basic study guide

7/23/2019 basic study guide

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PHYSICSOnly study guide for PHY308-K

(Atomic and Nuclear Physics)

M.L. Lekala

UNIVERSITY OF SOUTH AFRICAPRETORIA


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i PHY308K 1

c2010 University of South Africa

All rights reserved

Printed and published by theUniversity of South AfricaMuckleneuk, Pretoria

Page make-up by the Department

PHY308K/1/2010-2010


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1 Introduction iv

1.1 Prescrib ed Textb ook . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

1.2 Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

1.3 Assignments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

1.4 The Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

2 Basic Concepts 1

2.1 Learning Ob jectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2.3 Self-assessment Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Elements of Quantum Mechanics 4


3.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

4 Nuclear Properties 6


4.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6


5 The Force between Nucleons 10


5.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5.3 Self-assessment Questions . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6 Nuclear Models 17


6.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17


7 Nuclear Decay and Radioactivity 24

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7.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24


8 Nuclear Reactions 33


8.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

8.3 Self-assessment questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

9 Nuclear Energy 40


9.2 Study Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40


10 Evolution of Stars 48


10.2 S tudy guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

10.3 Self-Assessment questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50


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Chapter 1

Introduction

Welcome to the PHY308K module! In this module we teach Nuclear Physics. NuclearPhysics is an exciting and fast-developing field of physics, having fascinated mankind inthe 20th century with the awful destructive force of the nuclear bomb and radioactivefall-out, and with the exciting possibility of nuclear-fusion power. Through research ofthe structure of nuclei and nuclear reactions, we have even deliberated about the origins

of the universe, via nucleosynthesis. This is an opportunity for you to learn more aboutyour present existence, your origins, and possibly your future. I trust that you will putevery effort into this course and that you will find it instructive and enjoyable.

1.1 Prescribed Textbook

The prescribed textbook for the PHY308K module is the 1988 Edition of IntroductoryNuclear Physics by Kenneth S Krane, Published by John Wiley & Sons, Inc. Thisbook contains many problems at the end of each chapter and you can test your masteryof this module by attempting as many as possible of these questions. Additional problemsare given in this Study Guide. Thus this textbook was chosen to give you, the student,

a broad outline of nuclear physics, the basic mathematical background required for further studies in this field.

Although it is expected that you have a previous background in quantum physics, theimportant concepts of this field are reiterated in Chapter 2 of the textbook.

1.2 Study Guide

The Study Guide should be used in parallel with the prescribed textbook. The purposeof this study guide is to

assist you in highlighting and summarising material which is important for exami-nation purposes,

provide additional background and discussion on certain topics, and give a new perspective on the arrangement of the study material.

In the course of your study you should compile a summary to achieve the objectives ofeach chapter. In order to help you, each chapter of the Study Guide is organised as follows:

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Learning objectives : specified in order to facilitate systematisation of material tobe studied.

Study guidance : intended to give a special direction of the work to be studied. Worked examples : to show how problems are solved and thereby give problem-

solving skills.

Practice problems (where necessary) : for more problem-solving practice. Self-assessment questions : to check the extend to which the Learning Objectives

have been attained.

The Study Guide treats the Chapters in succession as they are given in the prescribedTextbook. However, large sections and few chapters of the Textbook are omitted. Theomitted sections and Chapters may be read for interest. They are not for examinationpurposes, although you may find a knowledge from them useful in the examination.

1.3 Assignments

To qualify for the examination a number of assignments have to be completed satisfactorily.The work that needs to be covered for each assignment is outlined in Tutorial Letter 101.In your first and second years of physics, you should have learned how to solve problemsin physics. Of course, the only way to do this successfully in your third year, is to makesure that you understand the theory first! This is my special plea to you to study therelevant sections in the textbook first, before attempting any of the prescribed problems!For those who still experience difficulty with problem-solving we hope that the hints andideas attached by the lecturer to the first assignment will resolve any difficulties.

Make life easier for yourself by attempting the assignments well in advance of the givendue dates. Submit your assignments in good time. After your assignment has beenmarked, you will receive solutions or guidelines for solving the prescribed problems. Thisshould not be filed away in some place until just before the examinations! Please workthrough these guidelines when you receive them, solving each problem in detail yourselfand paying particular attention to the problems with which you may have struggled. Alittle persuasion from your lecturer - remember that a number of the assignment problemsmay also appear in the examination, in modified form.

1.4 The Exam

Although we hope to impart the fascination and enjoyment that we get out of our subjectto our students, we, as ex-students, realise that ultimately all you want to achieve at thisstage is to pass the examination. To this end, I have included a Model Exam Paper inAppendix A of the study guide. This is to help you know what type of understandingand problem-solving ability will be required in the examination. Perhaps you would evenenjoygiving yourself a practice run on exam-writing by writing this paper under examconditions i.e. within 2 hours and with no assistance from the textbook or study guide -Good luck! A memorandum to the paper is provided in Appendix B.

Conclusion

Finally, I wish you all well and a great deal of fulfillment as you study Nuclear Physics. Feelfree to contact the lecturer in charge of this module should you experience any difficulties.

M.L. LEKALADEPARTMENT OF PHYSICS


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Part II

CHAPTERS


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Chapter 2

Basic Concepts

2.1 Learning Objectives

An appreciation of the development and discoveries in nuclear physics,

An understanding of some of the terminology p eculiar to this subject e.g. protons,

neutrons, atomic number, mass number etc., an outline of the properties of nuclei that will be important in the study of this

subject,

A feel for the dimensions and the units used in nuclear physics.

2.2 Study Guidance

Section 1.1: History and OverviewThe origins of nuclear physics can be traced back to the fifth century BC, when philoso-phers started to ask questions about the structure of matter. The more recent history ofnuclear physics is closely linked to the development of quantum physics. A few historical

milestones which might be of interest are given below, but bear in mind that the list isfar from complete.

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Section 1.3: Nuclear PropertiesProperties of importance are mass, radius, abundance, decay modes, half-lives, reactionmodes and cross sections, spin, magnetic dipole and electric quadrupole moments, andexcited states.

Section 1.4: Units and DimensionsLengths: Nuclear sizes range from 1 fm to 7 fm where 1 fm = 1015 m.Nuclear energies: Measured in MeV where 1 eV = 1.602 1019 J.Nuclear masses: Measured in terms of the unified atomic mass unit, u, where 1 u = 931.502MeV.

Worked Example

In the decay of6He to 6Li, the maximum amount of energy that can be lost is 3.5 MeV.Find the mass of 6He if the mass of 6Li is 6.01512 u.

Solution

Since energy and mass at the subatomic level are interchangeable, the mass that is lost is

3.5 MeV

931.502 MeV u1 = 0.00376 u

The mass of 6He is thus just the mass of 6Li plus the lost mass, i.e.

6.0152 u + 0.00376 u = 6.01888 u

Practice Problem

See if you understand this conversion by solving the following problem:In the decay of 242Cm to 238Pu, the maximum energy that can be lost is 3732 MeV. Ifthe mass of 238Pu is 238.05, find the mass of 242Cm.

2.3 Self-assessment Questions

1. Give four reasons why electrons cannot be found in the nucleus.

2. Use spin arguments to show that the proton-electron model of the nucleus is incor-rect.

Solution to Practice Problem

The mass lost is3732 MeV

931.502 MeV u1 = 4.00667 u

..

.The mass of 242

Cm is

238.05 + 4.00667 = 242.05667 u


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Chapter 3

Elements of Quantum

Mechanics


To give a background understanding of Quantum Mechanical concepts such as the wave behaviour of particles,

the Heisenberg uncertainty principle,

the Schrodinger equation and its solution for various potentials,

normalization,

expectation values,

operators,

the stationary states,

the reflection and transmission coefficients,

degeneracy.

3.2 Study Guidance

In this chapter, unlike the other chapters, we do not discuss section by section, but rathergive a brief overview of quantum mechanics. Sometimes physicists refer to the newphysics and the old physics. The new physics refers to quantum mechanics, whichbegan with Max Plancks theory of quanta in 1900, and relativity, which began with AlbertEinsteins special theory of relativity in 1905. The old physics is the physics of IsaacNewton, which he discovered about three hundred years ago.

Quantum mechanics forced itself upon the scene at the beginning of the 20th century. Aquantum is a quantity of something, a specific amount. Mechanics is the study of

motion. Therefore, quantum mechanics is the study of the motion of quantities. Quan-tum theory says that nature comes in bits and pieces (quanta), and quantum mechanics isthe study of this phenomenon. Quantum mechanics does not replace Newtonian physics,it includes it. Newtonian physics still is applicable to the large-scale world, but it doesnot work in the subatomic realm. Quantum mechanics resulted from the study of thesubatomic realm, that invisible universe underlying, embedded in, and forming the fabricof everything around us.

In the subatomic realm, we cannot know both the position and the momentum of a particlewith absolute precision. We can know both, approximately, but the more we know about

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the one, the less we know about the other. This is Werner Heisenbergs uncertaintyprinciple. As incredible as it seems, it has been verified repeatedly by experiment. Sincewe cannot determine both the position and momentum of subatomic particles, we cannotpredict much about them. Accordingly, quantum mechanics does not and cannot predictspecific events. It does, however, predict probabilities. Quantum theory can predict theprobability of a microscopic event with the same precision that Newtonian physics canpredict the actual occurrence of a macroscopic event.

Another peculiarity of physics at the subatomic level is that mass and energy changeunceasingly into each other. Particle physicists are so familiar with the phenomena of

mass becoming energy and energy becoming mass that they routinely measure the massof particles in energy units. Strictly speaking, mass, according to Einsteins special theoryof relativity, is energy and energy is mass.

The wave-particle duality was (and I suppose still is) one of the thorniest problems inquantum mechanics. Physicists could no longer accept the proposition that light is eithera particle or a wave because they had proved to themselves via experiment that it wasboth, depending on how they looked at it.

Niels Bohr, in 1924, suggested that the waves in question were probability waves i.e.mathematical entities by which physicists could predict the probability of certain eventsoccurring or not occurring. The unfolding of these probabilities occurs according to theSchrodinger wave equation. In a nutshell, the Schrodinger wave equation governs thedevelopment in isolation of the observed system (which is a photon in the case of light)

which is represented mathematically by a wave function.

Worked Example

Hopefully the heading of this section has enticed you to read this important comment.We have not included a worked example here as you do not have to study this chapterfor exam purposes. However, assignment problems may be set from this chapter as it isvitallyimportant that you understand all of the concepts described in this chapter. Aswith all the read only sections of the textbook, an understanding of these sections isvery necessary for understanding of other sections prescribed as study material for thismodule!


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Chapter 4

Nuclear Properties


To understand a description of the nucleus in terms of the followingstatic properties nuclear radius,

electric charge,

mass,

isotopes,

binding energy,

angular momentum,

parity,

magnetic dipole and electric quadrupole moments, and

energies of excited states.

4.2 Study Guidance

Section 3.1: The Nuclear RadiusThe size of the nucleus is determined by the balance between Coulomb repulsion of theprotons and the strong attraction between nucleons at short distances. Depending on themeasuring technique, one can differentiate between a charge density radiusand anucleonmass density radius. It is relatively natural to characterize the nuclear shape with twoparameters: the mean radius - where the density is half its central value, and the skinthickness - over where the density drops from near its maximum to near its minimum.The number of nucleons per unit volume is roughly constant, and therefore

A34

R3 constant

whereR is the mean nuclear radius, we have

R= R0A1/3 .

There are various techniques for determining the value of the proportionality constant R0(e.g. electron scattering measurement, K X ray isotope shifts, optical isotope shifts,muonicK X ray isotope shifts, direct measurement of the Coulomb energy differencesin mirror nuclei). All these techniques give essentially the same results, namely R0 =1.2 1.25 fm.

Section 3.2: Mass and Abundance of NuclidesTo master this section the following points need to be understood

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The measurement of nuclear masses is done by means ofmass spectrometry. Massspectrograph is the instrument used to do this.

The fixed point on the atomic mass scale is 12C, which is taken to have a massof exactly 12 u. By means of the mass doublet method, the small differences, ,between close-lying masses can be measured.

The Q value of a reaction is the difference in mass between the reactants andproducts. In the nuclear reaction x+X y + Y, the Q value of the reaction isdetermined as follows:

Q= [m(x) + m(X) m(y) m(Y)]c2 .Section 3.3: Nuclear Binding EnergyIn Chapter 2, I mentioned that mass and energy are interchangeable at the subatomiclevel. Thus, I am going to refer to the mass energyof a nucleus. Themass energyof acertain nuclide is its atomicmass energy less the sum of the total mass energy of the Zelectrons (me) and the total electronic binding energy. The latter is usually very smalland can be ignored, unless otherwise instructed.

Mnucleus = mass of the atom ZmeWhat, therefore, is the binding energy(B) of a nucleus? It is the mass (energy) of all thefree nucleons less the mass (energy) of the nucleus, i.e

B = {Zmp+ Nmn Mnucleus}c2

= {Zmp+ Nmn (mass of the atom Zme)}c2

= {Zm(1H) + Nmn mass of the atom}c2

where

m(1H) = mp+ me.

Theneutron separation energySn and the proton separation energySp are the amount ofenergy required to remove a neutron and a proton respectively from the nucleus.

Atomic mass tables often give the mass defect (), which is the difference between themass of the atom and the mass number ( A), i.e.

= (m

A)c2

A useful value to remember is the average binding energy per nucleon (i.e. B/A) of mostnuclei. It is approximately 8 MeV per nucleon.

(The remainder of this section in the textbook is devoted to nuclear models which we willreturn to in Chapter 5 of this study guide).

Section 3.4: Nuclear Angular Momentum and ParityFor a specificnucleonthe orbital angular momentum couples with the spin s to give totalangular momentum j. The total angular momentum of the nucleus is called the nuclearspin I. In a magnetic field the state Isplits up into 2I+ 1 substates, the so-calledZeemaneffect. There is a restriction on the allowed values ofI:

odd-A nuclei: I= half-integraleven-A nuclei: I= integral

Along with the nuclear spin, the parity(behaviour of the wave function when r r)is also used to label nuclear states; = + or =, i.e. if the wave function remainspositive when r is changed tor, we say that it has even parity ( = +). If it becomesnegative, the parity is odd (= ).

Section 3.5: Nuclear Electromagnetic MomentsAny distribution of electric charges and currents (as in the nucleus) produces electric andmagnetic fields. These fields consist of various multipole moments (depending on theirspatial dependence) e.g. monopole, dipole and quadrupole moments. Due to the symmetryof the nucleus, the lowest order non-vanishing moments are the electric monopole moment(Ze), the magnetic dipole moment() and the electric quadrupole moment(eQ).


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Section 3.6: Nuclear Excited StatesExcited states of the nucleus are obtained by moving individual nucleons to higher energyorbits. This can be achieved by adding energy to the core. These states are unstable anddecay rapidly to the ground state. Properties observed for each excited state are: energyof excitation, lifetime and mode(s) of decay, spin and parity, magnetic dipole moment andelectric quadrupole moment.

Worked Examples

Problem 1

a) Compute the difference in binding energy between11B and 11C ifm(11B) = 11.009305 uand m(11C) = 11.011433 u.

Solution

B(11B) = {5m(1H) + 6mn m(11B)}c2

= {5(1.007825 u) + 6(1.008664 u)(11.009305 u)}(931.502 MeV u1)

= (0.08181 u)(9315.502 MeV u1)

= 76.21 MeV

B(11C) = {6m(1H) + 5mn m(11C)}c2

= (0.078842 u)(931.502 MeV u1)

= 73.44 MeV

... B = B(11B) B(11C) = 2.77 MeV

b) Assuming that this difference arises from the difference in Coulomb energy, computethe nuclear radius of 11B and 12C.

Solution

Using equation (3.17) in the textbook, we have

B= 3

5

e2

40R (2Z 1)

Solving for Rwe get (Z= 6):

R = 3

5

e2

40

2Z 1

B

= 3

5 (1.44 MeV fm)

(2(6) 1)2.77 MeV

= 3.43 fm

(Note: we have used the well known constant e/40= 1.44 MeV fm)

Problem 2

Compute the mass defect of 133Cs.

Solution

For 133Cs:

= [m(133Cs) 133]c2


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= (132.905429 u 133 u)(931.502 Mev u1)= (0.09458 u)(931.502 MeV u1)= 88.10MeV

Problem 3

Evaluate the proton separation energy of 32S.

Solution

For 32S:

Sp = [m(A1Z1XN), (AZXN) + m(1H)]c2

= [m(31P) m(32S) + m(1H)]c2

= (30.973762 31.972071 + 1.007825) u (931.502 MeV u1)= (0.009516 u)(931.502 MeV u1)

= 8.86 MeV

Problem 4

The spin-parity of 17O and 17F are both 5+

2 . Assuming in both cases that the spin

and parity are characteristic only of the odd nucleon, show how it is possible to observespin-parity of 18F (1+)).

Solution

For 17O, we expect the valence (unpaired) neutron to have total angular momentumjn = 52 with n = +. For

17F, the valence proton has jp = 52 with p = +. The vectorcoupling of jn to jp yields the values 0, 1, 2, ... jn + jp = 5. Total parity pn = +.Therefore the possible spin-parity states of 18F are 0+, 1+, 2+, ..., 5+, with 1+ being theobserved value.


1. What is meant by the charge radius and the matter radius of a nucleus?

2. Describe what is meant by the abundance of stable isotopes of a particular element.

3. Define separately the binding energy, the neutron separation energy and the protonseparation energy of a nucleus.

4. What is meant by the parity of a nuclear state?


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Chapter 5

The Force between Nucleons


To understand of the deuteron and its properties, To understand why we use the deuteron to discuss nuclear forces, To understand nucleon-nucleon scattering as an instrument for understanding the

nucleon-nucleon interaction,

To understand concepts involved in scattering experiments, such as- impact parameter,

- potential; square-well potential,

- phase shift,

- boundary conditions,

- differential cross section,

- total cross section

- triplet and singlet states,

- spin dependence,

- scattering length,

- effective range approximation.

To understand the properties of nuclear forces, such as- an attractive central potential,

- spin dependence,

- a tensor potential,

- charge symmetry,

- charge independence,

- repulsiveness at short distances,

- velocity dependence.

To understanding the basic nucleon-nucleon interactions.

5.2 Study Guidance

Section 4.1: The Deuteron

This section aims to assist you in making your own summary about the deuteron and itsproperties. Include the following sub-headings in your summary:

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a) Description of a deuteron- it is the nucleus of the deuterium atom, consisting of aneutron and a proton.

b) Excited states- none exist. (Why?)

c) Binding energy - the binding energy of the deuteron is 2.224 MeV. (What threemethods are used to determine the binding energy?)

d) Strength of the nucleon-nucleon potential- the potential is approximately 35 MeV.(Write out the mathematical calculations that arrive at this result).

e) Spin- the total measured spinfor the deuteron is I= 1, where

I=sn+ sp+ and sn, sp are the neutron and proton intrinsic spins and is the orbital angularmomentum of the nucleons.

f) Parity- experimentally we know that the deuteron has even parity, i.e. the wavefunction does not change sign if we replace r with -r. (Explain, as described in thetextbook, why the deuteron can have = 0 and = 2 possibilities for the orbitalangular momentum).

g) Magnetic Dipole Moment- The deuteron wave function is given by

= as(= 0) + ad(= 2).

The magnetic dipole moment is therefore

m= a2s(= 0) + a2s(= 2).

h) Electric Quadrupole Moment- due to the d-state admixture, this moment is non-zero.

Section 4.2: Nucleon-Nucleon Scattering

In the Learning Objectives, a set of points under the heading of scattering experimentswas given. See if, after reading through this section, you can write down a definitionfor each of these concepts. Now that you have done this, it may be easier to follow themathematics outlined in this section:

Make sure that you understand how the author comes to the conclusion that thetotal cross-section must be constant at low energies (see top of Page 92 for example).

Follow the discussion on page 92 of the textbook. Do you understand why it canbe concluded that the nuclear force must be spin dependent?

Section 4.4: Properties of the Nuclear Force

i) The interaction between two nucleons consists mainly of an attractive central po-tential.

ii) It is strongly spin dependent.

iii) It includes a noncentralterm - a tensor potential.

iv) It is charge symmetric.

v) It is nearlycharge independent.

vi) It becomes repulsiveat short distances.

vii) It may depend on the relative velocityor momentumof the nucleons.

Section 4.5: The Exchange Force Model

According to this model, two nucleons interact by exchanging something between them.One of the greatest developments in physics was the introduction of the concept offields,according to which one object establishes throughout space a force field and the secondobject interacts only with the field, and not directly with the first object. A later devel-opment was that all exchanges of energy must occur in bundles of discrete size i.e. thefield must be quantized. These quanta exist only for fleeting instants (due to the uncer-tainty principle) and are known as virtualparticles. The exchange particles that carry thenuclear force are called mesons. The lightest meson, the -meson or pion, is responsiblefor the major portion of the longer range part of the nucleon-nucleon potential.


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The Basic Nucleon-Nucleon Interactions

These interactions are mentioned only briefly in sections of the textbook. IT IS THERE-FORE IMPORTANT THAT YOU STUDY THE FOLLOWING SECTION DIRECTLYFROM THE STUDY GUIDE.

1. The Strong InteractionThe high binding energy/nucleon of 8 MeV cannot be explained by either gravityor the electromagnetic force and is due to the so-called strong interaction betweennucleons. The existence of the strong interaction took some time to be discovered,due to the short range of the interaction which limits it to the dimensions of thenucleus. The strong force is spindependent, but it is independent of charge to agood approximation. This means that in terms of the strong interaction the neutronand proton are the same particle, as long as we do not forget Paulis principle.

In analogy to the photon, quantum of the electromagnetic field, Yukawa predictedthe existence of the -meson (or pion), quantum or carrier of the strong inter-action field. According to Heisenbergs uncertainty principle, the -meson shouldhave a mass of 140 MeV. In 1947 Powell et aldiscovered a particle with the re-quired characteristics in cosmic radiation. To transmit forces between neutrons andprotons, three varieties of pions are called for to conserve electric charge, namely a+, and 0 meson.

In higher and higher energy experiments (>1 GeV), more and more particles were

observed as a result of the strong interaction b etween pions and nucleons. Theseparticles are called strange particles, as they always appear in pairs and do notdecay back rapidly into pions and nucleons as would b e expected. Together withthe -mesons, more than 200 members of a strongly interacting or hadron familyare already known. Hadrons can be divided into mesons, which are bosons, andbaryons which are fermions and which include protons and neutrons.

Presently fundamental particles are classified by postulating the existence of quarks,elementary particles from which all hadrons are made. For example the proton andneutron would be made up of three distinct quarks. One striking feature of themodel is that quarks should have fractional charge of (2/3)e and(1/3)e.Yukawa also postulated for the strong interaction field the existence of a potentialwith the asymptotic form:

V(r) f er/c/rwith c = hmc and f relating to the strength of the strong interaction in thesame way ase, the elementary charge, relates to the strength of the electromagneticreaction. Therefore just as e2/hc = = 1/137 is an indication of the strength ofthe E/M force, f2/hc is an indication of the strength of the strong interaction.A value of approximately 14.5 was determined for f2/hc experimentally, which isroughly 1000 times greater than e2/hc.

The form of the strong interaction potential can be deduced from nucleon-nucleonscattering experiments as shown in the figure below.


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r(fm)

V(r

)

3210

For distances greater than 2 fm, the potential is practically zero. The asymptoticform (above 1 fm) is described by Yukawas potential. It is attractive and staysattractive to around 0.3 fm, below which it becomes strongly repulsive. To ob-

tain better agreement with experiment, one needs additions to the central staticpotential of Yukawa namely

a) a non-static central potential which also depends on the relative momentumof the nucleons,

b) a non-central term dependent on spin, also known as the tensor potential and

c) an exchange term, resulting from pion exchange.

The complete form of the nucleon-nucleon force is still unknown at the presenttime.

2. The Weak InteractionIn decay processes it was noticed that the electron spectrum was continuous andnot discrete, as would be expected from conservation of energy and momentum.Pauli resolved this anomaly by introducing the massless particle, the neutrino,

which could carry away some of the momentum in the reaction. The neutrinocould at first not be detected and many scientists, inter alia N Bohr, doubted itsexistence.

Another peculiarity of decay was that it was much slower than other decays forexampledecay. This indicated that another type of interaction, the so-called weakinteraction, was involved in the emission of an electron-neutrino pair. The relativeweakness of the reaction also explains why it took so long before the neutrino waseventually detected.

During the search for Yukawas particle (-meson) a particle with mass 200e wasobserved in cosmic radiation. However, its interaction probability appeared to b etoo small for it to be considered a quantum of the strong interaction. It was laterfound to be a -meson (or muon), a particle formed in the decay of the -mesonthrough the weak interaction:

+

+ + (t

2.5

108 s)

+ (t 2.5 108 s)where is a particle resembling the neutrino and its anti-particle.

NOTE: 0 decays through the electromagnetic interaction, mainly according tothe modes:

0 + (98.8%)0 e+ + e + (1.2%) } (t 1016s)


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4. Classification of Particles

Worked Example

The square-well form of potential for s-wave neutron-proton scattering is defined as

V(r) = + r < Rcore= V0 Rcore r R= 0 r > R,

a) Write down the solutions to the Schrodinger equation in each of the three regionswhere the potential is defined;

b) Write down the continuity conditions for these wave functions at r = Rcore andr= R; and hence show how Rcore can be determined, if we let 0= 0.

Solution

a) The radial part of the wave function that satisfies the Schrodinger equation for the= 0 (s-wave) is:

u(r) = A sin(kr+ 0), r > R

= B sin(kr+ ), Rcore < r < R

= 0, r < Rcore

b) The continuity conditions at r = R are

i) A sin(kR+ 0) = B sin(kR+ )

ii) kA cos(kR+ 0) =kB cos(kR+ )

Dividing these two equations yields

Kcot(kR + 0) = k cot(kR+ ) (5.1)

A continuity condition at r = Rcore is

B sin(kRcore+ ) = 0

Thus

kRcore+ = 0

or = kRcore (5.2)


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16 PHY308K 1

Substituting (2) into (1) yields

k cot(kR+ 0) = k cot[k(R Rcore)]

Solving for Rcore with 0= 0, we get

cot(kR) = k

k cot[k(R Rcore]

... cot[kR kRcore] = kk

cot(kR)

... kR kRcore = cot1

k

kcot(kR)

... Rcore = R+ 1

kcot1

k

kcot(kR)


1. Discuss the general properties of the nucleon-nucleon interaction and in particularcharge independence, spin dependence and charge symmetry.

2. Discuss the relation between the range of an interaction and the mass of the ex-change particle.

3. Why is there a slight difference in the mass of the three pions?

4. Use binding energy considerations to show that the proton is not a + emitter.

5. How is it possible for a massless neutrino to carry away momentum?

6. What are the properties of neutrinos and anti-neutrinos?

7. Discuss the properties of the six leptons.

8. Classify the following as b osons or fermions:proton, neutrino, +-meson, W+ particle, photon, electron, graviton, neutron,positron, muon.

9. Define leptons, baryons, hadrons and mesons.

10. Discuss the properties of the four basic interactions.


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Chapter 6

Nuclear Models


Understanding the liquid-drop model - a semiclassical (particle) model,

Understanding the shell model - a quantum mechanical (wave) model, and

Understanding the collective model.

6.2 Study Guidance

The Liquid-drop Model

The liquid-drop model was proposed by Niels Bohr and is the oldest model of the nucleus.The essential assumptions are (by analogy to a drop of water)

1. The nucleus consists of incompressible matter so that the radius (R) is proportional

to A1/3

(mass number).2. The nuclear force is identical for every nucleon, and in particular does not depend

on whether it is a neutron or a proton.

3. The nucleus has constant density at its centre.

4. There is a definite surface tension.

5. Decay of a nucleus by emission of particles is analogous to the evaporation of watermolecules from the surface of a liquid.

This model is used in the explanation of

nuclear fission,

binding energy (semi-empirical mass formula),

nuclear reactions where an intermediate compound nucleus is formed.

Semi-empirical mass formula

Bethe and Weizsacker derived the semi-empirical mass formula for the binding energy ofa nucleus containing A nucleons, of which Zare protons, as

M(A, Z) =Zm(1H) + Nmn B(A, Z)/c2

17


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18 PHY308K 1

where

B(A, Z) =avA asA2/3 aC Z2

A1/3 asym (A 2Z)

2

A

The first three terms are explained in terms of the liquid-drop model:

Thevolume term,avA, is derived from the fact that B /Ais constant. It is positiveas it adds to the binding energy.

The surface term, - asA2/3, is a correction proportional to the surface area of thenucleus. This term reduces the binding energy, as nucleons at the surface are lessbound. By assuming a nuclear drop to be a sphere its surface area is given by

A = 4r2 = 4(r0A1/3)2

= 4r20A2/3

yielding a correction of - asA2/3 with as a constant proportional to av and the

surface thickness.

TheCoulomb term, -aCZ2/A1/3, reflects the repulsive interaction between protonsand its contribution will decrease the binding energy. It is derived by assuming thateach proton interacts with all the other protons and by again using R = r0 A

1/3.

The Symmetric term, - asym(Z A/2)2/A, is derived using the Fermi gas model.This term reflects the property of nuclei to be more stable when Z= (A Z) = N,the number of neutrons. It is zero when Z = N, but will decrease the bindingenergy when Z=N.

The pairing energy term, - , is due to the coupling of identical nucleons in pairs.It will increase the binding energy when Z and N are even, have no effect whenA= odd, and decreases the binding energy when Zand Nare odd.

The Shell Model

The liquid-drop model discussed in the previous section, is a phenomenological modelwhich

describes the average behaviour of the nucleons and

approximates the mass of the nucleus well.

For the correct prediction of

energy, spin, and magnetic moment,

a microscopic model is needed which describes the nucleus in terms of all its individualnucleons.

Thenuclear shell modelis similar to theatomicshell model in that we fill the shells with

nucleons (instead of electrons) in order of increasing energy, consistent with the require-ment of the Pauli principle. The energies of the subshells are calculated by solving theSchrodinger equation for a potential which represents the interaction between individ-ual nucleons. This model is sometimes called the independent particle model, because itassumes that due to the short range of the strong interactions each nucleon moves inde-pendently of all other nucleons and is acted upon by an average nuclear field produced bythe action of all the other nucleons.

There is experimental evidence that supports the existence of nuclear shells.


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19 PHY308K 1

i) It was, for example, noted that nucleon numbers 2, 8 and 20 were associated withparticular stability. These magic-number nuclei were more tightly bound thannon-magic nuclei.

ii) Sharp discontinuities in the separation energies occur for certain proton and neutronnumbers - these so-called magic numbers (Zor N= 2, 8, 20, 28, 50, 82 and 126) i.e.more energy is required to separate a proton or neutron from nuclei which containthese magic numbers of nucleons. These discontinuities correspond to a filling ofmajor shells.

As explained earlier, theenergiesof these subshells are obtained by solving the Schr odingerequation. Different forms of p otentials are used e.g. the infinite well and harmonic os-cillator potentials. For the harmonic oscillator the energy level of the shells are given by(equation 2.65 in Krane):

EN = h(N+ 3/2)

whereN= 0, 1, 2,... . The number of nucleons that can be put in each level is (2(2 +1))where can at most be equal to Nand takes on only even or odd values as N is even orodd.

Level scheme of harmonic oscillator:

N degeneracy no. protons/neutrons

0 0 2 2

1 1 6 82 0 2

2 10 12 203 1 6

3 14 20 404 0 2

2 10 30 704 18

5 1 63 14 42 1125 22

As you can see from the table above, only the first three magic numbers (shell closures) arepredicted by using a harmonic oscillator p otential. To get a better correspondence withthe magic numbers, we use the Woods-Saxonpotential as given by equation 5.1 in Krane.This potential includes a strong interaction between the orbital angular momentum andthe intrinsic spinangular momentum of each nucleon, the so-called spin-orbit coupling. Ifa strong spin-orbit interaction exists, a different energy is associated with the j = + 1

2

and j = 12

states, i.e. the j = 12

degeneracy is removed, and the correct magicnumbers are predicted. The potential has the form

V =VWS+ VC+ VS0si i

The Collective Model

The collective model of A Bohr and Mottelson combines features from both the shellmodel and the liquid-drop model. It is an outgrowth of the shell model which assumesa deformed potential instead of a spheric symmetric one. The deformations arise frombulges in the nuclear drop caused by a nucleon moving in an orbit near the surface,similar to the effect of a moon on the tides. The deformed nuclei give rise to additionalenergy levels by virtue of their rotation and vibration.


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20 PHY308K 1

Worked Examples

Problem 1

Derive an expression for the atomic number Zof the most stable isobar A by optimizingthe binding energy (semi-empirical mass formula) for the strong interaction (symmetryterm) and the Coulomb repulsion (Coulomb term).

Solution

B(A, Z) =avA asA2/3 ac Z2

A1/3 asym (A 2Z)

2

A

To optimize, we take the derivative with respect to Zand set it equal to 0.

B(A, Z)

Z = 2ac

A1/3Z+ 4asym

(A 2Z)A

= 0

... 2acA1/3

Z 8asymA

Z+ 4asym = 0

... Z

2acA1/3

+8asym

A = 4asym

... Z = 4asym2acA1/3

+ 8asym

A

= 4asym2acA2/3+8asym

A

= A

2 + (ac/2asym)A2/3

Problem 2

The mass excesses for 15C and 15O are respectively:

(15C) = 9.87 MeV; (15N) = 0.1 MeV; (15O) = 2.86 MeV.

a) Use the formula derived in 5.4.1 above to determine which of these nuclei are stable.

b) Calculate the binding energy difference between 15N and 15O, bearing in mind thatthe proton-neutron mass difference is about 1.3 MeV.

c) Show that this difference comes mainly from the Coulomb term in the semi-empiricalmass formula.

Solution

a) A = 15. We choose ac= 0.72 MeV and asym= 23 MeV, as in the textbook (Krane

page 68).

... Z = 15

2 + (0.72/(2 23)152/3

= 7.16

7... 157 N would be the most stable nucleus.


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21 PHY308K 1

b) The binding energy of a nucleus is given by the formula

B(A, Z) = {Zm(1H) + (A Z)mn Mat(A, Z)}c2

For 157 N we have

B(15, 7) = (7m(1H) + 8mn 15.0001)c2,

and for 158 O we have

B(15, 8) = (8m(1H) + 7mn

15.0031)c2.

The binding energy difference is thus given by (neglecting the electron mass)

B(15, 8) B(15, 7) = {(m(1H) mn) 0.003}c2

= 1.3 MeV (0.003 u)c2

= 1.3 MeV 0.003 u 931.5 MeVu

= (1.3 2.795) MeV= 4.095 MeV

c) The Coulomb term in the semi-empirical mass formula:

C(A, Z) = ac Z2A1/3

For 157 N we have

C(15, 7) = 0.72MeV

72

151/3

= 14.305 MeV,and for 158 O we have

C(15, 8) = 0.72

82

151/3

= 18.685 MeV.The difference is thus

C(15, 8) C(15, 7) = 18.685 (14.305)= 4.38 MeV.

Comparing this to the solution of (b) above, we see that the difference comes mainlyfrom the Coulomb term.

Problem 3

Use the semi-empirical mass formula to derive an expression for the mass difference be-tween two mirror nuclei for which NandZdiffer by 1. Use the atomic mass of two mirrornuclei (e.g. 2311Na and

2312Mg) to determine the value for aC.

Solution:The semi-empirical mass formula:

B(A, Z) =avA asA2/3 ac Z2

A1/3 asym (A 2Z)

2

A


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22 PHY308K 1

For a mirror mucleus with Z =Z 1,

B(A, Z 1) = avA asA2/3 ac (Z 1)2

A1/3

asym (A 2(Z 1))2

A

... B(A, Z 1) B(A, Z) = acZ2

A1/3 ac(Z 1)

2

A1/3

+asym (A 2Z)2

A asym (A 2(Z 1))2

A

= ac

A1/3{zZ2 (Z2 2Z+ 1)}

+asym

A {A2 4AZ+ 4Z2 (A 2Z+ 2)2}

= ac

A1/3 (2Z 1) asym

A (4A 8Z+ 4)

Now

B(23, 11) B(23, 12) = (11 12)m(1H) + (12 11)mn 22.989768 + 22.994124= 1.007825 + 1.008665 22.989768 + 22.994124

= 0.005196 u

= 4.84 MeV

Ifasym= 23MeV. Z= 12, A= 23,

B(23, 11) V(23 12) = ac231/3

(23) 2323

(4(23) 8(12) + 4)

= 8.088 ac 0... 8.088ac = 4.84

... ac = 0.60 MeV

Problem 4

Differentiate the semi-empirical mass formula with respect to Zand show that

a) if the symmetry term is absent the most stable nucleus would consist entirely ofneutrons and

b) if the Coulomb term is absent the most stable nucleus would contain equal numbersof protons and neutrons.

Solution:

B(A < Z) = avA asA2/3 acZ2

A1/3 asym (A 2Z)

2

A

B (A, Z)

Z =

1acZ

A1/3 + 4

asym(A

2Z)

A

a) If the symmetry term is absent

B(A, Z)

Z =

2acZ

A1/3

For the most stable nucleus, this derivative must be zero

... 2acZ

A1/3 = 0

... Z = 0


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23 PHY308K 1

Thus the most stable nucleus would have 0 protons, and will consist entirely ofneutrons.

b) If the Coulomb term is absent

B(A, Z)

Z = 4

asym(A 2Z)A

... 4asym(A 2Z)

A = 0

... A

2Z = 0

... Z = A/2

Now,

N = A Z= A A/2= A/2

Thus, the most stable nucleus would contain equal numbers of protons and neu-trons. (Explain why this would be so!)


1. Explain the origin of each term in the semi-empirical mass formula and show howit affects the mass of the nucleus.

2. What is meant by magic numbers?

3. Explain how the shell model accounts for these magic numbers.

4. Determine the energy level scheme for the harmonic oscillator potential.

5. Show how the inclusion of a spin-orbit p otential can remove the j = 12

degen-eracy.


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Chapter 7

Nuclear Decay and

Radioactivity


To understand what is meant by radioactive decay, To derive the radioactive decay laws, To apply the decay law in solving various decay problems, To understand the following concepts

i) decay constant,

ii) half-life,

iii) mean lifetime,

iv) activity,

v) partial decay constant,

vi) branching ratios,

vii) partial half-lives,

To consider and understand a series or chain of decays, To be able to give examples of alpha, beta and gamma decay, To understand the various units used for measuring radiation.

7.2 Study Guidance

This chapter and the next in the study guide will be concerned with a discussion of thedynamic, or time-varying, properties of nuclei: radioactive decay and nuclear reactions.Both of these are characterized by a transition from some initial system to some final sys-

tem, occurring either spontaneously (radioactive decay) or artificially (nuclear reaction).

In this chapter we consider radioactive decay. It has been found that naturally occurringradioactive nuclides emit one or more of three types of radiation, namely alpha (), beta() and gamma () rays. The initial nuclide in any decay is called theparent and the(heavy) product nuclide is called the daughter.

24


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25 PHY308K 1

RADIOACTIVE DECAY

Section 6.1: The Radioactive Decay Law

Radioactive decay obeys an exponential law such that

N(t) =N0et,

where N0 = original number of nuclei present at t = 0, = decay constant, N(t) =

number of parent nuclei present after time t.

The number of daughter nuclei at time t will be

ND(t) = N0(1 et).

Thehalf-lifet 12

gives the time taken for half of the nuclei to decay:

t 12

=0.693

.

Themean lifetimeof a nucleus is defined as

= 1

.

The number of decays that occur between times t and t + t:

|N| =N0et(1 et).

Defining the activityA to be the rate at which decay occurs within a sample,A(t) = N(t).

It can happen that a given initial nucleus decays in two or more different ways, endingwith two different final nuclei. Ifa and b represent the partial decay constants for thesetwo modes, then the total decay constant is given by

t= a+ b.

Section 6.3 Production and Decay of Radioactivity

Radioactive isotopes can be produced artificially by employing nuclear reactions. For ex-ample a radioisotope can be produced by bombarding a sample of material with neutrons.TherateR (per unit time) at which these radioisotopes are produced is

R= N0 I,

where

N0 = number of target atoms/cm2

=

xNA

MA

= x(g cm2)NA(at mol1)

MA(g mol1)

= cross section of reaction (in cm2),

I = incoming flux (atoms/s).

Theyieldthen of these radioisotopes is

N(t) =Rt,


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26 PHY308K 1

where t = irradiation time (s).

The radioisotope decays at a rateN where N is the number of radioactive nucleipresent, and = the decay constant. The net rate of change ofN is therefore

dN

dt =R N

Solving this equation yields

N(t) = R

(1

et).

Section 6.4 Growth of Daughter Activities

It is common for a radioactive decay to result in a product nucleus that is also radioactive.One can therefore have series or chains of radioactive decays. In other words a parent 1decays with a decay constant 1 and produces a daughter 2. Then the daughter 2 decaysin turn with a decay constant 2 producing a granddaughter 3, and so on. For the simplecase where the granddaughter 3 is stable, let N1, N2 and N3 be the respective numbersof radioactive nuclei present at any given time t, then we find

dN1dt

= 1N1 (7.1)

dN2dt

= 1N1 2N2 (7.2)

dN3dt

= 2N2 (7.3)

In deriving equation (2) we need only to remember that every decaying nucleus 1 producesone nucleus 2. Since 2 is also radioactive, it also decays. The number of parent nuclei caneasily be found by integrating equation (1):

N1 =N0e1t

Substitution into equation (2) yields

dN2dt

+ 2N2 = 1N0et (7.4)

If you know anything about solving differential equations (and now is the time to do

so), you will know that the complete solution of this inhomogeneous differential equationconsists of a general solutionof the homogeneous equation

dN2dt

+ 2N2 = 0 (7.5)

plus any particular solutionof the original equation (4).

A general solution of equation (5) is

N2= ce2t.

A particular solution of equation (4) is

N2Ke1t.

Substitution into equation (4) yields

K = N01

2 1

... N2 = N01

2 1 e1t + Ce2t.

If initially no nuclei 2 are present (N2(0) = 0), we can evaluate Cand find

N2 = N012 1 (e

1t e2t)


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27 PHY308K 1

Substituting into equation (3) and integrating (using N3 = 0 at t = 0) we get

N3 =N0122 1

1 et

1 1 e

2t

2

It is vitally important that you verify this for yourself!

Section 6.5: Types of Decay

In - and -decay processes, an unstable nucleus emits an or particle as it tries tobecome a more stable nucleus. In -decay processes, an excited state decays towards theground state without changing the nuclear species.

-decay: In this process a nucleus of helium, 42He2, is emitted from the parent nucleus.

-decay: In this process a proton can be converted into a neutron or vice versa. This canoccur in three ways:

decay - emission of an electron,+ decay - emission of a positron,electron capture.

Recall that in chapter 4 we described -decay as a weak interaction and that in all threeprocesses mentioned above a neutral particle called the neutrino is also emitted. Note

also that in -decay a particle is created whereas in -decay the particle emitted alreadyexisted within the parent nucleus.

-decay: In this process, an excited state decays to a lower excited state or to the groundstate by the emission of a photon.

Spontaneous fission:In this process, a heavy nucleus splits spontaneously into two roughly equal lighter nuclei.

Nucleon emission:In this process, unstable nuclei decay by means of nucleon emission.

Branching ratios and partial half-lives:Often nuclei decay through various competing modes. The relative intensities of thesemodes is specified by the branching ratios. The decay constant for each mode is called thepartial decay constant from which the partial half-lifecan be calculated.

Section 6.6: Natural Radioactivity

Most of the elements produced during the formation of the Earth were radioactive, buthave since decayed to form stable nuclei. A few of the radioactive elements have half-livesthat are long compared with the age of the Earth, and form thus a major portion of ournatural radioactive environment. There are also other natural sources of radioactivity ofrelatively short half-lives that are being formed continuously today.

Section 6.7: Units for Measuring Radiation

You need to read through this section carefully, as there are various units defined here thatwill be used in solving problems later on. Make sure that you understand the following:

curies exposure roentgen absorbed dose rad gray relative biological effectiveness (RBE)


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28 PHY308K 1

quality factor (QF) dose equivalent (DE) rem sievert

To make life a little easier for you, we mention below a few of these units.

The unit for measuring activity is the curie(Ci):

1 Ci = 3.7 1010 decays/s.

Nuclear radiations are able to ioniseatoms, i.e. the radiated particles are able to knockout electrons from the atoms with which they interact. The total electric charge Q on theions produced in a given mass mof air is called the exposureX:

X= Q

m

and is measured in Coulomb per kilogram.

The exposure resulting in an ionisation charge of 1 electrostatic unit in 1 cm 3 of air at 0

and 760 mm pressure is called the roentgenR:

1R= 2.58 104

C kg

1

.

ALPHA DECAY

Section 8.1: Why-decay occurs

Alpha radioactivity has been investigated for a long time because the naturally radioactivesubstances which led to the discovery of radioactivity (Becquerel, 1896) were found to bealpha emitters (Curie, Rutherford). Alpha particles (4He nuclei) are emitted because theCoulomb repulsion within the nucleus becomes large for large nuclei (A > 140).

Section 8.2: Basic-decay processes

You may be asked to show that the alpha-particle kinetic energy T is always less thanthe decay energy Q. In order to do this, you will have to derive the relation

T= A 4

A Q.

Make sure that you can do this by using the conservation of energy as described in thetextbook. This is known as the energeticsof a alpha-particle decay.

BETA DECAY

Section 9.1: Energy release in-decay

Beta decay is the most common type of radioactive decay. The process consists of theemission of an electron directly from a nucleus.

Both positive and negative electrons can be emitted. However, these electrons cannot beemitted in isolation, as this would result in violation of conservation of energy, angularmomentum and linear momentum. Make sure that you understand from the textbookhow these difficulties concerning the conservation laws were overcome by the neutrinohypothesisof Pauli in 1931.

You must understand how to demonstrate -decay energetics by considering each of thethree processes:


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29 PHY308K 1

-decay, +-decay, electron capture.

We have, for example,

Q = Te + E ,

Q = [m(AX) m(AX)]c2,

Q+ = Te++ E,

Q+ = [m(AX) m(AX) 2m0]c2.

Section 9.2: Fermi theory of-decay

Fermis theory of-decay begins with Fermis Golden Rule for the transition rate:

=2

h|Vfi |2(Ef)

The matrix element Vfi is the integral of the interaction Vbetween the initial and final

states,Vfi =

f V idv.

In your summary, you need to look at the interaction Vand the final state f.

The factor (Ef) is the density of final states. Write it down.

What is the allowedapproximation?

Show, how, using this approximation you can obtain an expression for the partial decayrate and hence the momentum and energy distribution.

N(p) = c

c2 p2(Q

p2c2 + m2ec4 + mec

2)2

N(Te) = ccs

(T2e + 2Temec2) 12 (Q Te)2(Te+ mec2)

GAMMA DECAY

A nucleus can be brought to an excited state by a variety of means. An excited nucleuscan always decay to a lower-energy state by

emission of electromagnetic radiation or internal conversion.

Emission of electromagnetic radiationor in other words a photon can occur when a nucleonmoves from a higher to a lower state. This is gamma decay.

Internal conversionoccurs when the nuclear energy released (Ei Ef) in the transitionof a nucleon from a higher to a lower state is transferred directly to an atomic electron,which is ejected with a kinetic energy

Te= Ei Ef EB

whereEB is the binding energy of the electron in the atomic shell form which it has beenejected.


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30 PHY308K 1

We will concern ourselves here with gamma decay.

Section 10.1: Energetics ofdecay

If the initial excited nucleus has a rest mass M0 and the final state has a rest mass M0,conservation of energy and momentum require

M0 c2 = M0c

2 + E+ TR

0 = PR+ P

where

E, P = energy and momentum of gamma decay

TR, PR = recoil energy and momentum of final nucleus

The recoil speed of the nucleus is so small that nonrelativistic formulae may be used tocomputeTR.

TR = P2R

2m0

= P 2

2M0

= E

2

2M0c2

Thus, using

E = Ei Ef, we get

E = E+ E22M0c2

Worked Examples

Problem 1

3H has a half-life of 12.26 years. If we start with 1 kg 3H, how long will it take before 1 gis left?

Solution:The exponential law of radioactive decay gives the number of nuclei left after time t:

N(t) =N0 et

Firstly, we need the ratioN(t)

N0

Now, N(t) can be calculated as follows:

number of moles (n) = mass

molar mass of 3H nucleus

... number of nuclei(N(t)) = NA n

= NA 1 g

molar mass of 3H

whereNA is Avogadros number.


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31 PHY308K 1

Similarly:

N0 = NA 1000 gmolar mass of 3H

... N(t)

N0=

1

1000

Secondly, we need , the decay constant:

= 0.693

t 12

= 0.693

12.26 years

= 0.0565 years1

We can thus find how long it takes to have 1 g left:

et = N(t)

N0

n N(t)

N0= t

... t = 1

n

N0N(t)

= 1

0.0565 years1

= 122.26 years.

Problem 2

Compute the Q-values for the following reactions

a) Alpha decay of 230Th: 230Th226Ra +b) Beta () decay of 193Os: 193Os193Ir

Solution:a)

Q = [m(230Th) m(226Ra) m(4He)]c2

= (230.033128 u 226.025403 u 4.002603 u)(931.5 MeV u1)= 4.771 MeV.

b)

Q = [m(193Os) m(193Ir)]c2

= (192.964138 u 192.962917 u)(931.5 MeV u1)= 1.137 MeV.

Problem 3

(The next worked example may confuse you a little, as it comes from the section aboutgamma decay that is supposed to be read only. However, we feel it is important that youunderstand the modes of gamma decay.)

Lets explain what is meant by modes of gamma decay, before posing the problem. When aphoton is emitted, it carries a certain angular momentum (Lr) and the parity of the initial


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32 PHY308K 1

and final state of the nucleon is affected. Conservation of angular momentum requires that

Ii= If+ Lr

The parity change r is directly related to Lr, and

For electric multipole radiation r= (1)LrFor magnetic multipole radiation r = (1)Lr+1 . Now, the question is determine thedominant polarities of the following transitions:

a) 2+ 0+b) 1+ 0+c) 1

2

12

+

d) 2+ 2+e) 9

2

+ 12

Solution:

a)

2+ 0+ r = if = +Lr = 2

... Since r = (1)Lr = +, this is an electric dipole transition and is written asELr or E2 in this case.

b)

1+ 0+ r = +Lr = 1

... Since r= (1)Lr+1 = +, this is an M1 transition.c)

1

2

12

+

r = +

Lr = 1

... Since r= (1)1 = , this is an E1 transition.d)

2+ 2+ r = +Lr = 1

... Since r= (1)Lr+1 , this is an M1 transition.e)

9

2

+

12

r =

Lr = 9

21

2 = 4

... Since r= (1)Lr+1 = , this is an M4 transition.


1. Explain what is meant by the half-life of a nucleus.

2. Describe briefly the five modes of radioactive decay.


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Chapter 8

Nuclear Reactions


To know all the types of nuclear reactions listed in the textbook, To understand the conservation laws, To apply the conservation of linear momentum and energy to obtain the energetics

of nuclear reactions,

To understand isospin, To be able to define the following

- cross section,

- differential cross section,

- doubly differential cross section,

To understand and be able to derive the differential cross section for Rutherfordscattering,

To understand the optical model, compound-nucleus reactions, direct reactions andresonance reactions.

8.2 Study Guidance

Section 11.1 Types of Reactions and Conservation Laws

A typical nuclear reaction is written as

a + x Y + b

or

x(a, b)Y.

Reactions can be classified into the following groups by the mechanism that governs theprocess:

Radioactive capture reactions: In these reactions one of the products (b) is a raye.g. 14N(p, )15O.

Photonuclear reactions: In these reactions one of the reactions (a) is a ray e.g.14N(, p)13C.

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Scattering reactions: The incident and outgoing particles are the same in these reactionse.g.

14N(p, p)14N : elastic scattering,

14N(p, p)14N : inelastic scattering.

Knockout reactions: These reactions cause another nucleon to be ejected, so that thereare 3 particles in the final state e.g. 3He(e, ep)2H.

Transfer reactions:e.g. 3He(d, p)4He.

Here the neutron is transferred from the deuteron (d) to the 3He nucleus.

Compound reactions: These reactions occur via a series of intermediary steps. Theincoming and target nuclei merge briefly, sharing energy.

Direct reactions: These reactions take place immediately without the formation of acompound nucleus.

Resonance reactions: These reactions are intermediate reactions in which the incomingparticle forms a quasibound state with the target.

In nuclear reactions (particularly at low energies) we can apply the following conservation

laws:

Conservation: of total energyoflinear momentumofproton and neutron numberofangular momentumofparity.

Section 11.2 Energies of Nuclear Reactions

Since I, so kindly, summarized Section 11.1 for you, it is up to you to summarize the rest.Of course, this study guide is there to assist you in this task.

Using conservation of energy and linear momentum, you can write down the equations

governing these quantities in a nuclear reaction.

We define the reactionQ valueas the initial mass energy minus the final mass energy.

IfQ >0, the reaction is exothermic, and ifQ


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fm2. Thedifferential cross section(d/d) represents the number of scatterings per unittime and unit solid angle for one incident particle per unit time and one target particleper unit area. Its measurement gives us important information on theangular distributionof the reaction products. Integrating the differential cross section over all angles gives usthe reaction cross section.

Of course, you also need to know the definition of the doubly differential cross section.Find this in the textbook and write it down below:

The variation of the cross section with energy, (E), is also known as the excitationfunction. At high energies (E) will simply be equal to the geometrical cross section.

Section 11.6: Coulomb Scattering

Coulomb scattering occurs as a result of the electric charge distribution of the nucleus.This scattering may be either elastic or inelastic. Elastic Coulomb scattering is calledRutherford scattering. Rutherford used scattering of alpha particles from gold foil toinvestigate the nature of the atom. He derived a formula to predict the yield of scatteredparticles per solid angle expected at a given angle.

Thedistance of closest approach(in a head-on collision) is given by

d= zZe2

r0Ta

whereTa= 12

m20 .

Thefraction of particleswith impact parameters less than b, or that are scattered at anglesgreater than is

f=nxb2

where

n = target nuclei per unit volume

x = target thickness

b = impact parameter.

The relationship between band is given by

b= b

2cot

2

The Rutherford cross sectionis thus given by

d

d =

zZe2

40 1

4Ta2 1

sin4 2

.

To simplify calculations with e4 we introduce the numerical constant:

e2/(40) = 1.44 MeV.fm

(Notice that the above constitutes a brief outline of the derivation of the Rutherford crosssection. You need to fil l in the missing steps.)

Section 11.7 Nuclear Scattering


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A deviation from the Rutherford formula is caused by interference between nuclear andCoulomb scattering of charged particles.

Section 11.9 The Optical Model

The approach used in the optical model is very similar to the approach used in the shellmodel. The Schrodinger equation is solved for a potential which accounts for elasticscattering in the presence of absorptive effects. This potential is complex

U(r) = V(r) + iW(r)

The real part is responsible for the elastic scattering and the imaginary part is responsiblefor the absorption.

In practice the optical model consists of the sum of central, spin-orbit and Coulombpotentials.

Section 11.10 Compound-Nucleus Reactions

The optical model is used to describe average trends in the cross section. To explain theresonances observed in excitation functions at low energies the concept of a compoundnucleus is introduced.

The most important feature of the compound nucleus model is the following:

1. The projectile is absorbed by the target to form a compound nucleus. The life-time of the compound nucleus ( 1017s) is long compared to the time neededto transgress the nucleus ( 1021s). This means that the projectile and targetnucleus form a new nucleus which exists long enough to forget how it was formed.

2. A prerequisite for the formation of a compound nucleus is that the projectile hasenough energy to overcome the Coulomb barrier.

3. The projectile loses its energy to other nucleons in the target atom and becomesindistinguishable from them. The compound nucleus is thus formed in an excitedstate. As the bombarding energy increases, the number of energy levels availablefor occupation increases and the resonances are more closely spaced.

4. The compound nucleus loses energy by emitting energy radiation or particles. Twodecay modes are always possible, namely photon radiation and compound-elastic

scattering. At higher energies other reaction channels become available.

Section 11.11 Direct Reactions

At still higher energies (>20 MeV) an immediate interaction occurs between the pro-jectile and target without the formation of a compound nucleus. Features of this type ofreaction mechanism are:

1. Direct processes occur very rapidly; compound-nuclear processes take much longer.

2. Enhancement in cross section for forward angles (angular distribution is peaked inforward direction).

3. No definite resonances in excitation function.

4. A very large cross section.

Two specific processes are stripping reactions, in which target removes one or more nucle-ons from projectile (e.g. (d, p) reaction), and pickup reactions, in which projectile picksup a nucleon from target (e.g. (p, ) reaction).

From Fermis Golden Rule for the transition probability (see Section 9.2 of the textbook),you should have realized that the transition rate or amplitude is governed by the nuclearmatrix element:

M=

y

b V xa dv


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where we have to consider the direct reaction

X+ a Y + b.

In the plane-wave Born approximation, a and b are treated as plane waves.

To obtain a slightly more realistic result, wee use the optical model to account for the factthat the incoming and outgoing plane waves are changed (or distorted) by the nucleus.This gives the distorted wave Born approximationor DWBA.

Section 11.12 Resonance Reactions

For certain incident energies the exterior and interior wave functions match exactly, withthe result that the incident particle penetrates easily and the cross section rises to a max-imum. This is called a resonance reaction. The Breit-Wigner formuladescribes the shapeof a single, isolated resonance. It is important to note that direct scattering without theformation of a resonant state (potential scattering) and resonant scattering both contributeto the elastic scattering amplitude.

You must be able to derive the Breit-Wigner formula for the shape of a single, isolatedresonance:

For the reaction a + x a + X,

=

k2

g (ax)

2

(E ER)2

+ 2

/4

,

and for the reaction a + x b + Y,

=

k2 g

(ax)(by)

(E ER)2 + 2/4 ,

by following the steps in the textbook.

Worked Examples

Problem 1

Determine the distance of closest approach for an alpha particle incident on a gold targetat incident energy of 10 MeV.

Solution:

d = zZe2

40Ta

= zZ

Ta e

2

40

= (2)(79)

10 MeV (1.44 MeV fm)

= 22.8 fm

= 2.28 1012 cm.

Problem 2

Derive equation (11.10) in the textbook by making use of the conservation of linear mo-mentum (equations (11.4)) and the difference between the final and initial kinetic energies(Q) (equation (11.3))


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Solution:Equations (11.4) give

pa = pbcos +pycos ,

0 = pbsin pysin .We can write these equations as

mava = mbvbcos + myvycos ,

0 = mbvbsin myvysin .In order to eliminate , substitute mv = (2mT)

12 for each particle and rewrite the equa-

tions:

(maTa)12 (mbTb)

12 cos = (myTy)

12 cos ,

(mbTb)12 sin = (myTy)

12 sin ,

Squaring both of these equations and adding:

maTa 2(maTambTb)12 cos + mbTb(cos

2 + sin2 ) = myTy(cos2 + sin2 )

or

maTa 2(maTambTb)12

cos + mbTb(cos2

+ sin2

) = myTy

Now, using equation (11.3):Ty = Q Tb+ Ta+ Tx

we can eliminate Ty to get:

myQ myTbmyTa+ myTx= maTa 2(maTambTb)12 cos + mbTb

... myQ = Tb(my+ mb) Ta(my ma) 2(mzTambTb)12 cos

... Q = Tb(1 +mbmy

) Ta(1 mamy

) 2 mamb

mymy TaTb12

cos

(as required).

Problem 3

Protons of energy 6 MeV are Coulomb scattered by a silver foil of thickness 3.5 106m. What fraction of the incident protons is scattered at angles greater than 90? (Thedensity of the silver foil is 10.6 g cm3.)

Solution:The fraction of protons scattered at an angle greater than is given by equation (11.19)in the textbook:

f = nxb2

where

n = NA

A

= (10.5 g cm2)(6.02 1023 atomsmol1)

107.9 g mol1

= 5.86 1022 atoms cm2


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To calculate busing equation (11.26) we need d. From equation (11.16) we have that

d = zZe2

40T

= (1)(47)(1.44 MeV fm

6 MeV

= 11.28 fm

At = 90,

b =

d

2

cot

2

= 11.28

2 cot45

= 5.64 fm

... f = nxb2

= (5.86 1022 atoms cm3)(3.5 104 cm)()(5.64 1013 cm)2

= 2.05 105

8.3 Self-assessment questions

1. Give the 3 categories into which nuclear reactions can be divided.

2. Define the Q value of a reaction.

3. Explain the meaning of endothermic and exothermic reactions.

4. What is isospin?

5. Define differential cross section, total reaction cross section and excitation function.

6. What is the difference between elastic and inelastic scattering?

7. Derive an expression for the differential cross section for charged particles at lowenergies.

8. Show that a complex potential (in the optical model) gives rise to absorption.

9. Describe the propagation of a compound nucleus reaction.

10. It is possible to have direct and compound nucleus processes both contribute to agiven reaction. Describe the two principal differences that can be observed experi-mentally.

11. Derive the Breit-Wigner formula for isolated resonances.


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Chapter 9

Nuclear Energy


To b e able to distinguish between fission and fusion reactions,

To understand why nuclei split (or fission),

To understand the characteristics of fission such asi) mass distribution of the fragments produced in the fission process,

ii) number of emitted neutrons,

iii) radioactive decay processes, and

iv) cross sections,

To be able to derive the energetics of fission, To describe the factors that must be considered in the design of a fission reactor, To be able to classify fission reactors according to

i) type of fuel used,

ii) average neutron energy,

iii) moderators used,

iv) form of energy extraction, and

v) the purpose of the reactor,

To identify safety measures, To understand how fusion occurs, To understand the characteristics of fusion, To describe solar fusion.

9.2 Study Guidance

At closer inspection it will be clear that all our energy is basically nuclear in origin. Wewill concentrate on two self-sustaining processes which constitute enormous amounts ofenergy, namely nuclear fission and nuclear fusion.

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NUCLEAR FISSION

Section 13.1: Why Nuclei Fission

A reactionx(a, b)yis usually called fission, ifb and yhave comparable masses (not normallyequal). Usually fission is produced only if sufficient energy is given to a nucleus by captureof a slow neutron or by bombardment with n; p; d; ... particles of gamma rays. As faras we know, the fission process always proceeds through a compound-nucleus stage. Thecompound nucleus breaks up into two parts with some prompt neutron emission. The two

main nuclear components, called fission fragments, do not have equal masses because ofenergetics. We consider two fission processes below viz. spontaneous fission and inducedfission.

1. In the spontaneous fission of a heavy nucleus into two fragments, the binding energyper nucleon increases, with the result that energy is released - about 200 MeV pernucleus. Due to the Coulomb barrier this is a very slow process. The height ofthe Coulomb barrier is roughly equal to the energy released. Spontaneous fissionoccurs when the energy released puts the two fragments just below the Coulombbarrier. A rough indicator that spontaneous fission will occur can be obtained fromthe semi-empirical mass formula and is

Z2

A >47.

2. Absorption of a relatively small amount of energy, from a neutron or photon, formsan intermediate step that is at or just above the barrier so that induced fissionoccurs. Thus, by bombarding heavy nuclei with neutrons, the reaction is sped upconsiderably and a fission chain reaction is established due to the release of anaverage of 2.5 neutrons per fission. This chain reaction can be controlled, utilizingthe energy for power-generating purposes, or b e left uncontrolled, leading to anexplosion, e.g. an atom bomb.

Section 13.2: Characteristics of Fission

You need to summarize this section under the heading described in the textbook andoutlined below:

Mass distribution of fragments: Here you should note that fission of a specific nucleus (e.g.235U) into equal or nearly equal fragments is not highly probable.

Number of emitted neutrons: Neutrons are shedded by the fission fragments either atthe instant of fission or after a short delay These are called prompt neutronsor delayedneutrons respectively.

Radioactive decay processes: The initial fission products are highly radioactive and decayby emitting and radiations. These radioactive products constitute the waste matterof nuclear reactors.

Fission cross sections: Take note of the various regions in the cross section.

Section 13.3: Energy in Fission

When a nucleus captures a neutron to form a compound state, that state has a certainexcitation energy. The energy needed to overcome the fission barrier of a nucleus is calledtheactivation energy. If the excitation energy is greater than the activation energy, zeroenergy neutrons can be used to fission the nucleus.

Section 13.6: Fission Reactors

A device in which a self-sustaining reaction takes place is called a reactor. A self-sustainedreaction is obtained when at least one neutron released in a fission reaction induces anotherfission reaction. Factors which inhibit this process are


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i) absorption of neutrons by non-fissile nuclei in the fission fuel (e.g. 238U).

ii) Absorption of neutrons by materials used in the construction of reactor core.

iii) Leakage of neutrons from reactor core.

These factors have to be considered in the design of a reactor.

Reactors can be classified according to the following five properties.

a) Type of fuel:

i) Natural uranium (0.72% 235U)

ii) Enriched uranium (>0.72% 235U)

iii) Plutonium (239Pu)

iv) Uranium-233 (233U).

b) Average neutron energy:

i) Fast neutrons (10 keV - 10 MeV)

ii) Intermediate neutrons (partially moderated to 100 eV - 10 keV)

iii) Thermal neutrons (moderated to 0.025 eV).c) Moderators (in intermediate and thermal reactors)

i) Graphite (C)

ii) Deuterium (D2O)

iii) Hydrogen (H2O)

d) Energy extraction

i) Gas

ii) Water, D2O

iii) Liquid metals e.g. molten sodium

e) Purpose of reactor

i) Research

ii) Power generation

iii) Breeding

iv) Converting.

The following safety measures can be identified:

a) Nuclear fuel is kept in containers which provide sufficient radiation protection whenfuel elements are exchanged.

b) The reactor core is enveloped in reinforced concrete. The high Z value materialscreens neutrons and the low Zmaterial screens -radiation.

c) The reactor is cooled. Usually the cooling material and moderator are the samee.g. water. In pressure water reactors the reactor core is placed under high pressureto prevent water from boiling below 600 K. This pressurized water is also used insteam turbines for generation of power.

d) The value of the multiplication factork is controlled by insertion and withdrawalof neutron absorbers e.g. control rods made of boron or cadmium.

e) Reactors also have safety factors inherent to their design. The popular conceptof reactors being controlled nuclear b ombs is wrong. In a nuclear explosion fastneutrons are needed, while most reactors work with slow neutrons. In nuclearfission - 0.7% of neutrons are released a few seconds after fission has occurred.These delayed neutrons are included in the calculation ofk with the result that thetempo of energy release is slowed down to prevent overheating of reactor.


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In pressure water reactors overheating will cause the water (H2O or D2O) whichis used as moderator, to evaporate. Vapour is a poor moderator, with the resultthat the neutrons have higher energy. The probability for leakage is enhanced,causing k to become smaller than 1 and the reactor to become sub-critical. Inthermal reactors the high energy neutrons will of course be less likely to causefission (smaller cross section), in effect halting the chain reaction.

PS: Damage from nuclear weapons are not restricted to the immediate slaughter. Long-time biological and ecological effects result from the fall-out produced by bombtests.

NUCLEAR FUSION

Section 14.1: Basic Fusion Processes

Theadvantagesof using fusion reactions as a source of energy are:

1. Due to the strong binding of the 4He nucleus, fusion of lighter nuclei to produce4He will yield great amounts of energy; in theory up to about six times as much asthe yield from fission of heavy nuclei,

2. the seemingly inexhaustible supply of fuel (deuterium) and

3. the absence of radioactive waste.

Adisadvantageis that to initiate these reactions, the nuclei must get close enough to inter-act (


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Reaction rate: From the dependence of < v > on temperature, you should be ableto understand why at extremely high temperatures the D-T reaction may become lessfavourable than others.

Section 14.3: Solar Fusion

The sun is a prototype of a self sustaining thermonuclear reactor. The basic process inthe sun is the fusion of hydrogen into helium:

p +p d + e+

+ p+ d 3He + 3He +3 He 4He +p +p

p p I chain

This complete process is known as the proton-proton cycle. The net reaction is the con-version of 4 protons into helium:

41H 4 He+ 2e+ + 2

An alternative fate of the 3He is to encounter an particle and for the sequence to proceedin one of the following two ways:

p-p II chain (T3 107 K)4H

3H

7Be + 4He + 3He

7Be +

7Be + e 7Li + 7Be + p 8Be + e+ +7Li + p 8Be + 8B 8Be + e+ +8Be 4He + 4He 8Be 4He + 4He

At temperatures below 3 107K the decay of 7Be to 7Li is the fastest process and thep-p II chain dominates. When the temperature is 3 107K the protons overcome theCoulomb barrier of 7Be faster than the half-life of electron capture and the p-p III chainwould win.

If carbon, nitrogen and oxygen are present in the star, it is possible that these elementscan also play the role of catalyst in the 4p 4He synthesis, as shown in the sequencebelow. (This is known as the carbon or CNO cycle):

12C+p 13N+13N 13C+e+ + 13C+p 14N+14N+p 15O+15O 15+e+ + 15N+p 12C+4He

It is estimated that 56% of the 4He formed at the centre of the sun comes from the p-p Ichain, 40% from the p-p II chain, 0.05% from the p-p III chain and 3.2% from the CNOcycle.

Once a star has exhausted its hydrogen fuel, helium fusion reactions can take place:

34

He 12

C.

Other reactions involving fusion of light nuclei and particle capture can continue torelease energy until the process ends near 56Fe, beyond which there is no energy gain incombining nuclei.


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Worked examples

Problem 1

Calculate the energy released in the fission reaction:

235U + n 79 GE +152 ND + 4n

SolutionThe energy released is the Q-value of the reaction:

Q = [m(235U) m(79Ge) m(52Nd) m(3n)]c2

= (235.043924 u 78.920946 151.924680 3(1.008665 u))(931.502 MeV u1

= (1.172299 u)(931.502 MeV u1)

= 1092 MeV.

Problem 2

244

Pu decays by spontaneous fission with a half-life of 8.1107

years. If the energy re-leased is about 230 MeV per fission, determine how much (in g) of244Pu would be neededto produce a total fission power of 0.1 W.

Solution

From the power required, we can determine the number of fissions per second required;since

Power = Energy per fission fissions per second

... Fissions per second = power

energy per fission

= 0.1 W

(230 MeV/fission)(1.6

1013 J/MeV)

= 2.72

The number of fissions per second is also known as the activity (A) of the isotope:

A= N

is the decay constant:

= 0.693

(8.1 107 years)(3.15 107 s/years)

= 2.71 1016 s1

... N = A

= 2.72 fissions/s

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