad/s, 5.5 rad/s

63. 14 rad/s

65. 1.5 � 10�19 rev/day

67. and . The satellite closer to

earth has the greater speed. and

. The satellite closer to earth has the

smaller angular momentum.

69. �4.3 � 10�22 rad/s2, 3.5 � 1016 N·m, 2.6 � 1012 W

71. �7�

73. a) 5.0 m/s; b) �0.009 rad/s; c) �6860� or 19 rev

75.

77.

79. 0.37 rad/s

mv30

g132,

mv30

g12

m2v2

2g (m � 13M )(m � 1

2M )

L2 � m1r2GME

L1 � m1r1GME

v2 � BGME

r2

v1 � BGME

r1


ANSWERS A-45

39. 7.2 m/s2

41. a) 8.8 m/s2; b) 5.4 m/s2; c) 3.8 m/s2

43. 2mg;

45. tan�1 (2s)

47.

49. 1580 N; 1340 N

51. 400 N

53. 240 N

55. 0.010 micrograms resolution; 0.20000 milligrams max load

57. 736 N

59. a) b)

c)

61. 8.9

63. 393

65. 36

67. 2 � 10�3 m

69. 3.5 � 104 m

71. a) 4 � 10�3 m; b) 3.5 � 103 N

73. 0.057 m

75. 0.033%

77. 4.3 � 10�6 m

79. 0.52 cm

81. 1.5�

83. 3.96 � 108 N/m2

85. 624 rad/s

87. 425 rad/s

89. 360 N

91. 1.23 � 104 N; 2.13 � 104 N

93. 2400 N�m

95. 0.577 Mg; 0.289 Mg

97. 490 N

99. 1200 N

101. 0.71 mm

103. 5.0 � 103 m

105. 4.85 � 10�5 N

Chapter 15

1. a) 3.0 m; 0.318 Hz; 2.0 rad/s; 3.14 s;

b) tmidpoint � 0.785 s; tturn.point � 1.57 s

3. a) 0.83 Hz; 5.2 rad/s; b) 0.20 m; c) 0.30 s; 0.40 s;

d) 1.05 m/s; 0.91 m/s

5. a) 251 m/s; b) 251 m/s

7. 211 N; 4.2 m/s

F�

F�

2l

R1 � R2

F � 12 mg

(R1 � R2)

l;1

2 mg (R1 � R2);

T1 �t

Ra 1

emsp � 1b and T2 �

t

Ra 1

1 � e�mspb

mg

13 ;

mg

13

81. The instantaneous change in angular momentum opposes

the original direction of the angular momentum and makes

the tilt worse

83. a) 110 kg·m2/s; b) 34 N·m; c) 34 N·m

85. 1.6 � 108 kg·m2/s, east or west, 1.2 � 104 N·m, 1.0 � 104 N

87. 3100 N·m

89. 3100 J

91. a) 76.2�; b) 290 N, �250 N

93. a) 4.10 � 10�5 kg·m2; b) 0.12 rad/s2; c) 4.8 � 10�6 N·m;

d) 0.0023 J

95. 420 N

97. 8.50s

99. a) 2.0 � 10�10; b) 4.9 � 109; c) 4.9 � 109 rev/month

or 1860 rev/s

101. 1.3 � 1014 kg·m2/s, north

Chapter 14

1. 590 N

3. 5.5 � 106 N; 5.1 � 106 N

5. 5200 N

7. 8 cm; 18 cm

9. 5.88 kg

11. 3500 N; 6800 N

13. 51 N; 29 N

15. 1420 N; 2500 N

17. 30�

19. 7.65 m

21. Proof required.

23.

25. a) 9 � 103 N; b) 2.6 � 103 N

27. 0.408 Mg

29.

31. 0.62 mg

33. a); b) 26.6�; c) 56 J

35. 1.04 L

37. 17.5 m/s

10°

U

θ20° 30° 40° 50° 60°

53 J

1/2 × 53 J

T �Mg a L � R

2(L � R)2 � R2b ; N �Mg

R

2(L � R)2 � R2

F � Mg R

2R2 � (R � h)2F � Mg

2R2 � (R � h)2

(R � h);


75. a) ;

b)

77. 1.0 � 103

79. 92; 0.32 W

81. 30

83. 395

85. 21

87. 1.5 cm; 66.7 Hz

89. a) midpoint at 2 s; 6 s, 10 . . .; turning point at 0 s, 4 s, 8 s, . . .;

b) midpoint at 0 s, 4 s, 8 s, . . .; turning point at 2 s, 6 s, 10 s, . . .

91. 26.7 m/s; 1.68 � 104 m/s2; 2.0 � 104 N

93. 2.12 Hz

95. 1.25 Hz

97. 18.75 J; 3.54 m/s

99. 0.375 m

103. 0.35 Hz

Chapter 16

1. 4.3 � 1014 Hz to 7.5 � 1014 Hz (violet)

3. 2.08 cycles/hour, 356 km

5. a) 4.4 m/s; b) 205 m/s

7. a) 10.8 hour; b) 2.5 /hour

9. a) vmax � 0.27 m/s as it passes through equilibrium between

crests; b) � amax � � 6.2 m/s2 at the wave crests.

11. a) 0.02 m; b) 1.4 Hz; c) 10 m

13. a) 0.2 sec, 5.0 Hz, 31 rad/sec, 5.2 m�1;

b) y � 0.020 cos (5.2 x �31.4 t)

15. wavelength decreases by 20 cm

17. � amax � � 2.41 m/s2; wavelength � 156 m

19. 0.45 m/s

21. 0.97 sec

23. 2.0 kg/m

25. 250 m/s

27. 1280 N

29. 184 km

31. v � v

33. 0.017 sec

35.

37. ; Proof required.

41. a) ; b) 0.20 m

43. 6.0 m, .6 m, 0.80 Hz; �3.2 m

45. 1.14 m, 6.28 m

p

2

mV 2

2pR

22l /g

Bm

m

p

2 B

l

g ; mg (1 � A2)

mg B1 �A2

2�

3A2

2 sin2 aB

g

l t bR

A-46 ANSWERS

9.

11. 3.98 � 10�6 m

13.

15. a) x � 0.292 cos(6� t � 0.815); b) 0.043 s; �103 m/s2

17. 2.8 � 106 N/m; 3.16 Hz

19. 1.9 � 104 N/m

21. 1.13 � 1014 Hz

23. 0.20 m; 7.3 rad/s; 0 rad; 1.46 m/s; 10.7 m/s2

25. 5.51 � 103 Hz; 1.0 � 10�4 g; 6.7 � 10�6 cm

27. x � 0.27 cos 6t, with the axis chosen so the initial position

of the unstretched spring is at x � 0.27 m.

29.

31.

33. 3.6 J

35. 2.12 Hz; E same; A same; 2.0 m/s; 26.7 m/s2

37. 0.34 m

39.

43. 10.4 s

45. 34.8 s

47. 0.188 Hz

49. a) 0.73 min/day; b) 1.0 mm

51. 24.8 m

53. 3.0 s

55. a) b) �0 c) �0 � 1 radian

57. 2.09

59. a) 1.26 � 10�3 J; b) 0.145 m/s

61. 1.64 s

63. 9.8 � 10�3 m/s2

65. 1.6 s

67.

69. 3.6 rad/s; 3.3 m/s

71.

73. 2pB4L

5g

m �k

g

2p BL

g am1 � m2

m1 � m2

b

BL

g

B2k

MR2 ;B

2k

MR2 ;

Fd �kd2

2

f �1

2p B

k

(m � 6M)

mg sin u

k ;

1

2p B

k

m

d �3p

2

A � Bx20 �

v20

v2 ; d � �tan�1 a v0

vx0

b


47. a) 0.0194 m; b) �0.0179 m

49. a) y � 0.0060 cos(400�t) � 0.0040 cos(1200�t); b)

51. 0.028 is the fractional increase or decrease. We cannot tell

which from the given information.

53. 392 Hz, 588 Hz, 784 Hz, 980 Hz

55. 1.58 Hz, 3.16 Hz

57. 28 Hz

59. 7.07 m, 66.6 m/s, 628 m/s2

61. 9.3 ms, down, 54 Hz

63. 8.16 Hz, 16.3 Hz, 24.5 Hz, 32.7 Hz, 40.8 Hz

65. 71 N

67. 1.3 � 10�7 m, 2.5 � 10�7 m

69. 3.7 m/s, 6.8 � 103 m/s2

71. a) 4.82 � 103 N; b) 8.2 Hz

73. y � (0.20 mm) sin(2�x � 880�t) � (0.20 mm)

sin(2�x � 880�t), A � 0.20 mm, v � 440 m/s

75. a) ; b) ;

c)

77. Large amplitude is at . Smallest amplitude is at

where n is an integer.

79. a) 0.030 m; b) 5.2 m; c) crests: 0 m, 5.2 m, 10.4 m, . . .;

troughs: 2.6 m, 7.9 m, 13.5 m, . . .

81. 2.1 m/s2, 4.9 m/s

83. 13 m/s, 7.9 � 103 m/s2

85. 0.0731 kg/m, 261 m/s

87. a)

b) v1 � 7.2 m/s, v2 � 8.6 m/s

89. 12 Hz

91. a) , 14 m/s, 9.9 m/s, 0 m/s; b) 2.9 secv(x) � 2g (L � x)

T1 �22

1 � 23 T, T2 �

2

1 � 23 T ;

x �(2n � 1)p

k

x �np

k

f �n

2L B

F

A � BL

l(x) �1

f B

F

A � Bxv(x) � B

F

A � Bx

0.002

0.003

0.001 0.004 0.005 st

y

0

0.005

–0.005

–0.010

0.010 m

ANSWERS A-47

93. , . . ., where 4L is the longest possible

wavelength

Chapter 17

1. 17 m (20 Hz) to 1.7 cm (20 kHz)

3. 765 m; 166 m

5. about 9 cm

7. 1.9 mm and 0.10 mm

9. 6.8 m/s

11. D–D# 1.9 cm, D#–E 1.8 cm, E–F 1.7 cm, F–F# 1.6 cm,

F#–G 1.5 cm, G–G# 1.5 cm, G#–A 1.3 cm, A–A# 1.3 cm,

A#–B 1.2 cm, B–C 1.1 cm, C–C# 1.1 cm, C#–D 1.0 cm

13. 1.0 � 103 W/m2

15. D#, 6 octaves above the one listed in Table 17.1

17. �3.0 dB

19. 83 dB

21. 0.11 W

23. 130 times (intensity measured in W/m2), 21 dB

25. 9.1 sec

27. 249 m

29. a) f � 3.0 � 10�5 Hz; T � 3.3 � 104 s (about 9 h); b) It’s

possible because the period of the first overtone is close to

1/4 of the tidal period.

31. a) 3.0 � 10�5 s; b) glass

33. 272 Hz, 3.9%

35. about 4000 Hz

37. 2.0 km

39. 3.3 km in sea water

41. 2.8 km

43. 92.4 m

45. 337 m/s

49. a) 0.632 m; b) C–C# 3.5 cm, C#–D 3.4 cm, D–D# 3.2 cm,

D#–E 3.0 cm, E–F 2.8 cm, F–F# 2.7 cm, F#–G 2.5 cm,

G–G# 2.4 cm, G#–A 2.2 cm, A–A# 2.1 cm, A#–B 2.0 cm,

B–C 1.9 cm

51.

53. 619 Hz

55. 21.5 m/s, 0.215 Hz

57. 381 m/s

59. 2.63 m/s

61. 30�

63. 405 Hz

65. 594 Hz, 595 Hz

67. 476 Hz

69. 481 Hz

71. 29.4�

fn �nv

2L, n � 1, 2, 3, . . .

ln �2L

n

l � 4L, 4L

3,

4L

5,

4L

7


57. 12.6 cm

59. a) Proof required; b) �2g; c) "

61. 4.4 cm

63. a) Proof required; b) 0.094 Hz

65. 1.13 � 105 Pa

67.

69. 1.9 � 104 N

71.

73. 0.013 m/s

75. a) 332 N; b) Average rate � 4.3 kW; Peak rate � 8.6 kW

77. 12.4 hp

79. 2.7 m/s; 190 liters/s

81. 7.3 � 103 N

83. 1.12 � 1012 Pa

87. 8.06 mm-Hg

89. 1.21 � 105 Pa; 1.01 � 105 Pa; 2 � 103 N

91. 0.73 cm

93. 2.5 � 105 kg

95. 5.3 � 10�3 m3

97. 1.07 � 103 kg/m3

99. 2.0 � 10�3 m3/s

101. The pressure inside increases by 209 Pa; Smaller

Chapter 19

1. 32�F, � 380�F, � 423.4�F, � 452.2�F, � 454�F, � 459.67�F

3. 5.3 � 1021 atoms

5. 78% N2, 22% O2

7. The frequency decreases by 14 Hz

9. 1.9 atm

11. a) pO2 � 7.5 � 104 Pa, pN2 � 8.6 � 104 Pa; b) 1.6 � 105 Pa

13. 3.4 atm

15. �i/�0 � 1.05

17. 1.4 � 10�9 Pa

19. 12 kg/m3

21. 4.3 atm

23. 500 kg/m3

25. 100 kg/m3

27. 1.3 cm

29. 4.5 � 107 Pa

31. 96.3 g

33. 3150 kg, 2.8 � 103 m3

35. a) Water rises 1.2 m; b) 6.8 kg at 2.6 � 105 Pa

37. 29 g/mol

39. Differentiating p � p0 � ��gy yields dp � ��gy. Using the

Ideal-Gas Law and .n �m

M, r �

m

V�

pM

RT

v � B2ptank

r

B2gh � 2 patm

r

A-48 ANSWERS

73. a) Proof required; b) 165 m/s

75. 0.15 mm

77. 1.5 mm, 0.33 mm

79. 5.0 m/s, 3.2 � 107g

81. 3 women

83. 3.0 dB

85. a) 3.0* 10�3 sec; b) The bat will think distances are 0.77

times the real distances.

87. a) 33.5�; b) 30.2 sec

89. a) 660 Hz; b) 691 Hz; c) 723 Hz

Chapter 18

1. In the hose: 1.39 m/s; 2.8 m/s; 4.2 m/s; In the nozzle:

22.3 m/s; 25.1 m/s; 23.9 m/s

3. 7.23 � 105 W

5. 12 m/s

7. 8.84 cm/s; 8.84 m/s

9. a) 11.5 m; b) 8.1 cm; 11.7 cm

11. 84 m

13. 1370 lbf � 6090 N

15. 132 cm2

17. 2.34 � 105 Pa

19. 5.08 � 104 N

21. 48.6 cm2

23. 2.0 � 104 Pa; 7.5 � 106 Pa

25. a) 360 N; b) 330 N

27. 3.56 � 105 Pa; 3.60 � 105 Pa; 4 � 103 N

29. 0.85 m

31. 10.3 m

33. 3.3 � 104 Pa

35. 2.94 � 105 Pa

37. 2.1%

39. 3.1 � 108 N

41. a) Proof required; b) 5.0 � 109 Pa

43. a) ;

;

b) ;

45. a) 4.7 � 107 m3; b) 4.3 � 1010 kg

47. Yes

49. 0.32 m

51. 61 m/s2

53. 113 kg

55. 31 m

Ftotal � �prghQR2

1 � R1R2 � R22R

3

Fb � rghp R21; F� � �

prgh(R2 � R1)(2R1 � R2)

3

Ftotal � �prghQR2

1 � R1R2 � R22R

3

Fb � rghp R22; F� �

prgh(R2 � R1)(R1 � 2R2)

3


Substituting � into the differential equation:

or,

41. 615 m/s

43. 1200 K

45. 4100 m/s, 1.4 � 10�20 J

47. 0.12 m/s

49. 5.65 � 10�21 J

51. For O2, vrms � 428 m/s, For O3, vrms� 349 m/s, For O2,

(translational) K � 4.87 � 10�21 J, for O3, (translational)

K � 4.85 � 10�21 J

53. 0.43%

55. 9.7 � 106 K

57. 0.47 m/s

59. Using the hint, the volume swept out per molecule with an

effective radius 2R0 going a distance l is cylindrically shaped

with volume V/N � � (2R0)2 l. Solving for l yields the

desired result.

61. a) for 1 atm: � 0.091%; b) for 1000 atm: � 91%

63. 1.9 � 105 J, translational 0.6, rotational 0.4

65. A 7% increase in kinetic energy by changing temperature,

no change in the kinetic energy by changing the pressure.

67. 1.0 � 105 J

69. 291 J

71. 1.88 � 1025 molecules

73. 1.3 � 103 N

75. 1.1 � 1023 nitrogen molecules, 2.9 � 1022 oxygen mole-

cules, 1.4 � 1023 total

77. a) 1.8 � 1032 particles/m3 3.7 � 1016 Pa; b) 9.0 � 1031

particles/m3, 1.9 � 1016 Pa; c) 4.5 � 1031 particles/m3,

9.3 � 1015 Pa

79. From the Ideal-Gas Law at constant temperature

pV � p�V �. So,

.

This can be rearranged to where �V is the

decrease in volume and �p is the corresponding increase in

pressure.

Furthermore, and

for �p �� p.

Using the specifics of the problem, Vc � Vs � V or Vs �

Vc � ¢V a p

¢pbVc � V �

V � ¢V a1 �p

¢pb � ¢V a p

¢pb

p�

¢p�

p � ¢p

¢p� 1 �

p

¢p

V � ¢V a p�

¢pb ,

¢Z

V�

V � V�

V� 1 �

V �

V � 1 �

p

p��

p� � p

p��

¢p

p�

dp

p� �

Mg

RT dy.

dp � �pMg

RT dy

ANSWERS A-49

81. 615 m/s

83. a) ; b) 1; c) 1; d) 0.5

85. 375 K

Chapter 20

1. 540 s

3. 0.28�C

5. 1.6 km

7. 750 s

9. 8500 steps

11. 0.17�C

13. 1.7�C

15. 1.7 � 10�4 �C/km

17. The heat produced from electric power 2.6% of the inci-

dent solar heat.This is enough to slightly increase the local

temperature.

19. 1.1 � 10�3 m3/s

21. a) 1.7 � 103 N�m; b) 2.3 � 103 W; c) 4.0 � 10�4 �C

23. 27�C

25. 136�C

27. 0.67�C

29. 38�C

31. 0.18 liter

33. 0.028 J of work done by iron, 2.7 � 107 J of heat absorbed

by iron, amount of work is 1.0 � 10�6 times the heat

absorbed

35. 4.9 � 10�4 m, 17 N

37. a) 3.8 � 10�4, 1.9 � 10�4; b) 16 s

39. Proof required.

41. 100.28�C

43. 23000 W, the rate through window is 13 times greater

than the rate through the wall

45. a) 2.4 m2�s��C/J; b) 13.6 ft2�h��F/BTU

47. 4.2 � 103 W

49. The solution is a proof.

51. 11 W, 79�C


55. 0.51 cm/h

57. a) 4.26 � 1014 J; b) 5.3 bombs

59. 270 g

61. 1.1 � 109 J, 1.2 � 104 W

63. 3.9 kg

65. a) 2.0 � 1011 kg; b) 1.1 � 1017 cal; c) 2.9 � 1015 J � 7.0 �

1014 cal; d) 1.0 � 1013 J � 2.4 � 1012 cal. The kinetic

energy is smaller than the potential energy due to fric-

tional losses with the air.

67. 0.092 kg

69. 4.3 km3/h

71. 41�C

22


31. 8.2 � 106 J/s, 0.19 kg/s

33. 0.999 999 997

35. 1.4 � 105 J

37. 44%, 1100 W

39. 75 W

41. 19.5

43. 8.5 � 103 J, 3.4

45. a) �1.3 � 103 J; b) Heat is absorbed during step 2 and

rejected in step 3; c) heat is rejected by the system in step 1,

d) 0.39

47. a) 0.067; b) 1.39 � 107 W; c) 180 kg/s

49. a) 48 W; b) 20 times

51. a) eturbine � 0.34, eengine � 0.42; b) 0.62, the two efficien-

cies are the same

53. 9.5 � 103 J/K

55. 3.0 W/K

57. 12 400 W/K

59. 3 � 103 J/K

61. �SAl � 430 J/K, �SFe � 150 J/K, �SAg � 80 J/K, �SHg �

47 J/K. The change in entropy seems to decrease with

increasing atomic number. Largest is aluminum; smallest

mercury.

63. 41 W/K

65. 9.5 � 106 W/K

67. 0.94 W/K

69. 5.8 J/K

71. a) Proof required; b �S � 780 J/K

73. 37 J/K

75. Proof required.

77. 120 K

79. a) 4.16 � 105 Pa, 2.27 � 105 Pa; b) W � �Q � �3.0 �

102 J

81. 24%, 4.8 � 104 W

83. 4.0 � 102 W

85. a) Beginning with the point at the upper left, the gas under-

goes an isobaric expansion in step 1 as the volume increases,

followed by a isovolumetric reduction of pressure in step 2

as the temperature is reduced. The gas is then compressed

isobarically in step 3 by reducing the volume, before an

isovolumetric increase in pressure in step 4 by increasing

the temperature.

b) W1 � 2100 J, W2 � 0 J, W3 � �700 J, W4 � 0 J;

c) Q1 � 5260 J, Q2 � �3160 J, Q3 � �1750 J, Q4 � 1050 J;

d) 44%

87. 48%, 1 � 107 W

89. �SN � 2600 J/K, �SO � 2300 J/K, �SH � 22 500 J/K.

Hydrogen is largest and oxygen smallest.

91. a) �S � 1.1 J/K; b) Q2 � 340 J, �S � 1.1 J/K; c) �S �

0 J/K

Q �myou(9.81 m/s2) (3.0 m)

0.055 a 1 kcal

4187 Jb

A-50 ANSWERS

73.

The gas with the highest specific heat per kilogram is

helium; and that with the lowest is argon.

75. 971 m/s

77. �V � 3.7 � 10�2 m3, W � 3.7 � 103 J

79. Cp � 26.3 J/(mol�K), CV � 18.0 J/(mol�K)

81. 110 kcal/h

83. 1 K

85. a) 0.072 m3; b) 0.42 m3, 145 K

87. 214 K

89. 3.6�C

91. 160 liters/h

93. 0.33 m, 0.050 m, 46 m2

95. 2.3 � 10�5 kg/s

97. 0.52 kg

99. 880 J, 1500 J

101. a) 1700 J; b) 1200 J; c) 0.029 m3, 1.7 � 104 N/m2;

d) 5.0 � 102 J, �2.4 � 103 J

Chapter 21

1. Q � 1.9 � 103 J, �E � 1.1 � 103 J

3. a) W � 0, Q � �E � 610 J; b) W � 810 J, Q � 2.0 � 103 J,

�E � 1.2 � 103 J

5. �470 J

7. a) 4.29 moles; b) W � 1010 J, �E � 2490 J; c) 5/2, diatomic

9. a) W � �9.19 J, �E � 3.34 � 105 J; b) � 18.4 J, the heat

of vaporization remains unchanged

11. W � �37.9 J, �E � �37.1 J

13. 4.87 � 104 J, 2.56 J

15. 4.3 � 104 J

17. 0.014 m3, 7.2 � 105 Pa

19. 2.43 � 106 J/kg

21. 35%

23. 14%

25. 60%, 1.2 � 107 J

27. 44.5%

29. 5.5%, W � myou (9.81 m/s2) (3.0 m),

GAS CV (J/(kG.K)

He 3.12 � 103

Ar 3.13 � 102

N2 7.42 � 102

O2 6.50 � 102

CO 7.39 � 102

NH3 1.60 � 103

CH4 1.69 � 103


ANSWERS A-51

Chapter 22

1. 5.8 � 105 N

3. 58 N, 3.5 � 1028 m/s2

5. 51 N

7. 1.6 � 1020 electrons

9. 2.39 � 10�7 N

11. Fg � 6.7 � 10�13 N, Fe � 2.3 � 10�8 N

13. 9.63 � 104 C

15. 4.7 � 1013 electrons

17. 3.2 � 1019 N

19. 1.3

21. 99.9 %

23. 2.9 � 10�9 N/m

25. (�2.3 � 10�5 N) i � (�3.5 � 10�5 N) j, (2.3 � 10�5 N) i

� (3.5 � 10�5 N) j

27. 6.81 � 1032 electrons on Earth and 1.9 � 1032 electrons

on Moon

29. 1.0 � 10�9 at 1 m, 1.0 � 10�5 at 1 � 104 m

31. 3.8 � 10�39 C, ratio � 2.8 � 106 (attractive)

33. �x direction

35. �(1.9 � 10�7 N) i�(1.7 � 10�7 N) j

37. 1.0 � 10�7 C

39. �1.35k i�1.35k j ��1.35k (i�j)

41. �(3.1 � 10�15 N) i � (6.9 � 10�16 N) j

43. 1.2 � 10�4 kg or 0.12 g

45.

47.

49. (1.116 i�1.75 j�0.5k)

51. 0.35d

53. p � p → n � n � π+, p � p → n � p � π0, p � p → n �

p � π0 + π�

55. 1

57. 5.6 � 1021 electrons

59. 1.9 � 10�9 kg

61. any negative charge

63. 3.6 � 10�8 N, 6.9 � 10�2 m/s2

65. 1.0 � 10�6 C, 6.5 � 1012 electrons

67. 0 C, �e, �e, 0 C

k Q2

a2

kQq c 1

x2�

4

(d � 2x)2d

212 Q132 � 1R2 � 9.85

Q2

L2

Q2

L2

Q2

L2

2kqxQ

�4d2

� x2�3/2,

Chapter 23

1. F � 5.4 � 10�14 N, a � 6.0 � 1016 m/s2

3. � � �20�

5. E � �2.1 � 105 N/C j

7. Fe � 0.6 Fg

9. 6.3 � 10�7 m

11. E � 5.1 � 1011 N/C

13. E � 5.1 � 1012 N/C

15. 28 N/C

17.

19. EP � 9.5 � 103 N/C i � 2.8 � 104 N/C j

21. E(x) � 2 � 8.99 � 109

23. E ≈

25. a) Emax is at y � �R

b) The field distribution for the ring is the same as that for

two positive charges rotated around the y axis.

27. E0.5 � 7.2 � 10�4 N/m, E1.0 � 3.6 � 10�4 N/m,

E1.5 � 2.4 � 10�4 N/m

29. E � 2.6 � 1010 (i � j)

31. E � (cos 30�) j

33. Ex �

35. EA � 1.13 � 105 N/C j, EB � �1.13 � 105 N/C j,

EC � �3.39 � 105 N/C j, ED � �1.13 � 105 N/C j

37. E � 2.4 � 107 N/C directed at an angle of 45� with respect

to each sheet

39.Ez � �

zs

2�02z2 � R2

1

4p�0

c �l

2x2 � y2d , Ey �

l

4p�0yc1 �

x

2x2 � y2d

1

p�0

l

2d

L

Nm2

C2ba

122

Q

2p�0x2

£ (�40 C)(10000 m)

�x2 � (10000 m)2�3/2�

(30 C)(4000 m)

�x2 � (4000 m)2�3/2§

Nm2

C2ba

ED � � a1.15 � 1010 Nm2

C2b Q

L2j

EC � � a1.15 � 1010 Nm2

C2b Q

L2i,

EB � a1.15�1010 Nm2

C2b Q

L2j,

EA � a1.15 �1010 Nm2

C2b Q

L2i,


A-52 ANSWERS

41.

43. a)

c) When x �� l, the electric field resembles that of a point

charge and goes to zero as x goes to infinity.

45.

47.

49.

51. The solution is a sketch of the electric field.

53. The solution is a sketch of the electric field.

+2q –3q

E � 0.0609 l

�0R

� B 1y ¢1 �

x

(x2 � y2)1/2≤ �

1

(x2 � y2)1/2R jr

E � �l

4p�0

b B1x

¢1�y

(x2�y2)1/2≤ �

1

(x2 � y2)1/2R i

E �l22

p�0l i �l22

p�0l j

b) Ep �1

2p�0

�Q

l 2 B1

2 ln ¢x � l

x≤2

� x

x � l �1R i,

l0 �2Q

l,

b) EP � � 1

2pe0

Q

y

1

2l 2 � 4y2 j

a) EP � 1

4p�0

Q

l ¢1

x �

1

x � l≤ i, 55. The solution is a sketch of the electric field.

57. E � 3.1 � 105 N/C

59. v0 � 1.4 � 105 m/s

61. v0 � 110 m/s

63. 1085 electrons

65. p � 2.0 � 105 Cm

67. a) p � 1.6 � 10�29 Cm, b) The dipole moment is reduced

due to the motion of electrons.

69. E � 520 N/C

71. E � 1.1 � 107 N/C, � � 3� with respect to the y-axis

73. E � 1.21 � 104 N/C, � � 74� with respect to the y-axis

75. (a)

(c) The magnitude of the force is the same as in (b), but its

direction is opposite that of the rod on the charge.

77.

79.

Chapter 24

1. 1.1 � 1012 Nm2/C

3. 0.16 Nm2/C

5.

7. �E,1 � �E,6 � 0, �E,2 � ��E,4 � 2.0 Nm2/C, �E,3 � ��E,5

� �3.5 Nm2/C, �total � 0

9.

11. 1.4 � 103 Nm2/C

13. 0.038 Nm2/C

15. (a) (b) (c) F �sq

2�0

, E �s

2�0

sq

2�0

q

2�0

f �q

�0

�23

8 s

�0

a2, 23

24 s

�0

a2

E �1

4p�0

Q

y2

E �Q

2p�0

£ 1

y2 �

1

(y2 � d 2)32

≥ j

E �1

4p�0

¢ld≤ , (b) F �

q

4p�0

¢ld≤ ,

–

++


ANSWERS A-53

17. (a) (b)

19. �45 Nm2/C, 5.5 � 10�8 C/m

21. 2.3 � 102 Nm2/C

23. �G � 4�Gminside

25. Imagine a small cube of volume dV . If the cube itself con-

tains no charge and it’s located in a uniform electric field,

the net flux through the six faces of the cube must be zero,

no matter how the cube is oriented in the field. If the cube

contains charge, then the flux through it cannot be zero

and the field cannot be uniform. If the field is uniform,

Gauss’ law tells us that the charge density inside the cube,

which is the charge inside divided by the volume dV, must

be zero.

27. 160 N

29.

since A � 0, E � 0

31. in direction perpendicular to axis

33. r a E � 0, a r b

r � b

35. r a E � 0, a r b

r � b

37. (a) where q ��e. where

. Substituting yields

where the minus sign indicates this is a restoring force and

the magnitude of that force is . (b) 7.2 � 1015 Hz

39. (a) (b) (c)

41. (a) (b) (c)

(d) if

43. For For

45. f � �q

�qEy2pRdx � 2pR�

q

�qEydx �

q�

�0

E �Cd 3

24�0

�x� �d

2E �

Cx3

3�0

�x� d

2,

Q � qn � �3,E � k>2n � �1,E �

kr n�1

�0(n � 3)Q �

4pk

n � 3rn�3

E �Q

4p�0r2E �

Q

4p�0R2C �

Q

2pR2

F �

e2r

4p�0R3

FS

� � e2r

4p�0R3 rQ � �e a r3

R3b

E �Q

4p�0r2 rF � qE

E �Q

4p�0r2

E �Q

4p�0r2 (r3 � a3)

(b3 � a3) ,

E �r

3�0

(b3 � a3)

r2

E �r

3�0

(r3 � a3)

r2,

lr

2pR2 �0

E A �qenc

�0

1 qenc � 0 ‹ E A � 0

s

4�0

q

4�0‹

‹

47.

49.

51. and

where

‹ ‹

53. 0 C,

55.

57.

59.

61. (a)

(b) (c)

(d)

63.

65. (a) (b)

67. (a) (b)

(c) (d) 0

69. where the minus sign indicates

the field points towards the center of the sphere. r � R�

E � �rl

3pR2�0

r R�

1.7 � 1012 Nm2/C

�1.1 � 1012 Nm2/C1.1 � 1012 Nm2/C

f �3q

4�0

f �q

4�0

fcurved �q

2�0

fbase �q

2�0

,

E �Q1 � Q2

4p�0r2Qlarge,outer � Q1 � Q2

Qlarge,inner � �Q1,E �Q1

4p�0r2

Qsmall,outer � �Q1Qsmall,inner � �Q1,

8.85 � 10�10 C/m2

sb �q

4pb2sa � �

q

4pa2,E �

q

4p�0r2,

r � bE � 0,a � r � bE �q

4p�0r2,r � a

E � 0E �l

2p�0x,

1.1 � 105 C1.1 � 105 C,

E �qd

4p�0R3Etotal �

q d

4p�0R3

Etotal � Ebigsphere � Esmallspherer� � d� r

Esmallsphere �qr�

4p�0R3Ebigsphere �

qr

4p�0R3

ES

�rd

�0

c a 1

2�

1

3025b i �

1

1525j dr

S�

1

2di � d j

E �47rd

96�0

i ,r � 2d iES

�rd

6�0

i ,r �1

4di

E �Q

4p�0

° 1

r2�

1

8 a r �R

2b 2¢ r

�q

�qFydt �

q

v

q�

2p�0R

q

v �q

�qEydx �

q

v

q�

2p�0R�q

�qEydx �

q�

2pR�0


Take any point (x,y,z) on the surface of the sphere. Then

x2 � y2 � z2 � R2

r1 and r2 as shown

Therefore,

Therefore, potential constant is at 0 on entire surface.

39. (a)

(b)

(c)

(d)

41.

47.

Ey � �2p

bcos

2px

asin

2py

bcos

2pz

c;

Ex � �2p

asin

2px

acos

2py

bcos

2pz

c;

Q

4p�0

a 3

c�

1

a�

1

b�

r2

a3�

1

2ab

3Q

4p�0c�

Q

4p�0r�

Q

4p�0b;

3Q

4p�0c;

3Q

4p�0r;

Vd �Q

4p�0R1

�Q

4p�0rR0 � r � R,

Vc �Q

4p�0R1

�Q

4p�0rR � r � R1,

Vb � 0R1 � r � R2,

r � R2, Va � 0

1

(h2 � 2zh � R2)1�2d � 0

c 1

(R2 � 2zh � h2)1�2��

Q

4p�0

1

h�r (R2 � 2R2z�h � R4�h2)1�2d

c 1

(R2 � 2zh � h2)1�2��

Q

4p�0

1

(x2 � y2 � z2 � 2R2z�h � R4�h2)1�2d

c 1

(x2 � y2 � z2 � 2zh � h2)1�2�

R

h�

Q

4p�0

�R

h

1

Bx � y � aR2

h� z b 2 ¥

V(x,y,z) �Q

4p�0

≥ 1

2x2 � y2(h � z)2

r2 � Bx2 � y2 � aR2

h� z b 2

r1 � 2x2 � y2 � (h � z)2

V �1

4p�0 cQr1

��Q(R/h)

2d

A-54 ANSWERS

where again, the minus sign indicates

the field points towards the center of the sphere.

71.

73.

75.

77.

Chapter 25

1.

3. 60,000 V

5.

7. a) b)

9.

11.

13.

15.

17.

19.

21.

23.

25. a) b)

27.

29.

31.

33.

35. for the wire and the same for the cylinder.

37. V(r) �4k

�01r Volt

1.16 � 104 V/m

l

2p�0

c a2 ln (b�a)

b2 � a2�

1

2d

� ln a l � 2x2 � l2

xb � ln a x � l � 2l2 � (x � l )2

x � 2x2 � l 2b d

� ln a l � 2l2 � (x � l )2

l � xbQ

l

1

4p�0

c ln a l � x

xb

Q

l

1

4p�0

ln a 2 � 23

2 � 23b

�23 V

4.0 � 107 m/s5.5 � 10�12 J;

lu

4p�0

Q

ap�0

a1 �2

25�

1

3b

5.8 � 10�12 J

3.86 � 10�12 J

7.36 � 10�8 C

2.05 � 105 m/s

9.2 � 102 V

2.7 � 107 m/s

1.45 � 106 m/s2.1 � 106 m/s;

6.9 � 106 m/s

2.4 � 106 J

6.6 � 1020 Hz

1.1 � 10�34 kgm2/s,2.3 � 10�13 J,1.2 � 107 m/s,

E � 0r � b

E � �l

2p�0r b2 � r2

b2 � a2,a r bE � �

l

2p�0r,r a

ES

total �2l

5p�oRiS

E �l

2p�0R,

m �4pR3rs

6g�0

E � �R�3l

3p�0R2r2


49.

On the z axis: On the x axis:

51. a)

b)

53.

55.

57.

59. a) b)

61.

63.

65.

67.

69.

73.

75. b)

77.

79. a) b)

81.

83. 0 V

85.

87. a) 2Q

3p�0R2a2R2 � x2 � B

R2

4� x2 b ;

3.1 � 107 m/s

2.5 � 105 V9.96 � 10�2 m;

l2

4p�0

a 1

4� ln

b

abl2

8p2�0r2;

4.1 � 10�7 J.

Ugrav

Urest mass

�(3�5)(GM2�R)

Mc2�

3

5 GM

Rc2 � 1.9 � 1011

Ugrav �3

5 GM2

R � 1.24 � 1029 J

8.6 � 105 eV

Q2

8p�0

a 7

10Rb

�50 eV

5.8 � 106 eV

Q2

4p�0d(4 � 22)

1.8 � 109 J/m3

2.3 � 1011 J4.4 � 10�8 J�m3;

3 � 10�2 J

6.3 � 10�4 J

�8.1 � 10�18 J

4l

2l 2 � 4lx � 4(R2 � x2)Q�l � 2x � 2l 2 � 4lx � 4(R2 � x2) R

�Q

4p�0

Q

l

1

4p�0

c ln a x � l�2 � 2R2 � (x � l�2)2

x � l�2 � 2(x � l�2)2 � R2b d ;

Ez � �p

2p�0x3Ex � Ey � 0,

Ez �p

2p�0z3;Ex � Ey � 0,

Ez �p

4p�0

c 3z2

(x2 � y2 � z2)5�2�

1

(x2 � y2 � z2)3�2d ;

Ey �p

4p�0

3yz

(x2 � y2 � z2)5�2;

Ex �p

4p�0

3xz

(x2 � y2 � z2)5�2;

Ez � �2p

ccos

2px

acos

2py

bsin

2pz

c

ANSWERS A-55

b)

89.

91.

Chapter 26

1.

3.

5.

7.

9.

11.

13.

15.

17.

19.

21.

23.

25.

27. The only arrangement of capacitors to give the same net

capacitance is two pairs of two capacitors in series, con-

nected in parallel.

29.

31.

33.

35.

37.

39. a) b)

c)

41. a) b)

43.

45. a)

sbound,outside � 1.7 � 10�6 C/m2,

sbound,inside � � 2.6 � 10�6 C/m2,

¢Vempty � 4.5 V¢Vfilled � 1.8V,Q � 9.0 � 10�6 C,

C � 2.9 � 10�11 FC0 � 2.0 � 10�11 F,

sbound �2�0¢V

da 1

1 � kb

E �2¢V

da 1

1 � kb ,Efree �

2¢V

da k

1 � kb ,

C � (k1 � k2) a e0A

2db

k � 1.7

k � 5100

C � 4p�0kR

¢VPP � � 68 VQ3 � 4.8 � 10�4 C,

Q2 � 4.1 � 10�4 C,Q1 � 3.0 � 10�4 C,

¢V5.0mF � 2.3 V¢V2.5mF � 4.5 V,Q � 1.1 � 10�5 C,

¢V � 3.5 VQ � 12.0 mC,

Cnet �2

3 C

Q � 1.7 � 10�3 C

C � 9.9 mF

C � 8.0 � 10�12 F/m

Q � 1.8 � 10�18 CC � 1.8 � 10�17 F,

Q � 125 C

Amax � 6.8 � 10�3 m2Amin � 5.6 � 10�4 m2,

n � 4.5 � 1014 electrons

Q � 1.3 � 10�8 CC � 1.1 � 10�9 F,

Q � 1.1 � 10�6 CC � 1.1 � 10�11 F,

V1 � 3V2Q2 � 3Q1,

Utotal �Q2

8p�0 b(b3 � a3)2a 6

5b6 � 3a3b3 �

9

5a5b b

V �Q2

4p�0l2 c xln

(x � 2l )x

(x � l )2� 2l ln

(x � 2l )

(x � l )d

�2Q

3p�0R2± x

2R2 � x2�

x

BR2

4� x2

≤ i


13.

15.

17.

19.

21.

23.

25.

27.

29.

31.

33.

35.

37.

39.

41.

43.

45.

47.

49. In series: in parallel: in combi-

nations of series and parallel: Req �

51.

53. a)

55. d � 1.5 km from point A

57.

59.

61. a) b)

c)

63.

65.

67.

69.

71.

73.

75. Connecting two sets of two resistors in parallel gives

77.

79.

81. a) b) harmless,

harmless, fatal

Chapter 281.

3. 6.9 � 106 J

largest battery: 1.9 � 10�2 kWhkg � 6.7 � 104

J

kg

smallest battery: 3.3 � 10�3 kWhkg � 1.2 � 104

J

kg;

I3 � 0.18 A,I2 � 88 mA,

I1 � 9.2 mA,Req � 1.3 � 106 �,

j � 3.3 � 106 A/m2, vd � 2.4 � 10�4 m/s

I � 0.30 A

Rnet � 1.0 �

R � 5.2 � 10�4 �, ¢V � 0.31 V

¢R

R0

� 0.089 or 8.9%

E � 2300 V/m

Q � 1.4 � 105 C, n � 9.0 � 1023 electrons

R � 2.73 �

¢V � 36 V

¢V1 � 3.6 V, ¢V2 � ¢V3 � 4.4 V

I1 � 1.8 A, I2 � 1.1 A, I3 � 0.7 A,

I � 1.8 A, Rnet � 4.4 �,

R � 5.4 � 10�2 �/m

R � 3.2 � 10�3 �

ICu � 210 A, b) Irubber � 2.7 � 10�19 A

I � 27 A

2.2 �2.0 �,1.6 �,

5.2 �,4.3 �,3.7 �,

Req � 0.92 �,Req � 9 �,

Iiron � 1.7 A, Ibrass � 4.3 A

I2 � 1.7 A, I3 � 1.3 A

Req � 2.2 �, I � 5.4 A, I1 � 2.4 A,

R � 4.4 � 1010 �, I � 6.8 � 10�9 A

I � 8.0 � 10�5 A

d � 0.16 cm

mCu � 380 kg, mAl � 190 kg

¢V � 9.9 V

T � 22�C

s � 5.9 � 107(�m)�1

j � 5.7 � 106 A/m2, E � 0.097 V/m

¢R � 0.92 �

r � 5.7 � 10�7 �m

R � 0.87 �

E � 0.069 V/m

R � 0.40 �

Rnet � R>32

I � 1.3 A, n � 8.2 � 1018 electrons/s

R � 3.0 �

A-56 ANSWERS

b)

c)

47.

49.

51. a) b)

c) d)

e)

53.

55.

57. (52 times more charge in

the supercapacitor), (3 times

more energy in the supply capacitor)

59. a)

b)

c)

d)


63.

65.

67.

69. a) b)

71.

73. a) b)

75.

77.

79.

81 . a ) b ) c )

d) Compare the answers from parts a) and c).

Chapter 271.

3.

5.

7.

9.

11. t � 3.8 � 10�14 s, vd � 0.054 m/s

I(0.0s) � 1.0 A, I(1.0s) � 0.25 A, Q(2.0s) � 0.67 C

I � 7.5 A, Q � 15 C

I � 4.0 A, E � 6.0 V/m

t � 1.2 � 10�4 s

Q � 1800 C/h, n � 1.1 � 1022 electrons/h

¢U �Q2

2�0A¢d,W � �

Q2

2A�0

¢d,F �Q2

2A�0

,

¢V � 4.7 � 10�2 V

C

l� 2p�0° k1k2

k2ln a c

ab � k1ln a b

cb¢

U2 � 3.6 � 10�5 JU1 � 7.2 � 10�5 J,C2 � 8.0 mF,

Volume � 2.7 � 10�4 m3

u � 4.8 � 104 J/m3,U � 13 J,

k � 2.0

Cnet � 8.0 � 10�7 FCnet � 8.0 � 10�6 F,

U3 � 0.24 JU2 � 0.077 J,U1 � 0.115 J,

Q3 � 1.2 � 10�3 C,Q1 � Q2 � 9.6 � 10�4 C,

F � 11 N

U3 � 2.5 � 10�3 JU2 � 1.9 � 10�3 J,

U1 � 3.1 � 10�2 J,Q3 � 2.0 � 10�4 C,

Q2 � 1.5 � 10�4 C,Q1 � 3.5 � 10�4 C,

¢V � 4.0 VQ � 8.5 � 10�9 C,C � 2.1 � 10�9 F,

¢V � 36 V,Q � 2.5 � 10�8 C,C � 7.1 � 10�10 F,

¢U � 1.0 � 10�7 J,

¢V � 12 V,Q � 2.5 � 10�8 C,C � 2.1 � 10�9 F,

U � 5.1 � 10�8 J,

¢V � 12V,Q � 8.5 � 10�9 C,C � 7.1 � 10�10 F,

Usupply � 66 JUsuper � 21 J,

Qsupply � 0.33 CQsuper � 17 C,

¢V � 2000 V

U � 2000 JQ � 0.2 C,

U � 1.1 � 10�5 J

u � 0.025 J/m3,E � 7.5 � 104 V/m,

¢V � 380 V,C � 1.6 � 10�10 F,

C � 2p�0 a 1

R1

�1

R2

b�1

(ktop � kbottom)

F � 0.050 N

Efree � 3.0 � 105 V/m

Eouter � 1.1 � 105 V/m,Einner � 1.6 � 105 V/m,


5.

7. 0.38 A,

9. 6.0 A,

11. current through R1 does not change, but it increases through

the other two resistors

13. (a) (b)

15. 115 V, 4.8 V

17. (a) 1.7 A, (b) 0.86 A through resistors 2, 3, and 4

19. 0.49 V

21. 0.11 A

23. 31.0 A, out of the junction

25.

27.

29.

31. 16 V

33. The current through is 2.58 A, through is 1.71 A,

through is 1.28 A, through is 2.15 A, through is

0.85 A.

35.

37. $0.36

39.

41. 1100 W

43. 4.9 A

45.

47. 6.7 h

49. 58 h

51.

53. 10%

55.

57. 0.50 W


61. (a) 3.3%, (b) 33%, lower current is more efficient

63. (a) 180 V, (b)

65. 19 liters/min,

67.

69. 3.958 A

71.

73.

75.

77. 0.011 s

79.

81. (a)

(b)

83. 0.020 s

Q(t) � E1C � (E2 � E1)Ce�t/R1CA � E1C, B � C (E2 � E1), t � R1CI,

Q(t) � E2C � (E1 � E2)Ce�t/R2CA � E2C, B � C (E1 � E2), t � R2C,

¢V �ER1

(R1 � R2) (e�t>(R1�R2)C)

Q � Ce a1 � e�t>(R1�R2)C b ,

4.0 � 10�4 s

3.2 � 10�4 s

RCu � 5.0 �, Rconstantan � 5.6 �

7.0 � 102 �

1.8 � 106 W

R115 � 13.23 �, R230 � 52.90 �

1.2 � 106 W

6.3 � 1012 protons/s, 7.0 � 102 W

3.7 � 10�3 A

6.0 � 10�4 W

V � 2.6 V

R5R4R3

R2R1

I1 � 1.25 A, I2 � 1.75 A, I3 � �0.50 A

I5 � �20 A, R5 � 4.0 �

I1 � �15 A, I2 � 25 A, I3 � �5 A, I4 � 15 A,

8.0 �

62 �,

4.8 � 10�6 V2.4 � 10�4 A;

I2 � I1 � 3.0 A

4.0 � 103 J

1.3 � 105 J

ANSWERS A-57

85. 5.3 s,

87. (a) (b) 430 A, (c) 2670 A

89.

91.

93.

95. (a) 0.024 m, (b)

97. (a) 47 A,

99. 2.2%

101. (a) (b) (c)

Chapter 29

1. opposite the direction of the current

3.

5.

7.

9. pointing downward

11.

13.

15. j

17. and points in direction above the

horizontal in North direction.

19. a)

b)

21.

23.

25.

27.

29.

31. a) b)

33.

35. pointing into the page.

37.

39. For

for

for

for y � 0, z � 0, �q � x � �q : B � �m0s i

y � 0, z � 0, �q � x � �q :B � 0,

y � 0, z � 0, �q � x � �q : B � m0s i,

y � 0, z � 0, �q � x � �q : B � 0,

r � r3 : B � 0.B �m0I

2pr r2

3 � r2

r23 � r2

2

,

r2 � r � r3 :B �m0I

2pr,r1 � r � r2 :B �

m0Ir

2pr21

,r � r1 :

B �m0I

2R�m0I

2pR,

B �m0

2p I

d 3

222

u � 16�B � 5 � 10�6 T,

a � 1.5 � 1013 m/s2

�

B ds �m0I

R

B � 5.03 � 10�3 T

B � 4 � 10�8 T, F � 1.92 � 10�18 N

B � 0.19 T

B �m0I

2pr�m0lv

2pr�m0v

2pr2p�0Er � �0m0vE

I � lv,

16�F � 9.98 � 10�18 N

� 3.33 � 10�18 k) N

F � (�2.98 � 10�18 i � 6.8 � 10�19F � 8.2 � 10�16 N

Bearth � 4.2 BB � 1.44 � 10�5 T,

B � 4.18 T,

u � 19.5�

�F � � 1.38 � 10�24 N

a � 1.14 � 1014 m/s2�F � � 1.04 � 10�16 N,

�F � � 1.07 � 10�16 N,

6.0 � 10�2 A1.2 � 10�4 C,7.6 � 10�5 C,

7.1 � 103 m

3.4 � 106 W

P1 � 5.4 W, P2 � 14 W

I1 � 0.45 A, I2 � 1.31 A, I3 � 0.85 A,

I1 �ER�i

Ri R� � RR� � RRi

, I2 �E

R�i � R aR�

Ri

� 1 b,

3.0 � 101 W, 1.7 � 101 W

1380 �,

5.0 � 10�6 A


63. both directed

opposite the current.

65. For currents in the same direction, for

currents in the oposite direction,

67. the angle between B and the straight wire

is

69. for for

71. directed in a plane parallel to the plane of the

wires and perpendicular to the current, with the exception

along the edges.

73. directed into the paper.

75. directed out of the paper.

Chapter 30

1.

3.

5.

7.

9.

11.

13.

15. (a) The electron will follow a circular path that spirals in

the direction of the magnetic field;

(b)

(c)

17.

19. (a) (b)

21.

23. Consider an element dl of the long, straight wire. As the

current I flows through the loop, the magnetic field B pro-

duced by the loop exerts a magnetic force F on the element

d l given by This force is perpendicular to

element d l. Applying Newton’s third law of motion, the

element d l exerts a force that is directed oppositely to d F,

which is also perpendicular to element d l.

25.

27.

29.

31. (a) sinusoidal behavior;

(b) W � 2.25 � 10�26 J

t � (1.1 � 10�26 sin u) Nm,

t � 2.4 � 10�7 Nm

I � 1.0 � 106 A

F � �13 N j

d F � I d l ? B.

F � 6.7 � 10�5 N

up � 3.3 � 10�3 radvp � 1.6 � 103 m/s;

B � 3.2 T

x � 0.29 m.

f � 1.4 � 107 Hz, T � 7.2 � 10�8 s;

I � 0.39 A

r12

r14

� 1

B � 0.036 T

p � 3.8 � 10�19 kgm/s

B � 3.3 T

p � 3.4 � 10�17 kgm/s

p � 1.1 � 10�17 kgm/s

B �m0I

2Ra 1

p�

3

8b ,

B �m0I25

2pL,

B �m0nI

2p,

r � R2, B � 0.

B � m0NI ;R1 � r � R2,For r � R1, B � 2m0nI;

u � 17.7�

�B � � 1.05m0I

R,

B � 1.05 � 10�5 T.

B � 7.33 � 10�5 T,

F � 4.8 � 10�17 N, a � 2.87 � 1010 m/s2,

A-58 ANSWERS

41.

43. For for

for

for

for

for

45.

47.

49.

51.

53.

55. directed into the paper



61. a)

b) At any point z, the fields produced by the coils are

Their first derivatives are

Both of these derivatives cancel each other at

The second derivatives are

Both of these derivatives cancel each other at z � R>2.

d 2Btop

dz2�

3m0IR2 SR2 �4(R � z)2T2 S(R � z)2 � R2T7�2

d 2Bbottom

dz2� �

3m0IR2(R2 � 4z2)

2(z2 � R2)7>2

z � R>2.

dBtop

dz�

3m0I

2

(R � z)R2

[(R � z)2 � R2]5>2

dBbottom

dz� �

3m0I

2

zR2

(z2 � R2)5>2

Btop �m0I

2

R2

[(R � z)2 � R2]3>2

Bbottom �m0I

2

R2

(z2 � R2)3>2

BP �8m0I

2125R,

B � 0.11m0I

L,

B �m0I

2pR�m0I

8R,

B � 7.9 � 10�5 T,

B �4m0I

22pL

B � m0nn�(r2 � r)I for r1 � r � r2

B � m0nn�(r2 � r1)I for r � r1;

Bmax � 3.34 T, Bmin � 1.43 T

B � 1.26 � 10�2 T

I � 26.5 A

Bmax � B3(z � 0) �m0I

pRB6 �

m0IR

p(z2 � R2),

z � �2R,B5 �m0I

2pa 1

R � z�

z � R

R2b ,

�2R � z � �R,B4 � �m0I

2pa 1

R � z�

R � z

R2b ,

�R � z � 0,B3 � �m0I

2paR � z

R2�

1

z � Rb ,

0 � z � R,B2 �m0I

2pa z � R

R2�

1

z � Rb ,

R � z � 2R,B1 �m0IR

p(z2 � R2),z � 2R,

B �m0

p I

btan�1

a b

2zb


33.

35. The magnetic and electric fields surrounding

the spinning paper disk are illustrated as follows:

37. (a) (b)

39. (a) (b)

41. (a) (b)

43.

45.

47. (a) (b)

49.

51. (a) (b) due north; (c)

due east; (b) due east.

53.

55. (a)

(b)

57.

59.

Chapter 31

1. left side is positive and right side is negative

3.

5.

7. current flows counterclockwise,

to the left

9.

11. �

13. 22 rev/sec

15. patient does not need to be pushed more

slowly

17.

19. 0.047 V

21.

23. (a) (b)

25.

27. 3.0 � 102 A

t � 5.59 � 10�3 NmI � 0.565 A,

7.4 � 10�6 V/m6.6 � 10�6 V/m;

3.5 � 10�7 Tm2

W � 146 JI � 0.606 A,�E� � 6.88 V,

9.7 � 10�3 V;

5.5 � 10�5 Tm2

9.8 � 10�3 V

F � 10 Nv � 0.48 m/s,

2.2 � 10�3 V

0.5 m/s

3.0 � 10�3 V;

a � 3.1 � 103 rad/s2

B �4mv

qd

y �eEL2

2mv, therefore

e

m�

2yE

B2L2

y � 12ayt

2 � 12 a F

mb a L

v0x

b 2

�eEL2

2mv2,

B � 0.010 T, f � 1.6 � 105 Hz

F � 0.16 N

F � 0.18 NF � 0.18 NF � 0 N;

RH � �4.7 � 10�10 m3/C

¢VH � 1.6 � 10�5 Vvd � 2.7 � 10�3 m/s;

B � 2.5 � 10�2 T

¢VH � 3.1 mV

m � 0x � �1;

m

V� 2.0 � 10�23 Am2M � 1.7 � 106 A/m;

f � 0.47 Hzt � (1.71 � 10�7 sin u) Nm;

m � 14Q�R2

m � 0.023 Am2

ANSWERS A-59

29. 0.63 A

31. At at

33. 0.010 C

35. 0.40 H

37. (a) 0.5 H; (b) 10 V

39.

41.

43. the shape of the coil wire does not matter

45.

49.

51.

53. B-field (T)

45

8

2

1.5

55.

57. 211 J/km

59.

61. satisfies the differential

equation if

63. (a) (b) (c)

65. (a) 2.0 A; (b)

67. (a) (b)

69. (a) (b)

71.

73.

75.

77.

79. (a) (b) 0; (c)

81. (a) 0.0188 T/s; (b) (c)

83. 102 V

85. (a) (b) percentage of

energy that

Chapter 32

1. (a) (b)

Pmin � 0 W

Pmax � 2400 W,Irms � 10.4 A, Imax � 14.8 A;

remains � 25%

Umax � 576 J,Imax � 48 A, t � 14 s;

6.4 � 10�3 V6.4 � 10�3 V;

9.2 � 10�15 C/m22.6 � 10�3 V;

7.5 � 10�5 V

0.014 J/m3

2.4 � 10�4 V

R � 0.125 �, L � 0.0396 H

IL2� 0.33 AIL1

� 0.17 A,

IResistor � 0.50 A,dI2

dt� 1.5 A/s;

dI1

dt� 0.75 A/s,

L

R1 � R2

L

R2

;

�3.0 � 103 V

9.51 � 104 J1.58 � 104 W;3.75 �;

� L�R

I �E

Re�t>tIR � L

dI

dt� 0;

1.0 � 10�3 J

U � 5.29 � 104 J; V � 1.64 � 10�3 m3

9.0 � 105

1.6 � 106

2.5 � 107

8.1 � 108

4.0 � 1011103

4.0 � 1021108

u (J/m3)

1.0 J/m3

1.2 � 10�5 J

1.1 � 10�7 H

M � 200m0npR2;

�1.9 V

1.1 � 104 A/s

e0 � 1100 VE0 � 117 V/m

r � 1.5 m:e0 � 704 V;E0 � 140 V/mr � 0.80 m:


57. 4348 turns

59. Transformer 1: Transformer 2:

Transformer 3: Transformer 4:

61.

63. 36 W

65. 0.139 A

67. One-quarter cycle after the maximum

One-half cycle after maximum

three-quarters cycle after maximum

69. (a) (b) 29.3 W

71. (a) (b) First max. at

first min. at (c)

73. One quarter cycle later, One half

cycle later, Three-quarters of a cycle later,

75. (a)

(b)

77. for fully magnetic energy;

for fully electric energy.

79. (a) (b) (c) 78

81. 0.026 A at the power plant, 0.0012 A in the transmission line

83.

Chapter 33

1. (c)

3. (a) 4.0 A; (b)

5. (a) 0.10 A; (b)

9. (a) (b) 1.0 sec; (c)

11. (a) (b) (c)

13.

�B # ds � m0 a I � ke0

d£E

dtb

��E # ds � �d£B

dt,�B # dA � 0,�E # dA �

Q

k�0

,

B �m0�0rC

2prB �

m0�0rCr

2pR2;E �

rCt

pR2;

2.7 � 10�8 T1.8 � 10�8 T,8.9 � 10�9 T,

8.9 � 10�4 A,2.0 � 10�3 A;

6.3 � 10�3 A

4.5 � 1011 Vm/s

Idisp � e0

d£E

dt�

dQ

dt

E2 � 15 sin (3000pt) V

1.2 � 103 V,1.36 � 103 Hz,

1.6 � 10�3 s

1.4 � 10�3 J; 7.9 � 10�4 s

� (2.4 � 10�3 sin (6000pt cos (6000pt) W

PL � 180t � 90t cos (6000pt) � (4.8 � 10�3) sin (6000pt)

PR � 4.5 � 10�3 � 1.1 � 10�3cos 2(6000pt) W,

IL(t) � 60t � (1.6 � 10�3) sin (6000pt) A.

IR(t) � 1.5 � 10�3 A � (7.5 � 10�4 A) cos (6000pt),

I � 0.

I � �0.89 A.

I � 0.Imax � 0.89 A.

2.95 � 10�8 J

5.9 � 10�8 J,t � 0;t � 1>240 sec,

Q � 6.5 � 10�8 sin (120 pt) F;

I � (2.4 � 0.43cos 360t)A;

I � 0.

I � �1.63 A,I � 0.

Imax � 1.63 A.

Igenerator � 9.09 � 104 A, Iline � 5.0 � 103 A

N1

N2

� 34.8N1

N2

� 16.5,

N1

N2

� 7.6,N1

N2

� 0.044,

A-60 ANSWERS

3. (a)

(b)

5.

7.

9. 0.032 A, 0.064 A

11.

13. (a) (b)

(c)

15.

17. For an amplitude of 1.00 V:

t (s) (W)

0.001 8.17E-04

0.002 1.58E-04

0.003 6.19E-05

0.004 6.52E-04

19.

21.

23.

25. 0.033 A, 0.017 A

27. 0.049 A

29. 0.343 A/s, 0.00218 Hz

31. at 60 Hz, at 100 MHz

33.

35. (a)

(b)

37.

39. 379 Hz

41. 663 Hz

43. 2.11 Hz, 259 H

45.

47.

49. (a)

(b) From the impedance triangle,

(c) . No.

51. (a) (b) 12 V; (c) 24 W; (d)

53.

55.

� B aEmax

Rb 2

� aEmax

XC

�Emax

XL

b 2

Imax � 2I 2R � (IC � IL)2

� �R

LEmax>10;0.995Emax;

1.6 � 1021.9 � 103 V;

cos f � 0.642, f � ;50�

cosf �R

Z

P �1

2Imax

Emax aR

Zb

8.23 � 10�6 F, 6.25 W

2.76 � 10�8 F, 7.5 � 10�8 H

9.4 � 103 rad/s

C � 1.66 � 10�6 F, L � 6.6 � 10�2 H

Imax � 5.0 � 10�6 A,

3.6 � 103 Hz

3.0 � 104 �0.018 �

f � 1.6 � 103 Hz

1.0 � 105 A/s

5.6 � 10�5 F

0.0012cos2(600t)

1.0 � 10�9 F

I(t � 0) � 0 A, I(t � 4p>�) � �6.8 � 10�4 A

9.6 � 10�4 A;208 �;

2.0 � 106 �, 1.5 � 106 �

Irms � 31.1 A, Pave � 48.5 kW

Irms � 10.4 A, Imax � 14.8 A, R � 11.1 �

Pmax � 3.4 � 108 W, Pave � 1.7 � 108 W

Emax � 3.25 � 105 V, Imax � 1.05 � 103 A;

IC

IR

IL


15. E is to acceleration. B is to acceleration and

propagation.

17.

19. west

21.

23.

25.

27. 0.30 GHz, no

29. North,

31. 1/4

33. 1.5%

35.

37. (b) The field rotates

clockwise with a period of

39.

41.

43. AM: 566 m to 187 m; FM: 3.41 m to 2.78 m

45. (a) (b) ;

(c) (“electric” wave)

47. 0.027 m

49.

51.

53. 1.7 � 10�9 T down; 6.6 � 10�4 W/m2 north

9.5 � 10�6 W/m2

2.8 � 10�8 J/m3

5.0 � 106 m

1.0 cm (microwave radio)1.5 � 10�10 m (X-ray);

3.0 � 1011 Hz

I0 cos 2f

2(sin 2f cos

2� � cos 2f)

2p/�

u � 0�, u � 90�, u � 180�, u � 270�;

39�

2.0 � 10�9 T

3.3 � 10�9 s,

5.00 � 103 s (8.33 min)

3.3 � 10�11 s

1.1 � 10�5 A/m

1 2 3

BE

a

BE

��

ANSWERS A-61

55.

57.

59.

61. 0.065 V/m, 0.043 V/m

63.

65.

67.

69. 51

71.

73. (a) each; (b) and 0 T,

or 0 V/m and (c) one is 7.08 � 10�17

J/m3, the other is zero.

75. 80.0 m, minus z direction,

79.

81. 0.067 sec

83.

85.

87.

89.

91.

93. (a) (b)

Chapter 34

1. 40

3.

5.

7. 7

9.

11. 416 nm,

13. 1.09

15.

17.

19. Fused Quartz

21.

23.

25.

27.

29.

31.

33. 1.000 21

35.

37. 0.77 mm

36.9�

61.0�41.1�,

64.5�

55.6�, 56.1�, 57.7�

24.4�

42�

41.7�

39.014�

50�

1.97 � 108 m/s4.74 � 1014 Hz,

H � 0.90 m; W � 0.31 m

20�

90� � u

5.9 � 1017 m � 63 ly, 3600 stars3.4 � 10�15 W;

2.9 � 10�20 W, 3.6 � 1027 W

5.5 � 10�2 V/m, 1.8 � 10�10 T

9.60 � 103 V/m, 3.2 � 10�5 T

5.00 � 10�7 T north

0.31 W/m2

m0

2pa V0 sin �t

rR�e0pr

dV0 �cos �t b

V0 sin �t

R� e0

A

dV0 �cos �t,e0

A

d�V0 cos �t,V0 sin �t/R,

Ey � cB0 cos (kx � �t)

2.36 � 107 s�1,0.0785 m�1,

1.33 � 10�11 T;

4.0 � 10�3 V/m,1.77 � 10�17 J/m3

3.5 � 1022 N6.0 � 108 N,

8.0 � 103 m2

7.81 � 10�11 N1.49 W/m2,

1.67 � 10�10 J2.19 � 103 V/m,6.37 � 103 W/m2,

6.42 � 10�5 W/m2, 2.02 � 104 W

4.0 � 1026 W

2.8 � 1028 W


111.

lens object distance image characteristics

concave virtual, upright, and

reduced

virtual, upright, and

reduced

convex virtual, upright, and

magnified

real, inverted,

magnified

real, inverted, reduced

113.

115. 1.9 mm, 108

Chapter 35

1.

3. 127 nm for both

5. 244 m

7. 78 nm

9. yes

11.

13. (a) Only one reflected ray suffers a phase reversal;

(c) 1.22 mm

15. 0.257 mm

17. 5.76 mm/s

19. 1.000277

21. 7.4 mm

23. 1.9 mm

25. 0.994

27. measured angles: predicted

angles:

29.

31.

� (degrees) � (radians) I�Imax

1.0 0.0175 0.626

2.0 0.0349 0.064

3.0 0.0524 0.154

33.

39.

41. First order:

Second order:

Third order: no third order max for 700 nm

Second and third orders overlap

43.

45. 0.34 mm

1.38 � 10�5 rad

u400 � 45.1� ;u400 � 28.2�, u700 � 55.7�

u400 � 13.7�, u700 � 24.4�;

19.0�, 40.5�, 77.2�

5.4 � 10�7 m

0, ; 0.0074�, ; 0.0149�, ; 0.0223�, . . .

6.1�, 18.6�, 32.2�, 48.2�

5.5�, 20.5�, 35.5�, 51.5�;

2d �1

2l,

3

2l,

5

2l, . . .

2d �l

n2

, 2l

n2

, 3l

n2

, ;

3.2 � 108 m/s

�30 cm

s � 2f

f � s � 2f

s � f

s � f

s � f

A-62 ANSWERS

39.

41.

43. no dependence on n�

45. 1.4002, 40.25%.

47. 137�29�, 139�14�

49.

51.

53.

55. 16.7 cm

57. 120.0 cm, concave

59. 60.0 cm, 30.0 cm

61. 8.3 mm

63. 21 cm

65. 12 cm

67. virtual, upright, smaller, 0.57

69. 14 cm, 14

71. 45 cm, inverted, enlarged


75.


79. 475 cm behind the lens (also behind the mirror)


83. 3.9 cm to the right of the center of the ball

85. 25.8 cm to infinity

87. 8.9 � 10�2 s

89. diverging lens,


93. 1/64

95. 1340


99.

101.

103. 7.5 cm

105.

107.

mirror object distance image characteristics

concave virtual, upright, and

magnified

real, inverted, and

magnified

real, inverted, and

reduced

convex all virtual, upright, and

reduced

109. 0.24 m�1.7 m,

s � 2f

f � s � 2f

s � f

� � 38.7�

56�

50�31�,

�0.72 cm�22 cm,

�21 cm�60 cm,

�8.6 cm,

�2.5 cm

s � 12Rs � 1

2R,

�10.9 cm

48.8�,

1 n 22

53.1�


47.

ANSWERS A-63

75. (a) (b) tremor is to

eliminating tremor would improve

resolution somewhat

77. (a) (b) 0.34 m; (c) 0.0054 mm

79. cat barely distinguishes the mice

81. film is at least 120 nm thick

83. 454 nm

85.

87. (b) 95.9 MHz

89. 13 km

91. 4.2 cm, 2.1 cm

93. 5.7 km

95. (a) 1.95 km; (b)

Chapter 36

1. (a) 0, 60 km/s; (b) 30 km/s, 90 km/s

3.

5. 0.995c

7. 0.866c

9. 0.16 s.

11. 0.447c

13.

15.

17. 366 m/s; time dilation

19. the clock at the

North Pole will be ahead by in one year.

21. 0.866 m

23. 0.19 m

25. 0.866c

27. 0.44c

29.

31. Relative to the upper part of the belt, the lower belt seg-

ment is moving with a speed greater than V and appears

shortened even more than the base. Therefore the belt is

tightened in both reference frames.

35. 0.65c

37. (a) t � �1.56 �

108 s; (b) The light from the nova will arrive at earth before

the radio message.

x � 7.67 � 1016 m, y � 1.2 � 1017 m,

3.3 � 10�6 C/m3

3.7 � 10�5 s

time dilation factor � 1 � 1.17 � 10�12;

factor � 1 � 7.4 � 10�13

f �

f� 0.58; 0.42; 0.23

4.8 � 10�3c (1.4 � 106 m/s)

(2.60 � 108 m/s)

¢t1 � ¢t2 �2(L1 � L2)

cc2 �

3

2a V

cb 2 d

¢t2 �2

cc (L2 � L1) � aL2 �

L1

2b a V

cb 2 d ;

¢t1 �2

cc (L1 � L2) � aL1 �

L2

2b a V

cb 2 d ;

5.02 � 10�4 W/m2

1.63 � 10�3 radian;

;14.5, ;48.6�

2.24 � 10�6 radian;

1.5 � 10�4 radian,

9.7 � 10�5 radian9.6 � 10�5 radian;

Delta (radians)

I/I m

ax

1.00.0 2.0 3.0

0.5

0.0

1.0

E0 E0

E0

E0

E0

E0

E

E0

E0

E = 4E0

E = 3.41E0

E = 2E0

E = 0

2

2E0

E = 0.59E

2E0

2E0

2E02E0

= 0δ

=δ

δ = π

π

E0

E0

4=δ π

43

=δ π

49. (b) (c)

51. 2.5 cm

53. 0.042 mm

55.

57. (a) not resolved; (b) resolved; (c) colors smeared, not resolved

59.

61. (b) m interference maxima between adjacent diffraction

minima

63.

65. about 0.22 mm in diameter

67.

69. about 0.06 arcsec, about eight times better than from Earth

71. (a) (b) about 1 cm in diameter

73. it is possible to resolve

the moons but they can only be seen as points of light

minimum separation � 6 � 107 m;

6.7 � 10�5 rad;

94�

0.952(3p)

0.045Imax

47�

u � ff � 0� ;


39.

41.

43. 0.00243 nm

45. 0.0312 nm; 0.0336 nm

47. 0.37%; 3.3%; 49%

49. 0.016 nm

51. 62.5 eV

53.

55. 1.24 kV

57. 8.042 kV

59.

61. 0.031 nm

63. 0.872 Hz;

65. a) 1.07 mm, microwave; b)

67. photons/s

69. 0.0486 nm;

71.

73. 7.8 eV

75. 0.062 nm and 0.070 nm; only 0.070 nm will be present

Chapter 38

1. 121.568 nm, 102.573 nm, 97.254 nm, 94.975 nm

3. 4.34 m/s

5. a) 410.3 nm, 434.2 nm; 1.0034; 0.0023; b)

7. a) ; ; b) “Principal” is analo-

gous to Lyman series, “Diffuse” is analogous to Balmer

series.

9.

11. 3.14 � 10�26 m2; 10�3

13. a) atoms ; b) ; c)

15.

17.

19. 0.0283 mm;

21. 10.2 eV

23. From the n � 3 to the n � 1 state.

25.

27. n � 3, the first two lines of the Lyman series and the first

line of the Balmer series.

29. ;

; R � 4a0

33.

35. ;

37. 0.00614 nm;

39. If , then E2>E1 � 1>16l2>l1 � 4

1.1 � 10�5

4pa02pa0

4.98 � 10�6 eV

f �(n2 � n1)

2pa e

24pe0meR3b

E �nUe

24pe0me R3�

e2

4pe0

a 3

2Rb

7.2 � 10�3

�2.5 � 10�5 eV

2.2 � 106 m/s

4.3 � 1032§1.2 � 1061.2 � 10�41.2 � 1018

5.16 � 10�12 J

p � �0.00407d � �0.0013

1.0 � 106 m/s

1.7 � 10�8 V/m

2.12 � 107 m/s

2 � 1020

1.54 � 109 W

1.96 � 10�16 m

1.3 � 10�28 kgm/s

4.0 � 10�15 J

9.21 � 10�24 kgm/s2.76 � 10�15 J;

6.8 � 10�34 Js

A-64 ANSWERS

45.

deviation � 0.8%

47. 0.85c

49. 0.94c

53. 2.7 m/s; 0.016 m/s;

55.

61.

63.

65. 1.5 times the rest mass energy of the electron.

67. 0.828c

69. 9.33 light years from earth, message arrives 21 years after

departure

71. (a) To the ship observer, the 600 nm pulse is emitted

before the 400 nm pulse; (b) 133 nm instead of 400 nm,

1800 nm instead of 600 nm; (c) 1.2 km

73. Dimensions will be The area of

the two faces perpendicular to the direction of motion will

be The four faces whose planes are parallel to the

direction of motion will have The volume

of the cube will be

75. The deviation is 2%.

77. 0.98c if forward, 0.54c if backward

79. 7.74 metric tons

81. tons of TNT

83. 0.33 m/s

85. (a) 0.906c ; (b)

Chapter 37

1.

3.

5. 0.0259 nm; 0.0366 nm

9.

11.

13. 43 W; 0.2 W; liquid nitrogen

15.

17.

21. 44 K

23. a) ; b) 0.07 nm

25. photons

27. photons/s

29. photons/s

31. 1.77 eV; 3.10 eV

33. 0.29 eV; 0.60 V

35. 1.86 eV

37. Red light (700 nm): None; Blue light (400 nm): K; UV light

(280 nm): K, Cr, Zn

3.2 � 1018

1.9 � 1031

7 � 1024

7.16 � 1015 Hz

1.38 � 103 W>m2

6.1 � 10�9 K

5.8 � 10�13 m

1.60 � 10�6 m

5.3 � 10�22 J

3.3 � 10�34 J

�vmuon� � 0.98c, �vantimuon� � 0.67c

1.2 � 1017 J; 2.9 � 107

6.96 � 1020 J;

0.80 m3.

area � 0.80 m2.

1.0 m2.

1.0 m � 1.0 m � 0.8 m.

2.60 ms

1.26 � 10�13 J,

mthermal

msun

� 1.1 � 10�4%

1.02 � 106 eV

1.06 � 10�17 kgm/s

1.3 � 104 m/s.

KN � 5.00 � 10�3mc2,Krel � 5.04 � 10�3mc2,


41. 5.8 mm

43.

45.

47. n � 1 locations where the probability per unit length � 0,

at ; n locations where the proba-

bility per unit length � 1, at

49. 0.091

51. 0.0014 nm

55. a) ; b) ; c) ; d) 37.7 eV, 151 eV,

339 eV, 603 eV

57. 91.176 nm; 820.58 nm; 1458.8 nm

59. 122 nm

61. 0.212 nm; ; ;

63. 0.0454

65. ;

67. 12.8 eV; 97.3 nm; 0.661 eV;

69. ; ; n � 323; �5.9

� 10�17eV

71. ;

Chapter 39

1. n � 1, l � 0 only, for n � 2, l � 0 or 1, and for n � 3, l �

0, 1, or 2

3. or

5.

7. (a) 18 states, (b) argon

9. U and

11.

13. , where g is approximately equal to 2 for the spin

moment and equal to 1 for the orbital moment.

15.

17. and

19. In its ground state, two electrons occupy the n � 1 level with

l � 0, m � 0, and , two electrons occupy the n �

2 level with l � 0, m � 0, and , and the remain-

ing electron has n � 2, l � 1, m � 0, and .ms � �1>2ms � ;1>2

ms � ;1>2

Lx � Ly � UBl(l � 1) � m2

2

Lx2 � Ly

2 � l (l � 1)U2 � m2U2

mp � 2.79mN, mC � 0.703 mN

;msgmB

30�

ms � �1, 0, � 1

5

2U ,

3eU2me

4U�4U ,�3U ,�2U�U , 0, U , 2U , 3U ,

1 � 10�21 m/s5.3 � 10�23 kgm/s

En � �(GM)2m3

2n2U2rn �

n2U2

Gm2M

1.88 � 103 nm

4.87 � 106 m/s1.67 � 10�14 m

5.71 � 1021 m/s2

2.12 � 10�34 Js1.10 � 106 m/s

n2h2

8mL2

nh

2L

2L

n, n � 1,2,3, p

x �(2m � 1)L

2n, m � 1,2, p n

x �mL

n, m � 0,1,2, p n

2

L, 0,

2

L

2.7 � 10�20 m/s

ANSWERS A-65

21.

Number n l m ms of States

1 0 0 �1�2 2

0 0 �1�2 2

2 1 �1 �1�2 2

1 0 �1�2 2

1 �1 �1�2 2

23. K shell: n � 1, l � 0, m � 0, , �3/2, , and

�1/2 (4 states)

L shell: n � 2, l � 0, m � 0, , �3/2, , and

�1/2 (4 states) and l � 1, m � 0 and �1, , �3/2,

, and �1/2 (12 states)

In comparing the hypothetical periodic table with the known

periodic table, we see that the two additional spin states

cause significant changes. For example, the completed shells

would no longer be the present noble gases, but would be Be,

Mg, Ca, Sr, … Instead of two elements with n � 1 ground

states, there would be four: H, He, Li, Be. Instead of Li

through Ne (8 elements) having n � 2, there are now 16

elements, C through Ca.

25. ,

27. 30 (zinc), 60 (neodymium)

29. 0.128 nm

31. 74 (tungsten)

33.

35. (a) , (b) ,

,

37.

39. (a) 37.9 eV, (b) 0.0150 eV

41. , 0.089 nm

43.

45. Because of the steep slope of the curve, a small change in

voltage across the emitter and base produces a large

increase in voltage crossing the emitter-base junction.

This increased current appears in the current IC

leaving

the collector.

47.

1.1 � 10�6 m

9.26 � 10�47 kg # m2

1.60 � 1014 Hz

7.8 � 10�4 eV3.9 � 10�4 eV

1.3 � 10�4 eV5.43 � 10�46 kg # m29.27 � 1013 Hz

1.31 � 10�10 m9.48 � 103 eV

�1>2ms � �3>2

�1>2ms � �3>2�1>2ms � �3>2

++ +V1 V2 V3

R3R1

R2

E

C

B

E

C

n

p

n

p

n

p

B


27. ZTOT � 9 on both sides, NTOT � 9 on both sides

29. 1.64 MeV

31. (a)180 MeV; (b) for Kr;

for Br; 110 MeV for Kr; 70.1 MeV for Br

33. 212Pb; 235U

35. 22Ne; 64Ni

37. 0.820 MeV

39. 4.0 km/s

41. 0.782 MeV

43. 14N; 0.156 MeV

45. 140 yr

47.

49.

51. 271 days;

53. 11 J; 0.15 Gy; 0.26 Sv (26 rem)

55. 20 Gy; 2.3 Gy

57. 1.0 kg

59.

61. 123 MeV

63. 4.78 MeV

65. (a) ; (b) ; (c)

67. (a) ; (b)

69. 3.27 MeV; 4.03 MeV; 17.6 MeV; 18.4 MeV; 7.2

MeV/nucleon

71. 6.4 MeV

73. 0.157 MeV

75. 3.00 MeV

77. 9.91 MeV

79. 1.5 Bq

81.

83. 1.0087 u

85. (a) 16,000 reactors; (b) 42 yr.

Chapter 41

1. (b) 1.36 T

3. (a) 1.88 GeV, (b) 43.4 GeV, (c) 115 GeV

5. 12 leptons altogether

7. Time dilation factor � 9.0;

9. 0.294 kg

11. 118 MeV

13. 67.5 MeV, 1.84 � 10�14 m for each of the -rays

15. Baryon number is conserved; Strangeness is not conserved.

17. Strangeness is conserved in the first reaction; strangeness

is not conserved in the second and third reaction.

19. In the rest frame of the electron, the energy before the emis-

sion equals the rest mass energy of the electron. After the

emission, the total energy equals the rest mass energy of the

electron plus the energy of the photon.This violates the con-

servation of energy law.

g

t � 1.98 � 10�5 s

8.1 � 10�7 g

7.2 � 1010 yr6.6 � 1011kg/s

1.7 � 108 K7.10 � 10�14 J3.24 � 10�15 m

6.94 � 103 kg

1.44 � 10�6 g

2.2 � 108 Bq

7.7 � 1015 Bq

9.69 � 106 m/s1.51 � 107 m/s

A-66 ANSWERS

49. 0.060 W, 12%

51. (a) for n � 1, two electrons and for n � 2, eight electrons,

(b) two, six, and zero

53. , 0, or , , and

55. two electrons occupy the n � 1 level with l � 0, m � 0, and

, two electrons occupy the n � 2 level with l �

0, m � 0, and , and six electrons have n � 2, l �

1, m � 0 and , and .The remaining two elec-

trons have n � 3, l � 0, m � 0, and .

57. 5 (boron), 111 (transuranic elements)

59. 28 (nickel)

61. , , ,

, ,

63. (a) 1.32 eV/bond, (b) 5.12 eV

Chapter 40

1. 16O: Z � 8, A � 16, N � 8; 56Fe: Z � 26, A � 56, N � 30;238U: Z � 92, A � 236, N � 146

3.ISOTOPE Z A N � A - Z

24Na 11 24 1327Al 13 27 1452Cr 24 52 2852Mn 25 52 2763Cu 29 63 3463Zn 30 63 33124Xe 54 124 70138La 57 138 81

5.

Z � 8 13O 14O 15O 16O 17O 18O 19O 20O 21O 22O 23O

N � A - 8 5 6 7 8 9 10 11 12 13 14 15

7. First reaction is not possible (Z is OK, A is wrong); Second

reaction is not possible (Z is wrong, A is OK); Third reac-

tion is not possible (Z is wrong, A is OK)

9.

11. is correct; is not correct; is correct; is

not correct; ;

13. 13C; 16O

15.

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

No. of isobars 1 1 2 1 2 3 2 4 4 3 5 4 4 5 4 4 61H, 2H, 4He have no isobars

17.

19.

21. 28.3 MeV

23. 128 MeV for 16O, 112 MeV for 16F

25. 6.3 MeV/neutron vs 8.38 MeV/nucleon in 140Ce

1.46 � 10�8 u

4.3 � 10�15

19779Au89

39Y

20783Pb118

50Sn4422Sc31

15P

2.5 � 10�15 m; 3.2 � 10�15 m

1.60 � 10�4 m1.24 � 10�21 J2.40 � 10�4 m

8.27 � 10�22 J4.80 � 10�4 m4.14 � 10�22 J

ms � ;1>2ms � ;1>2;1

ms � ;1>2ms � ;1>2

45�90��45��1m � �1


21.left side right side left side right side

Reaction baryon # baryon # strangeness strangeness

K� � p S �� 0�1�1 1�0�1 �1�0��1 �1�0��1

K� � p S �0 � �0 0�1�1 1�0�1 �1�0��1 �1�0��1

K� � p S � � �0 0�1�1 1�0�1 �1�0��1 �1�0��1

K� � p S � � �� 0�1�1 1�0�0�1 �1�0��1 �1�0�0��1

23.

ANSWERS A-67

35.

37.

39. The Earth and the oldest globular clusters were older than the

estimated total age of the universe according to Hubble’s Law.

41. 0.789 MeV

43. Baryon number is not conserved in the first reaction, elec-

tric charge is not conserved in the second reaction, energy

is not conserved in the third reaction.

45. Electric charge is , the particle is .

47.

©��1

9.87 � 10�9 nm

t

x

v

w+

e–

p–

n–

25. ;

27. 129 GeV, greater than the energy required to achieve sym-

metry between W and photons.

29.

31. 30 quarks are created.

33. In table 41.2, no particle is its own antiparticle. In table

41.3, is its own antiparticle, is its own antiparticle,

is its own antiparticle. The latter three consist of a quark

and the corresponding antiquark.

�

J>cp0

uud

¢tZ � 7.23 � 10�27 s, dZ � 2.17 � 10�18 m

¢t0 � 4.70 � 10�24 s, d0 � 1.41 � 10�15 m

P

+

t

x



Owners Manual: p. xxv: Michael Kim/Corbis; p. xxvi: James L. Amos/Corbis; p. xxix:

NASA/Johnson S pace Center ; p. x x i x : Mark Bol ton/Corb i s ; p. x x i x : Rober t

Laberge/Allsport/Getty Images; p. xxix: Robert Laberge/Allsport/Getty Images; p. xxx:

Duomo/Corbis; p. xxxi: George Hall/Corbis.

Prelude:

100: John Markert/Junenoire Photography; 101: John Markert/Junenoire Photography; 102: John

Markert/Junenoire Photography; 103: Jim Mairs; 104: Aerial Imagery Courtesy of GlobeXplorer.com;

105: Aerial Imagery Courtesy of GlobeXplorer.com; 106: Aerial Imagery Courtesy of

GlobeXplorer.com; 107: Jacques Descloitres, MODIS Rapid Response Team, NASA/GSFC; 108:

NASA/Corbis; p.5: 109: NASA/ JSC; 1021: Dr. Fred Espenak/Photo Researchers, Inc.; 1022: David

Malin; 1023:The Hubble Heritage Team AURA/STScl/NASA; 1024: NASA/ESA/R.Thompson

(University of Arizona); 1026: Max Tegmark/SDSS Collaboration; 100: John Markert/Junenoire

Photography; 10�1: John Markert/Junenoire Photography; 10�2: Professor Pietro M. Motta/Photo

Researchers, Inc.; 10�3: OMIKRON/Photo Researchers, Inc.; 10�4: SPL/Photo Researchers,

Inc.; 10�5: G. Murti/Photo Researchers, Inc.; 10�6: DOE/Science Source.; 10�7: Kenneth

Eward/Photo Researchers, Inc..; 10�8: Eurelios/Phototake.

Part Openers: Part Opener 1: NASA/GRIN; Part Opener 2: L.Weinstein, NASA/Photo

Researchers, Inc.; Part Opener 3: Alfred Pasieka/Photo Researchers, Inc.

Chapter Opener 1: Roger Ressmeyer/Corbis; fig. 1.5: Photo by H. Mark Helfer/NIST; fig.

1.7: Photo by Barry Gardner; fig. 1.8: ©2004 Bruce Erik Steffine; table 1.1a: NASA/JSC; table

1.1b: Charles O Rear/Corbis; table 1.1c: U.S. Mint Handout/Reuters/Corbis; table 1.1d:

Courtesy of Robert G. Milne, Plant Virus Institute, National Research Council,Turin, Italy; fig.

1.9: NIST; fig. 1.10: Robert Rathe/NIST; fig. 1.11: BIPM (International Bureau of Weights

and Measures/Bureau International des Poids et Mesures), www.bipm.org; table 1.7a: NASA/JSC;

table 1.7b: Gene Blevins/LA Daily News/Corbis; table 1.7c: Royalty-Free/Corbis; table 1.7d:

Clouds Hill Imaging Ltd./Corbis; fig. 1.13: Reuters/Corbis; fig. 1.15: National Maritime

Museum, UK; fig. 1.19: John Brecher/Corbis.

Chapter Opener 2: Reuters/Corbis; table 2.1a: Corbis; table 2.1b: Randy Wells/Corbis; table

2.1c: David Muench/Corbis; fig. 2.19: James Sugar/Black Star; p.51: The Granger Collection,

Photo credits

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A-70 Photo Credits

New York; fig. 2.22: Tom Sanders/Photri/Microstock; fig. 2.23: Wally McNamee/Corbis;

fig. 2.26: Taxi/Getty Images; fig. 2.31: NASA/JPL.

Chapter Opener 3: NOAA/Hurricane Research Division; fig. 3.3: Ron Watts/Corbis; p.71:

(Pip) Galen Rowell/Corbis.

Chapter Opener 4: Dr. J. Alean (Stromboli); p.99: (Pip fig. 2) Photo Courtesy of Mark

Wernet/NASA/Glenn Research Center, Cleveland Ohio; fig. 4.7: Krafft-Explorer/Photo

Researchers, Inc.; fig. 4.9: Richard Megna/Fundamental Photographs; fig. 4.19: Aero

Graphics/Corbis; fig. 4.15: Richard Megna/Fundamental Photographs; fig. 4.16: Wally

McNamee/Corbis; fig. 4.22: NASA/JSC; fig. 4.27: Roger Ressmeyer/Corbis; fig. 4.34: Fermilab;

fig. 4.35: Robert Harrington, www.BobQat.com.

Chapter Opener 5: Bob Krist/Corbis; fig. 5.1: Courtesy of John Markert; p.131: The Granger

Collection, New York; fig. 5.5: Courtesy DYNCorp/NASA/JSC; table 5.1a: NASA/JSC; table

5.1b: Mark Bolton/Corbis; fig. 5.6: Bettmann/Corbis; fig. 5.8: TEK Image/Photo Researchers,

Inc.; fig. 5.14: Tim Kiusalaas/Corbis; fig. 5.15: Jim Sugar/Corbis; fig. 5.16: Roger

Ressmeyer/Corbis; fig. 5.44: Science Museum, London.

Chapter Opener 6: Michael Kim/Corbis; p.174: James L. Amos/Corbis; fig. 6.17: Kai

Pfaffenbach/Reuters/Corbis; p.189: (Pip fig. 2) Photodisc Green/Getty Images; p.189: (Pip

fig. 3) Courtesy Beckman Coulter, Inc.; fig. 6.25: Duomo/Corbis; fig. 6.26: George Hall/Corbis;

fig. 6.27: October 2001 Physics Today (Volume 54, Number 10, p.39), Courtesy of John Yasaitis,

Analog Devices, Inc.; fig. 6.28: Brian Erler/Corbis; fig. 6.39: Duomo/Corbis.

Chapter Opener 7: Lester Lefkowitz/Corbis; p.207: Hulton Deustch/Corbis; fig. 7.17: Jim

Cummins/Corbis; fig. 7.19: Courtesy Klockit, Inc.; p.221: Bettmann/Corbis; fig. 7.23: David

Cumming/Eye Ubiquitous/Corbis; fig. 7.32: Jim Cummins/Corbis; fig. 7.34: Lester

Lefkowitz/Corbis; fig. 7.37: Image Bank/Getty Images.

Chapter Opener 8: Courtesy Blenheim-Gilboa Pumped Storage Power Project/New York

Power Authority; p.236: The Granger Collection, New York; p.242: (Pip) Courtesy New York

Power Authority; fig. 8.11: Paul A. Souder/Corbis; p.248: Corbis; table 8.1a: Courtesy

International Dark Sky Association, Defense Meteorological Satellite Program (DMSP).table

8.1b: Royalty-Free/Corbis; table 8.1c: Mika/Zefa/Corbis; p.254: North Wind Picture Archives;

table 8.2a: Earth Observatory/NASA; table 8.2b: Stone/Getty Images; table 8.2c: Taxi/Getty

Images; fig. 8.17: Courtesy New York Power Authority; fig. 8.18: Corbis; fig. 8.21: Paul A.

Souders/Corbis; fig. 8.27: National Archives.

Chapter Opener 9: Bettmann/Corbis; fig. 9.5: Science Museum and Society Picture Library;

p.277: The Granger Collection, New York; p.279: The Granger Collection, New York; fig. 9.8:

NASA/Corbis; p.281: (Pip) NASA; table 9.1a: JPL/NASA; p.285: The Granger Collection,

New York; fig. 9.15: Bettmann/Corbis; fig. 9.28: NASA; fig. 9.30: ©2002 Calvin J. Hamilton;

fig. 9.36: National Astronomy Observatiries/AP Images; fig. 9.37: Dennis di Cicco/Corbis;

fig. 9.40: Photofest; fig. 9.41: JPL/NASA; fig. 9.42: Photographer: Mark Avino ©Copyright

1995 by Smithsonian Institution.


Photo Credits A-71

Chapter Opener 10: Reuters/Corbis; fig. 10.6: Richard Megna 1990 Fundamental Photographs;

fig. 10.13: ©2005 Estate of Alexander Calder/Artist Rights Society (ARS), New York; fig.

10.15: Roger Ressmeyer/Corbis; p.320: (Pip fig. 1) Macduff Everton/Corbis; p.320: (Pip fig. 2)

Right George D. Lepp/Corbis; fig. 10.18: Jason Reed/Reuters/Corbis; fig. 10.22: NASA/Roger

Ressmeyer/Corbis; fig. 10.25: Philip James Corwin/Corbis.

Chapter Opener 11: Tom Wright/Corbis; fig. 11.1: Daimlerchrysler; p.343: (Pip: top and

bottom) Robert Laberge/Allsport/Getty Images; fig. 11.6: Smithsonian Institute; fig. 11.11:

Charles D. Winters/Photo Researchers, Inc; fig. 11.17: Reuters/Corbis; fig. 11.20: Oxford

University Press, UK; fig. 11.21: Charles & Josette Lenars/Corbis.

Chapter Opener 12: Courtesy Sandia National Laboratories; fig. 12.1: Courtesy Palm Press, Inc.;

fig. 12.3c: Royalty-Free/Corbis; fig. 12.3b: Picture Arts/Corbis; fig. 12.3d: Hans C. Ohanian;

fig. 12.3d: Richard T. Nowitz/Corbis; fig. 12.3a: Hans C. Ohanian; fig. 12.20: Corbis; fig.

12.32: Neil Rabinowitz/Corbis.

Chapter Opener 13: Don Harlan/Gravity Probe B; fig. 13.6: Courtesy of Gravity Probe B

Photo Archive, Stanford University; fig. 13.11: Wally McNamee/Corbis; fig. 13.14: Nancy

Ney/Corbis; p.414: (Pip fig. 2) Ron Keller, N.M. Museum of Space; p.414: (Pip fig. 1) Science

& Society Picture Library; fig. 13.25: Lawrence Lucier/Getty Images; fig. 13.34: Reuters/Corbis;

fig. 13.37: NASA/JSC.

Chapter Opener 14: Kroll Cranes A/S - Denmark; fig. 14.5: Corbis; fig. 14.29: Tom Pantages;

fig. 14.50: Tim De Waele/Isosports/Corbis.

Chapter Opener 15: NASA/JSC; fig. 15.6: Courtesy of David Hammond; fig. 15.8: Courtesy

of John Markert; fig. 15.15: Loren Winters/Visuals Unlimited; fig. 15.19: National Maritime

Museum; fig. 15.23: Bibliotheque Nationale de France; p.493: (Pip fig. 3) Courtesy John Markert;

fig. 15.26: P.B. Umbanhowar, F. Melo, and H. L. Swinney, “Localized excitations in a vertically

vibrated granular layer,” Nature 382, 793-796 (1996); fig. 15.27: Jim Craigmyle/Corbis; fig.

15.36: Courtesy of John Markert; fig. 15.37: NASA/JSC.

Chapter Opener 16: Vladimir Smolyakov/Stolichnaya Vechernyaya Gazeta /AP Images; fig. 16.1:

Richard Megna Fundamental Photo; fig. 16.17: Richard Megna Fundamental Photo; fig. 16.18:

Keystone/Getty Images; fig. 16.19: Royalty-Free/Getty Images; fig. 16.20: David Nock of

British Car Specialists; fig. 16.21: Royalty-Free/Corbis; fig. 16.27: Francois Gohier/Photo

Researchers, Inc.

Chapter Opener 17: Stone/Getty Images; fig. 17.1: Aaron Horowitz/Corbis; fig. 17.2 Engineering

Applications of Lasers and Holography by Winston E.Kock, Plenum Publishing Co., New York

1975; fig. 17.5: William B. Joyce; fig. 17.7: Courtesy of C.F. Quate and L. Lam Hansen

Laboratory; p.546: (Pip fig. 1) www.777life.com/free photo stock; p.546: (Pip fig. 3) John Ross

Buschert, Goshen College, IN; p.549: AIP Emilio Segre Visual Archives; fig. 17.19: Gary S.

Settles/Photo Researchers, Inc.; p.552: The Granger Collection, New York; fig. 17.20: Museum

of Flight/Corbis; fig. 17.21: John Shelton Photography; fig. 17.22: PSSC Physics 2nd edition

1965 DC Health & Company and Educational Development Center, Inc., Newtown, MA; fig.


A-72 Photo Credits

17.23: PSSC Physics 2nd edition 1965 DC Health & Company and Educational Development

Center, Inc., Newtown, MA; fig. 17.27: Tim Bird/Corbis.

Chapter Opener 18: Woods Hole Oceanographic Institution fig. 18.3: Hans C. Ohanian; fig.

18.8: AD Moore from Introduction to Electromagnetic; fig. 18.9: D.C. Hazen and R.F. Lehnert,

Subsonic Aerodynamics Laboratory, Princeton; fig. 18.10: D.C. Hazen and R.F. Lehnert,

Subsonic Aerodynamics Laboratory, Princeton; fig. 18.11: Royalty-Free/Corbis; fig. 18.13:

Image Design by Nature; fig. 18.14: Paul Edmondson/Corbis; fig. 18.17: Tom Kleindinst,

Woods Hole Oceanographic Institution; p.574: Stefano Bianchetti/Corbis; fig. 18.20: AC

Hydraulic; p.579: (Pip fig. 1) Anonia Reeve/Photo Researchers, Inc.; fig. 18.26: DW Stock

Picture Library/S. Drossinos; fig. 18.28: Stan White Photography; p.581: Archivo Iconografico,

S.A./Corbis; p.585: Bettmann/Corbis; fig. 18.34: Columbia University Physics Department;

fig. 18.37: Principals of Physics, 1994; fig. 18.38: Peter Finger/Corbis; fig. 18.40: Roger

Ressmeyer/Corbis; fig. 18.41: Bettmann/Corbis; fig. 18.48: Royalty-Free/Corbis; fig. 18.49:

Roger Ressmeyer/Corbis; fig. 18.50: AFP/Hyundai Heavy Industries.

Chapter Opener 19: Vince Streano/Corbis; p.604: AIP Emilio Segre Visual Archives; p.606:

Chemical Heritage Foundation Collection; p.608: Bettmann/Corbis; fig. 19.7: Royalty-

Free/Corbis; fig. 19.8: Tom Pantages; fig. 19.9: Crown copyright 1999, Reproduced permis-

sion; fig. 19.10: Courtesy of Cole-Parmer Instrument Company; fig. 19.12: Liquid Crystal

Resources, Glenview, IL; table 19.1a: Julian Baum/Photo Researchers, Inc.; table 19.1b: Visuals

Unlimited; table 19.1c: Dr. Arthur Tucher/Photo Researchers, Inc.; fig. 19.17: Courtesy of

Worthington Cylinders; fig. 19.18: Dr. Kimberly Strong, University of Toronto; fig. 19.20:

National Institute of Standards and Technology.

Chapter Opener 20: Peter Arnold, Inc./Alamy; p.629: (bio) Burnstein Collection/Corbis; fig.

20.1: Bettmann/Corbis; fig. 20.6: Norbert Wu; fig. 20.9: Visuals Unimited; fig. 20.10: AP

Images; table 20.4a: David Taylor/Corbis; table 20.4b: David Pollack/Corbis; table 20.4c:

D.Winters/Photo Researchers, Inc.; fig. 20.19: Lowell Georgia/Corbis; fig. 20.20: Alfred

Pasieka/Photo Researchers, Inc.; fig. 20.22; Private Collection; fig. 20.24: Stanford University.

Chapter Opener 21: The Image Bank/Getty Images; fig. 21.5: Science & Society Picture

Library; fig. 21.6: Courtesy of BMW World; fig. 21.8: Jim Cummins/Corbis. p.667: Bridgeman

Art Library; fig. 21.17: Tom Pantages; fig. 21.19: Tom Pantages; p.678: Bettmann/Corbis;

p.681: Bettmann/Corbis; fig. 21.23: ©2005 The M.C. Escher Company - Holland. All rights

reserved; fig. 21.27: Picture Arts/Corbis; fig. 21.32: Inga Spence/Visuals Unlimited.

Part Opener 4: Andrew Syred/Photo Researchers, Inc.; Part Opener 5: NASA/Marshall Space

Flight Center.

Chapter Opener 22: Canon Inc.; fig. 22.3: Departamento de Física Faculdade de Ciências e

Tecnologia Universidade de Coimbra; p. 698: Art Resource, NY; fig. 22.10a: Corbis; fig. 22.10c:

Department of Physics, University of Oslo, Norway; fig. 2210d: IBMRL/Visuals Unlimited; fig.

22.9a: Jun Yang,Ting-Jie Wang, Hong He, Fei Wei, Yong Jin. Particle size distribution and mor-

phology of in situ suspension polymerized toner. Industrial & Engineering Chemistry Research.

2003, 42 (22): 5568-5575; fig. 22.10b: Wolfson Nanometrology Laboratory at the University

of Strathclyde by Gregor Welsh; fig. 22.13: Chuck Doswell/Visuals Unlimited; fig. 22.14:


H. David Seawell/Corbis; p. 710: North Carolina Museum of Art/Corbis; fig. 22.15:

Lightingmaster, Streamer Delaying Air Terminal; fig. 22.17: Larry Stepanowicz/Visuals Unlimited.

Chapter Opener 23: Kent Wood/Photo Researchers, Inc.; p. 735: (Pip fig. 2): PPC Industries;

fig. 23.22: Courtesy of Ionoptika Ltd., England.

Chapter Opener 24: Trustees of Princeton University; p. 763: American Institute of Physics; fig.

24.22: Trustees of Princeton University; fig. 24.46: Tom Pantages.

Chapter Opener 25: Hank Morgan/Photo Researchers, Inc.; p. 793: Bettmann/Corbis; p. 805:

(Pip fig. 3) IT Stock Int’l/indexphoto.com; p. 805: (Pip fig. 4) Euclid Garmet/Georgia Power;

fig. 25.32: Stanford University.

Chapter Opener 26: (left and right) National Ignition Facility; fig. 26.18: Edward Kinsman/Photo

Researchers, Inc.; fig. 26.23: Edward Kinsman/Photo Researchers, Inc.

Chapter Opener 27: William Taufic/Corbis; fig. 27.2: American Journal of Physics 30, 19,

1963; fig. 27.3: American Journal of Physics 30, 19, 1963; Table 27.1a: National Center for

Atmospheric Research; Table 27.1b: Patrick Bennett/Corbis; Table 27.1c: Brownie Harris/Corbis;

Table 27d: Courtesy of Bosch; Table 27e: Corbis; p. 866: Bettmann/Corbis; fig. 27.9: © Crown

copyright 1999. Reproduced by permission of the Controller of HMSO and the Queen’s Printer

for Scotland; fig. 27.21: John Wilkes Studio/Corbis; fig. 27.26: Principals of Physics by Hans

Ohanian, 1994; fig. 27.28: Courtesy of Superpower, Inc.

Chapter Opener 28: Martyn F. Chillmaid/Photo Researchers, Inc.; fig 28.6b: Loren

Winters/Visuals Unlimited; fig. 28.7b: NASA/Johnson Space Center; fig. 28.7c: AP Photos;

fig. 28.22: Loren Winters/Visuals Unlimited; fig. 28.34: Mark C. Burnett/Photo Researchers,

Inc.; fig. 28.56: Kim Kulish/Corbis.

Chapter Opener 29: © 1990 Richard Megna, Fundamental Photographs, NYC; fig. 29.5: Tom

Pantages; p. 930: American Institute of Physics, Emilio Segrè Visual Archives.; p. 934: Science

Museum/Science and Society Picture Library; Table 29.1a: Maximilian Stock Ltd./Photo

Researchers, Inc.; Table 29.1b: John Chumack/Photo Researchers, Inc.; Table 29.1c: Adrianna

Williams/zefa/Corbis; Table 29.1d: NASA/HST/ASU/J. Hester et al; fig. 29.15: Loren

Winters/Visuals Unlimited; fig. 29.17: Loren Winters/Visuals Unlimited; p. 940: Science

Museum/Science and Society Picture Library; fig. 29.24: © 1990 Richard Megna, Fundamental

Photographs, NYC; p. 943: © 1990 Richard Megna, Fundamental Photographs, NYC; p. 950:

Science Museum/Science and Society Picture Library.

Chapter Opener 30: Lawrence Berkley Laboratory; fig. 30.1: Science Museum/Science and

Society Picture Library; fig. 30.4: © 2002 Richard Megna, Fundamental Photographs, NYC;

fig. 30.21: Neil Borden/Photo Researchers, Inc.; fig. 30.22: © 1986 Richard Megna, Fundamental

Photographs, NYC; p. 978 (Pip fig. 1): Up The Resolution (uptheres)/Alamy; p. 979: (Pip fig.

2) Tek Image/Photo Researchers, Inc.; p. 979: (Pip fig. 3) Courtesy of John Markert; p. 981:

AIP Emilio Segrè Visual Archives; fig. 30.24a: Appl. Phys. Lett., Vol. 61, No. 16, 19 October

1992; fig. 30.24b: Appl. Phys. Lett., Vol. 61, No. 16, 19 October 1992; fig. 30.27: LBNL/Photo

Researchers, Inc.

Photo Credits A-73


Chapter Opener 31: Porterfield-Chickering/Photo Researchers, Inc.; p. 998: Hulton-Deutsch

Collection/Corbis; Table 31.1: Phototake Inc./Alamy; Table 31.2: Image courtesy of Indigo

(R) Instruments, www. indigo.com; Table 31.4: AllOver Photography/Alamy; p. 1011:

Bettmann/Corbis; fig. 31.31: Maximilian Stock Ltd./Photo Researchers, Inc.; fig. 31.45: Marty

Snyderman/Visuals Unlimited.

Chapter Opener 32: (left) Lester Lefkowitz/Getty Images; (right) Chris Knapton/Photo

Researchers, Inc.; fig. 32.3: David Michael Zimmerman/Corbis; fig. 32.18: Loren Winters/Visuals

Unlimited; fig. 32.35: Steve Callahan/Visuals Unlimited; fig. 32.42: Tom Pantages.

Chapter Opener 33: The Imagebank/Corbis; p. 1080: American Institute of Physics p. 1086:

Library of the Academy of Science, Paris. Courtesy of AIP, Niels Bohr Library; fig. 33 18: © 1992

Diane Hirsch, Fundamental Photographs, NYC; fig. 33 20: Philip Bailey/Corbis; fig. 33.21a:

Corbis; fig. 33.21b: Food pix/Getty Images; fig. 33.29: Jim Sugar/Corbis; fig. 33.31: Courtesy

of Hans Ohanian; fig. 33.32: Maximilian Stock Ltd./Photo Researchers, Inc.; fig. 33.33: Steve

Percival/Photo Researchers, Inc.; fig. 33.34: Courtesy of Hans Ohanian.

Chapter Opener 34: NASA/Science and Society Picture Library; fig. 34.1: Marli Miller/Visuals

Unlimited; fig. 34.2: Digital Vision/Getty Images; fig. 34.6: © 1997 Richard Megna, Fundamental

Photographs, NYC; fig. 34.11: Courtesy of Hans Ohanian; fig. 34.20: © 1990 Richard Megna,

Fundamental Photographs, NYC; fig. 34.22 : Hugh Turvey/Photo Researchers, Inc.; p. 1124:

(Pip fig. 1) Hank Morgan/Photo Researchers, Inc.; p. 1124: (Pip fig. 2): PHT/Photo Researchers,

Inc.; fig. 34.24: Junenoire Photography; fig. 34.28: Will/Demi McIntyre/Photo Researchers, Inc.;

fig. 34.31: © 1987 Ken Kay, Fundamental Photographs, NYC; fig. 34.41b: Courtesy of John

Markert; fig. 34.43: David Parker/Photo Researchers, Inc.; fig. 34.44: David Parker/Photo

Researchers, Inc.; fig. 34.61: Reuters/Corbis; fig. 34.66a: Roger Ressmeyer/Corbis; fig. 34.66b:

McDonald Observatory; fig. 34.67:The Lady and the Unicorn: ‘Sight’ (tapestry), French School,

(15th century)/Musée National du Moyen Age et des Thermes de Cluny, Paris/The Bridgeman

Art Library; fig. 34.68: Courtesy of Hans Ohanian; fig. 34.69: © 1990 Paul Silverman,

Fundamental Photographs, NYC; fig. 34.70: Point Reyes National Seashore; fig. 34.88: DK

Limited/Corbis; fig. 34.90: Roger Ressmeyer/Corbis; fig. 34.95: Bo Zaunders/Corbis.

Chapter Opener 35 (left): Steve Percival/Photo Researchers, Inc.; (right): Peter Steiner/Alamy;

fig. 35.1: Peter Aprahamian/Photo Researchers, Inc.; p. 1170: The Granger Collection, NY;

fig. 35.7: Sciencephotos/Alamy; fig. 35.10: Bureau of International des Poids et Mesures, Sevres,

France; p. 1175: The Granger Collection, NY; fig. 35.16: Erich Schrempp/Photo Researchers,

Inc.; fig. 35.17: Courtesy of Chris C. Jones; fig. 35.21: Courtesy of Chris C. Jones; fig. 35.23:

Roger Ressmeyer/Corbis; fig. 35.28: Principals of Physics by Hans Ohanian, 1994; fig. 35.29:

© 1987 Ken Kay, Fundamental Photographs, NYC; p. 1191: Stefano Bianchetti/Corbis; fig.

35.33: Courtesy of Chris C. Jones; fig. 35.38: Courtesy of Chris C. Jones; fig. 35.39: Courtesy

of Chris C. Jones; fig. 35.40: Courtesy of Chris C. Jones; fig. 35.41: Denis Scott/Corbis; fig. 35.42:

Stephanie Maze/Corbis; p. 1198: Mary Evans Picture Library/Alamy; fig. 35.43: Courtesy of

Michael Lockwood, University of Illinois Urbana-Champaign.; fig. 35.45: Courtesy of Chris

C. Jones; fig. 35.46: Northwestern University/Photo Researchers, Inc.; fig. 35.47: Eye of

Science/Photo Researchers, Inc.; fig. 35.57: Principals of Physics by Hans Ohanian, 1994.

Part Opener 6: NASA/Marshall Space Flight Center.

A-74 Photo Credits


Chapter Opener 36: ESA/CE/Eurocontrol/Photo Researchers, Inc.; p. 1217: Hulton

Archive/Getty Images; p. 1235: Science Photo Library/Photo Researchers, Inc.

Chapter Opener 37: Larry Landolfi/Photo Researchers, Inc.; fig. 37.1: Simon Lewis/Photo

Researchers, Inc.; fig. 37.2: Edward Kinsman/Photo Researchers, Inc.; p. 1260: AIP Emilio

Segrè Visual Archives; fig. 37.8: Photo Researchers, Inc.; p. 1268: (Pip fig. 2): Courtesy of Burle

Industries; p. 1269: AIP Emilio Segrè Visual Archives; fig. 37.15: Science Museum/Science

and Society Picture Library; fig. 37.16: © Mediscan/Visuals Unlimited; fig. 37.17: © Science

VU/Visuals Unlimited; fig. 37.21a-c: Courtesy of E.R. Huggins; fig. 37.21d: Courtesy of Chris

C. Jones; p. 1277: AIP Emilio Segrè Visual Archives; p. 1278: AIP Niels Bohr Library.

Chapter Opener 38: (top) Courtesy of Eastman Kodak, Inc. (bottom) Courtesy of Hans C.

Ohanian; fig. 38.1: Prof. C.K. Shih, Department of Physics, The University of Texas at Austin;

fig. 38.3: Adam Jones/Visuals Unlimited; p. 1289: Courtesy of Eastman Kodak; fig. 38.5:

Courtesy of Hans C. Ohanian; fig. 38.6: Courtesy of Hans C. Ohanian; p. 1294: Cavendish

Laboratory, Cambridge; p. 1295: Princeton University/American Institute of Physics/Science

Photo Library; p. 1301: AIP Emilio Segrè Visual Archives; p. 1304: AIP Emilio Segrè Visual

Archives; p. 1310: (Pip fig. 1b): Richard J. Green/Photo Researchers, Inc.; p. 1310: (Pip fig.

2b): Andrew Syred/SPL/ Photo Researchers, Inc.; p. 1311: (Pip fig. 3b): © IBM Research; p.

1311: (Pip fig. 4): Delft University of Technology/Photo Researchers, Inc.

Chapter Opener 39: Professor C.K. Shih, Department of Physics, The University of Texas at

Austin; fig. 39.3a-d: A.F. Burr and A. Fisher, New Mexico State University; p. 1325: AIP Emilio

Segrè Visual Archives; p. 1326: Granger Collection, NY; p. 1332: Courtesy of The University

of Oxford, Museum of the History of Science; fig. 39.10: Courtesy of John Markert; fig. 39.24:

Ton Kinsbergen/Photo Researchers, Inc.; fig. 39.25: Astrid & Hanns-Frieder Michler/Photo

Researchers, Inc.; fig. 39.29: NASA/Corbis; fig. 39.30: Reuters/Corbis; fig. 39.31: Corbis; fig.

39.32: Roger Ressmeyer/Corbis.

Chapter Opener 40: U.S. Department of Energy/Photo Researchers, Inc.; fig. 40.7: John

Cockcroft, Cambridge, UK; p. 1366: Jean-Loup Charmet/Science Photo Library; p. 1373:

Physics Today Collection/AIP/Science Photo Library; p. 1376: (Pip fig. 1): Reuters/Corbis; p.

1377: Ullstein bild/The Granger Collection, NY; p. 1380: AIP Emilio Segrè Visual Archives;

fig. 40.18: Corbis; fig. 40.21: Corbis; p. 1384: Corbis; fig. 40.22: Roger Ressmeyer/Corbis; fig.

40.23: National Ignition Facility (NIF).

Chapter Opener 41: CERN, P. Lopez/Photo Researchers, Inc.; fig. 41.1: Fermilab/Photo

Researchers, Inc.; fig. 41.2: Fermilab/Photo Researchers, Inc.; fig. 41.3: David Parker/Photo

Researchers, Inc.; fig. 41.4a: CERN/SPL/ Photo Researcher, Inc.; fig. 41.4b: CERN/SPL/

Photo Researcher, Inc.; fig. 41.6: Stanford Linear Accelerator Center/SPL/Photo Researchers,

Inc.; fig. 41.7: CERN/Photo Researchers, Inc.; fig. 41.8: LBNL/Photo Researchers, Inc.; p.

1410: Harvey of Pasadena/American Institute of Physics/Science Photo Library; fig. 41.19a: Jean-

Charles Cuillandre/CFHT/Photo Researchers, Inc.; fig. 41.19b: NOAO/Photo Researchers, Inc.;

fig. 41.20a: Dr. Jean Lorre/Photo Researchers, Inc.; fig. 41.20b: Celestial Image Co./Photo

Researchers, Inc.; fig. 41.20c: C. Butler/Photo Researchers, Inc.; fig. 41.21a: NASA; fig. 41.23:

Courtesy of Jeff Hester; fig. 41.22: NASA; fig. 41.23: NOAO/AURA/NSF/Photo Researchers,

Inc.; fig. 41.24: Roger Ressmeyer/Corbis.

Photo Credits A-75



Index

A-77

aberration:chromatic, 1144spherical, 1128, 1144

absolute acceleration, 132absolute motion, 1217absolute temperature scale, 604absolute thermodynamic temperature scale, 609,

686absolute zero, entropy at, 680absorbed dose, 1375Acapulco, divers, 688accelerated charge:

electric field of, 1075–79, 1077radiation field of, 1075–76, 1076, 1077

acceleration, 39–54absolute, 132angular, see angular accelerationaverage, 39, 39t, 40, 60average, in three dimensions, 101–2average, in two dimensions, 96of center of mass, 323centripetal, 113–14, 113, 114, 132, 184–90,

195, 371, 372components of, 95–98, 101as derivative of velocity, 41formulas for, 39, 41instantaneous, 40–41, 41instantaneous, components of, 97instantaneous, in two dimensions, 96–97motion with constant, 42–49, 43, 63, 102–4,

103, 104, 122motion with variable, 54–56negative, 39positive, 39–40standard g as unit of, 52tangential, 371–72translational, 402vectors, 100–101

acceleration of free fall, 49–54, 64universality of, 49, 49

acceleration of gravity, 52–53, 64, 274–75measurement of, 52–53variation of, with altitude, 274–75

accelerators:linear, 1413, 1415for particles, 1363, 1363, 1398–99

acceptor impurities, 1339

accidents, automobile, 339, 343, 355AC circuits, 1030–67AC current, 1031

hazards of, 913–14acoustic micrograph, 539action and reaction, 144–51, 144, 145, 146action-at-a-distance, 274, 722action-by-contact, 722, 723action-reaction pairs, 144–51, 144, 145, 146, 149AC voltage, 1004Adams, J. C., 272addition law for velocity, Galilean, 1218addition of vectors, 72–76, 72, 73, 74, 89

commutative law of, 74by components, 78–79

addition rule for velocities, 115–16, 117adiabatic equation, for gas, 649adiabatic expansion, 668, 689adiabatic process, 647–49air, composition by element and mass, 620, 623,

657air bag, 343air conditioner, 672, 673airfoil, flow around, 570, 582–83airplane:

motion, pitch, roll, yaw, 366propeller, 392

air resistance, 49, 51, 61, 181, 181in projectile motion, 111

Al’Aziziyah, Libya, hottest temperature, 621Alpha Centarui, A and B, xli, 296alpha decay, 1365–67alpha particles, 1363–64

scattering of, 1293–94alpha rays, 1365alternating current, 1031

hazards of, 913–14alternating emf, 1004, 1032–33, 1035,

1046–53Alvin, DSV, 565, 565, 574, 574, 577, 582ammeter, 905, 916Amontons, Guillaume, 174Ampère, André Marie, 941amperes, 697Ampère’s Law, 939–40

displacement current and, 1073–74electric flux and, 1073

modified by Maxwell, 1071, 1073, 1074, 1080,1096, 1097, 1097

amplitude:of motion, 470of wave, 511

Analytical Mechanics (LaGrange), 236analyzer, 1085Andromeda Galaxy, xlivAngers, France, bridge collapse at, 491, 491angle:

elevation, 109, 111, 111of incidence and of reflection, 1115–16, 1115

angle in radians, 368angular acceleration:

average, 370constant, equations for, 374instantaneous, 370rotational motion with constant, 374–76time-dependent, 376–78torque and, 400

angular frequency, 470–71, 471of simple harmonic oscillator, 477of wave, 512, 513–16

angular magnification:of magnifier, 1147of microscope, 1149of telescope, 1150

angular momenta, some typical values, 407tangular momentum, 284, 407t

for circular orbit, 409in elliptical orbit, 291–92orbital, 409quantization of, 1322–23spin, 409torque and, 410–16

angular momentum, conservation of:in planetary motion, 284in rotational motion, 406–10

angular-momentum quantum number, 1296,1322, 1324–26

angular momentum vector, 411, 411angular motion, 375angular position, for time-dependent angular

velocity, 376–77angular resolution, of telescope, 1196–99angular velocity, 369t, 471

average, 369

Page numbers in italics refer to biographies. Page numbers in boldface refer to figures. Page numbers followed by “t” refer to tables.

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A-78 Index

angular velocity (continued)instantaneous, 369for time-dependent angular acceleration,

376–77annulus of sheet metal, 381antibaryons, 1404antielectron, see positronantimatter-matter annihilation, 706antimesons, 1404antineutrino, 1368–70antinodes and nodes, 520–21, 521, 544–45,

544–45antiparticles, 706, 1403–4antiquarks, 1413aphelion, 282, 284, 409

of planets, xl, 285apogee:

of artificial satellites, 286of moons, xxxixof planet, 286

Apollo 16, xxxixApollo astronaut, 295apparent weight, 187–88, 187apple, chemical energy of, 632, 652Archimedes, 581Archimedes’ Principle, 580–82, 599area, 13areas, law of, 283–84Arecibo radiotelescope, 1198–99, 1198argon:

compression, 658Lennard-Jones potential, 263monatomic kinetic energy, 616thermal window of, 656

artificial satellites, 271–72, 281, 286–87, 1344,1344, 1421

apogee of, 286perigee of, 286

astigmatism, 1147astrology, 295astronaut:

weightlessness training, 589see also Apollo; International Space Station;

Skylab missionastronomical unit (AU), 24Atlas rocket, guidance system, 414atmosphere, 573atmospheric electric field, 806–8atmospheric pressure, 577–78

gauge, 573atom, 1397

electron configuration of, 1328–32electron distribution in, 695nuclear model of, 1294, 1294nucleus of, see nucleusquantum structure of, 1320–40stationary states of, 1299structure of, 695, 1287–95

atomic clock, Cesium, 9atomic force microscope (AFM), 475, 475, 1311,

1311atomic mass, 11–12atomic mass unit, 11, 20, 1355atomic number, 1356atomic standard of mass, 11atomic standard of time, 9atomic states, quantum numbers of, 1328–32atomic structure, 695, 1287–95atom smashers, 1397

attractors, 492Atwood’s machine, 403, 421automobile battery, 707, 890–91, 891, 892automobiles:

collisions, 339, 343, 355crash tests of, 339, 340, 355efficiency of, 674, 674electric fields and, 805energy conversions, 674engine cycle, 674impact speed, 343t

automobile stopping distances, 45, 46, 47, 47average acceleration, 39, 39t, 40, 60

formula for, 39in three dimensions, 101–2in two dimensions, 96

average angular acceleration, 370average angular velocity, 369average power, 253average speed, 29–31, 30taverage velocity, 32–35, 33, 101–2

in two dimensions, 95Avogadro’s number, 11, 20, 607axis of symetry, 380–82, 382t

back emf, 1012balance, 136–37

beam, 136–37, 137Cavendish torsion, 277Coulomb’s, 698, 700spring, 136, 136, 151watt, 11, 11

ballistic curve, 111ballistic pendulum, 349–50, 350balloons:

hot air, 126, 581, 594, 602, 605, 612, 618, 622Raven S-66A, 593research, 622

Balmer, Johann, 1291Balmer series, 1291, 1291tbanked curve, 186–87, 187barometer, mercury, 577, 577baryon, 1403, 1404, 1404t, 1405, 1406, 1422baryon number, 1406, 1407

conservation law for, 1406base units, 13bathyscaphe, 589, 589battery:

automobile, 707, 890–91, 891, 892dry cell, 891, 891internal resistance of, 895–96lead-acid, 707, 890–91, 891, 892, 893

Bay of Fundy, 531, 531, 559–60beam balance, 136–37, 137beam dump, 659beat frequency, 518beats, of a wave, 518becquere (Bq), 1375Becquerel, Antoine Henri, 1365, 1366Bell Laboratories, 1421, 1421Bernoulli, Daniel, 585Bernoulli’s equation, 582–85, 586, 587, 598, 599beta decay, 1368–70beta rays, 1365bicycle:

rounding curve, 456suspended, 431upright, 433

Big Bang, xliv, 626, 1420–21

Big European Bubble Chamber (BEBC), CERN,1399

bimetallic strip thermometers, 610, 610, 636, 637binary star system, 297

resolution of telescope and, 1197binding energy of nucleus, 1359–65

curve of, 1361, 1377binoculars, 1156Biot, Jean Baptiste, 950Biot-Savart Law, 948–50, 948, 949blackbody, spectral emittance of, 1259blackbody radiation, 1255–58, 1259–61black holes, 299block-and-tackle, 443–44, 444blood pressure, 579blood vessels, xlviiblowhole, 546blue, 1414–16body-mass measurement device, 134, 134, 468,

468, 478, 482, 490Bohr, Niels, 1295, 1296, 1321Bohr magnetron, 976Bohr radius, 1297Bohr’s postulates, 1296boiling points, common substances, 642tBoltzmann, Ludwig, 608Boltzmann’s constant, 607bomb, hydrogen, 1380bomb calorimeter, 250, 250bonds, interatomic, 1333bones as lever, 442, 442boom, sonic, 552–53, 552Born, Max, 1277, 1277, 1302Bose-Einstein condensate, 623boson, 1403bottom quark, 1415boundary conditions, 522, 522bound charges, 838bound orbit, 245Boyle, Robert, 606Boyle’s Law, 606Brackett series, 1292Brahe, Tycho, 285brake, hydraulic, 575–76brake, power, 456breeder reactor, 1383Bremsstrahlung, 1090, 1273–74Brewster’s Law, 1124bridge, 433, 433

thermal expansion and, 637, 637bridge collapse:

at Angers, France, 491, 491at Tacoma Narrows, 523–24, 524

British system of units, 6–7, 12British thermal unit (Btu), 630Brown Mountain hydroelectric storage plant,

242–43, 242, 243, 249, 257–58, 258bubble chamber, 1396, 1399–1400, 1400, 1402,

1409bulk modulus, 447–48, 447tbullet:

impact on block, 350measuring speed of, 356

bungee jumping, 246–47, 246, 247buoyant force, 580–81

cable, superconducting, 883cable capacitance, 843Cailletet, liquify oxygen, 658


Index A-79

calculus (review), A10–21antiderivative in, A14approximation of small values in, A18–19chain rule for derivatives in, A12derivatives in, 38, A10–12, A11tintegral, A12–17, A15tintegration rules in, A15–16partial derivatives in, A12Taylor series in, A18uncertainties, propagation of, A19–21,

application to Ohm’s Law of, A20–21Calder, Alexander, mobile, 317Caledonian Railway wheel set, 454California Speedway, 343calorie, 250, 630, 631calorimeter, bomb, 250, 250camera, photographic, 1144–45, 1144camera, ultrasonic range finder in, 560Canes Venatici, 1421capacitance, 811–16, 829–38

of earth, 830of single conductor, 829

capacitative reactance, 1036capacitor microphone, 833capacitors, 811–16, 829–38

circuit with, 1035–38energy in, 844–47guard rings for, 848, 848multiplate, 851, 851, 852, 852in parallel, 834parallel-plate, 831–32, 851, 851, 852, 852in series, 834–35

Caph, spectrum of, 1288carbon:

isotopes of, 1355–59, 1356mass of, 1358

Carnot, Sadi, 667Carnot cycle, 668–69, 669, 671–73, 675–76Carnot engine, 667–73

efficiency of, 671–73Second Law of Thermodynamics and, 676

Carnot’s theorem, 675–76carrier, 1410Cartesian diver, 589, 589Cavendish, Henry, 277, 698Cavendish torsion balance, 277, 277cavity radiation, 1257ceiling fan, 367, 368, 371–72cell, triple-point, 609, 609cello, notes available on, 562cells (of eye), rods, cones, xlviiCelsius temperature scale, 611, 612Centaurus, xlicenter of force, 240center of mass, 313–23, 320

acceleration of, 323of continuous mass distribution, 316gravitational force acting on, 430–33motion of, 323–27velocity of, 323–24, 348

centrifugal compressor, 99centrifugal force, 188–89, 189centrifuge, 114, 114, 365, 365, 373, 383centripetal acceleration, 113–14, 113, 114, 132,

184–90, 195, 371, 372Newton’s Second Law and, 185

centripetal force for circular motion, 185centroid, 316Cerenkov counters, 1399

CERN (Organisation Européenne pour laRecherche Nucléaire), 1225, 1238, 1398–99,1402, 1402, 1411

Cesium atomic clock, 9Cesium standard of time, 9cgs system of units, 713Chadwick, James, 1355chain reaction, 1378–79, 1378Chamonix waterfall, 652Champlain Canal, 334changes of state, 642–43chaos, 492–93characteristic spectrum, 1274characteristic time, 1016–17

of RC circuit, 909characteristic X rays, 1303charge, electric, 698–702, 729–30

bound, 838–39bound, in dielectrics, 838–39conservation of, 706–7of electron, 696–97, 698of electron, measurement of, 747of elementary particles, 1403t, 1404t, 1414of particles, 696–97, 698, 706, 1414point, 699of proton, 696–97, 698quantization of, 706SI unit of, 972static equilibrium of, 774–75surface, on dielectric, 840–41

charge distribution, electric field of, 732Charles’ Law, 606charm, of quarks, 1415chemical elements, Periodic Table of, 1328–32,

1329tchemical reactions, conservation of charge in,

707–8Chernobyl, 1383Chicago (Sears Tower), 653, 654chromatic aberration, 1144chromatic musical scale, 539, 539chronometer, 21, 21circuit, electric:

AC, 1030–67with capacitor, 1035–38DC, 1031, 1032frequency filter, 1037with inductor, 1038–41LC, 1041–46loop method for, 898multiloop, 897–900RC, 907–12with resistor, 1031–35RL, 1015, 1015, 1018single-loop, 893–97

circular aperture:diffraction by, 1196–1999, 1196minimum in diffraction pattern of, 1196

circular motion:centripetal force for, 185translational speed in, 374

circular orbits, 278–82, 278, 1321angular momentum for, 409energy for, 290–91in magnetic field, 966

circular polarization, 532, 532clarinet, sound wave emitted by, 538classical electron radius, 1315classical mechanics, quantum mechanics vs., 1287

Clausius, Rudolph, 678Clausius statement of Second Law of

Thermodynamics, 676Clausius’ theorem, 678clock:

Cesium atomic, 9grandfather, 219, 219pendulum, 487, 487, 495, 495synchronization of, 4, 5, 133n, 1220–23

Coast and Geodetic Survey, U.S., 500coaxial cable, 843Cockroft-Walton accelerator, 1363coefficient of kinetic friction, 175–78, 175tcoefficient of linear thermal expansion, 633, 637coefficient of performance (heat), 673coefficient of restitution, 358coefficient of static friction, 175t, 179–80coefficient of volume thermal expansion, 634–35coefficients of friction, 174–81, 175tcoherence of light, 1177“cold resistance,” 869Collider Detector, Fermilab, 1399, 1399colliding beams, 1398–99collisions, 338–64

automobile, 339, 355impulsive forces and, 339–44

collisions, elastic, 342–47conservation of energy in, 344–45, 351–52, 353conservation of momentum in, 344–45, 351–52,

353in one dimension, 344–47one-dimensional, speeds after, 345–47in three dimensions, 351–53in two dimensions, 351–53

collisions, inelastic, 348conservation of energy in, 351–52, 353conservation of momentum in, 351–52, 353in three dimensions, 351–53totally, 348in two dimensions, 351–53

color:of quarks, 1414–15of visible light, 1092

“color” force, 1405color-strip thermometer, 610, 610Coma Berenices, xliiicombination principle, Rydberg-Ritz, 1293comets, 291

Hale-Bopp, 299, 299Halley’s, 291, 298, 298perihelion of, 291period of, 291Shoemaker-Levy, 299

communication satellites, 271–72, 281, 290–91commutative law of vector addition, 74compact disc, 367compass needle, 927, 927components, of vectors, 77–86, 78, 95–98, 97, 99,

101formulas for, 77

compression, 446, 448–49compressor, centrifugal, 99Compton, Arthur Holly, 1269, 1269Compton effect, 1269–72Compton wavelength, 1315concave spherical mirror, 1128, 1130Concorde SST, 553, 557, 563

sonic boom of, 553concrete, thermal expansion of, 637


A-80 Index

condenser, 833conduction band, 1338conduction of heat, 638–42conductivity, thermal, 638–41, 639tconductors, 708, 708, 709–10, 1337

in electric field, 774–76insulators vs., 871potential energy of, 812–13

conservation laws, 205for baryon number, 1406elementary particles and, 1406–7for mass number, 1406

conservation of angular momentum:in planetary motion, 284in rotational motion, 406–10

conservation of electric charge, 706–7in chemical reactions, 707–8

Conservation of Electric Charge, Law of, 710conservation of energy, 205, 207, 223, 235–70,

290, 797in analysis of motion, 223general law of, 248, 249, 252, 662in inelastic collision, 351–52, 353law of, 790in one-dimensional elastic collision, 345in rotational motion, 397in simple harmonic motion, 483in two-dimensional elastic collision, 351–52, 353in two-dimensional inelastic collision, 351–52,

353conservation of mass, 205, 252conservation of mechanical energy, 238

equation for, 239law of, 221–23, 221, 222, 223, 238

conservation of momentum, 307–12, 310, 345,348

in elastic collisions, 344–45, 351–52, 353in fields, 723in inelastic collisions, 351–52, 353law of, 309

conservative electric field, 804–5conservative force, 236–43, 238

gravity as, 288potential energy of, 236–43

constant, fine-structure, 1315constant angular acceleration, equations for, 374constant force, 205, 208constant-volume gas thermometer, 609–10, 609Constitution, USS, 331constructive and destructive interference, 517,

517, 1169in Michelson interferometer, 1175–76for wave reflected by thin film, 1169–73

contact force, 142–43, 143“contact” forces, 697continuity equation, 570control rod, in nuclear reactor, 1381convection, 641conversion factors, 17–19, 20conversion of units, 16–17, 18convex spherical mirror, 1129cooling, evaporative and laser, 624Coordinated Universal Time (UTC), 9coordinate grid, 3–4, 115coordinates, Galilean transformation of, 1218,

1234, 1241coordinates, origin of, 3, 3, 4, 44, 45, 46, 47Copernicus, Nicholas, 279corner reflector, 1116

Corona Borealis, 59corona discharge, 710, 710corona wire, 709cosecant, A8–10cosine, 19, 473–74, 486, A8–10

formula for derivatives of, 473law of cosines, A10

cosmic background radiation, 1421Cosmological Principle, 1419cosmology, 1416–23cotangent, A8–10Coulomb, Charles Augustin de, 700coulomb (C), 696–97, 972Coulomb constant, 699Coulomb potential, 794–95Coulomb’s balance, 698, 700Coulomb’s Law, 698–99, 703, 711, 762, 790, 810,

1074in vector notation, 699

crane (tower), see K-10000 tower cranecritical angle, for total internal reflection, 1123cross product, 83–86, 84, 85, 410–11, 743

of unit vectors, 85crystals, atomic arrangement of, xlixCurie, Marie Sklowdowska, 1373Curie, Pierre, 1373current, electric:

AC, 1031AC, hazards of, 913–14in capacitor circuit, 1035–38DC, 859–63, 862t, 887–925, 1031DC, hazards of, 913–14displacement, 1073–74generating magnetic field, 939–40, 940, 948in inductor circuit, 1038–41“let-go,” 913SI unit of, 972time-dependent, 907–12

current density, 869current loop:

potential energy of, 975–76right-hand rule for, 942torque on, 972–76

current resistor circuit, 1031–35curve:

ballistic, 111banked, 186–87, 187of binding energy, 1361, 1377of potential energy, 244–47, 244

cutoff frequency, 1275cutoff wavelength, 1274–75cyclic motion, 469cyclotron, 964, 967–68, 968, 979–80cyclotron emission, 1090cyclotron frequency, 967Cygnus X-1, 298cylindrical symmetry, 767–68

da Costa, Ronaldo, 14damped harmonic motion, 489damped oscillations, 488–91, 491damped oscillator:

driving force on, 490harmonic, 1045resonance of sympathetic oscillation of, 489,

490–91, 491dark energy, xlv, 1423dark matter, 1423daughter material, 1366

da Vinci, Leonardo, 174–75, 174Davisson, C. J., 1303day:

mean solar, 9sidereal, 294solar, 9

DC-3 airplane:efficiency, 686engines, 267take off speed, 266

DC-10 airliner, accident (Orly, France), 591DC current, 887–925, 1031, 1032

hazards of, 913–14instruments used in measurement of,

905–7DC voltage, 1004de Broglie, Louis Victor, Prince, 1302, 1303de Broglie wavelength, 1302, 1336decay constant, 1373decay of particles, 1365–70decay rates of radioisotopes, 1374–76deceleration, 40decibel, 541–42, 542tdees, 967–68degree absolute, 604density, 13, 316–17, 566

of field lines, 739of fluid, 566–67, 567tof nucleus, 1359

depth finder, 557depth of field, 1145derivative, of the potential, 806–8derivatives, rules for, 38derived unit, 13–14, 20, A20–21, A21tdestructive and constructive interference, 517,

1169in Michelson interferometer, 1175–76for wave reflected by thin film, 1169–73

determinant, 86deuterium, 498dialectic strength, 841diamagnets, 977diatomic gas, energy of, 617–18diatomic molecule, 244dielectric, 829, 838–47

electric field in, 838–40energy density in, 844–45Gauss’ Law in, 842linear, 838

dielectric constant, 840, 841tdiffraction, 553–55, 1190–98

at a breakwater, 553by circular aperture, 1196–99, 1196by a single slit, 1190–96of sound waves, 554of water waves, 553

diffraction pattern, 554, 554of single slit, 1194

diffuse series, 1314dimensional analysis, 16dimensionless quantities, 17dimensions, 16diodes, 1340–42dioxyribonucleic acid (DNA), xlviiidip angle, 954dip needle, 954dipole, 742–44, 756

potential energy of, 743torque on, 743


Index A-81

dipole moment, 743–44magnetic, 973permanent and induced, 887–925

Dirac, Paul Adrien Maurice, 1302, 1325, 1326direct current, 887–925, 1031, 1032

hazards of, 913–14Discoverer II satellite, 297discus thrower, 109, 109dispersion, of light, 1125displacement current, 1073–74displacement vector, 69, 70–72, 70, 71, 88, 96dog, hearing, 558dominoes, toppling, 525donor impurities, semiconductors with, 1339door (swinging), 367, 396Doppler, Christian, 549Doppler shift, 547–53, 547, 548, 561

of light, 1228–29dot product:

in definition of work, 208–9of vector components, 82–83, 82, 86of vectors, 81–83, 81, 83, 86, 208–9

double-well oscillator, 492, 493“doubling the angle on the bow,” 89, 89down quark, 1413–14drag forces, 180–81drift velocity, of free-electron gas, 864driving force, on damped oscillator, 490dry cell battery, 891, 891dumbell, rotation, 411, 412dynamics, 29, 130–72

fluid, 582–87of rigid body, 394–428

dynodes, 1267, 1269

Earth, xxxix–xl, 285t, 286, 286angular momentum of, 409, 427capacitance of, 830coldest and hottest temperature, 621densities and pressures in upper atmosphere, 621tescape velocity from, 292mass distribution within, 390tmoment of inertia of, 388, 389, 390–91perihelion of, 295polar ice cap melting, 425reference frame of, 132rotational motion of, 120rotation of, 9, 132, 476translational motion of, 120

earthquake, Tangshan, China, 533echolocation, 536, 536Echo satellites, 1421Eddington, Arthur, 1276eddy currents, 1003efficiency:

of automobiles, 674of Carnot engine, 671–73of engines, 666–67

Eiffel Tower, 654eigenfrequencies, 523, 545, 546Einstein, Albert, 251, 394, 1217, 1217, 1264–65,

1361photoelectric equation of, 1266–67

elastic body, 182, 445elongation of, 445–49

elastic collision, 342–47conservation of energy in, 344–45, 351–52, 353conservation of momentum in, 344–45, 351–52,

353

in one dimension, 344–47speeds after one-dimensional, 345–47in three dimensions, 351–53in two dimensions, 351–53

elasticity of materials, 445–49elastic moduli, some values, 447telastic potential energy, 236–37, 238electrical measurements, 905–7electric charge, 698–702, 729–30

bound, 838bound vs. free, in dielectrics, 838conservation of, 706–7of electron, 696–97, 698of electron, measurement of, 747of elementary particles, 1403t, 1404t, 1414of particles, 696–97, 698, 706, 1414point, 699of proton, 696–97, 698quantization of, 706SI unit of, 972static equilibrium of, 774–75surface, on dielectric, 840–41

Electric Charge, Law of Conservation of, 710electric circuit, see circuit, electricelectric constant, 699electric dipole, 725, 742–44, 756electric energy:

in capacitors, 844–47of spherical charge distribution, 813

electric energy density in dielectric, 844–45electric field, 721–55, 722, 724t, 1078

of accelerated charge, 1075–79, 1077atmosphere, 806–8of charge distribution, 732conductors in, 774–76conservative, 804–5definition of, 722–23in dielectric, 838–40electric dipole in, 742–44electric force and, 723, 724energy density of, 815as fifth state of matter, 722of flat sheet, 734–36, 759of harmonic traveling wave, 1098induced, 994motion in, 740–44of plane wave, 1079–80, 1079of point charge, 723–24, 728superposition of, 772–73, 773of thundercloud, 725–28in uniform wire, 860

electric field lines, 736–39made visible, with grass seeds, 859of point charge, 736–37, 737, 739, 936sources and sinks of, 739–40

electric flux, 757–61Ampère’s Law and, 1073

electric force, 695–99compared to gravitational force, 696, 699Coulomb’s law for, 698–99electric field and, 723, 724in nucleus, 1355, 1359qualitative summary of, 695–97superposition of, 703and xerography, 709

electric fringing field, 848electric generators, 892electric ground, 803electricity:

frictional, 695, 711Gauss’ Law for, 1074

electric motor, 974electric resistance thermometer, 610, 610electric shock, 913–14electrolytes, 709electromagnet, 946electromagnetic flow meter, 996electromagnetic force, 191, 695, 1405, 1406, 1406telectromagnetic generator, 1000, 1003–4electromagnetic induction, 993–1029electromagnetic interactions, 1405, 1406, 1406telectromagnetic radiation:

kinds of, 1088–91wavelength and frequency bands of, 1090–91,

1091electromagnetic wave, 1070–1110, 1071, 1075–76,

1076, 1077energy flux in, 1093–94generation of, 1088–91momentum of, 1094–96right-hand rule for, 1078–79, 1079speed of, 1080

electromagnetic wave pulse, 1080–83, 1082electromagnetism, 1411electromotive force, see emf (electromotive force)electron:

charge of, 696, 698distribution of, in atoms, 695elliptical orbit of, 1321–22free, 708, 866–67free of gas, 711, 863–65magnetic moment of, 1325, 1355mass of, 137, 137t, 1356measurement of charge of, 748neutron and proton vs., 1356quantum behavior of, 1302–9, 1320–43spin of, 1324–25, 1328–29, 1355–56states of, 1327t

electron-attracting wire, 709electron capture, 1390electron configuration:

of atoms, 1328–32of solids, 1337–39

electron field, 1409–10electron-holography microscope, 694electron microscopes, see microscope, electronelectrons, of neon, xlix-lelectron scanning microscope, 1310, 1310electron-volts (eV), 248, 796, 1302, 1359,

1397–98electrostatic equilibrium, 774electrostatic force, 695–97electrostatic induction, 711electrostatic potential, 790–98

calculation of, 798–803electrostatic precipitators, 735electroweak force, 1411elektron, 695elementary particles, 1396–1430

collisions between, 342–44conservation laws and, 1406–7electric charges of, 1403t, 1404t, 1414masses of, 1403t, 1404, 1404tspins of, 1403t, 1404t

elements, chemical:age of, 1421during Big Bang, 1421transmutation of, 1363


A-82 Index

elephant, rumble (communication), 558elevation angle, 109, 111, 111elevator with counter weight, 154–55, 154, 155,

157ellipse, xxxix, 283

major axis of, 282semimajor axis of, 282, 285, 291

elliptical orbits, 282–86, 282angular momentum in, 291–92of electron, 1321–22energy in, 291–92vs. parabolic orbit for projectile, 287

elongation, 445–49, 445emf (electromotive force), 887–93

alternating, 1004, 1032–33, 1035, 1046–53induced, 994, 998–1007motional, 993–97power delivered by, 901in primary and secondary circuits of trans-

former, 1053–54sources of, 890–92steady, 1004

emission, stimulated, 1090Empire State Building, 63, 227energy, 204–23

alternative units for, 248–49in capacitor, 844–47for circular orbit, 290–91conservation of mechanical, 221–23, 221, 222,

223, 238, 239dark, xlv, 1423of diatomic gas, 617–18electric, see electric energyin elliptical orbit, 291–92equivalent to atomic mass unit, 1361gravitational potential, 218–23, 219, 220, 238,

288–93gravitational potential, of a body, 321of ideal gas, 616–19internal, 616kinetic, see kinetic energylaw of conservation of, 790in LC circuits, 1042Lorentz transformations for, 1234–37magnetic, 1012magnetic, in inductor, 1013–15and mass, 1242–44mass and, 251–53mechanical, see mechanical energymomentum and, relativistic transformation for,

1242–44in orbital motion, 288–93of photon, 1264–65of point charge, 796potential, see potential energyrate of dissipation of, 264rest-mass, 1242rotational, 1334of rotational motion of gas, 617–18sample values of some energies, 249tin simple harmonic motion, 480–83of stationary states of hydrogen, 1299of system of particles, 327–28thermal, 248, 616, 629threshold, 355total relativistic, 1243vibrational, 1333in wave, 1092–96zero-point, 1307

energy, conservation of, 205, 207, 223, 235–70,290, 797

in analysis of motion, 223general law of, 248, 249, 252, 662in inelastic collision, 351–52, 353in one-dimensional elastic collision, 345in rotational motion, 397in simple harmonic motion, 483in three-dimensional elastic collision, 351–52,

353in two-dimensional elastic collision, 351–52, 353

energy bands, 1337energy banks, in solids, 1336–40energy density:

in dielectric, 844–45in electric field, 815in magnetic field, 1014

energy flux, in electromagnetic wave, 1093–94energy level, 244

of hydrogen, 1299in molecules, 1333–36

energy-level diagram, 1298–99, 1299, 1334, 1337,1337

energy quantization, of oscillator, 1259–60energy quantum, 1258–63energy-work theorem, 215, 236, 400engine:

automobile, 665, 674Carnot, 667–73efficiency of, 666–67flowchart, 667heat, 665steam, 665, 671, 671

enriched uranium, 1381entropy, 678

at absolute zero, 680change, in isothermal expansion of gas, 668disorder and, 680irreversible process, 679–80negative, 683

equation of motion, 151–53, 151, 174integration of, 54–56of simple harmonic oscillator, 477of simple pendulum, 485see also Newton’s Second Law

equilibrium:electrostatic, 774of fluid, 575of mass, 155nuetral, 432, 432static, see static equilibriumunstable, 432

equilibrium point, 245, 245equilibrium position, 476–77equipartition theorem, 617–18equipotential surface, 808–9escape velocity, 292

from Earth, 292from Sun, 292

Escher, M. C., waterfall, 662ether, 1218–19ether wind, 1219evaporative cooling, 624evaporative loss, Mediterranean, 657excited states, 1299Exclusion Principle, 1321, 1328–32, 1337expansion, free, of a gas, 664, 668expansion, thermal:

of concrete, 637

linear, coefficient of, 634, 637of solids and liquids, 633–37, 633of water, 635, 635

expansion, volume, 634expansion joints, bridge, 636expansion of railroad rails, 636Explorer I, 286, 298–99Explorer III, 286Explorer X, 298exponential function, 910external field, 742external forces, 311eye:

astigmatic, 1147compound, 1202, 1202insect, 1202, 1202myopic and hyperopic, 1146nearsighted and farsighted, 1146, 1146as optical instrument, 1145–47

eye, components of:cone cells, xlviiiris, xlviretina, xlviirod cells, xlvii

Fahrenheit temperature scale, 611, 612fallout, from nuclear accident, 1383farad (F), 830Faraday, Michael, 994, 998Faraday cage, 804–5Faraday’s constant, 715Faraday’s Law, 997–1001, 1005, 1006, 1008, 1009,

1071, 1071, 1080, 1096–97, 1097farsightedness, 1146, 1146Fermi, Enrico, 1358, 1380fermi (fm), 1358Fermilab, 125, 422Fermi National Accelerator Laboratory

(Fermilab), 1398–99, 1398, 1415Collider Detector, 1399, 1399Tevatron, 1398–99, 1398

fermion, 1403Ferris, George, 388, 421Ferris wheel, 388, 421, 527ferromagnetic materials, 947ferromagnets, 977Feynman, Richard P., 1410, 1410Feynman diagram, 1410, 1410fibrillation, 913field, 722

depth of, 1145electric, see electric fieldelectron, 1409–10as fifth state of matter, 722magnetic, see magnetic fieldquanta and, 1409–11, 1409tradiation, of accelerated charge, 1075–76, 1076,

1077field lines, 736–39

density of, 739of point charge, 736–37, 737, 739sources and sinks of, 738–39

fifth state of matter, 722filter, polarizing, 1084–86, 1084, 1086fine-structure constant, 1315fire extinquisher, 595fire hose:

pressure within, 596rate of water flow, 590t


Index A-83

First Law of Thermodynamics, 662–64first overtone, 522, 522fission, 252, 1355, 1377–79

in alpha-decay reaction, 1366, 1366as source of nuclear energy, 1242

flat sheet, electric field of, 734–36, 759flip coil, 1019flow:

around an airfoil, 570, 582–83of free electrons, 866–67of heat, 638–39, 638incompressible, 569laminar, 569methods for visualizing, 570in a nozzle, 586from source to sink, 570steady, 569streamline, 569turbulent, 571, 571velocity of, 566, 568around a wing, 571, 584

flowmeter, Venturi, 585–86, 596, 596fluid, 566

density of, 566–67, 567equilibrium of, 575incompressible, 569nuclear, 1360static, 575–79, 575

fluid dynamics, 582–87fluid mechanics, 565–99flute, 519, 546, 560, 561flux:

electric, 757–61magnetic, 997–1001, 999

f number, 1144, 1152focal length, of spherical mirror, 1128focal point:

of lenses, 1136of mirror, 1128

foot, 6–7force, 133, 133, 135, 135t, 136

buoyant, 580–81calculated from potential energy, 241center of, 240centrifugal, 188–89, 189“color,” 1405, 1414–15conservative, 236–43, 238, 288conservative vs. nonconservative, 238–39contact, 142–43, 143, 697as derivative of potential energy, 241electric, see electric forceelectromagnetic, 191, 695, 1405, 1406, 1406telectromotive, see emf (electromotive force)electrostatic, 695–97electroweak, 1411external, 311fundamental, 191gravitational, see gravitational forceimpulsive, 339–44internal and external, 311inverse-square, 240moment arm of, 400motion with constant, 151–59net, 138–40, 138, 139normal, 143, 143, 144, 147nuclear, 1355power delivered by, 255between quarks, 1414–15restoring, 182–84, 183

resultant, 138of a spring, 182–84, 183, 184“strong,” 191, 1355, 1359–65, 1405, 1406,

1406t, 1414torque and, 395–97units of, 135–36, 141“weak,” 191, 1369, 1405, 1406, 1406twork done by constant, 205, 208work done by variable, 211–13, 211, 212, 213,

214force, magnetic, 695, 697, 927–31, 927

due to magnetic field, 931magnitude of, 933on moving point charge, 928–30right-hand rule for, 934, 935, 936vector, 933on wire, 969–72

forced oscillations, 488, 490–91, 1046Fornax constellation, xlivFoucault pendulum, 499Fourier’s theorem, 519, 519, 538fractal striations, 492, 493frames of reference, 3, 4, 20, 114, 115

in calculation of work, 208of Earth, 132freely falling, 142, 142inertial, 132, 132, 133, 1219for rotational motion, 366

Franklin, Benjamin, 710, 710Fraunhofer diffraction pattern, 1194Fraunhofer Lines, 1290“free-body” diagram, 146, 146, 147, 148, 153,

157, 158, 159, 176, 180, 181, 184, 186, 187,188

for automobile tire, 403for backbone as lever, 442for box on truck, 440for box titled, 439for bridge, 433for ladder, 438for pulley, 402, 443for string-bob system, 484for tower crane, 435–36

free charges, 844free-electron gas, 711

friction in, 863–65free electrons, 708

flow of, 866–67free expansion of a gas, 664, 689

entropy change in, 668free fall, 49–54, 51, 52, 53, 64, 141, 142, 142,

495–96formulas for, 49, 50universality of, 49, 49weightlessness in, 142, 142

French Academy, speed of sound, 560French horn, 546freon, 672–73frequency, 369–70

beat, 518normal, 523proper, 523resonant, 1043of simple harmonic motion, 470–71threshold, 1266of wave, 510–11

frequency bands, of electromagnetic radiation,1090–91

frequency filter circuit, 1037

Fresnel, Augustin, 1191Fresnel lens, 1156friction, 171–81

air, 49, 51, 61, 181, 181coefficient of kinetic, 175–78, 175tcoefficients of, 174–81, 175tof drag forces, 180–81equation for kinetic, 175equation for static, 179in flow of free–electron gas, 863–65heat produced by, 248kinetic (sliding), 174–78, 175, 176, 177, 190loss of mechanical energy by, 238–39microscopic and macroscopic area of contact

and, 174–75as nonconservative force, 238–39static, 178–80, 179, 180, 190static, coefficient of, 175t, 179–80of viscous forces, 181

frictional electricity, 695, 711fringes, 1170fringing field, 792, 848fuel cells, 892–93

on Skylab, 892fuel consumption, 263tfuel rods, of nuclear reactor, 1381function, oscillating, 1047, 1051fundamental forces, 191

strength of, 191fundamental frequency, 523fundamental mode, 522, 522fusion:

heat of, 642, 642tnuclear, 1421

g, 40, 52–53, 189measurement of, 52–53, 277–78NASA centrifuge, 592standard, 52–53

Gagarin, Yuri, 299gain factor, 1343galaxies, 1417–18

Milky Way, 1417, 1417Galilean addition law for velocity, 1218Galilean coordinate transformations, 1218, 1234,

1241Galilean telescope, 1142Galilean transformation:

for momentum, 1239for velocity, 1218

Galilean velocity transformation, 116Galilei, Galileo, 51, 131, 495

claim on isochronous pendulum, 495, 501experiment on free fall, 495experiments on universality of free fall by, 495isochronism of pendulum and, 501pendulum experiments by, 495tide theory of, 120

Galvani, Luigi, 793galvanometers, 975gamma emission, 1369–70gamma rays, xlix, 1090, 1365, 1421gas:

adiabatic equation for, 649diatomic, energy of, 617, 617, 620, 625diatomic, molar heat, 645, 646tdistribution of molecular speeds in, 615energy of ideal, 616–19entropy change in isothermal expansion of, 668


A-84 Index

gas (continued)of free electrons, 711, 863–65free expansion of, 664, 664, 668ideal, see ideal gasLaw of Boyle for, 606Law of Charles and Gay-Lussac for, 606monatomic, energy of, 616–18, 620monatomic, molar heat, 645, 646tpolyatomic and non-linear, 618, 620polyatomic and non-linear, molar heat, 646troot-mean-square speed of, 614–15specific heat of, 644–47

gas constant, universal, 604gas thermometer, constant-volume, 609–10, 609gauge, pressure, 578gauge blocks, 6, 6gauss (G), 934Gaussian surface, 762Gauss’ Law, 274, 757, 762–72, 790, 1422

in dielectrics, 842for electricity, 1074for magnetic field, 937, 940for magnetism, 1074

Gay-Lussac’s Law, 606Gedankenexperiment, 287Geiger, H., 1294Gell-Mann, Murray, 1413General Relativity theory, 394general wavefunction, 511generator:

electric, 892electromagnetic, 1003–4homopolar, 1004–5

geometric optics, 1112geometry (review), A3–27

angles, A3areas, perimeters, volumes, A3

geostationary orbit, 271–72, 281geostationary satellite, 280, 281, 290–91geosynchronous orbit, 271–72geothermal power plant (Waiakei, NZ), 690Germer, L., 1303GeV, 1398–99, 1411gimbals, 414Glashow, Sheldon Lee, 1411Global Positioning System (GPS), 25, 1216, 1220,

1228gluon, 1409, 1411golf ball, club impact, 347Goudsmit, Samuel, 1325gradient, of the potential, 806–10grandfather clock, 219, 219grand unified theory (GUT), 1411graphite, xlixgrating, 1184–85, 1184

principal maximum of, 1184–85reflection, 1185resolving power of, 1186

gravitation, 271–303law of universal, 131, 272–76, 278

gravitational constant, 273measurement of, 277–78

gravitational force, 191, 272–76, 1405, 1406, 1406tacting on center of mass, 430–33compared to electric force, 696, 699

gravitational interactions, 1405, 1406, 1406tgravitational potential energy, 218–23, 219, 220,

238, 288–93of a body, 321

graviton, 1409, 1411, 1415gravity:

action and reaction and, 146, 146as conservative force, 288of Earth, see weightgalaxies and, 1420work done by, 207, 207

gravity, acceleration of, 52–53, 64,274–75

measurement of, 52–53variation of, with altitude, 274–75

Gravity Probe B satellite, 394, 394, 401, 401, 406,414

gray, 1375Great Pyramid (Giza), 318, 334Greek alphabet, A1green, 1414–16Greenwich time, 9Griffiths-Joyner, Florence, 59ground, electric, 803ground state, 1299guard rings, for capacitors, 848, 848Guericke, Otto von, 161, 591guitar, 523, 526, 530, 531

lowest note available, 558gyrocompass, 414gyroscope, 394, 394, 401, 406, 414, 414

precession of, 415–16

hadron, 1403Hahn, Otto, 1377Hale-Bopp comet, 299, 299half-life, 1372–73Hall, Edwin Herbert, 983Hall coefficient, 990Hall effect, 980–83Halley’s comet, 298, 298Hall sensors, 982Hall voltage, 981hammer in free fall, 366harmonic function, 470harmonic motion, damped, 489

see also simple harmonic motionharmonic oscillator, 477, 492harmonic wave, 510–13, 1098harmonic wavefunction, 513–16Hawking radiation, 1281hearing:

dog, 558threshold of, 538

hearing trumpet, 559heat, 248, 628–60

as energy transfer, 662–63of fusion, 642, 642tlatent, 642mechanical equivalent of, 631–32specific, see specific heattemperature changes and, 630–31transfer, by convection, 641transfer, by radiation, 641transfer; see also R valueof transformation, 642of vaporization, 642, 642t

heat capacity, specific, 630heat conduction, 638–42

equation of, 638–39two-layer, 640, 640

heat current, 638heat engine, 655, 665

heat flow, 638–39, 638across electric insulator, 656

heat pump, 672, 672, 673heat reservoir, 665height, maximum, of projectiles, 109–11Heisenberg, Werner, 1278, 1278, 1302Heisenberg’s uncertainty relation, 1278heliocentric system, 279helium:

ion collision with O2, 352and hydrogen, abundance of in universe, 1421monatomic kinetic energy, 616

helium liquid, 624helium nuclei, 1366Helmholtz, Hermann von, 248henry, 1010Henry, Joseph, 1011hermetic chamber, 626Hertz, Heinrich, 1080, 1080, 1265hertz (Hz), 471high-voltage transmission line, power dissipated

in, 904Hobby-Eberly telescope, 1152, 1152“holes,” 982

in semiconductors, 1338–39, 1340–42homopolar generator, 1004–5Hooke’s law, 182–84, 445–46, 476Hoover Dam, 993horizontal velocity, 103–8, 103, 106, 107, 108horse, work efficiency, 689, 689horsepower (hp), 254, 254, 256house walls, heat flow in, 639–40, 652, 659Hubble, Edwin, 1416–17, 1418Hubble’s Law, 1418–19, 1420Hubble Space Telescope, 1111, 1130, 1135, 1152,

1198, 1198, 1209human:

aorta, average blood flow, 590chemical energy conversion, 685–86circulatory system, 571, 571ear, audible frequencies, 558heart pressure, 591muscle striation, 685power ed bicycling, 666, 666powered flight (over English Channel), 264venous presssure, 597, 597voice, pitch, 559

human body:lever-like motion of, 442, 451, 458–59mean tissue density, 598, 626speed of sound in, 558temperature, 612ultrasound frequencies used in, 558, 562

hurricane, barometric pressure, 598Huygens, Christiaan, 221, 495Huygens’ Construction, 1113–14, 1113Huygens-Fresnel Principle, 1190–91Huygens’ tilted pendulum, 495, 495Huygens’ wavelets, 1113, 1120Hyades Cluster, xliiiHyatt Regency hotel, collapse of “skywalks” at,

451, 451hydraulic brake, 575–76, 576hydraulic car jack, 576hydraulic press, 575, 575hydroelectric pumped storage, see Brown

Mountainhydrogen:

atom, stationary states of, 1298, 1300


Index A-85

atomic mass, xlixelectron orbit, 423energy levels of, 1300and helium, abundance of in universe, 1421interstellar gas, density and temperature, 621isotopes of, 1357quantum numbers for stationary states of, 1328spectral series of, 1291–93, 1299, 1324

hydrogen bomb, 1380hydrogen molecule:

diatomic gas, 625oscillations of, 483, 483vibration frequency, 498

hydrogen spectrum, 1291–92, 1291tproduced by grating, 1184

hydrometer, 594hyperbola, 291hyperbolic orbit, 291hyperopic eye, 1146

ice:density of, 581, 581density v. liquid water, 634

iceberg, 593, 593ideal gas, 604

energy of, 616–19kinetic theory and, 602–27

Ideal Gas Law, 603–8, 627, 646, 670ideal-gas temperature scale, 609–12ideal particle, 3, 20ideal solenoid, 943–44, 943image:

real, 1133virtual, 1116

image charges, 727image orthicon, 1269impact parameter, 1294, 1294impedance, for RLC circuit, 1050impulse, 339–40impulsive force, 339–44impurities, acceptor, 1339incidence, angle of, 1115–16, 1115inclined plane, 153, 157incompressible flow, 569incompressible fluid, 569, 576indeterminacy relation, 1278index of refraction, 1118, 1119t, 1125–26, 1125induced dipole moments, 744induced electric field, 994, 996induced emf, 994, 998–1007induced magnetic field, 1071

in capacitor, 1072, 1082inductance, 1008–12

mutual, 1009–10self-, 1011

induction:electromagnetic, 993–1029electrostatic, 711Faraday’s Law of, 997–1001

induction furnace, 1024induction microphone, 1000–1001inductive reactance, 1039–40inductor:

circuit with, 1038–41current in, 1038–41

inelastic collision, 348conservation of energy in, 351–52, 353conservation of momentum in, 351–52, 353in three dimensions, 351–53

totally, 348in two dimensions, 351–53

inertia:law of, 132, 132moment of, see moment of inertia

inertial reference frames, 132, 132, 133infrared radiation, 1090initial speed, 111instantaneous acceleration, 40–41, 41

components of, 97in two dimensions, 96–97

instantaneous angular acceleration, 370instantaneous angular velocity, 369instantaneous power, 253instantaneous velocity, 35–39, 36, 61

components of, 96as derivative, 38formulas for, 37, 38graphical method for, 37numerical method for, 37–39as slope, 35–37in two dimensions, 96, 96

instantaneous velocity vector, 98, 98insulating material, see dielectricinsulators, 708

conductors vs., 871electron configurations of, 1338resistivities of, 871

integrals, for work, 212–13integrated circuit, 1345integration, of equations of motion, 54–56INTELSAT, 281, 281intensity:

of sound waves, 538, 540–43, 542t, 543for two-slit interference pattern, 1179

interaction:“color,” 1405electromagnetic, 1405, 1406, 1406tgravitational, 1405, 1406, 1406t“strong,” 1405, 1406, 1406t“weak,” 1369, 1405, 1406, 1406t

interatomic bonds, 1333interference, 1169–86

constructive and destructive, 517, 1169,1169–75

maxima and minima, for multiple slits, 1184,1187–89

maxima and minima for, 1179from multiple slits, 1183–89in thin films, 1169–73two-slit, pattern for, 1180, 1180from two slits, 1177–83, 1177

interferometer, Michelson, 1174–77, 1175interferometry, very-long-baseline, 1202internal energy, 616internal forces, 208, 311internal kinetic energy, 348internal resistance, of batteries, 895–96International Bureau of Weights and Measures,

1175International Space Station, 389, 468international standard meter bar, 5–6, 5International System of Units (SI) 5, 14, 972

see also system of units (SI)interstellar hydrogen gas, density and temperature,

621invariance of speed of light, 1220–21inverse Lorentz transformation equations, 1236inverse-square force, 240

inverse-square law, 739Io, 64, 64ion gun, 740–41, 740ionosphere, temperature and density, 620ions, 697

in electrolytes, 711iris, of eye, xlviirreversible process, 678

entropy change in, 679–80isobars, 1389isochronism, 477, 501

of simple pendulum, limitations of, 486isospin, 1407isothermal compression of gas, 689isothermal expansion of gas, entropy change in, 668isotopes, 1355–59

of carbon, 1355–59, 1356chart of, 1357tof hydrogen, 1357radioactive, 1372–76of uranium, 1362, 1383

J/� meson, 1415Jodrell Bank radiotelescope, 1209Jordan, P., 1302Joule, James Prescott, 207, 629, 652joule ( J), 206, 254Joule heat, 902–5Joule’s experiment and apparatus, 631–32, 631Jupiter, 285t, 286

moons Europa, Ganymede, Io, 296

K-10000 tower crane, 429, 429, 435–37, 435–37,448, 448

Keck Telescope, 1151kelvin, 604Kelvin, William Thomson, Lord, 458, 604Kelvin-Planck statement of Second Law of

Thermodynamics, 675–76Kelvin temperature scale, 604, 609–10Kepler, Johannes, 285Kepler’s Laws, 282–86

of areas, 282–84First, 282limitations of, 288for motion of moons and satellites, 286–88Second, 282–84Third, 285–86

kilocalorie, 248–49, 250kilogram, 5, 11, 13, 14

multiples and submultiples of, 13t, 134standard, 11, 134

kilometers per hour (km/h), 30kilowatt-hours, 248, 250kinematics, 29kinetic energy, 214–17, 216, 238

equation for, 215of ideal monatomic gas, 616–18internal, 348relative examples of, 216trelativistic, 1240–41of rotation, 378–84in simple harmonic motion, 480–83of a system of particles, 327–28

kinetic friction, 174–78, 175, 176, 177, 190coefficient of, 175–78, 175tequation for, 175

kinetic pressure, 613–16kinetic theory, ideal gas and, 602–27


A-86 Index

Kirchhoff, Gustav Robert, 894Kirchhoff ’s current rule, 897–900Kirchhoff ’s rule, 1031, 1032, 1038, 1044, 1046,

1054Kirchoff ’s voltage rule, 894, 898, 900, 908Knot meter, 568krypton, monatomic kinetic energy, 616

Lagrange, Joseph Louis, Comte, 236lamda particle, 1400, 1400laminar flow, 569La Paz, Bolivia, airport, 620Large Magellanic Cloud (galaxy), xlivlaser, 1090

stabilized, 6, 6laser cooling, 624Laser Interferometer Gravitational Observatory

(LIGO), 1175laser light, 1178laser printers, 694, 709latent heat, 642launch speed, 111Lave, Max von, 1273Lave spots, 1273law of areas, 283–84Law of Boyle, 606Law of Charles and Gay-Lussac, 606law of conservation of angular momentum, 407–9Law of Conservation of Electric Charge, 710Law of Conservation of Energy, 662, 790Law of Conservation of Energy, general, 248, 249,

252law of conservation of mechanical energy, 221–23,

221, 222, 223, 238, 584law of conservation of momentum, 309law of inertia, 132, 132law of Malus, 1086law of radioactive decay, 1372–76law of reflection, 1114–15law of refraction, 1121law of universal gravitation, 131, 272–76, 278Lawrence, Ernest Orlando, 968laws of planetary motion, 282–86Lawson’s criterion, 1391LC circuit, 1041–46

energy in, 1042freely oscillating, 1041–46natural frequency of, 1042

lead-acid battery, 707, 890–91, 891, 893leaf spring oscillator, 492length, 5–8

precision of measurement of, 6standard of, 5–6, 5, 6

length contraction, 1230–32visual appearance and, 1230–32

Lennard-Jones potential, 262, 263lens:

focal point of, 1136magnification produced by, 1139objective, 1149–50ocular, 1149–50thin, 1135–39

lens equation, 1138lens-maker’s formula, 1135–36Lenz’ Law, 1001–2, 1006lepton, 1403, 1403t, 1406lepton number, 1406, 1407“let-go” current, 913Leverrier, U. J. J., 272

levers, 441–45, 445, 458–59human bones acting as, 442, 442, 451, 458–59mechanical advantage and, 441, 441, 444

Lichtenstein, R., 294lift, 584light:

coherent, 1177dispersion of, 1125Doppler shift of, 1228–29from laser, 1178polarized, 1126–27pressure of, 1094–95quanta of, 1254–55reflection of, 1112, 1114–17refraction of, 1112, 1117–27spectral lines of, 1127spectrum of, 1126, 1184, 1291–93ultraviolet, 1090unpolarized, 1084, 1084visible, 1090–91

light, speed of, 6, 1218–20in air and water, 527invariance of, 1220–21in material medium, 1118measurement of, 1080universality of, 1220–21, 1238

light-emitting diode (LED), 1343–44lightning, 710, 721, 805

distance of, 560lightning rod, 710–11light waves, 1079–1110

coherent, 1090incoherent, 1090

light-year, xlv, 24linear dielectric, 838linear magnification, 1141–42linear polarization, 532, 532lines of electric field, 736–39

made visible, 859, 859liquid drop model of nucleus, 1360liquids:

bulk moduli for, 447–48thermal expansion of, 633–37, 633

locomotive steam engine, 661, 666, 671Loki (volcano on Io), 64Long Island, xxxviiilongitudinal wave, 508, 509, 509long wave, 1088looping the loop, 187–88, 187loop method, for circuits, 898Lorentz, Hendrik Antoon, 1235Lorentz force, 933Lorentz transformations, 1232–38

for momentum and energy, 1234–37Lyman series, 1292

Mach, Ernst, 552Mach cone, 552, 552Mach number, 562macroscopic and microscopic parameters, 603Magdeburg hemispheres, 161, 161magic nuclei, 1389magnet, permanent, 977magnetic constant, 929magnetic dipole moment, 973magnetic energy, 1012

in inductor, 1013–15magnetic field, 931–38, 935t

circular motion in, 965–69

circular orbit in, 966energy density in, 1014Gauss’ Law for, 937generated by current, 939–40, 940, 948induced, in capacitor, 1072, 1072made visible with iron filings, 937, 942magnetic force and, 931of plane wave, 1081–83, 1082, 1097of point charge, 936of point charge, represented by field lines,

936–37principle of superposition for, 938right-hand rule for, 932, 932, 933of solenoid, 943–44of straight wire, 940–41, 941

magnetic field lines:of circular loop, 942, 942made visible with iron filings, 942, 942,

943magnetic flux, 997–1001, 999magnetic force, 695, 697, 927–31, 927

magnetic field and, 931nmagnitude of, 933on moving point charge, 928–30right-hand rule for, 934, 935, 936vector, 933on wire, 969–72

magnetic levitation, 1008–9magnetic moment, 1355

of electron, 1325spin, 976

magnetic permeability, 977magnetic quantum number, 1321, 1328magnetic recording media, 978–79, 978–79magnetic resonance imaging (MRI), 656, 977, 977magnetism, Gauss’ Law for, 1074magnetization, 977, 984magneto, 1019magnification, 1139magnifier, 1147–49major axis of ellipse, 282Malus, Étienne, 1086Malus, law of, 1086mandolin, 546

frequencies, 530, 531, 558manometer, 578, 578marker point, 29, 29Mars, xl, 285t, 286, 286Mars Climate Orbiter, 16, 16Marsden, E., 1294mass, 11–13

atomic, 11–12atomic standard of, 11of carbon, 1358center of, see center of massconservation of, 205, 252definition of, 134–35of electron, 137, 137t, 1356of elementary particles, 1403t, 1404tand energy, 1242–44energy and, 251–53equilibrium of, 155molecular, 11moment of inertia of continuous distribution of,

379–80of neutron, 137, 137t, 1356of proton, 137, 137t, 1355relative examples of, 12tstandard of, 134–35


Index A-87

of universe, 1422–23, 1423tweight vs., 141

mass defect, 1362mass-energy relation, 1242–44, 1361mass number, 1356

binding energy per nucleon vs., 1361conservation law for, 1406

mass spectrometer, 986, 1393mathematics (review), A1–27

algebra, A3–25arithmetic in scientific notation, A2–23equation with two unknowns, A5exponent function, A5–27logarithms, common (base–10) and natural,

A5–27powers and roots, A1–22quadratic formula, A5symbols, A1see also calculus; geometry; trigonometry; uncer-

tainties (propagation of )matter, dark, 1423matter-antimatter annihilation, 706maximum height, of projectiles, 109–11Maxwell, James Clerk, 1071, 1080, 1411

formula for speed of light by, 1080Maxwell-Ampère’s Law, 1071, 1073, 1074, 1080,

1096, 1097, 1097Maxwell distribution, 615, 615Maxwell’s demon, 683Maxwell’s equations, 1071–72, 1074–75, 1112Mayer, Robert von, see von Mayer, Robertmean free path, 624mean lifetime, 1373mean solar day, 9mechanical advantage, 441, 441, 443–44mechanical energy, 238

law of conservation of, 221–23, 221, 222, 223,238

loss of, by friction, 238–39mechanical equivalent of heat, 631–32mechanics, 29

classical vs. quantum, 1287fluid, 565–99

Mediterranean, evaporative loss, 657medium, wave propagation, 508medium wave, 1088Meissner effect, 989Meitner, Lise, 1377, 1377melting points, common substances, 642tMercedes Benz automobile crash test, 340Mercury, xl, 284, 285t, 286, 286mercury barometer, 577mercury-bulb thermometers, 610, 610, 636, 636merry-go-round, 367meson, 1403, 1404, 1405, 1405t, 1406, 1415metal:

resistivities, 870temperature dependence of resistivity of,

868–69Meteor Crater (Arizona), 362meteoroid incidents, 301meter, 5–6, 13–14

cubic, 13cubic, multiples and submultiples of, 14tsquare, 13, 13square, multiples and submultiples of, 14t

meters per second (m/s), 30meters per second squared (m/s2), 39metric system, 5

MeV, 1359, 1361–62, 1364, 1397, 1415Michelson, Albert Abraham, 1175, 1219, 1220Michelson interferometer, 1174–77, 1175Michelson-Morley experiment, 1175–76, 1219,

1220microelectromechanical system (MEMS), 503,

503microfarad, 830micrograph, acoustic, 539microscope, 1149

angular magnification of, 1149atomic-force (AFM), 475, 475, 1311, 1311electron-holographic, 694scanning capitance, 849

microscope, electron:scanning (SEM), xlvii, 1310, 1310scanning tunneling (STM), xlviii, xlix, 1311,

1311, 1320transmission scanning (TEM), xlviii

microscopic and macroscopic parameters, 603microwaves, 1084, 1090Midas II satellite, 296middle C, 539, 539Milky Way Galaxy, xliii, xliv, 1417, 1417Millikan, R. A., 1267Millikan’s experiment, 747mirror:

focal point of, 1128image formed by, 1116–17magnification produced by, 1139plane, 1116reflection by, 1116–17spherical, 1128–35in telescope, 1151

mirror equation, 1131mirror nuclei, 1388mode, fundamental, 522moderator, in nuclear reactor, 1381modulation, of wave, 518modulus, see bulk modulus; elastic moduli; shear

modulus; Young’s modulusmole, 11, 20, 604–8molecular mass, 11molecular speeds:

distribution of, in gas, 615Maxwell distribution, 615most probable speed, 615root mean square, 614

molecule, diatomic, 244molecules:

energy levels in, 1333–36quantum structure of, 1321, 1333–36rotational energy of, 1334vibrational energy of, 1333water, 389, 743

moment, magnetic dipole, 973moment arm, 400moment of inertia:

of continuous mass distribution, 379–80of Earth, 388, 389, 390–91of nitric acid molecule, 388of oxygen molecule, 388of system of particles, 378–84of water molecule, 389

momentum:angular, see angular momentumof an electromagnetic wave, 1094–96energy and, relativistic transformation for,

1242–44

Galilean transformations for, 1239Lorentz transformations for, 1234–37of a photon, 1270rate of change of total, 312relativistic, 1239–40of a system of particles, 306–13, 324

momentum, conservation of, 307–12, 310, 345,348

in elastic collisions, 344–45, 351–52, 353in fields, 723in inelastic collisions, 351–52, 353law of, 309

monatomic gas, kinetic energy of ideal, 616–18Moon, xxxixmoons of Saturn, 296, 297tMorley, E. W., 1175, 1219, 1220most probable speed, 615, 615motion:

absolute, 1217along a straight line, 28–68, 32amplitude of, 470angular, 375of center of mass, 323–27circular, translational speed in, 37with constant acceleration, 42–49, 43, 63with constant acceleration, in three dimensions,

102–4, 103, 104, 122with constant force, 151–59cyclic, 469energy conservation in analysis of, 223equation of, see equation of motionfree-fall, 49–54, 51, 52, 53, 64, 141, 142, 142harmonic, 489Newton’s Laws of, 130–72one-dimensional, 28–68, 32parabolic, 108–9, 109periodic, 469planetary, 282–86position vs. time in, 32–33, 33, 34, 35, 35, 36,

60of projectiles, 104–12, 122, 124of projectiles, formulas for, 104as relative, 31, 115–18, 1217of rigid bodies, 366–67rotational, see rotational motionsimple harmonic, see simple harmonic motionthree-dimensional, 95with time-dependent angular acceleration,

376–78translational, 29, 95, 95, 120, 366, 404two-dimensional, 94–129uniform circular, 112–15, 112, 113, 125, 184–90in uniform electric field, 740–44with variable acceleration, 54–56wave, 508–9

motional emf, 993–97Mount Fuji, 334Mt. Everest, descent, 127Mt. Palomar telescope, 1151, 1151Mt. Pelée volcano, 622multiloop circuits, 897–900multiplate capacitor, 851, 851, 852, 852multiwire chambers, 1401–2, 1402muon, 1403musical instruments, standing waves, 546,

546mutual inductance, 1009–10muzzle velocity, 287, 331myopic eye, 1146


A-88 Index

nanoelectromechanical system (NEMS), 503NASA centrifuge, 592NASA Spacecraft Center, 114, 114NASA weightlessness training, 589National Ignition Facility (NIF), 828, 845National Institute of Standards and Technology, 9,

500, 609National Ocean Survey buoys, 527National Research Council of Canada, speed of

sound, correction, 560navigation, vectors in, 71near point, 1147nearsightedness, 1146, 1146negative acceleration, 39negative vectors, 75, 75negative velocity, 39negentropy, 683neon, 696

atomic structure of, xlixnucleus of, l–li

Neptune, 285t, 286discovery of, 272

Nernst, Walther Hermann, 681net force, 138–40, 138, 139neutral equilibrium, 432, 432neutrino, 1368–69, 1403, 1423neutron, 696, 1355, 1397

beta-decay reaction and, 1368isotopes and, 1356mass of, li, 137, 137t, 1356nuclear binding energy and, 1359–62in nuclear reactor, 1381quark structure of, 1414

Newton, Isaac, 131, 131, 287, 1358on action-at-a-distance, 722bucket experiment of, 588experiments on universality of free fall by, 496pendulum experiments by, 496

newton (N), 135, 141newton-meter, 396newton per coulomb, 724Newton’s cradle (pendulum), 355Newton’s Laws:

angular momentum and, 400First, 131–33, 131of Motion, 130–72in rotational motion, 395Third, 144–51of universal gravitation, 131, 272–76, 278

Newton’s rings, 1173Newton’s Second Law, 133–38, 171, 205, 214,

400, 413, 514, 604centripetal acceleration and, 185empirical tests of, 135see also equation of motion

Newton’s theorem, 274New York City, xxxvii–xxxviii, 386

electric power, 653Public Library, xxxvii

NGC 2997 (spiral galaxy), xlivNiagara Falls, 266, 688nitric acid molecule:

distance and angles between component atoms,333

moment of inertia of, 388nitrogen molecule: diatomic kinetic energy, 617,

617nodes and antinodes, 520–21, 521, 544–45noise, white, 538

noise reduction, 557normal force, 143, 143, 144, 147normal frequencies, 523normalization condition, 1308n-p-n junction transistor, 1342, 1342n-type semiconductor, 1338–42, 1341, 1343–45nuclear bomb, 1355nuclear density, 1359nuclear explosion, overpressure of blast wave,

590nuclear fission, see fissionnuclear fluid, 1360nuclear force, see “strong” forcenuclear fusion, 1421nuclear model of atom, 1294, 1294nuclear power plant, 1382–83nuclear radius, 1358nuclear reactions, 1371nuclear reactor, 1381–82, 1381

containment shell of, 1382, 1382control rods in, 1381fission, 1380–81fuel rods in, 1381heavy water, 358mass and, 1293moderator in, 1381neutrons in, 1381plutonium (Pu) produced in, 1383rods in, 1381water-moderated, 1383

nucleons, 1355, 1355t, 1359see also neutron; proton

nucleus, 1355binding energy of, 1359–65, 1359, 1360density of, 1359electric force and, 1355, 1359of helium, 1366liquid-drop model of, 1360magic, 1389mirror, 1388stable and unstable, 1360

Oak Ridge National Laboratory, 387object, 1131objective lens, 1149–50observable universe, 1420ocean, energy extraction, 687ocean waves:

amplitude of, 558diffraction of, 553see also seiche; tides; tsunami

octave, 539ocular lens, 1149–50Oerlikon Electrogyrobus, 391Oersted, Hans Christian, 930Ohm, Georg Simon, 866ohm (�), 868ohmmeter, 916, 916Ohm’s Law, 866, 1034oil pipeline, lateral loops, 651oil tanker:

cross section, 591see also supertanker

ommatidia, 1202opera house acoustics, European, 557optical fiber, 1124optical pyrometer, 610, 610optics:

geometric, 1112

geometric vs. wave, 1169wave, 1169

orbit:bound, 245circular, see circular orbitsof comets, 290–91elliptical, 282–86, 282, 287, 291–92, 1321–22geostationary, 271–72, 281geosynchronous, 271–72hyperbolic, 291parabolic, 287, 291period of, 279planetary, 279, 286planetary, data on, 285–86, 285tsynchronous, 271–72unbound, 245

orbital angular momentum, 409orbital motion, energy in, 288–93orbital quantum number, 1321, 1328organ, pipe, 546, 546, 561Organisation Européenne pour la Recherche

Nucléaire (CERN), 1225, 1238, 1398–99,1402, 1402, 1411

origin of coordinates, 3, 3, 4, 44, 45, 46, 47orthicon, 1269oscillating beads, 497oscillating function, 1047, 1051oscillating mass on spring, 473oscillations, 468–506

forced, 488, 490–91, 1046oscillator, 493

damped, see damped oscillatordouble-well, 492energy quantization of, 1259–60harmonic, 492leaf spring, 492simple harmonic, 476–79, 481torsional, 500

overpressure, 578nuclear blast wave, 590

overtones, 522Oxygen, liquify, 658oxygen molecule:

collision with He, 352diatomic kinetic energy, 617, 617moment of inertia of, 388

palladium, atoms, xlixparabola, 291parabolic motion, 108–9, 109parabolic orbit, 291

vs. elliptical orbit for projectile, 287parallel-axis theorem, 382–83parallel-plate capacitor, 831–32, 851, 851, 852,

852paramagnetism, 977parbuckle, 459parent material, 1366parity, 1407parsec, 24particle:

accelerators for, 1363, 1363, 1398–99alpha, 1293–94center of mass of system of, 313–23decays of, 1365–70electric charges of, 696–97, 698, 706, 1414elementary, see elementary particlesideal, 3kinetic energy of system of, 327–28


Index A-89

moment of inertia of system of, 378–84momentum of system of, 306–13, 324in quantum mechanics, 1302–9scattering of, 1294stable, 1404system of, see system of particlesunstable, 1404W, 1409–11w�, 1409, 1411, 1415W�, 1409, 1411, 1415wave vs., 1276–79Z, 1409–11Z0, 1409, 1411, 1415

pascal, 573Pascal, Blaise, 574, 588Pascal’s Principle, 575, 575, 599Paschen series, 1292Pauli, Wolfgang, 1325, 1328–32, 1329t, 1368pendulum, 476, 495, 496, 500

ballistic, 349–50Foucault, 499Huygens’ tilted, 495physical, 487, 501see also Foucault pendulum; Huygens’ tilted pen-

dulum; Newton’s cradle; simple pendulumpendulum clocks, 487, 654–55, 655penetration depth, 989Pentecost Island “land divers,” 356Penzias, Arno A., 1421perigee:

of artificial satellites, 286of comets, 291of Earth, 295of Moon, xxxixperihelion, 282of planets, 286

perihelion, 282, 284, 409of comets, 291of Earth, 295of planets, xl, 285

period, 369of comets, 291of orbit, 279of planets, 285tof simple harmonic motion, 470–71of wave, 510, 510

periodic motion, 469Periodic Table of Chemical Elements, 1328–32,

1329tperiodic waves, 509–16permanent dipole moments, 743permanent magnet, 977permeability constant, 929permittivity constant, 699perpetual motion machine:

of the first kind, 662hypothetical, 662M. C. Escher, 662of the second kind, 662

Perseus Cluster, xliiiphase, 471phase constant, 471phase �, RLC current of, 1050phasor, 1047, 1047, 1049, 1187Phobos, moon of Mars, 424photocopiers, 694photoelectric effect, 1264–69photoelectric equation, 1266photographic camera, 1144–45, 1144

photomultiplier, 1268photomultiplier tube, 1266–67photons, 1255, 1264–69

in Compton effect, 1270–71emitted in atomic transition, 1300–1301energy of, 1264–65momentum of, 1270virtual, 1410

physical pendulum, 487, 501piano:

frequencies, 531notes avaiable on, 558

picofarad, 830pion, 1400, 1401, 1404, 1414pitch, 366Pitot tube, 596, 596planar symmetry, 770–72Planck, Max, 675, 1259–61Planck’s constant, 976, 1259, 1403Planck’s Law, 1259–60plane mirror, 1116planetary motion:

conservation of angular momentum in, 284Kepler’s laws of, 282–86

planetary orbit, 279aphelion of, 285data on, 285–86, 285tperihelion of, 285periodicity of, 492period of, 285t

plane wave, 537electric and magnetic fields of, 1081–83, 1082,

1097plane wave pulse, electromagnetic, 1080–83, 1082plasma, 710Pleiades Cluster, xliii“plum-pudding” model, 1293Pluto, xli, 285t, 286

xli, mass of, 294plutonium (239Pu), produced in nuclear reactor,

1383in chain reaction, 1379

p-n junctions, 1340–42, 1341, 1343–45, 1343point charge, 699

electric field lines of, 736–37, 737, 739, 936electric field of, 723–24, 728energy of, 796magnetic field of, 936moving, magnetic force and, 928–30potential of, 794–99

Poisson spot, 1202, 1202polarization, 839–40, 1083, 1126–27

circular and linear, 532, 532polarized light, 1126–27polarized plane waves, electric and magnetic fields

of, 1081–83, 1082polarizing filter, 1084–86, 1084, 1086Polaroid, 1085, 1086polyatomic and non-linear molecules, energy in,

618, 620position, time vs., 32–33, 33, 34, 35, 35, 36, 60position vector, 76–77positive acceleration, 39–40positive velocity, 39positron, 1369positronium, 1316potential:

derivatives of, 806–8electrostatic, 790–98, 798–803

gradient of, 806–10of point charge, 794–99of spherical charge distribution, 799–800

potential difference, 793potential energy, 218

of argon (Lennard-Jones potential), 263of conductor, 812–13of conservative force, 236–43of current loop, 975–76curve of, 244–47, 244of dipole, 743in double-well oscillator, 492elastic, 236–37, 238of force, equation for, 239force calculated from, 241gravitational, 218–23, 219, 220, 238, 288–93gravitational, of a body, 321in simple harmonic motion, 480–83of a spring, 236–37, 237, 241turning points and, 244–45

potentiometers, 872, 906pound (lb), 12, 134pound-force (lbf ), 136pound-force per square inch, 574Powell, Asafa, 31power, 253–58

average, 253delivered by force, 255delivered by source of emf, 901delivered by torque, 397dissipated by resistor, 1033dissipated in high-voltage transmission line, 904instantaneous, 253sample values of somepowers, 257ttime-average, 1033transported by a wave, 516

power brake, 456power cable, 444

copper contraction, 461Poynting, John Henry, 1093Poynting vector, 1093–94precession, 415

of a gyroscope, 415–16pressure, 448, 566, 567, 573–75, 574t, 590

atmospheric, 577–78barometric in hurricane, 598blood, 579gauge, 578human vein, 597, 597in incompressible fluid, 576kinetic, 613–16standard temperature and, 604, 607–8in a static fluid, 575–79

primary, of a transformer, 1053–54principal maximum, of grating, 1184–85principal quantum number, 1321, 1328–29principal rays, 1129

of lens, 1137Principia Mathematica (Newton), 131, 287principle of relativity, 1220–23principle of superposition, 138, 703

for magnetic field, 938for waves, 516–20, 518

prism, 1123, 1126probability interpolation of wave, 1277problem-solving, guidelines for, 50projectile:

maximum height of, 109–11parabolic vs. elliptical orbit for, 287


A-90 Index

projectile (continued)range of, 109, 111, 111, 287time of flight of, 109–10trajectory of, 105, 106, 107, 111, 111

projectile motion, 104–12, 122, 124with air resistance, 111

propagation of uncertainties, see uncertaintiesproper frequencies, 523proton, 1355

charge of, 696, 698collision of, 361isotopes and, 1356mass of, li, 137, 137t, 1356nuclear binding energy and, 1359–62quark structure of, 1414

Proxima Centauri, xli, 52p-type semiconductor, 1338–42, 1341, 1343–44pulleys, 443–44, 458, 459

mechanical advantage and, 443–44pulsar (neutron star), rotation, 391, 427pupil, of eye, xlvip-V diagram, 668–69, 678

four step, 687, 690one step, 689three step, 686

pyrometer, optical, 610, 610

Q, or quality factor, of oscillation, 489–91, 502–3quality factor (Q), 1045quantization:

of angular momentum, 1296, 1322–23of electric charge, 706of energy, 1260

quantum:of energy, 1260fields and, 1409–11, 1409tof light, 1254–55

quantum Hall effect, 982quantum jump, 1296quantum mechanics, 1287, 1302–9quantum numbers, 1260

angular-momentum, 1296, 1322, 1324–26of atomic states, 1328–32electron configuration and, 1328magnetic, 1328orbital, 1321, 1328principal, 1321, 1328spin, 1324, 1328for stationary states of hydrogen, 1328

quantum structure, 1320–53quark, 1397, 1404t, 1412–16

bottom, 1415charmed, 1415charm of, 1415color of, 1414–15down, 1413–14electric charges of, 706, 1414tforce between, 1414–15strange, 1413–14top, 1415up, 1413–14up and down, within proton, li

quark structure of proton and neutron, 1414Quito, Ecuador, 386

rad, 1375radar “guns” (Doppler), 558radians, position angle, 368radiation, 641

blackbody, 1255–58, 1259–61cavity, 1257electromagnetic, 1090–91electromagnetic, wavelength and frequency

bands of, 1090–91, 1091Hawking, 1281heat transfer by, 641infrared, 1090thermal, 1255, 1256–57, 1256

radiation field, of accelerated charge, 1075–76,1076, 1077

radioactive dating, 1376, 1421radioactive decay, law of, 1372–76radioactive series, 1367–68radioactivity, 1365–72radioisotopes, decay rates of, 1374–76radio station WWV, 9radio telescope, 1186

at Arecibo, 1198–99, 1198at Jodrell Bank, 1209Very Large Array, 1186

radio waves, 1079–1110, 1089, 1094railroad tracks, thermal expansion and, 637, 637range, of projectiles, 109, 111, 111, 287rate of change, of momentum, 312Rayleigh, John William Strutt, Lord, 1198, 1258Rayleigh’s criterion, 1197, 1209rays, 1115, 1115

alpha, 1365beta, 1365gamma, 1090, 1365principal, 1129, 1137

rays, X, 1090, 1303RC circuit, 907–12, 907, 911, 912reactance:

capacitative, 1036inductive, 1039–40

reaction and action, 144–51, 144, 145, 146, 149reactor, nuclear, 1381–82, 1381

breeder, 1383containment shell of, 1382, 1382control rods in, 1381fission, 1380–81fuel rods in, 1381moderator in, 1381neutrons in, 1381rods in, 1381water-moderated, 1383

real image, 1133recoil, 309–10rectangular coordinates, 3, 3rectifier, 1340–42red, 1414–16red-shift, 1418reference circle, 472reference frames, 3, 4, 20, 114, 115

in calculation of work, 208, 208of Earth, 132freely falling, 142, 142inertial, 132, 132, 133for rotational motion, 366

reflection, 1112, 1114–17, 1114angle of, 1115–16, 1115critical angle for total internal, 1123law of, 1114–15by mirror, 1116–17polarization by, 1086, 1125total internal, 1123

reflection grating, 1185

reflector, corner, 1116refraction, 1112, 1117–27, 1126

index of, 1118, 1119t, 1125–26, 1125law of, 1121

refrigerators, use of freon and, 672–73, 673relative biological effectiveness (RBE), 1375relative measurement, 4relativistic kinetic energy, 1240–41relativistic momentum, 1239–40relativistic total energy, 1243relativistic transformation, for momentum and

energy, 1242–44relativity:

of motion, 31, 115–18, 1217principle of, 1220–23of simultaneity, 1220–22special theory of, 1216–53of speed, 31, 31of synchronization of clocks, 1222–23

Relativity, General, theory of, 394rem, 1375resistance, 866

air, 49, 51, 61, 111, 181, 181in combination, 872–76of human skin, 913internal, of batteries, 895–96of wires connecting resistors, 875–76

resistance thermometers, 880resistivity, 865, 866t

of insulators, 871tof materials, 868–72of materials, temperature dependence of, 869of metals, 870of semiconductors, 871ttemperature coefficient of, 880

resistors, 872–76, 872, 873tcircuit with, 1031–35in parallel, 874power dissipated by, 1033power dissipated in, 902resistance of wires connecting, 875–76in series, 874

resolution, angular, of telescope, 1196–99resolving power, of grating, 1186resonance, 523, 1404

of bridge (Tacoma, WA), 523, 524of damped oscillator, 490–91, 491in musical instruments, 546tides, 531

resonant frequency, 1043rest-mass energy, 1242restoring force, 182–84, 183resultant, of vectors, 72, 74resultant force, 138retina, of eye, xlviireversible process, 667Reynolds number, 595rheostats, 872, 872right-hand rule, 84, 84, 85, 743

for a current loop, 942for electromagnetic waves, 1078–79, 1079for magnetic field, 932, 932, 933for magnetic force, 934, 935, 936for magnetic moment, 974for solenoids, 944

right triangle, 19rigid body:

dynamics of, 394–428kinetic energy of rotation of, 378–84


Index A-91

moment of inertia of, 378–84motion of, 366–67, 366–67parallel-axis theorem for, 382–83rotation of, 365–93some moments of inertia for, 382tstatics of, 430–33translational motion of, 366

RLC circuit, 1044–45, 1044, 1045impedance Z of, 1050phase f of, 1050

RL circuit, 1015–18, 1015, 1046roll, 366root-mean-square (rms) speed, of gas molecules,

614–15root-mean-square voltage, 1033rotation:

of the Earth, 9, 132, 476frequency of, 369–70kinetic energy of, 378–84period of, 369of rigid body, 365–93

rotational energy, of molecule, 1334rotational motion:

analogies to translational motion, 374t, 407tconservation of angular momentum in, 406–10conservation of energy in, 397with constant angular acceleration, 374–76of Earth, 120equation of, 399–406about a fixed axis, 367–73of gas, 617–18Newton’s laws in, 395reference frame for, 366torque and, 405work, energy, and power in, 395–99

roulette wheel, 367Rumford, Benjamin Thompson, Count, 629, 629Rutherford, Sir Ernest, 1287, 1293–95, 1294,

1355, 1363, 1365Rutherford backscattering, 357Rutherford scattering, 1293R value, 640, 641, 655

three layers, 659Rydberg constant, 1292, 1300–1301Rydberg-Ritz combination principle, 1293

Sagittarius, constellation, xliiiSalam, Abdus, 1411Sandia National Laboratory centrifuge, 365, 365satellites:

artificial, 271–72, 281, 286–87, 1344, 1344, 1421communication, 271–72, 281, 281, 290–91geostationary, 280, 281, 290–91Kepler’s laws for motion of moons and, 286–88see also specific satellites

Saturn, 285t, 286moons of, 296, 297t

scalar, 72scalar (dot) product, of vectors, 81–83, 81, 83, 86,

208–9scale, chromatic musical, 539, 539scanning capitance microscope, 849scanning tunneling microscope, 1311, 1311, 1320scattering, of alpha particles, 1294scattering cross section, 1315Schlieren photograph (of bullet), 552Schrödinger, Erwin, 1302, 1303Schrödinger equation, 1303–6Schwinger, Julian, 1410

scientific notation, xxv, 14–15, A2–23scissorjack, 460Sears Tower, 653, 654seat belt, 343secant, A8–10second, 5, 9–10, 13, 14

multiples and submultiples of, 10tsecondary, of a transformer, 1053–54second harmonic, 523Second Law of Thermodynamics, 675–80second overtone, 522, 522seiche, 558seismic waves (P and S), 533, 558seismometer, 558, 558selection rule, 1333self-inductance, 1011semiconductors, 871, 1340–45

with donor and acceptor impurities, 1339electron configurations of, 1338–39, 1339“holes” in, 1338–39, 1340–42n-type, 1338–42, 1341, 1343–45p-n junctions of, 1340–42, 1343–44p-type, 1338–42, 1341, 1343–45resistivities of, 871ttemperature dependence of resistivity of, 1338

semimajor axis of ellipse, 282, 285related to energy, 291

series limit, 1291Sèvres, France, 5shear, 445–46, 447t, 449shear modulus, 447–48, 447tsheerlegs, 463shells, 1330, 1330tship collison, Andrea Doria and Stockholm, 361shock wave, of bullet, 552Shoemaker-Levy comet, 299short wave, 1088sidereal day, 294sievert, 1375significant figures, 14–15, 18silicon:

nanoparticle, 264wafer, 654

silicon solar cells, 893silicon structures, micromachined, 195silk, 711silo, grain, 588simple harmonic motion, 469–76

conservation of energy in, 483frequency of, 470–71kinetic energy in, 480–83period of, 470–71phase of, 471potential energy in, 480–83

simple harmonic oscillator, 476–79angular frequency of, 477equation of motion of, 477as timekeeping element, 479

simple pendulum, 484–88, 484equation of motion for, 485isochronism of, 486, 501

simultaneity, relative, 1220–22sine, 19, 473–74, 486, A8–10

formula for derivatives of, 473law of sines, A10

single-loop circuits, 893–97single slit:

diffraction by, 1190–96diffraction pattern of, 1191

minima in diffraction pattern of, 1191–92sinks, of field lines, 738–39siphon, 595, 596Sirius, star, xliiiSI units, see system of units (SI)skater, figure, 408skiing, speed record, 267skin, human, resistance of, 913sky divers, 53Skylab mission, 134, 134, 229, 295, 424, 504

body-mass measurement device on, 134, 134fuel cell on, 892

sliding friction, 174–78, 175, 176, 177, 190slope, 17, 18, 32–33

instantaneous velocity as, 35–37slug, 136nsmall angle approximations, 485Small Mass Measurement Instrument, 504, 504Snell’s Law, 1121, 1124sodium gas, 624solar cells, 893, 1343–45

silicon, 893solar day, 9solar heat collector, 653Solar System, xli, 282

data on, 285–87solenoid, 943–46, 946

magnetic field of, 943–44self-inductance of, 1012

solid:compression of, 446, 448elasticity of, 445–49elastic moduli for, 447telectron configuration of, 1337–39elongation of, 445–49, 445energy banks in, 1336–40quantum structure of, 1321, 1336–45shear of, 445–46, 448thermal expansion of, 633–37, 633ultimate tensile strength of, 447t

sonic boom, 552–53, 552sonography, 536, 536, 544sound:

intensity level of, 542tspeed of, 543–45, 544t, 559speed of (French Academy), 560speed of correction (NRC, Canada), 560speed of in air, 559–60speed of in freshwater, 527, 559speed of in human body, 558theoretical expression for speed in air, 543theoretical expression for speed in gas, 657

sound waves:in air, 538–39diffraction of, 554intensity of, 538, 540–43, 543longitudinal, 538, 538maximum speed in air, 558

sources:of electromotive force, 890–92of field lines, 738–39

Space Shuttle, 125, 142, 280, 280, 290–91, 426Space Telescope, 1209Special Relativity theory, 1216–53specific heat:

of common substances, 630tat constant pressure, 644at constant volume, 644of a gas, 644–47, 646t


A-92 Index

specific heat capacity, 630spectra, color plate, 1288spectral band, 1335spectral emittance, 1257

of blackbody, 1259spectral lines, 1127, 1286, 1287–88, 1289, 1290,

1290splitting of, 1324–26

spectral series, of hydrogen, 1291–93, 1299, 1324spectrum, 1288, 1290

of Caph, 1288of hydrogen, 1291–92, 1291tproduced by grating, 1184, 1184produced by prism, 1126

speed:average, 29–31, 30tof bullet, 356in circular motion, 371of electromagnetic wave, 1080initial, 111launch, 111of light, see speed of lightof light in air and water, 527molecular, in gas, 615most probable, 615, 615after one-dimensional elastic collision, 345–47as relative, 31, 31of sound, 543–45, 559, 560standard of, 6terminal, 53unit of, 30velocity vs., 33–34, 96of wave, 510of wave in Bay of Fundy, 559–60of waves, on a string, 513–16

speed of light, 6, 1218–20invariance of, 1220–21in material medium, 1118measurement of, 1080universality of, 1220–21, 1238

spherical aberration, 1128, 1144spherical charge distribution:

electric energy of, 813potential of, 799–800

spherical mirrors, 1128–35concave, 1128, 1130convex, 1129focal length of, 1128

sphygmomanometer, 579spin:

of electron, 1324–25, 1328–29, 1355–56of elementary particles, 1403t, 1404t

spin angular momentum, 409spin magnetic moment, 976spin quantum number, 1324, 1328Spirit of America, 136, 136spokes of wire wheels, 389, 392, 426, 526, 531spring balance, 136, 136, 151spring constant, 183springs, 476–79

force of, 182–84, 183, 184potential energy of, 236–37, 237, 241

spring tides, 296Sputnik I, 286, 299, 301Sputnik II, 286Sputnik III, 286stabilized laser, 6, 6stable equilibrium, 432, 432stable particle, 1404

standard g, 52–53standard kilogram, 11, 134standard meter bar, international, 5–6, 5standard model, 1415standard of length, 5–6, 5, 6standard of mass, 11, 134–35standard of speed, 6standard of time, 9standard temperature and pressure (STP), 604,

607–8standing wave, 520–24

of bridge (Tacoma WA), 523, 524mode and overtone, 522–23, 522, 545in a tube, 545, 545

Stanford Linear Accelerator (SLAC), 1411, 1413,1415

beam dump, 658, 659star, 55 Cancri system, 297star, binary system:

Cyguns, 298Kruger 60, 297PSR 1913+16, 297

states of aggregation (solid, liquid, gas), xxxvistates of atoms, stationary, 1299states of matter, fifth, 722static equilibrium, 430–41

condition for, 430of electric charge, 774–75examples of, 432–41, 432

static fluid, 575–79, 575static friction, 178–80, 179, 180, 190

coefficient of, 175t, 179–80equation for, 179

statics, 174of rigid body, 430–33

stationary state, 1296of hydrogen, 1298, 1300

steady emf, 1004steady flow, 569steam engines, 671, 671steel, maximum tensile stress and density, 528steel rods, deformation, 449Stefan-Boltzmann Law, 1261–62step-down transformer, 1054step-up transformer, 1054stimulated emission, 1090stopping distances, automobile, 45, 46, 47, 47stopping potential, 1266STP (standard temperature and pressure), 604,

607–8straight line, motion along, 28–68, 32strangeness, 1407strange quark, 1413–14Strassmann, Fritz, 1377streamline flow, 569streamlines, 569–71, 569, 570, 584, 585

velocity along, 584stream tube, 569–70stress, thermal, 461stringed instruments, 546

see also specific stringed instrumentsstring theory, li, 1416“strong” force, 191, 1405, 1406, 1406t, 1414

nuclear binding energy and, 1359–65, 1359,1360

in nucleus, 1355strong interactions, 1405, 1406, 1406tstrontium, radioactive decay of, 1372–76sublimation, 642

suction pump, 592Sun, xl

age of, 1420escape velocity from, 292

sunglasses, Polaroid, 1086superconducting cable, 883superconductivity, 858, 861, 870superconductors, high-temperature, 870superposition:

of electric fields, 772–73, 773of electric forces, 703

superposition principle, 138, 703, 729, 734for magnetic field, 938for waves, 516–20, 518

supersonic aircraft, sonic boom and, 552–53, 552supertanker:

empty and loaded, 593Globtik Tokyo, 266Seawise Giant, 659see also oil tanker

surface tension, 566symmetry, axis of, 380–82, 382tsymmetry, in physics:

cylindrical, 767–68planar, 770–72

synchronization of clocks, 4, 5, 133n, 1220–23relativity of, 1222–23

synchronous (geostationary) orbit, 271–72synchrotron, 1398Syncom communications satellite, 271–72,

290–91system of particles, 305–37

center of mass in, 313–23energy of, 327–28kinetic energy of, 327–28moment of inertia of, 378–84momentum of, 306–13, 324

system of units (SI) 5, 14, 20base and derived, A21–23of current, 972of electric charge, 972for radioactive decay rate, 1374–75

Tacoma Narrows, 523–24, 524Tampa Bay, Fla., 99tangent, 19, A8–10tangent galvanometer, 954, 954tangential acceleration, 371–72telescope, 1149–52

angular magnification of, 1150angular resolution of, 1196–99Arecibo radio-, 1198–99, 1198Galilean, 1142Hobby-Eberly, 1152, 1152Jodrell Bank radio-, 1209Keck, 1151mirror, 1151Mt. Palomar, 1151, 1151radio-, 1186, 1198–99, 1198, 1209Space, 1209Very Large Array radio–, 1186

television cameras, 1269Telstar satellites, 1421temperature:

coefficient of resistivity, 880on Earth, hottest and coldest, 620some typical values, 611tstandard pressure and, 604, 607–8at Sun’s surface and center, 622, 625


Index A-93

temperature scales:absolute, 604absolute thermodynamic, 609Celsius, 611, 612comparison of, 611, 612Fahrenheit, 611, 612ideal-gas, 609–12Kelvin, 604, 609–10

Tennessee River, frictional losses, 267, 653tensile strength, of solids, 447ttension, 149–50, 149, 151, 155terminal speed, 53terminal velocity, 53Tesla, Nikola, 935tesla (T), 934Tethys, moon of Saturn, 297TeV, 1398Tevatron, Fermilab, 1398–99, 1398thermal conductivity, 638–41, 639tthermal energy, 248, 616, 629thermal engine, efficiency of, 666–67thermal expansion, 637

of concrete, 637linear, coefficient of, 634, 634t, 637of solids and liquids, 633–37, 633volume, coefficient of, 634tof water, 635, 635

thermal radiation, 1255, 1256–57, 1256thermal resistance (R value), 640, 641, 655thermal stress, 461thermal units, 249thermocouples, 610, 610thermodynamics, 661–91

calculation techniques, 672First Law of, 662–64Second Law of, 675–80Third Law of, 680

thermodynamic temperature scale, absolute, 609thermograph, 628, 652thermometer:

bimetallic strip, 610, 610, 636, 637color-strip, 610, 610constant-volume gas, 609–10, 609electric resistance, 610, 610mercury-bulb, 610, 610, 636, 636resistance, 880thermocouple, 610, 610

thermonuclear bomb, 260thermos bottle, 1257thickness monitor, 498thin films, interference in, 1169–73thin lenses, 1135–39third harmonic, 523Third Law of Thermodynamics, 680Thomson, Sir Joseph John, 1293thread of a screw, 22threshold energy, 355threshold frequency, 1266threshold of hearing, 538thundercloud, electric field of, 725–28tidal flow, 99tides, 120, 530

height at Bay of Fundy, 531, 531height at Pakhoi, 530resonance, 531spring, 296

tightrope walker, 418time:

atomic standard of, 9

Cesium standard of, 9Coordinated Universal, 9position vs., 32–33, 33, 34, 35, 35, 36, 60standard of, 9unit of, 9–10velocity vs., 61

time constant, 1373of RC circuit, 909

time-dependent angular acceleration, 376–78time dilation, 1224–28time of flight, of projectiles, 109–10time signals, 9TNT, 263Tomonaga, Sin-Itiro, 1410top, 415top quark, 1415tornado, air pressure within, 590toroid, 945, 945torque, 395–99

angular acceleration and, 400angular momentum and, 410–16on a current loop, 972–76on dipole, 743equation of, 400power delivered by, 397rotational motion and, 405static equilibrium and, 430–41

torque vector, 410, 411torr, 578Torricelli, Evangelista, 578Torricelli’s theorem, 585torsional oscillator, 500torsion balance, 277, 277total internal reflection, 1123totally inelastic collision, 348tower crane, see K–10000 tower cranetracking chambers, 1401–2traction apparatus, 149, 149train à grande vitesse (French TGV), 563, 563train box car collisions:

elastic, 346inelastic, 349

trajectory of projectiles, 105, 106, 107, 111, 111transfer of heat, 638–41, 641transformation, heat of, 642transformation equation:

Galilean, 1218, 1234, 1239, 1241Lorentz, 1232–38

transformation of coordinates, 1218, 1234, 1241transformer, 1010, 1053–57, 1055, 1056

step-down, 1054step-up, 1054

transistors, 1340, 1342–43, 1343translational acceleration, 402translational motion, 29, 95, 95, 366, 404

analogies to rotational motion, 374t, 407tof Earth, 120

transmission line, power dissipated in, 904transmutation of elements, 1363transverse electric field, 1078transverse radiation field of accelerated charge,

1076–77transverse wave, 508, 508triangle, right, 19Triangulum Galaxy, xlivTrieste, 589, 589trigonometry (review), A8–10

functions of (sine, cosine, tangent, secant, cose-cant, cotangent), A8–10

identities, A9–10laws of cosines and sines, A10of right triangle, 19, A8small angle approximation, 486

triode, 1340, 1342–43triple-point cell, 609, 609triple point of water, 609, 609trombone, 546trumpet, 546

sound wave emitted by, 538tsunami, height (Mexico), 526, 527tuning fork, 530tunneling, 1311turbulent flow, 571, 571turning points, 244–45

of motion, 470TV waves, 1088twin paradox, 1226two-slit interference, 1177–83, 1177

pattern for, 1180, 1180

UA1 detector, CERN, 1402, 1402UFOs, 557Uhlenbeck, George, 1325ultimate tensile strength, 447tultrahigh vacuum, 624, 624ultrasound, 536, 538, 539

frequencies used in human body, 558, 562ultraviolet catastrophe, 1259ultraviolet light, 1090unbound orbit, 245uncertainties:

application to Ohm’s Law of, A20–21propagation of, A19–21

uncertainty relation, 1278unified field theory, 1411unified theory of weak and electromagnetic forces,

1411uniform circular motion, 112–15, 112, 113, 125,

184–90Unionville, Md. rainfall, 332unit, derived, see derived unitunit of length, 5–8units:

consistency of, 16, 18, 20conversions of, 16–17, 18prefixes for, A21t

Units, International System of (SI), 5, 14of current, 972of electric charge, 972

units of force, 135–36, 141unit vector, 79–80, 79

cross product of, 85universal gas constant, 604Universal Gravitation, Law of, 131, 272–76,

278universality of acceleration of free fall, 49, 49universality of free fall, 49, 49universality of speed of light, 1220–21, 1238universe:

contraction of, 1422expansion of, xlv, 1419–23mass of, 1422–23, 1423tobservable, 1420

unpolarized light, 1084, 1084unstable equilibrium, 432, 432unstable particles, 1404ununquadium, atomic mass of, xlixup quark, 1413–14


A-94 Index

uranium, 1355alpha decay of, 1365–67, 1366enriched, 1381fission of, 1377–79, 1377isotopes of, 1362as nuclear fuel, 1242

uranium isotopes (235U, 238U, UF6), 623–24Uranus, 272, 285t, 286Ursa Major, 1417

valence band, 1338van der Waals equation of state, 624Vanguard I, 286vaporization, 642, 642tvariable force, 211–13, 211, 212, 213, 214Vasa (Swedish ship), 320vector, magnetic force, 933vector addition, 72–76, 72, 73, 74, 89

commutative law of, 74by components, 78–79

vector product, 83–86, 84, 85vectors, 69–93

acceleration, 100–101addition of, see addition of vectorscomponents of, 77–86, 78, 95–98, 97, 99, 101cross product of, 83–86, 84, 85definition of, 72displacement, 69, 70–72, 70, 71, 88, 96dot (scalar) product of, 81–83, 81, 83, 86,

208–9instantaneous velocity, 98, 98multiplication of, 75, 81–86in navigation, 71negative, 75, 75notation of, 71position, 76–77Poynting, 1093–94resultant of, 72, 74subtraction of, 75three–dimensional, 79–81, 79, 80unit, 79, 79velocity, 98–100

vector triangle, 73velocity:

acceleration as derivative of, 41addition rule for, 115–16, 117along streamlines, 584angular, 369t, 376–77, 471angular, average and instantaneous, 369average, 32–35, 33average, in three dimensions, 101–2average, in two dimensions, 95of center of mass, 323–24, 348components of, 95–98, 197, 199escape, 292of flow, 566, 568of galaxies, 1422Galilean addition law for, 1218horizontal, 103–8, 103, 106, 107, 108instantaneous, see instantaneous velocitymagnitude of, 96muzzle, 287, 331negative, 39positive, 39speed vs., 33–34, 96terminal, 53time vs., 61transformation equations for, 1218vectors, 98–100

Velocity Peak, Colorado, speed skiing, 267velocity transformation, Galilean, 1218venous pressure, 597, 597Venturi flowmeter, 585–86, 596, 596Venus, 285t, 286, 286, x.Verne, Jules, 300Very Large Array (VLA) radiotelescope, 1186very-long-baseline interferometry (VLBI),

1202vibrational energy, of molecules, 1333violin, 546, 546

frequencies, 530, 531, 558sound wave emitted by, 538vibration of standing wave, 546wavefront emitted, 538

Virgo Cluster (galaxies), xliv, xlvvirtual image, 1116virtual photon, 1410viscosity, 566, 595viscous forces, 181, 196visible light, 1090–91

colors and wavelengths of, 1090–91, 1091volcanic bombs, 94, 100, 100, 105volt (V), 792Volta, Alessandro, Conte, 793voltage, DC, 889–90, 1004

AC, 1004voltmeter, 905, 906, 916, 917volume, 13

compression of, 446volume expansion, 634volume expansion coefficient, 634, 634tVon Mayer, Robert, 629, 653vortex, 546, 546, 571Vostok, Antartica, coldest temperature, 621Voyager spacecraft, 358

W� particle, 1409, 1411, 1415Wairakei, New Zealand (geothermal plant), 690Warsaw Radio tower, 634water:

thermal expansion of, 635, 635triple point of, 609, 609volume as function of temperature, 634

waterfall:Chamonix, 652hypothetical (M. C. Escher)see also Niagara Falls

water-moderated nuclear reactor, 1383water molecule, 743

distance and algle between component atoms,332

moment of inertia for, 388, 389motion, 566

water park, 507waterwheel, 215–16, 216, 217

overshot, 267undershot, 267, 653

Watt, James, 254watt (W), 254, 540watt balance, 11, 11wave, 507–64

amplitude of, 511angular frequency of, 512, 513–16beats of, 518constructive and destructive interference for,

517crests, 509–10deep water, 526

electromagnetic, see electromagnetic waveenergy in, 1092–96equation, 513freak (North Atlantic), 533frequency of, 510–11harmonic, 510–13, 511intensity of, 540, 1093–94light and radio, 1079–1110long, 1088longitudinal, 508–9, 509, 538, 538measurement of momentum and position of,

1277–78medium, 1088modulation of, 518motion, 508–9number, 511ocean, 553ocean wavelength and speed, 528particle vs., 1276–79periodic, 509–16, 510period of, 510, 510, 513plane, 537, 537, 1081–83, 1082, 1097pool, 507, 511power transported by, 516probability interpolation and, 1277pulse, 508–9, 514, 522radio, 1079–1110, 1094shallow water, 515–16, 527, 528short, 1088sound, see sound wavesspeed, 510speed of, on a string, 513–16standing, 520–24, 524standing, in a tube, 545, 545superposition of, 516–20, 518transverse, 508, 508trough, 509–10, 510TV, 1088water, 515–16, 526–28

wave equation, 1096–99wave fronts, 537, 541–42, 541, 1113, 1113, 1115

circular, 537plane, 537spherical, 537

wavefunction, 511harmonic, 511, 513–16

wavelength, 510–13, 510, 511of visible light, 1090–91, 1091

wavelength bands, of electromagnetic radiation,1090–91, 1091

wavelength shift, of photon, in Compton effect,1270–71

wavelets, 1113, 1120, 1276wave mechanics, 1309, 1322wave optics, 1169wave pulse, electromagnetic, 1080–83, 1082waves, radio, 1089wavicle, 1255, 1276

measurement of, 1277–78“weak” force, 191, 1369, 1405, 1406, 1406t“weak” interactions, 1369, 1405, 1406, 1406tWeber, Franz, speed skiing, 267weber (Wb), 997weight, 11, 141–42, 141

apparent, 187–88, 187mass vs., 141

weightlessness, 142, 142simulated, 589

Weinberg, Stephen, 1411


Index A-95

whale:breaching, 263, 264song, 558

Wheatstone bridge, 906–7wheel pottery, 409white noise, 538Wien’s Law, 1261Wilson, Robert W., 1421winches, 441, 441, 458–59, 458

geared, 459wind instruments, 546wing, flow around, 571, 584wire:

magnetic force on, 969–72straight, magnetic field of, 940–41,

941uniform, electric field in, 860

work, 205–18calculation of, 218definition of, 205done by constant force, 205, 208done by gravity, 207done by variable force, 211–13, 211, 212, 213,

214dot product in definition of, 208–9frame of reference in calculation of, 208,

208integrals for, 212–13internal, in muscles, 208in one dimension, 205–8, 206in rotational motion, 395–99in three dimensions, 208–10, 209, 210zero, 238–39

work-energy theorem, 215, 236, 400

work function, 1266W�particle, 1409, 1411, 1415WWV (NIST radio station), 9, 23

xerography, 709X-ray, scattering, xlviiiX rays, 1090, 1269, 1273–75, 1273, 1274, 1421

characteristic, 1303

yaw, 366Young, Thomas, 1169, 1170, 1177Young’s modulus, 447–48, 447t, 460, 479

zero-point energy, 1307zero work, and conservative force, 238–39Z0 particle, 1409, 1411, 1415Zweig, G., 1413


PREFIXES FOR UNITS

FACTOR PREFIX SYMBOL

exa E

peta P

tera T

giga G

mega M

kilo k

hecto h

10 deka da

deci d

centi c

milli m

micro

nano n

pico p

femto f

atto a10�18

10�15

10�12

10�9

m10�6

10�3

10�2

10�1

102

103

106

109

1012

1015

1018

FUNDAMENTAL CONSTANTS

(See Appendix 7 for More)

Speed of light

Planck’s constant

Gravitational constant

Permeability constant

Permittivity constant

Coulomb constant

Electron charge

Electron mass

Proton mass

Neutron mass

Bohr radius

Compton wavelength

Bohr magneton

Boltzmann constant

Avogadro’s number

Universal gas constant R � NAk � 8.31 J/mole·K

NA � 6.02 � 1023�mole

k � 1.38 � 10�23 J�K

mB �e U

2me

� 9.27 � 10�24 J� T

h�mec � 2.43 � 10�12 m

4p�0-U2�mee

2 � 5.29 � 10�11 m

mn � 1.675 � 10�27 kg

mp � 1.673 � 10�27 kg

me � 9.11 � 10�31 kg

�e � �1.60 � 10�19 C

1�4p�0 � 8.99 � 109 m�F

�0 � 8.85 � 10�12 F�m

m0�4p � 1.00 � 10�7 H�m

m0 � 1.26 � 10�6 H�m

G � 6.67 � 10�11 N�m2�kg2

U � h�2p � 1.05 � 10�34 J�s

h � 6.63 � 10�34 J�s

c � 3.00 � 108 m�s

MISCELLANEOUS PHYSICAL CONSTANTS

Standard acceleration

of gravity

Molecular mass of air 28.98 g/mole

Density of dry air

Speed of sound in air 331 m/s

Density of water

Heat of vaporization 2.26�106 J/kg � 539 kcal/kg

of water

Heat of fusion of ice 3.34�105 J/kg � 79.7 kcal/kg

Mechanical equivalent

of heat

Solar constant

Index of refraction 1.33

of water

1.4 kW>m2

1 cal � 4.187 J

1000 kg>m3

(0�C, 1 atm)

(0�C, 1 atm)1.29 kg�m3

1 g � 9.81 m�s2 � 32.2 ft�s2

SPECIAL UNITS AND CONVERSION FACTORS (See Appendix 8 for More)

QUANTITY UNIT SYMBOL CONVERSION

length inch in.

length angstrom Å 1 Å

length light year — 1 light year � 9.461 � 1015 m

volume cubic centimeter

volume liter liter, l

time year y

mass pound lb

mass atomic mass unit u

energy electron-volt eV

force pound-force lb-f 1 lb-f � 4.448 N

pressure atmosphere atm

pressure torr torr

temperature Celsius scale

temperature Fahrenheit scale

angle radian rad 1 rad � 57.30�

�F � 95(�C) � 32�F

�C � K � 273.15�C

1 torr � 1 atm>760

1 atm � 1.013 � 105 Pa

1 eV � 1.602 � 10�19 J

1 u � 1.661 � 10�27 kg

1 lb � 0.453 6 kg

1 y � 365.25 d � 3.56 � 107 s

1 l � 103 cm3 � 10�3 m3

1 cm3 � 1 ml � 10�3 lcm3, cc

� 10�8 cm � 10�10 m

1 in. � 2.540 cm

GK023-1423G-EPV2[51-54].qxd 07/02/2007 8:41 AM Page 51 PMAC-291 27B:GK023:Chapters:End paper Vol 2: TechBooks [PPG -QUARK]

SOLAR SYSTEM DATA

EARTH

Mass

Equatorial radius

Polar radius

Mean density

Surface gravity

Period of rotation

Moment of inertia:

about polar axis

about equatorial axis

Mean distance from Sun

Period of revolution (period of orbit)

Orbital speed 29.8 km/s

1 year � 365 days 6 h � 3.16 � 107s

1.50 � 1011 m

I � 0.329MER2E

I � 0.331 MER2E

1 sidereal day � 23 h 56 min 4 s � 8.616 � 104 s

g � 9.81 m�s2 � 32.2 ft�s2

5520 kg�m3

RE � 6.357 � 106 m

RE � 6.378 � 106 m

ME � 5.98 � 1024 kg

MOON

Mass

Radius

Mean density

Surface gravity

Period of rotation 27.3 days

Mean distance from Earth

Period of revolution

SUN

Mass

Radius

Mean density

Surface gravity

Period of rotation � 26 days

Luminosity 3.9 � 1026 W

274 m�s2

1410 kg�m3

6.96 � 108 m

MS � 1.99 � 1030 kg

1 sidereal month � 27.3 days

3.84 � 108 m

1.62 m�s2

3340 kg�m3

1.74 � 106 m

7.35 � 1022 kg

MEAN DISTANCE PERIOD OF EQUATORIAL SURFACE PERIOD OF PLANET FROM SUN REVOLUTION MASS RADIUS GRAVITY ROTATION

Mercury 0.241 year 2 439 km 0.38g 58.6 days

Venus 108 0.615 6 052 0.91 243

Earth 150 1.00 6 378 1.00 0.997

Mars 228 1.88 3 393 0.38 1.026

Jupiter 778 11.9 71 398 2.53 0.41

Saturn 1430 29.5 60 000 1.07 0.43

Uranus 2870 84.0 25 400 0.92 0.65

Neptune 4500 165 24 300 1.19 0.77

Pluto 5890 248 1 500 0.045 6.391.5 � 1022

1.03 � 1026

8.70 � 1025

5.67 � 1026

1.90 � 1027

6.42 � 1023

5.98 � 1024

4.87 � 1024

3.30 � 1023 kg57.9 � 106 km


3Online

ConceptTutorial

Online Concept Tutorials www.wwnorton.com/physics( chapter sec t ion )

1 Unit Conversion 1.5, 1.6

2 Significant Digits 1.6

3 Acceleration 2.4, 2.5, 2.6

4 Vector Addition and Vector Components 3.1, 3.2, 3.3

5 Projectile Motion 4.4

6 Forces 5.4

7 “Free-Body” Diagrams 5.3, 5.5, 5.6

8 Friction 6.1

9 Work of a Variable Force 7.1, 7.2, 7.4

10 Conservation of Energy 8.1, 8.2, 8.3

11 Circular Orbits 9.1, 9.3

12 Kepler’s Laws 9.4

13 Momentum in Collisions 11.1, 11.3

14 Elastic and Inelastic Collisions 11.2, 11.3

15 Rotation about a Fixed Axis 12.2

16 Oscillations and Simple Harmonic Motion 15.1

17 Simple Pendulum 15.4

18 Wave Superposition 16.3, 16.4

19 Doppler Effect 17.4

20 Fluid Flow 18.1, 18.2, 18.6

21 Ideal-Gas Law 19.1

22 Specific Heat and Changes of State 20.1, 20.4

23 Heat Engines 21.2

24 Coulomb’s Law 22.2

25 Electric Charge 22.1, 22.5

26 Electric Force Superposition 22.3

27 Electric Fields 23.1, 23.3

28 Electric Flux 24.1

29 Gauss’ Law 24.2, 24.3

30 Electrostatic Potential 25.1, 25.2, 25.4

31 Superconductors 27.3

32 DC Circuits 28.1, 28.2, 28.3, 28.4, 28.7

33 Motion in a Uniform Magnetic Field 30.1

34 Electromagnetic Induction 31.2, 31.3

35 AC Circuits 32.1, 32.2, 32.3, 32.5

36 Polarization 33.3

37 Huygens’ Construction 34.1, 34.2, 34.3

38 Geometric Optics and Lenses 34.4, 34.5

39 Interference and Diffraction 35.3, 35.5

40 X-Ray Diffraction 35.4

41 Special Relativity 36.1, 36.2

42 Implications of Special Relativity 36.2, 36.3

43 Bohr Model of the Atom 38.1, 38.2, 38.4

44 Quantum Numbers 39.1, 39.2

45 Radioactive Decay 40.3, 40.4

Problem-So lv ing Techn iques(chapter sec t ion; page)

Units and Significant Figures (1.6; 18)

General Guidelines (2.6; 50)

Vector Addition and Subtraction (3.3; 79)

Dot Product and Cross Product of Vectors (3.4; 86)

Projectile Motion (4.4; 108)

“Free Body” Diagrams (5.6; 158)

Friction Forces and Centripetal Forces (6.3; 190)

Calculation of Work (7.3; 218)

Energy Conservation in Analysis of Motion (7.4; 223)

Energy Conservation (8.1; 238)

Conservation of Momentum (10.1; 310)

Center of Mass (10.2; 320)

Conservation of Energy and Momentum in Collisions

(11.4; 353)

Angular Motion (12.3; 375)

Torques and Rotational Motion (13.2; 405)

Conservation of Angular Momentum (13.3; 410)

Static Equilibrium (14.2; 437)

Bernoulli’s Equation (18.6; 586)

Ideal-Gas Law (19.2; 608)

Temperature Units; Thermal Expansion (20.2; 636)

Thermodynamic Calculations (21.2; 672)

Electric Fields (23.1; 728)

Electric Field of Charge Distribution (23.2; 732)

Gauss’ Law for Charge Distributions with Symmetry (24.5; 777)

Energy Conservation and Motion of Point Charge (25.1; 797)

Electrostatic Potential and Field (25.4; 810)

Combinations of Capacitors (26.2; 837)

Combinations of Resistors (27.4; 878)

Kirchhoff ’s Rules and Multiloop Circuits (28.4; 900)

Direction of Magnetic Force (right-hand rule) (29.2; 936)

Ampère’s Law for Current Distributions with Symmetry

(29.4; 947)

Faraday’s Law; Lenz’ Law (31.3; 1005)

Sign Conventions for Mirrors (34.4; 1132)

Images of Spherical Mirrors (34.4; 1134)

Sign Conventions for Lenses (34.5; 1138)

Images of Lenses (34.5; 1143)

Thin-Film Interference (35.1; 1174)

Two-slit Interference (35.4; 1181)

Nuclear Reactions (1371; 40.3)


TRIGONOMETRIC IDENTITIES (See Appendix 3 and 4 for More)

sin � � y�r � cos(90� � �)

cos � � x�r � sin(90� � �)

tan � � y�r � sin ��cos �

cot � � x�y � 1�tan � � tan(90� � �)

sec � � r�x � 1�cos �

csc � � r�y � 1�sin �

cos2 � � sin2 � � 1

sec2 � � 1 � tan2 �

csc2 � � 1 � cot2 �

sin 2� � 2 sin � cos �

cos 2� � 2 cos2 � � 1

In the following, u is in radians:

In the following, ku is in radians:

�cos (ku) du � �1

k sin (ku)

�sin (ku) du � �1

k cos (ku)

d

du csc u � �cot u csc u

d

du sec u � tan u sec u

d

du tan u � sec2

u

d

du cos u � �sin u

d

du sin u � cos u

θ

θ90° –r

y

x

B

C

A γ

β

α

Phys ics in Prac t i ce(chapter sec t ion; page number )

Stopping Distances (2.5; 47)

Vectors in Navigation (3.1; 71)

Velocity Vectors (4.2; 99)

Elevators (5.6; 157)

Ultracentrifuges (6.3; 188)

Hydroelectric Pumped Storage (8.1; 242)

Communications Satellites and Weather Satellites (9.3; 281)

Center of Mass and Stability (10.2; 320)

Automobile Collisions (11.1; 343)

Gyrocompass (13.4; 414)

Efficiency of Automobiles (21.2; 674)

Chaos (15.5; 492)

Musical Instruments (17.3; 546)

The Sphygmomanometer (18.4; 579)

Xerography (22.5; 709)

Electrostatic Precipitators (23.2; 735)

Electric Shielding (25.3; 804)

Capacitor Microphone (26.1; 833)

Fuses and Circuit Breakers (28.5; 903)

Magnetic Recording Media (30.4; 978)

Magnetic Levitation (31.3; 1008)

Frequency Filter Circuits (32.2; 1037)

AM and FM Radio (33.4; 1089)

Optical Fibers (34.3; 1124)

Photomultiplier (37.3; 1266)

Ultramicroscopes (38.5; 1310)

Radioactive Dating (40.4; 1376)

Math He lp ( chapter sec t ion; page number )

See also Appendices 2, 3, 4, and 5.

Trigonometry of the Right Triangle (1.6; 19)

Differential Calculus; Rules for Derivatives (2.3; 38)

Integrals (7.2; 213)

Ellipses (9.4; 283)

Derivatives of Trigonometric Functions (15.1; 473)

Small-angle Approximations for Sine, Cosine, and Tangent

(15.4; 486)

The Exponential Function (28.7; 910)

sin(� � �) � sin � cos � � cos � sin �

cos(� � �) � cos � cos � � sin � sin �

C2 � A2 � B2 � 2AB cos�

sin �

A�

sin �

B�

sin g

C


Documents

Physics for Engineers and Scientists, Volume III, 3E