Ch34 ISM 070623

Chapter 34 Wave-Particle Duality and Quantum Physics Conceptual Problems

1 • [SSM] The quantized character of electromagnetic radiation is observed by (a) the Young double-slit experiment, (b) diffraction of light by a small aperture, (c) the photoelectric effect, (d) the J. J. Thomson cathode-ray experiment. Determine the Concept The Young double-slit experiment and the diffraction of light by a small aperture demonstrated the wave nature of electromagnetic radiation. J. J. Thomson’s experiment showed that the rays of a cathode-ray tube were deflected by electric and magnetic fields and therefore must consist of electrically charged particles. Only the photoelectric effect requires an explanation based on the quantization of electromagnetic radiation. ( )c is

correct.

2 •• Two monochromatic light sources, A and B, emit the same number of photons per second. The wavelength of A is λA = 400 nm, and the wavelength of B is λB = 600 nm. The power radiated by source B (a) is equal to the power of source A, (b) is less than the power of source A, (c) is greater than the power of source A, (d) cannot be compared to the power from source A using the available data.

B

Determine the Concept Since the power radiated by a source is the energy radiated per unit area and per unit time, it is directly proportional to the energy. The energy radiated varies inversely with the wavelength ( λhcE = ); i.e., the longer the wavelength, the less energy is associated with the electromagnetic radiation. ( )b is correct.

3 • [SSM] The work function of a surface is φ. The threshold wavelength for emission of photoelectrons from the surface is equal to (a) hc/φ, (b) φ/hf, (c) hf/φ, (d) none of above. Determine the Concept The work function is equal to the minimum energy required to remove an electron from the material. A photon that has that energy also has the threshold wavelength required for photoemission. Thus, φ=hf . In

addition, λfc = . It follows that φλ =thc , so φλ hc=t and ( )a is correct.

1

Chapter 34

2

4 •• When light of wavelength λ1 is incident on a certain photoelectric cathode, no electrons are emitted, no matter how intense the incident light is. Yet, when light of wavelength λ2 < λ1 is incident, electrons are emitted, even when the incident light has low intensity. Explain this observation. Determine the Concept We can use Einstein’s photoelectric equation to explain this observation. Einstein’s photoelectric equation is: φ−= hfKmax

where φ , called the work function, is a characteristic of the particular metal.

Because f = c/λ: φ

λ−=

hcK max

We’re given that when light of wavelength λ1 is incident on a certain photoelectric cathode; no electrons are emitted independently of the intensity of the incident light. Hence, we can conclude that:

01

<− φλhc

⇒ φλ

<1

hc

When light of wavelength λ2 < λ1 is incident, electrons are emitted, electrons are emitted independently of the intensity of the incident light. Hence, we can conclude that:

02

>− φλhc

⇒ φλ

>2

hc

Thus, photons with wavelengths greater than the threshold wavelength tt fc=λ , where ft is the threshold frequency, do not have enough energy to eject an electron from the metal. 5 • True or false: (a) The wavelength of an electron’s matter wave varies inversely with the

momentum of the electron. (b) Electrons can undergo diffraction. (c) Neutrons can undergo diffraction.

(a) True. The de Broglie wavelength of an electron is given by ph=λ . (b) True (c) True

Wave-Particle Duality and Quantum Physics

3

6 • If the wavelength of an electron is equal to the wavelength of a proton, then (a) the speed of the proton is greater than the speed of the electron, (b) the speeds of the proton and the electron are equal, (c) the speed of the proton is less than the speed of the electron, (d) the energy of the proton is greater than the energy of the electron, (e) Both (a) and (d) are correct. Determine the Concept If the de Broglie wavelengths of an electron and a proton are equal, their momenta must be equal. Because mp > me, vp < ve. ( )c is correct.

7 • A proton and an electron have equal kinetic energies. It follows that the wavelength of the proton is (a) greater than the wavelength of the electron, (b) equal to the wavelength of the electron, (c) less than the wavelength of the electron. Determine the Concept The kinetic energy of a particle can be expressed, in terms of its momentum, as mpK 22= . We can use the equality of the kinetic energies and the fact that me < mp to determine the relative sizes of their de Broglie wavelengths. Express the equality of the kinetic energies of the proton and electron in terms of their momenta and masses:

e

2e

p

2p

22 mp

mp

=

Use the de Broglie relation for the wavelength of matter waves to obtain:

2ee

2

2pp

2

22 λλ mh

mh

= ⇒ 2ee

2pp λλ mm =

2e

2p λλ < and pe λλ >

and ( )c is correct.

Because me < mp:

8 • The parameter x is the position of a particle, <x> is the expectation value of x, and P(x) is the probability density function. Can the expectation value of x ever equal a value of x for which P(x) is zero? If yes, give a specific example. If no, explain why not. Determine the Concept Yes. Consider a particle in a one-dimensional box of length L that is on the x axis on the interval 0 <x L. The wave function for a particle in the n = 2 state (the lowest state above the ground state) is given

by ( ) ⎟⎠⎞

⎜⎝⎛=

Lx

Lx πψ 2sin2

2 (see Figure 31-13). The expectation value of x is L/2

and P(L/2) = 0 (see Figure 31-14).

Chapter 34

4

9 •• It was once believed that if two identical experiments are done on identical systems under the same conditions, the results must be identical. Explain how this statement can be modified so that it is consistent with quantum physics. Determine the Concept According to quantum theory, the average value of many measurements of the same quantity will yield the expectation value of that quantity. However, any single measurement may differ from the expectation value. 10 •• A six-sided die has the numeral 1 painted on three sides and the numeral 2 painted on the other three sides. (a) What is the probability of a 1 coming up when the die is thrown? (b) What is the expectation value of the numeral that comes up when the die is thrown? (c) What is the expectation value of the cube of the numeral that comes up when the die is thrown? Determine the Concept The probability of a particular event occurring is the number of ways that event can occur divided by the number of possible outcomes. The expectation value, on the other hand, is the average value of the experimental results. (a) Find the probability of a 1 coming up when the die is thrown: ( )

21

631 ==P

(b) Find the average value of a large number of throws of the die:

5.16

2313=

×+×=n

(c) The expectation value of the cube of the numeral that comes up when the die is thrown is:

5.46

2313 33

=×+×

=n

Estimation and Approximation 11 •• [SSM] During an advanced physics lab, students use X rays to measure the Compton wavelength, λC. The students obtain the following wavelength shifts λ2 – λ1 as a function of scattering angle θ :

θ 45º 75º 90º 135º 180º λ2 − λ1 0.647 pm 1.67 pm 2.45 pm 3.98 pm 4.95 pm

Use their data to estimate the value for the Compton wavelength. Compare this number with the accepted value.


5

)

Picture the Problem From the Compton-scattering equation we have

( θλλλ cos1C12 −=− , where cmh eC =λ is the Compton wavelength. Note that this equation is of the form y = mx + b provided we let y = λ2 − λ1 and x = 1 − cosθ. Thus, we can linearize the Compton equation by plotting

12 λλλ −=Δ as a function of θcos1− . The slope of the resulting graph will yield an experimental value for the Compton wavelength. (a) The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic FormA3 45 θ (deg) B3 1 − cos(A3*PI()/180) 1 − cosθ C3 6.47E^−13 λΔ = 12λ − λ

θ 1− cosθ λ2−λ1

(deg) 45 0.293 6.47E−1375 0.741 1.67E−1290 1.000 2.45E−12135 1.707 3.98E−12180 2.000 4.95E−12

The following graph was plotted from the data shown in the above table. Excel’s ″Add Trendline″ was used to fit a linear function to the data and to determine the regression constants.

0.0E+00

1.0E-12

2.0E-12

3.0E-12

4.0E-12

5.0E-12

6.0E-12

0.0 0.5 1.0 1.5 2.0

m 1003.1bm 1048.2

13

12

−

−

×−=

×=a

Δλ = a(1 − cosθ) + b

Δλ, m

1 − cosθ

Chapter 34

6

From the trend line we note that the experimental value for the Compton wavelength λC,exp is:

pm48.2expC, =λ

The Compton wavelength is given by:

2ee

C cmhc

cmh

==λ

Substitute numerical values and evaluate λC:

pm43.2eV1011.5nmeV1240

5C =×

⋅=λ

Express the percent difference between λC and λC,exp:

%21pm43.2pm48.2

1diff%exp

expC,

exp

expexpC,

≈−=

−=−

=λ

λλ

λλ

12 •• Students in a physics lab are trying to determine the value of Planck’s constant h, using a photoelectric apparatus similar to the one shown in Figure 34-2. The students are using a helium–neon laser that has a tunable wavelength as the light source. The data that the students obtain for the maximum electron kinetic energies are

λ 544 nm 594 nm 604 nm 612 nm 633 nm Kmax 0.360 eV 0.199 eV 0.156 eV 0.117 eV 0.062 eV

(a) Using a spreadsheet program or graphing calculator, plot Kmax versus light frequency. (b) Use the graph to estimate the value of Planck’s constant. (Note: You may wish to use a feature of your spreadsheet program or graphing calculator to obtain the best straight-line fit to the data.) (c) Compare your result with the accepted value for Planck’s constant. Picture the Problem From Einstein’s photoelectric equation we have

φ−= hfKmax , which is of the form bmxy += , where the slope is h and the Kmax-intercept is the work function. Hence we should plot a graph of Kmax versus f in order to obtain a straight line whose slope will be an experimental value for Planck’s constant. (a) The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic FormA3 544 λ (nm) B3 0.36 Kmax (eV)


7

C3 λ (m) A3*10^−19 /c λD3 3*10^8/C3

E3 B3*1.6*10^−19 Kmax (J)

λ Kmax λ f = c/λ Kmax(nm) (eV) (m) (Hz) (J) 544 0.36 5.44E−07 5.51E+14 5.76E−20594 0.199 5.94E−07 5.05E+14 3.18E−20604 0.156 6.04E−07 4.97E+14 2.50E−20612 0.117 6.12E−07 4.90E+14 1.87E−20633 0.062 6.33E−07 4.74E+14 9.92E−21


0.E+00

1.E-20

2.E-20

3.E-20

4.E-20

5.E-20

6.E-20

7.E-20

4.6E+14 4.8E+14 5.0E+14 5.2E+14 5.4E+14 5.6E+14

f (Hz)

Km

ax (J

)

J 1083.2sJ 1019.6

19

34max

−

−

×−=

⋅×=

+=

ba

bafK

(b) From the trend line we note that the experimental value for Planck’s constant is:

sJ1019.6 34exp ⋅×= −h

(c) Express the percent difference between hexp and h:

%6.6sJ1063.6sJ1019.61

1diff%

34

34

expexp

≈⋅×⋅×

−=

−=−

=

−

−

hh

hhh

Chapter 34

8

13 •• The cathode that was used by the students in the experiment described in Problem 12 is constructed from one of the following metals:

Metals Tungsten Silver Potassium Cesium Work function 4.58 eV 2.4 eV 2.1 eV 1.9 eV

Determine which metal composes the cathode by using the same data given in Problem 12. (a) Using a spreadsheet program or graphing calculator, plot Kmax versus frequency. (b) Use the graph to estimate the value of the work function based on the students’ data. (Note: You may wish to use a feature of your spreadsheet program or graphing calculator to obtain the best straight-line fit to the data.) (c) Which metal was most likely used for the cathode in their experiment? Picture the Problem From Einstein’s photoelectric equation we have

φ−= hfKmax , which is of the form bmxy += , where the slope is h and the Kmax-intercept is the work function. Hence we should plot a graph of Kmax versus f in order to obtain a straight line whose intercept will be an experimental value for the work function. (a) The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic FormA3 544 λ (nm)

Kmax (eV) B3 0.36 C3 λ (m) A3*10^−19

/c λD3 3*10^8/C3 E3 B3*1.6*10^−19 Kmax (J)

λ Kmax λ f = c/λ Kmax

(nm) (eV) (m) (Hz) (J) 544 0.36 5.44E−07 5.51E+14 5.76E−20594 0.199 5.94E−07 5.05E+14 3.18E−20604 0.156 6.04E−07 4.97E+14 2.50E−20612 0.117 6.12E−07 4.90E+14 1.87E−20633 0.062 6.33E−07 4.74E+14 9.92E−21


9


0.E+00

1.E-20

2.E-20

3.E-20

4.E-20

5.E-20

6.E-20

7.E-20

4.6E+14 4.8E+14 5.0E+14 5.2E+14 5.4E+14 5.6E+14

f (Hz)

Km

ax (J

)

J 1083.2sJ 1019.6

19

34max

−

−

×−=

⋅×=

+=

ba

bafK

(b) From the trend line we note that the experimental value for the work function φ is:

eV77.1

J101.602eV1J1083.2 19

19exp

=

×××= −

−φ

(c) The value of φexp = 1.77 eV is closest to the work function for cesium .

The Particle Nature of Light: Photons 14 • Find the photon energy in electron volts for light of wavelength (a) 450 nm, (b) 550 nm, and (c) 650 nm. Picture the Problem We can use E = hc/λ to find the photon energy when we are given the wavelength of the radiation. (a) Express the photon energy as a function of wavelength and evaluate E for λ = 450 nm:

eV76.2nm450

nmeV1240=

⋅==

λhcE

(b) For λ = 550 nm: eV25.2nm550

nmeV1240=

⋅=E

Chapter 34

10

(c) For λ = 650 nm: eV91.1nm650

nmeV1240=

⋅=E

15 • Find the photon energy in electron volts for an electromagnetic wave of frequency (a) 100 MHz in the FM radio band and (b) 900 kHz in the AM radio band. Picture the Problem We can find the photon energy for an electromagnetic wave of a given frequency f from E = hf where h is Planck’s constant. (a) For f = 100 MHz: ( )( )

eV1014.4

J101.602eV1J10626.6

s 10100sJ 10626.6

7

1926

1634

−

−−

−−

×=

×××=

×⋅×== hfE

(b) For f = 900 kHz: ( )( )

eV1072.3

J101.602eV1J10963.5

s 10900sJ 10626.6

9

1928

1334

−

−−

−−

×=

×××=

×⋅×== hfE

16 • What are the frequencies of photons that have the following energies (a) 1.00 eV, (b) 1.00 keV, and (c) 1.00 MeV? Picture the Problem The energy of a photon, in terms of its frequency, is given by E=hf. (a) Express the frequency of a photon in terms of its energy and evaluate f for E = 1.00 eV:

Hz1042.2

seV10136.4eV00.1

14

15

×=

⋅×== −h

Ef

(b) For E = 1.00 keV:

Hz1042.2

seV10136.4keV00.1

17

15

×=

⋅×= −f

(c) For E = 1.00 MeV:

Hz1042.2

seV10136.4MeV00.1

20

15

×=

⋅×= −f


11

17 • Find the photon energy in electron volts if the wavelength is (a) 0.100 nm (about 1 atomic diameter) and (b) 1.00 fm (1 fm = 10–15 m, about 1 nuclear diameter). Picture the Problem We can use E = hc/λ to find the photon energy when we are given the wavelength of the radiation. The energy of a photon as a function of its wavelength is given by:

λhcE =

(a) Substitute numerical values and evaluate E for λ = 0.100 nm:

keV4.12nm100.0

nmeV1240=

⋅=E

(b) Substitute numerical values and evaluate E for λ =1.00 fm = 1.00 × 10−6 nm:

GeV24.1nm1000.1

nmeV12406 =

×⋅

= −E

18 •• The wavelength of red light emitted by a 3.00-mW helium–neon laser is 633 nm. If the diameter of the laser beam is 1.00 mm, what is the density of photons in the beam? Assume that the intensity is uniformly distributed across the beam. Picture the Problem We can express the density of photons in the beam as the number of photons per unit volume. The number of photons per unit volume is, in turn, the ratio of the power of the laser to the energy of the photons and the volume occupied by the photons emitted in one second is the product of the cross-sectional area of the beam and the speed at which the photons travel, i.e., the speed of light. Express the density of photons in the beam as a function of the number of photons emitted per second and the volume occupied by those photons:

VN

=ρ

Relate the number of photons emitted per second to the power of the laser and the energy of the photons:

hcP

EPN λ

==

Express the volume containing the photons emitted in one second as a function of the cross sectional area of the beam:

AcV =

Chapter 34

12

Substitute for N and V and simplify to obtain: ( ) 222

422

4dhc

Pdhc

PAhc

Pπ

λλλρπ

===

where d is the diameter of the laser beam.

Substitute numerical values and evaluate ρ:

( )( )( )( ) ( )

31322834

m1006.4mm00.1m/s10998.2sJ10626.6

nm633mW00.34 −

−×=

×⋅×=

πρ

19 • [SSM] Lasers used in a telecommunications network typically produce light that has a wavelength near 1.55 μm. How many photons per second are being transmitted if such a laser has an output power of 2.50 mW? Picture the Problem The number of photons per second is the ratio of the power of the laser to the energy of the photons. Relate the number of photons emitted per second to the power of the laser and the energy of the photons:

hcP

hcP

EPN λ

λ

===

Substitute numerical values and evaluate N:

( )( )( )( )

116

834

s1095.1

m/s10998.2sJ10626.6m55.1mW50.2

−

−

×=

×⋅×=

μN

The Photoelectric Effect 20 • The work function for tungsten is 4.58 eV. (a) Find the threshold frequency and wavelength for the photoelectric effect to occur when monochromatic electromagnetic radiation is incident on the surface of a sample of tungsten. Find the maximum kinetic energy of the electrons if the wavelength of the incident light is (b) 200 nm and (c) 250 nm. Picture the Problem The threshold wavelength and frequency for emission of photoelectrons is related to the work function of a metal through tt λφ hchf == .

We can use Einstein’s photoelectric equation φλ

−=hcKmax to find the maximum

kinetic energy of the electrons for the given wavelengths of the incident light.


13

(a) Express the threshold frequency in terms of the work function for tungsten and evaluate ft:

Hz101.11Hz101.107

seV104.136eV4.58

1515

15t

×=×=

⋅×== −h

f φ

Using c = fλ, express the threshold wavelength in terms of the threshold frequency and evaluate λt:

nm271Hz101.107m/s10998.2

15

8

tt =

××

==fcλ

(b) Using Einstein’s photoelectric equation, relate the maximum kinetic energy of the electrons to their wavelengths and evaluate Kmax: eV1.62

eV4.58nm200

nmeV1240

max

=

−⋅

=

−=−=−= φλ

φφ hchfEK

(c) Evaluate Kmax for λ = 250 nm:

eV380.0

eV4.58nm250

nmeV1240max

=

−⋅

=K

21 • When monochromatic ultraviolet light that has a wavelength equal to 300 nm is incident on a sample of potassium, the emitted electrons have maximum kinetic energy of 2.03 eV. (a) What is the energy of an incident photon? (b) What is the work function for potassium? (c) What would be the maximum kinetic energy of the electrons if the incident electromagnetic radiation had a wavelength of 430 nm? (d) What is the maximum wavelength of incident electromagnetic radiation that will result in the photoelectric emission of electrons by a sample of potassium?

Picture the Problem (a) We can use the Einstein equation for photon energy to find the energy of an incident photon. (b) and (c) We can use the photoelectric equation to relate the work function for potassium to the maximum energy of the photoelectrons. (d) The maximum wavelength of incident electromagnetic radiation that will result in the photoelectric emission of electrons can be found from λ hct φ= . (a) Use the Einstein equation for photon energy to relate the energy of the incident photon to its wavelength:

eV4.13nm300

nmeV1240=

⋅==

λhcE

Chapter 34

14

(b) Using Einstein’s photoelectric equation, relate the work function for potassium to the maximum kinetic energy of the photoelectrons:

φ−= EKmax

Solve for and evaluate φ:

eV2.10

eV2.03eV13.4max

=

−=−= KEφ

(c) Proceeding as in (b) with

λhcE = gives:

eV0.78

eV2.10nm430

nmeV1240

max

=

−⋅

=

−= φλhcK

(d) Express the maximum wavelength as a function of potassium’s work function and evaluate λt:

nm590eV2.10

nmeV1240t =

⋅==

φλ hc

22 • The maximum wavelength of electromagnetic radiation that will result in the photoelectric emission of electrons from a sample of silver is 262 nm. (a) Find the work function for silver. (b) Find the maximum kinetic energy of the electrons if the incident radiation has a wavelength of 175 nm. Picture the Problem (a) We can find the work function for silver using

tλφ hc= and (b) the maximum kinetic energy of the electrons using Einstein’s photoelectric equation. (a) Express the work function for silver as a function of the maximum wavelength of the electromagnetic radiation that will result in photoelectric emission of electrons:

tλφ hc

=

Substitute numerical values and evaluate φ :

eV73.4nm262

nmeV1240=

⋅=φ


15

(b) Using Einstein’s photoelectric equation, relate the work function for silver to the maximum kinetic energy of the photoelectrons:

φλ

φ −=−=hcEKmax

Substitute numerical values and evaluate Kmax:

eV2.36

eV4.73nm175

nmeV1240max

=

−⋅

=K

23 • The work function for cesium is 1.90 eV. (a) Find the minimum frequency and maximum wavelength of electromagnetic radiation that will result in the photoelectric emission of electrons from a sample of cesium. Find the maximum kinetic energy of the electrons if the wavelength of the incident radiation is (b) 250 nm and (c) 350 nm. Picture the Problem We can find the minimum frequency and maximum wavelength for cesium using tt λφ hchf == and the maximum kinetic energy of the electrons using Einstein’s photoelectric equation. (a) Use the Einstein equation for photon energy to express the maximum wavelength for cesium:

φλ hc

=t

Substitute numerical values and evaluate tλ :

nm536eV90.1

nmeV1240t =

⋅=λ

Use c = fλ to find the maximum frequency: Hz104.59

nm653m/s10998.2

14

8

tt

×=

×==

λcf

(b) Using Einstein’s photoelectric equation, relate the maximum kinetic energy of the photoelectrons to the wavelength of the incident light:

φλ

−=hcKmax

Evaluating Kmax for λ = 250 nm gives:

eV06.3

eV90.1nm502

nmeV1240max

=

−⋅

=K

Chapter 34

16

(c) Proceed as in Part (b) with λ = 350 nm to obtain: eV64.1

eV90.1nm503

nmeV1240max

=

−⋅

=K

24 •• When a surface is illuminated with electromagnetic radiation of wavelength 780 nm, the maximum kinetic energy of the emitted electrons is 0.37 eV. What is the maximum kinetic energy if the surface is illuminated using radiation of wavelength 410 nm? Picture the Problem We can use Einstein’s photoelectric equation to find the work function of this surface and then apply it a second time to find the maximum kinetic energy of the photoelectrons when the surface is illuminated with light of wavelength 410 nm. Use Einstein’s photoelectric equation to relate the maximum kinetic energy of the emitted electrons to their total energy and the work function of the surface:

φλ

−=hcKmax (1)

The work function of the surface is given by:

maxmax KhcKE −=−=λ

φ

Substitute numerical values and evaluate φ:

eV22.1eV0.37nm807

nmeV1240=−

⋅=φ

Substitute for φ and λ in equation (1) and evaluate Kmax:

eV80.1

eV22.1nm104

nmeV1240max

=

−⋅

=K

Compton Scattering 25 • Find the shift in wavelength of photons scattered by free stationary electrons at θ = 60°. (Assume that the electrons are initially moving with negligible speed and are virtually free of (unattached to) any atoms or molecules.) Picture the Problem We can calculate the shift in wavelength using the Compton

relationship ( )θλ cos1e

−=Δcm

h .


17

The shift in wavelength is given by: ( )θλ cos1e

−=Δcm

h

Substitute numerical values and evaluate Δλ:

( )( )( ) pm1.2cos601m/s10998.2kg109.109

sJ106.626831

34

=°−××

⋅×=Δ −

−

λ

26 • When photons are scattered by electrons in a carbon sample, the shift in wavelength is 0.33 pm. Find the scattering angle. (Assume that the electrons are initially moving with negligible speed and are virtually free of (unattached to) any atoms or molecules.) Picture the Problem We can calculate the scattering angle using the Compton


−=Δcm

h .

Using the Compton scattering equation, relate the shift in wavelength to the scattering angle:

( )θλ cos1e

−=Δcm

h

Solving for θ yields: ⎟⎠⎞

⎜⎝⎛ Δ−= − λθ

hcme1 1cos

Substitute numerical values and evaluate θ :

( )( )( ) °=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅×××

−= −

−− 30pm0.33

sJ106.626m/s10998.2kg109.1091cos 34

8311θ

27 • The photons in a monochromatic beam are scattered by electrons. The wavelength of the photons that are scattered at an angle of 135° with the direction of the incident photon beam is 2.3 percent less than the wavelength of the incident photons. What is the wavelength of the incident photons? Picture the Problem We can calculate the shift in wavelength using the Compton


−=Δcm

h .

Express the wavelength of the incident photons in terms of the fractional change in wavelength:

023.0=λλΔ

⇒023.0λλ Δ

=

Chapter 34

18


( )θλ cos1e

−=Δcm

h

Substituting for Δλ yields: ( )θλ cos1023.0 e

−=cm

h

Substitute numerical values and evaluate λ:

( )( )( )( ) nm18.0135cos1m/s10998.2kg109.1090.023

sJ106.626831

34

=°−××

⋅×= −

−

λ

28 • Compton used photons of wavelength 0.0711 nm. (a) What is the energy of one of these photons? (b) What is the wavelength of the photons scattered in the direction opposite to the direction of the incident photons? (c) What is the energy of the photon scattered in this direction? Picture the Problem We can use the Einstein equation for photon energy to find the energy of both the incident and scattered photon and the Compton scattering equation to find the wavelength of the scattered photon. (a) Use the Einstein equation for photon energy to obtain:

keV17.4nm0.0711nmeV1240

1

=⋅

==λhcE

(b) Express the wavelength of the scattered photon in terms of its pre-scattering wavelength and the shift in its wavelength during scattering:

( )θλλλλ cos1e

112 −+=Δ+=cm

h

Substitute numerical values and evaluate λ2:

( )( )( ) nm0.0760cos1801m/s10998.2kg109.109

sJ106.626nm0.0711 831

34

2 =°−××

⋅×+= −

−

λ

(c) Use the Einstein equation for photon energy to obtain:

keV16.3nm0.0760nmeV1240

2

=⋅

==λhcE

29 • For the photons used by Compton (see Problem 28), find the momentum of the incident photon and the momentum of the photon scattered in the direction opposite to the direction of the incident photons. Use the


19

conservation of momentum to find the momentum of the recoil electron in this case. Picture the Problem Compton used X rays of wavelength 71.1 pm. Let the direction the incident photon (and the recoiling electron) is moving be the positive direction. We can use p = h/λ to find the momentum of the incident photon and the conservation of momentum to find its momentum after colliding with the electron. Use the expression for the momentum of a photon to find the momentum of Compton’s photons: m/skg109.32

pm71.1sJ106.626

24

34

11

⋅×=

⋅×==

−

−

λhp


( )θλλλ cos1C12 −+=

Substitute numerical values and evaluate λ2:

( )( ) pm76.0cos1801m102.43pm71.1 122 =°−×+= −λ

Apply conservation of momentum to obtain:

2e1 ppp −= ⇒ 21e ppp −=

Substitute for p1 and p2 and evaluate pe:

m/skg101.80pm76.0

sJ106.626m/skg109.32 2334

24e ⋅×=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅×−−⋅×= −

−−p

30 •• A beam of photons have a wavelength equal to 6.00 pm is scattered by electrons initially at rest. A photon in the beam is scattered in a direction perpendicular to the direction of the incident beam. (a) What is the change in wavelength of the photon? (b) What is the kinetic energy of the electron? Picture the Problem We can calculate the shift in wavelength using the Compton

relationship ( ) ( θλθλ cos1cos1 Ce

−=−=Δcm

h ) and use conservation of energy to

find the kinetic energy of the scattered electron.

Chapter 34

20

(a) Use the Compton scattering equation to find the change in wavelength of the photon:

( )( )( )

pm2.43

cos901m102.43

cos112

C

=

°−×=

−=Δ−

θλλ

(b) Use conservation of energy to relate the change in the kinetic energy of the electron to the energies of the incident and scattered photon:

21e λλ

hchcE −=Δ

Find the wavelength of the scattered photon:

pm8.43pm2.43pm6.0012

=+=Δ+= λλλ

Substitute numerical values and evaluate the kinetic energy of the electron (equal to the change in its energy since it was stationary prior to the collision with the photon):

keV60

pm8.431

pm6.001nmeV1240e

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅=ΔE

Electrons and Matter Waves 31 • An electron is moving at 2.5 × 105 m/s. Find the electron’s de Broglie wavelength. Picture the Problem We can use its definition to find the de Broglie wavelength of this electron. Use its definition to express the de Broglie wavelength of the electron in terms of its momentum:

vmh

ph

e

==λ

Substitute numerical values and evaluate λ: ( )( )

nm9.2

m/s102.5kg10109.9sJ106.626

531

34

=

××⋅×

= −

−

λ

32 • An electron has a wavelength of 200 nm. Find (a) the magnitude of its momentum and (b) its kinetic energy.


21

Picture the Problem We can find the momentum of the electron from the de

Broglie equation and its kinetic energy from nm226.1K

=λ , where K is in eV.

(a) Use the de Broglie relation to express the momentum of the electron:

m/skg103.31

nm200sJ106.626

27

34

⋅×=

⋅×==

−

−

λhp

(b) Use the electron wavelength equation to relate the electron’s wavelength to its kinetic energy:

nm226.1K

=λ

Solving for K gives:

eV1076.3

nm200nmeV1.226

5

221

−×=

⎟⎟⎠

⎞⎜⎜⎝

⎛=K

33 •• An electron, a proton, and an alpha particle each have a kinetic energy of 150 keV. Find (a) the magnitudes of their momenta and (b) their de Broglie wavelengths. Picture the Problem The momenta of these particles can be found from their kinetic energies. The values for the electron, however, are only approximately correct because at the given energy the speed of the electron is a significant fraction of the speed of light. The solution presented is valid only in the non-relativistic limit v << c. The de Broglie wavelengths of the particles are given by λ = h/p. (a) The momentum of a particle p, in terms of its kinetic energy K, is given by:

mKp 2=

Substitute numerical values and evaluate pe:

( )

sN1009.2sN10092.2

eVC101.602keV150kg10109.922

2222

1931

ee

⋅×=⋅×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××==

−−

−−Kmp

Chapter 34

22

Substitute numerical values and evaluate pp:

( )

sN1097.8sN10967.8

eVC101.602keV150kg10673.122

2121

1927

pp

⋅×=⋅×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××==

−−

−−Kmp

Substitute numerical values and evaluate pα:

sN1079.1sN10787.1

eVC101.602keV150

ukg10661.1u422

2020

1927

⋅×=⋅×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××⎟⎟

⎠

⎞⎜⎜⎝

⎛ ××==

−−

−−

Kmp αα

(b) The de Broglie wavelengths of the particles are given by:

ph

=λ (1)

Substitute numerical values in equation (1) and evaluate λe:

pm17.3

sN10092.2sJ10626.6

21

34

ee

=

⋅×⋅×

== −

−

phλ

Substitute numerical values in equation (1) and evaluate λp:

fm9.73

sN10967.8sJ10626.6

21

34

pp

=

⋅×⋅×

== −

−

phλ

Substitute numerical values in equation (1) and evaluate λα:

fm0.37

sN1079.1sJ1063.6

20

34

=

⋅×⋅×

== −

−

ααλ

ph

34 • A neutron in a reactor has kinetic energy of approximately 0.020 eV. Calculate the wavelength of this neutron. Picture the Problem The wavelength associated with a particle of mass m and

kinetic energy K is given by Kmc22nmeV1240 ⋅

=λ .


23

Substituting numerical values yields: ( )( )

nm20.0eV020.0MeV9402

nmeV1240=

⋅=λ

35 • Find the wavelength of a proton that has a kinetic energy of 2.00 MeV. Picture the Problem The wavelength associated with a particle of mass m and

kinetic energy K is given by Kmc22nmeV1240 ⋅

=λ .

Substituting numerical values yields: ( )( )

fm2.20

MeV00.2MeV9382nmeV1240

=

⋅=λ

36 • What is the kinetic energy of a proton whose wavelength is (a) 1.00 nm and (b) 1.00 fm?

Picture the Problem We can solve Equation 34-15 (Kmc22nmeV1240 ⋅

=λ ) for the

kinetic energy of the proton and use the rest energy of a proton mc2 = 938 MeV to simplify our computation. Solving Equation 34-15 for the kinetic energy of the proton gives:

( )22

2

2nmeV1240

λmcK ⋅

=

(a) Substitute numerical values and evaluate K for λ = 1.00 nm:

( )( )( )

meV820.0

nm00.1MeV9382nmeV1240

2

2

=

⋅=K

(b) Evaluate K for λ = 1.00 fm: ( )

( )( )MeV820

fm00.1MeV9382nmeV1240

2

2

=

⋅=K

37 • The kinetic energy of the electrons in the electron beam in a run of Davisson and Germer’s experiment was 54 eV. Calculate the wavelength of the electrons in the beam. Picture the Problem If K is in electron volts, the wavelength of a particle is

given by nm226.1K

=λ provided K is in eV.

Chapter 34

24

Evaluate λ for K = 54 eV: nm17.0nm54

1.226nm226.1===

Kλ

38 • The distance between Li+ and Cl– ions in a LiCl crystal is 0.257 nm. Find the energy of electrons that have a wavelength equal to this spacing.

Picture the Problem We can use nm226.1K

=λ , where K is in eV, to find the

energy of electrons whose wavelength is λ. Relate the wavelength of the electrons to their kinetic energy:

nm226.1K

=λ

Solving for K yields: 221eVnm226.1⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅=

λK

Substituting numerical values and evaluating K gives: eV8.22

nm257.0eVnm226.1

221

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅=K

39 • [SSM] An electron microscope uses electrons that have energies equal to 70 keV. Find the wavelength of these electrons. Picture the Problem We can approximate the wavelength of 70-keV electrons

using nm226.1K

=λ , where K is in eV. This solution is, however, only

approximately correct because at the given energy the speed of the electron is a significant fraction of the speed of light. The solution presented is valid only in the non-relativistic limit v << c. Relate the wavelength of the electrons to their kinetic energy:

nm23.1K

=λ

Substitute numerical values and evaluate λ:

pm6.4nmeV1070

226.13

=×

=λ

40 • What is the de Broglie wavelength of a neutron that has a speed of 1.00 × 106 m/s? Picture the Problem We can use its definition to calculate the de Broglie wavelength of a neutron with speed 106 m/s.


25

Use its definition to express the de Broglie wavelength of the neutron:

nnnn vm

hph

==λ

Substitute numerical values and evaluate λn: ( )( )

pm0.396

m/s1000.1kg101.675sJ106.626

627

34

n

=

××⋅×

= −

−

λ

A Particle in a Box 41 •• (a) Find the energy of the ground state (n = 1) and the first two excited states of a neutron in a one-dimensional box of length L = 1.00 ×10–15 m = 1.00 fm (about the diameter of an atomic nucleus). Make an energy-level diagram for this system. Calculate the wavelength of electromagnetic radiation emitted when the neutron makes a transition from (b) n = 2 to n = 1, (c) n = 3 to n = 2, and (d) n = 3 to n = 1.

Picture the Problem We can find the ground-state energy using 2

2

1 8mLhE = and

the energies of the excited states using . The wavelength of the electromagnetic radiation emitted when the neutron transitions from one state to

another is given by the Einstein equation for photon energy (

12EnEn =

λhcE = ).

(a) Express the ground-state energy:

2n

2

1 8 LmhE =

Substitute numerical values and evaluate E1:

( )( )( )

MeV205

MeV204.53J101.602

eV1m101.00kg101.67498

sJ106.6261921527

234

1

=

=×××

⋅×= −−−

−

E

( )

MeV818

MeV53.20442 12

2

=

== EE

Find the energies of the first two excited states:

and ( )

GeV84.1

MeV53.20493 12

3

=

== EE

Chapter 34

26

The energy-level diagram for this system is shown to the right:

1EE

0123

1

4

5

49

16

25

n

(b) Relate the wavelength of the electromagnetic radiation emitted during a neutron transition to the energy released in the transition:

EEhc

Δ⋅

=Δ

=nmeV1240λ

For the n = 2 to n = 1 transition: 11112 34 EEEEEE =−=−=Δ

Substitute numerical values and evaluate λ2→1: ( )

fm02.2

MeV53.2043nmeV1240

3nmeV1240

112

=

⋅=

⋅=→ E

λ

(c) For the n = 3 to n = 2 transition:

11123 549 EEEEEE =−=−=Δ


fm21.1

MeV53.2045nmeV1240

5nmeV1240

123

=

⋅=

⋅=→ E

λ

(d) For the n = 3 to n = 1 transition:

11113 89 EEEEEE =−=−=Δ


fm758.0

MeV53.2048nmeV1240

8nmeV1240

113

=

⋅=

⋅=→ E

λ

42 •• (a) Find the energy of the ground state (n = 1) and the first two excited states of a neutron in a one-dimensional box of length 0.200 nm (about the diameter of a H2 molecule). Calculate the wavelength of electromagnetic radiation emitted when the neutron makes a transition from (b) n = 2 to n = 1, (c) n = 3 to n = 2, and (d) n = 3 to n = 1.


27

Picture the Problem We can find the ground-state energy using 2

2

1 8mLhE = and

the energies of the excited states using . The wavelength of the electromagnetic radiation emitted when the neutron transitions from one state to

another is given by the Einstein equation for photon energy (

12EnEn =

λhcE = ).

(a) Express the ground-state energy:

2n

2

1 8 LmhE =


( )( )( )

meV11.5meV1133.5J101.602

eV1nm.2000kg101.67498

sJ106.62619227

234

1 ==××

⋅×= −−

−

E

( )

meV5.20

meV1133.542 12

2

=

== EE

Find the energies of the first two excited states:

and ( )

meV0.46

meV1133.593 12

3

=

== EE

(b) Relate the wavelength of the electromagnetic radiation emitted during a neutron transition to the energy released in the transition:

EEhc

Δ⋅

=Δ

=nmeV1240λ

For the n = 2 to n = 1 transition: 11112 34 EEEEEE =−=−=Δ


m8.80

meV1133.53nmeV1240

3nmeV1240

112

μ

λ

=

⋅=

⋅=→ E

(c) For the n = 3 to n = 2 transition:

11123 549 EEEEEE =−=−=Δ


m5.48

meV1133.55nmeV1240

5nmeV1240

123

μ

λ

=

⋅=

⋅=→ E

Chapter 34

28

(d) For the n = 3 to n = 1 transition:

11113 89 EEEEEE =−=−=Δ


m3.30

meV1133.58nmeV1240

8nmeV1240

113

μ

λ

=

⋅=

⋅=→ E

Calculating Probabilities and Expectation Values 43 •• A particle is in the ground state of a one-dimensional box that has length L. (The box has one end at the origin and the other end on the positive x axis.) Determine the probability of finding the particle in the interval of length Δx = 0.002L and centered at (a) 1

4 x = L, (b) x = 12 L, and (c) x = 3

4 L. (Because Δx is very small you need not do any integration.) Picture the Problem The probability of finding the particle in some range Δx is

dx2ψ . The interval Δx = 0.002L is so small that we can neglect the variation in ψ(x) and just compute ψ2Δx. Express the probability of finding the particle in the interval Δx:

( ) ( ) xxxxPP Δ=Δ= 2ψ

Express the wave function for a particle in the ground state:

( )Lx

Lx πψ sin2

1 =

Substitute for ( )x1ψ to obtain:

( )

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

Δ=

Lx

LLx

L

xLx

LP

π

π

π

2

2

2

sin004.0

002.0sin2

sin2

(a) Evaluate P at x = 4L: ( ) ( )

( ) 02sin004.024sin004.04

2

2

==

=

π

πLLLP

(b) Evaluate P at x = L/2: ( )

12

sin004.0

2sin004.02

2

2

=⎟⎠⎞

⎜⎝⎛=

=

π

πLLLP


29

(c) Evaluate P at x = 3L/4: ( )

002.04

3sin004.0

43sin004.043

2

2

=⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

π

πLLLP

44 •• A particle is in the second excited state (n = 3) of a one-dimensional box that has length L. (The box has one end at the origin and the other end on the positive x axis.) Determine the probability of finding the particle in the interval of length Δx = 0.002L and centered at (a) x = 1

3 L, (b) x = 12 L, and (c) x =

23 L.

(Because Δx is very small you need not do any integration.) Picture the Problem The probability of finding the particle in some range dx is

dx2ψ . The interval Δx = 0.002L is so small that we can neglect the variation in ψ(x) and just compute ψ 2Δx. Express the probability of finding the particle in the interval Δx:

( ) ( ) xxxxPP Δ=Δ= 2ψ

Express the wave function for a particle in its second excited state:

( )L

xL

x πψ 2sin22 =

Substitute for ( )x2ψ to obtain:

( )

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

Δ=

Lx

LL

xL

xL

xL

P

π

π

π

2sin004.0

002.02sin2

2sin2

2

2

2

(a) Evaluate P at x = L/3: ( )

003.03

2sin004.0

32sin004.03

2

2

=⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

π

πLLLP

(b) Evaluate P at x = L/2: ( )

( ) 0sin004.0

22sin004.02

2

2

==

⎟⎠⎞

⎜⎝⎛=

π

πLLLP

Chapter 34

30

(c) Evaluate P at x = 2L/3: ( )

003.03

4sin004.0

34sin004.032

2

2

=⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

π

πLLLP

45 •• A particle is in the first excited (n = 2) state of a one-dimensional box that has length L. (The box has one end at the origin and the other end on the positive x axis.) Find (a) x and (b) x 2 .

Picture the Problem We’ll use ( ) ( ) ( )∫= dxxxfxf 2ψ with ( )L

xnL

xnπψ sin2

= .

(a) Express ψ(x) for the n = 2 state: ( )

Lx

Lx πψ 2sin2

2 =

Express x using the n = 2 wave

function:

∫=L

dxL

xLxx

0

2 2sin2 π

Change variables by letting

Lxπθ 2

= . Then: θ

ππθθ

πdLdxdx

LdLx

2 ,2 ,

2===

and the limits on θ are 0 and 2π.

Substitute to obtain:

∫

∫

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

π

θθθπ

θπ

θθπ

2

0

22

0

2

sin2

2sin

22

dL

dLLL

xL

Use a table of integrals to evaluate the integral:

281

81

2

82cos

42sin

42

22

2

0

2

2

LL

Lx

=⎥⎦⎤

⎢⎣⎡ +−=

⎥⎦

⎤⎢⎣

⎡−−=

ππ

θθθθπ

π

(b) Express 2x using the n = 2

wave function: ∫=L

dxL

xLxx

0

22

2 2sin2 π


31

Change variables as in (a) and substitute to obtain:

∫

∫

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

π

θθθπ

θπ

θθπ

2

0

223

2

0

22

2

sin4

2sin

22

dL

dLLL

xL


22

2

3

3

22

0

23

3

22

321.08

131

234

442cos2sin

81

464

LL

LLx

=⎟⎠⎞

⎜⎝⎛ −=

⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

π

πππ

θθθθθπ

π

46 •• A particle in a one-dimensional box that has length L is in the first excited state (n = 2). (The box has one end at the origin and the other end on the positive x axis.) (a) Sketch ψ 2(x) versus x for this state. (b) What is the expectation value <x> for this state? (c) What is the probability of finding the particle in some small region dx centered at x = L/2? (d) Are your answers for Part (b) and Part (c) contradictory? If not, explain why your answers are not contradictory. Picture the Problem We’ll use ( ) ( ) ( )∫= dxxxfxf 2ψ with

( )L

xnL

xnπψ sin2

= . In Part (c) we’ll use ( ) ( )xxP 22ψ= to determine the

probability of finding the particle in some small region dx centered at x = ½L. (a) Express the wave function for a particle in its first excited state:

( )L

xL

x πψ 2sin22 =

Square both sides of the equation to obtain:

( )L

xL

x πψ 2sin2 222 =

The graph of ψ 2(x) as a function of x is shown to the right:

xL0

pro

bab

ilit

y d

ensi

ty

Chapter 34

32

(b) Express x using the n = 2 wave

function:

∫=L

dxL

xLxx

0

2 2sin2 π


Lxπθ 2

= . Then:

θπ

πθ

θπ

dLdx

dxL

d

Lx

2

and ,2

,2

=

=

=

and the limits on θ are 0 and 2π.

Substitute in the expression for x

to obtain:

∫

∫

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

π

θθθπ

θπ

θθπ

2

0

22

0

2

sin2

2sin

22

dL

dLLL

xL

Using a table of integrals, evaluate the integral:

281

81

2

82cos

42sin

42

22

2

0

2

2

LL

Lx

=⎥⎦⎤

⎢⎣⎡ +−=

⎥⎦

⎤⎢⎣

⎡−−=

ππ

θθθθπ

π

(c) Express P(x):

( ) ( )L

xL

xxP πψ 2sin2 222 ==

Evaluate P(L/2):

0sin22

2sin22

2

2

==

⋅=⎟⎠⎞

⎜⎝⎛

π

π

L

LLL

LP

Because P(L/2) = 0: 0

2=⎟

⎠⎞

⎜⎝⎛ dxLP

(d) The answers to Parts (b) and (c) are not contradictory. Part (b) states that the average value of measurements of the position of the particle will yield L/2 even though the probability that any one measurement of position will yield this value is zero.


33

47 •• A particle of mass m has a wave function given by ( ) axAex −=ψ , where A and a are positive constants. (a) Find the normalization constant A. (b) Calculate the probability of finding the particle in the region –a ≤ x ≤ a. Picture the Problem We can find the constant A by applying the normalization

condition and finding the value for A that satisfies this condition. As

soon as we have found the normalization constant, we can calculate the probability of the finding the particle in the region −a ≤ x ≤ a using

.

( )∫∞

∞−

= 12 dxxψ

( )∫−

=a

a

dxxP 2ψ

(a) The normalization condition is: ( )∫

∞

∞−

= 12 dxxψ

Substitute ( ) axAex −=ψ : ( ) ∫∫

∞−

∞

∞−

− =0

2222 dxeAdxAe axax

From integral tables: α

α 1

0

=∫∞

− dxe x

Therefore:

12

22 22

0

22 ==⎟⎠⎞

⎜⎝⎛=∫

∞− aAaAdxeA ax

Solving for A yields:

aA 1

=

(b) Express the normalized wave function:

( ) axea

x −=1ψ

The probability of finding the particle in the region −a ≤ x ≤ a is:

( )

865.012

12

2

0

2

0

22

=−==

==

−−

−

−

∫

∫∫

edxea

dxea

dxxP

aax

aax

a

a

ψ

48 •• A one-dimensional box is on the x-axis in the region of 0 ≤ x ≤ L. A particle in this box is in its ground state. Calculate the probability that the particle will be found in the region (a) Lx 2

10 << , (b) Lx 310 << , and (c) Lx 4

30 << .

Chapter 34

34

Picture the Problem The probability density for the particle in its ground state is

given by ( ) xLL

xP π2sin2= . We’ll evaluate the integral of P(x) between the limits

specified in (a), (b), and (c). Express P(x) for 0 < x < d: ( ) ∫=

d

xdxLL

xP0

2sin2 π

Change variables by letting xLπθ = .

Then: θ

π

πθθπ

dLdx

dxL

dLx

=

== and , ,

Substitute to obtain: ( )

∫

∫

=

⎟⎠⎞

⎜⎝⎛=

'

'

d

dLL

xP

θ

θ

θθπ

θπ

θ

0

2

0

2

sin2

sin2

Use a table of integrals to evaluate

: ∫'

0

2sinθ

θθ d

( )'

xPθθθ

π 042sin

22

⎥⎦⎤

⎢⎣⎡ −= (1)

(a) Noting that the limits on θ are 0 and π/2, evaluate equation (1) over the interval 0 < x < ½L:

( )

500.0

42

42sin

22 2

0

=

⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡ −=

ππ

θθπ

π

xP

(b) Noting that the limits on θ are 0 and π/3, evaluate equation (1) over the interval 0 < x < L/3:

( )

196.04

331

432sin

62

42sin

22 3

0

=−=

⎥⎦⎤

⎢⎣⎡ −=

⎥⎦⎤

⎢⎣⎡ −=

π

πππ

θθπ

π

xP


35

(c) Noting that the limits on θ are 0 and 3π/4, Evaluate equation (1) over the interval 0 < x < 3L/4:

( )

909.021

43

446sin

832

42sin

22 43

0

=+=

⎥⎦⎤

⎢⎣⎡ −=

⎥⎦⎤

⎢⎣⎡ −=

π

πππ

θθπ

π

xP

49 •• A one-dimensional box is on the x-axis in the region of 0 ≤ x ≤ L. A particle in this box is in its first excited state. Calculate the probability that the particle will be found in the region (a) 0 < x < 1

2 L, (b) 0 < x < 13 L, and

(c) 0 < x < 34 L.

Picture the Problem The probability density for the particle in its first excited

state is given by ( ) xLL

xP π2sin2 2= . We’ll evaluate the integral of P(x) between

the limits specified in (a), (b), and (c). Express P(x) for 0 < x < d: ( ) ∫=

d

xdxLL

xP0

2 2sin2 π


xLπθ 2

= . Then: θ

π

πθθπ

dLdx

dxL

dLx

2

and ,2 ,2

=

==

Substitute to obtain: ( )

∫

∫

=

⎟⎠⎞

⎜⎝⎛=

'

'

d

dLL

xP

θ

θ

θθπ

θπ

θ

0

2

0

2

sin1

2sin2

Use a table of integrals to evaluate

: ∫'

0

2sinθ

θθ d

( )'

xPθθθ

π 042sin

21

⎥⎦⎤

⎢⎣⎡ −= (1)

(a) Noting that the limits on θ are 0 and π, evaluate equation (1) over the interval 0 < x < ½L:

( )

500.0

21

42sin

21

0

=

⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡ −=

ππ

θθπ

π

xP

Chapter 34

36

(b) Noting that the limits on θ are 0 and 2π/3, evaluate equation (1) over the interval 0 < x < L/3:

( )

402.04

2331

434sin

621

42sin

21 32

0

=+=

⎥⎦⎤

⎢⎣⎡ −=

⎥⎦⎤

⎢⎣⎡ −=

π

πππ

θθπ

π

xP

(c) Noting that the limits on θ are 0 and 3π/2, evaluate equation (1) over the interval 0 < x < 3L/4:

( )

750.0

431

42sin

21 23

0

=

⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡ −=

ππ

θθπ

π

xP

50 •• The classical probability distribution function for a particle in a one-dimensional box on the x axis in the region 0 < x < L is given by P(x) = 1/L. Use this expression to show that x = 1

2 L and x 2 = 13 L2 for a classical particle in

the box. Picture the Problem Classically, ( )∫= dxxxPx and ( )∫= dxxPxx 22 .

Evaluate x with P(x) = 1/L:

LL

xdxLxx

LL

21

0

2

0 2=⎥

⎦

⎤⎢⎣

⎡== ∫

Evaluate 2x with P(x) = 1/L:

231

0

3

0

22

3L

Lxdx

Lxx

LL

=⎥⎦

⎤⎢⎣

⎡== ∫

51 •• A one-dimensional box is on the x-axis in the region of 0 ≤ x ≤ L. (a) The wave functions for a particle in the box are given by

ψ n x( )=

2L

sinnπx

L n = 1, 2, 3, . . .

For a particle in the nth state, show that x = 12 L and x 2 = L2/3 – L2/(2n2π2).

(b) Compare these expressions for x and x 2 , for n >> 1, with the expressions

for x and

x 2 for the classical distribution of Problem 50.

Picture the Problem We’ll use ( ) ( ) ( )∫= dxxxfxf 2ψ with

( )L

xnL

xnπψ sin2

= to show that 2Lx = and 22

222

23 πnLLx == .


37

(a) Express x for a particle in the

nth state:

∫=L

dxL

xnLxx

0

2sin2 π

Change variables by lettingL

xnπθ = .

Then:

θπ

πθθπ

dnLdxdx

Lnd

nLx === , ,

and the limits on θ are 0 and nπ.


∫

∫

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

π

π

θθθπ

θπ

θθπ

n

n

dn

L

dnL

nL

Lx

0

222

0

2

sin2

sin2


2

81

81

42

81

82cos

42sin

42

82cos

42sin

42

22

22

22

22

0

2

22

L

nn

Lnnnnn

L

nLx

n

=

⎥⎦

⎤⎢⎣

⎡+−=⎥

⎦

⎤⎢⎣

⎡+−−=

⎥⎦

⎤⎢⎣

⎡−−=

ππ

πππππ

θθθθπ

π

Express 2x for a particle in the

nth state:

∫=L

dxL

xnLxx

0

22

2 sin2 π

Change variables by letting L

xnπθ = .

Then:

θπ

πθθπ

dnLdxdx

Lnd

nLx === , ,

and the limits on θ are 0 and nπ.


∫

∫

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

π

π

θθθπ

θπ

θθπ

n

n

dn

L

dnL

nL

Lx

0

2233

2

0

22

2

sin2

sin2

Chapter 34

38


22

22

33

33

2

0

23

33

22

23

462

42cos2sin

81

462

π

πππ

θθθθθπ

π

nLL

nnn

Ln

Lxn

−=

⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

(b) For large values of n the result agrees with the classical value of L2/3 given in Problem 50. 52 •• (a) Use a spreadsheet program or graphing calculator to plot

x 2 as

a function of the quantum number n for the particle in the box described in Problem 48 and for values of n from 1 to 100. Assume L = 1.00 m for your graph. Refer to Problem 51. (b) Comment on the significance of any asymptotic limits that your graph shows.

Picture the Problem (a) From Problem 51 we have 2Lx = and

22

222

23 πnLLx −= . A spreadsheet program was used to plot the following graph

of <x2> as a function of n.

0.25

0.26

0.27

0.28

0.29

0.30

0.31

0.32

0.33

0.34

1 10 19 28 37 46 55 64 73 82 91 100

n

<x2 >


39

(b) 3

, As2

2 Lxn →∞→

53 •• The wave functions for a particle of mass m in a one-dimensional box of length L centered at the origin (so that the ends are at x = ± 1

2 L) are given by

ψ x( )=

2L

cosnπx

L n = 1, 3, 5, 7, . . .

and

ψ x( )=

2L

sinnπx

L n = 2, 4, 6, 8, . . .

Calculate x and

x 2 for the ground state (n = 1). Picture the Problem For the ground state, n = 1 and so we’ll evaluate

( ) ( ) ( )∫= dxxxfxf 2ψ using ( )Lx

Lx πψ cos2

1 = .

Because ( )x2

1ψ is an even function of x, is an odd function of x. It follows that the integral of

( )xx 21ψ

( )xx 21ψ

between −L/2 and L/2 is zero. Thus:

0=x for all values of n.

Express 2x : ∫

−

=2

2

222 cos2 L

L

xdxL

xL

x π

Change variables by letting Lxπθ = .

Then:

θπ

πθθπ

dLdxdxL

dLx === , ,

and the limits on θ are −π/2 and π/2.

Substitute for x and dx to obtain:

∫

∫

−

−

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

2

2

223

2

2

2

22

2

cos2

cos2

π

π

π

π

θθθπ

θπ

θθπ

dL

dLLL

x

Use a trigonometric identity to rewrite the integrand:

( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

−=

∫∫

∫

−−

−

2

2

222

2

23

2

2

2

223

22

sin2

sin12

π

π

π

π

π

π

θθθθθπ

θθθπ

ddL

dLx

Chapter 34

40

Evaluate the second integral by looking it up in the tables:

⎥⎦⎤

⎢⎣⎡ −=⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤+⎟⎟

⎠

⎞⎜⎜⎝

⎛−++

⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+=

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

−

−

22

3

3

2

23

23

3

2

2

2

23

3

2

2

2

233

3

22

21

121

4242

8cossin

81

1648

8cossin

81

16482

42cos2sin

81

462

42cos2sin

81

4632

πππ

π

πππππ

ππππππ

θθθθθπ

θθθθθθπ

π

π

π

π

LL

L

L

Lx

Remarks: The result differs from that of Example 34-8. Since we have shifted the origin by Δx = L/2, we could have arrived at the above result, without performing the integration, by subtracting (Δx)2 = L2/4 from 2x . 54 •• Calculate <x> and <x2> for the first excited state (n = 2) of the box described in Problem 53. Picture the Problem For the first excited state, n = 2, and so we’ll evaluate

( ) ( ) ( )∫= dxxxfxf 2ψ using ( )L

xL

x πψ 2sin22 = .

Because ( )x2

2ψ is an even function of x, is an odd function of x. It follows that the integral of

( )xx 22ψ

( )xx 22ψ

between −L/2 and L/2 is zero. Thus:

0=x

Express 2x : ∫

−

=2

2

222 2sin2 L

L

xdxL

xL

x π

Change variables by letting L

xπθ 2= .

Then:

θπ

πθθπ

dLdxdxL

dLx2

,2 ,2

===

and the limits on θ are −π and π.


41


∫

∫

−

−

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

π

π

π

π

θθθπ

θπ

θθπ

dL

dLLL

x

223

2

22

2

sin4

2sin

22

Evaluate the integral by looking it up in the tables:

⎥⎦⎤

⎢⎣⎡ −=⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤−+⎢

⎣

⎡−=

⎥⎦

⎤−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+

⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

−

22

3

3

2

33

3

2

23

23

3

2

23

3

22

81

121

234

46464

42cos2sin

81

46

42cos2sin

81

464

42cos2sin

81

464

πππ

π

πππππ

πππππ

ππππππ

θθθθθπ

π

π

LL

L

L

Lx

Remarks: The result differs from that of Example 34-8. Since we have shifted the origin by Δx = L/2, we could have arrived at the above result, without performing the integration, by subtracting (Δx)2 = L2/4 from 2x .

General Problems 55 • [SSM] Photons in a uniform 4.00-cm-diameter light beam have wavelengths equal to 400 nm and the beam has an intensity of 100 W/m2. (a) What is the energy of each photon in the beam? (b) How much energy strikes an area of 1.00 cm2 perpendicular to the beam in 1 s? (c) How many photons strike this area in 1.00 s? Picture the Problem We can use the Einstein equation for photon energy to find the energy of each photon in the beam. The intensity of the energy incident on the surface is the ratio of the power delivered by the beam to its delivery time. Hence, we can express the energy incident on the surface in terms of the intensity of the beam.

Chapter 34

42

(a) Use the Einstein equation for photon energy to express the energy of each photon in the beam:

λhchfE ==photon

Substitute numerical values and evaluate Ephoton:

eV3.10

eV3.100nm400

nmeV1240photon

=

=⋅

=E

(b) Relate the energy incident on a surface of area A to the intensity of the beam:

tIAE Δ=

Substitute numerical values and evaluate E:

( )( )( )

eV106.24eV106.242

J101.602eV1J0.0100

s1.00m101.00W/m100

1616

19

242

×=×=

××=

×=

−

−E

(c) Express the number of photons striking this area in 1.00 s as the ratio of the total energy incident on the surface to the energy delivered by each photon:

16

16

photon

1008.2

eV3.100eV106.242

×=

×==

EEN

56 • A 1-μg particle is moving with a speed of approximately 1 mm/s in a one-dimensional box that has a length equal to 1 cm. Calculate the approximate value of the quantum number n of the state occupied by the particle.

Picture the Problem The particle’s nth-state energy is 2

22

8mLhnEn = . We can find

n by solving this equation for n and substituting the particle’s kinetic energy for En. Express the energy of the particle when it is in its nth state: 2

22

8mLhnEn = ⇒ nmE

hLn 8=

Express the energy (kinetic) of the particle:

221 mvEn =

Substitute for En and simplify to obtain: h

mvLn 2=


43

Substitute numerical values and evaluate n:

( )( )( )

19

34

239

103

sJ106.626m10m/s101kg102

×≈

⋅××

= −

−−−

n

57 • (a) For the particle and box of Problem 56, find Δx and Δpx, assuming that these uncertainties are given by Δx/L = 0.01 percent and Δpx/px = 0.01 percent. (b) What is ( ) hpx xΔΔ ? Picture the Problem We can use the fact that the uncertainties are given by Δx/L = 0.01 percent and Δpx/px = 0.01 percent to find Δx and Δp. (a) Assuming that Δx/L = 0.01 percent, find Δx:

( ) ( ) m1m101010 244 μ===Δ −−− Lx

Assuming that Δpx/px = 0.01 percent, find Δp:

( )( )m/skg10

m/s10kg10101016

3944

⋅=

==−

−−−− mvpxΔ

(b) Evaluate ( ) hpx xΔΔ : ( )( )

11

34

16

102sJ 10626.6

m/skg10m1

×=

⋅×⋅

= −

−μh

px xΔΔ

58 • In 1987, a laser at Los Alamos National Laboratory produced a flash that lasted 1 × 10–12 s and had a power of 5 × 1015 W. Estimate the number of emitted photons, assuming they all had wavelengths equal to 400 nm. Picture the Problem We can estimate the number of emitted photons from the ratio of the total energy in the flash to the energy of a single photon. Letting N be the number of emitted photons, express the ratio of the total energy in the flash to the energy of a single photon:

photonEEN =

Relate the energy in the flash to the power produced:

tPE Δ=

Express the energy of a single photon as a function of its wavelength:

λhcE =photon

Chapter 34

44

Substituting for E and Ephoton and simplifying gives:

hctPN λΔ

=


( )( )( )( )( )

22834

91215

101m/s10998.2sJ106.626

m10400s101W105×≈

×⋅××××

= −

−−

N

59 • You cannot ″see″ anything smaller than the wavelength of the wave used to make the observation. What is the minimum energy of an electron needed in an electron microscope to ″see″ an atom that has a diameter of about 0.1 nm?

Picture the Problem We can use the electron wavelength equation nm,23.1K

=λ

where K is in eV to find the minimum energy required to see an atom. Relate the energy of the electron to the size of an atom (the wavelength of the electron):

nm23.1K

=λ ⇒( )

2

221 nmeV23.1λ

⋅=K

provided K is in eV.

Substitute numerical values and evaluate K:

( )( )

keV2.0nm1.0

nmeV23.12

221

≈⋅

=K

60 • A common flea that has a mass of 0.008 g can jump vertically as high as 20 cm. Estimate the de Broglie wavelength for the flea immediately after takeoff. Picture the Problem The flea’s de Broglie wavelength is ,ph=λ where p is the flea’s momentum immediately after takeoff. We can use a constant acceleration equation to find the flea’s speed and, hence, momentum immediately after takeoff. Express the de Broglie wavelength of the flea immediately after takeoff:

0mvh

ph

==λ

yavv Δ+= 22

02

or, since v = 0 and a = −g, ygv Δ= 20

Using a constant acceleration equation, express the height the flea can jump as a function of its takeoff speed:


45

Substitute to obtain: ygm

hΔ

=2

λ

Substitute numerical values and evaluate λ: ( ) ( )( )

m104

m0.2m/s9.812kg108sJ106.626

29

26

34

−

−

−

×≈

×

⋅×=λ

61 •• [SSM] A 100-W source radiates light of wavelength 600 nm uniformly in all directions. An eye that has been adapted to the dark has a 7-mm-diameter pupil and can detect the light if at least 20 photons per second enter the pupil. How far from the source can the light be detected under these rather extreme conditions? Picture the Problem We can relate the fraction of the photons entering the eye to ratio of the area of the pupil to the area of a sphere of radius R. We can find the number of photons emitted by the source from the rate at which it emits and the energy of each photon which we can find using the Einstein equation. Letting r be the radius of the pupil, Nentering eye the number of photons per second entering the eye, and Nemitted the number of photons emitted by the source per second, express the fraction of the light energy entering the eye at a distance R from the source:

2

2

2

2

2eye

emitted

eyeentering

4

4

4

Rr

Rr

RA

NN

=

=

=

ππ

π

Solving for R yields:

eyeentering

emitted

2 NNrR = (1)

Find the number of photons emitted by the source per second: photon

emitted EPN =

Using the Einstein equation, express the energy of the photons:

λhcE =photon

Substitute numerical values and evaluate Ephoton:

eV2.07nm600

nmeV1240photon =

⋅=E

Chapter 34

46

Substitute and evaluate Nemitted:

( )120

19emitted

s1002.3eVJ101.602eV2.07

W100

−

−

×=

⎟⎠⎞

⎜⎝⎛ ×

=N

Substitute for Nemitted in equation (1) and evaluate R:

km107

s20s103.02

2mm5.3

3

1

120

×≈

×= −

−

R

62 •• The diameter of the pupil of an eye under room-light conditions is approximately 5 mm. Find the intensity of light that has a wavelength equal to 600 nm so that 1 photon per second passes through the pupil. Picture the Problem The intensity of the light such that one photon per second passes through the pupil is the ratio of the energy of one photon to the product of the area of the pupil and time interval during which the photon passes through the pupil. We’ll use the Einstein equation to express the energy of the photon. Use its definition to relate the intensity of the light to the energy of a 600-nm photon:

tAE

API

Δ== photon1

photon1

Using the Einstein equation, express the energy of a 600-nm photon:

λhcE =photon1

Substitute for E1 photon to obtain: tA

hcIΔ

=λphoton1

Substitute numerical values and evaluate I1 photon:

( )( )( ) ( ) ( )

214

23

19

photon1 W/m102s1m105

4πnm600

J/eV10602.1nmeV1240 −

−

−

×≈

⎥⎦⎤

⎢⎣⎡ ×

×⋅=I

63 •• A 100-W incandescent light bulb radiates 2.6 W of visible light uniformly in all directions. (a) Find the intensity of the light from the bulb at a distance of 1.5 m. (b) If the average wavelength of the visible light is 650 nm, and counting only those photons in the visible spectrum, find the number of photons per second that strike a surface that has an area equal to 1.0 cm2, is oriented so that the line to the bulb is perpendicular to the surface, and is a distance of 1.5 m from the bulb.


47

Picture the Problem We can find the intensity at a distance of 1.5 m directly from its definition. The number of photons striking the surface each second can be found from the ratio of the energy incident on the surface to the energy of a 650-nm photon. (a) Use its definition to express the intensity of the light as a function of distance from the light bulb:

24 RP

API

π==

Substitute numerical data to obtain: ( )

2

22

mW/m29

mW/m96.19m1.54

W6.2

=

==π

I

(b) Express the number of photons per second that strike the surface as the ratio of the energy incident on the surface to the energy of a 650-nm photon:

photonEIAN =

where A is the area of the surface.

Use the Einstein equation to express the energy of the 650-nm photons:

λhcE =

Substituting for E yields: hc

IAN λ=


( )( )( )

( )4

19

242

100.3

eVJ101.602nmeV1240

nm650m101.0mW/m96.19×=

⎟⎠⎞

⎜⎝⎛ ×⋅

×=

−

−

N

64 •• When light of wavelength λ1 is incident on the cathode of a photoelectric tube, the maximum kinetic energy of the emitted electrons is 1.8 eV. If the wavelength is reduced to 1

2 λ1, the maximum kinetic energy of the emitted electrons is 5.5 eV. Find the work function φ of the cathode material. Picture the Problem The maximum kinetic energy of the photoelectrons is related to the frequency of the incident photons and the work function of the cathode material through the Einstein equation. We can apply this equation to the two sets of data and solve the resulting equations simultaneously for the work function.

Chapter 34

48

Using the Einstein equation, relate the maximum kinetic energy of the emitted electrons to the frequency of the incident photons and the work function of the cathode material:

φλ

φ

−=

−=hchfKmax

Substitute numerical values for the light of wavelength λ1:

φλ

−=1

eV8.1 hc (1)

Substitute numerical values for the light of wavelength 12

1 λ : φ

λφ

λ−=−=

1121

2eV5.5 hchc (2)

Solve equations (1) and (2) simultaneously for φ to obtain:

eV9.1=φ

65 •• An incident photon of energy Ei undergoes Compton scattering at an angle of θ. Show that the energy Es of the scattered photon is given by

( )( )θcos11 2ei

is −+

=cmEEE

Picture the Problem We can use the Einstein equation to express the energy of the scattered photon in terms of its wavelength and the Compton scattering equation to relate this wavelength to the scattering angle and the pre-scattering wavelength. Express the energy of the scattered photon Es as a function of their wavelength λs:

ss λ

hcE =

Express the wavelength of the scattered photon as a function of the scattering angle θ :

( ) λθλ +−= cos1e

s cmh

where λ is the wavelength of the incident photon.


49

Substitute for λ and simplify to obtain: ( )

( )

( )θ

θλ

λ

λθ

cos11

cos11

cos1

2e

i

i

2e

e

s

−⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

−+=

+−=

cmE

E

cmhc

hccm

hhcE

66 •• A particle is confined to a one-dimensional box. While the particle makes a transition from the state n to the state n – 1, radiation of 114.8 nm is emitted. While the particle makes the transition from the state n – 1 to the state n – 2, radiation of wavelength 147 nm is emitted. The ground-state energy of the particle is 1.2 eV. Determine n. Picture the Problem While we can work with either of the transitions described in the problem statement, we’ll use the first transition in which radiation of wavelength 114.8 nm is emitted. We can express the energy released in the transition in terms of the difference between the energies in the two states and solve the resulting equation for n. Express the energy of the emitted radiation as the particle goes from the nth to n – 1 state:

1−−=Δ nn EEE

Express the energy of the particle in nth state:

12EnEn =

Express the energy of the particle in the n – 1 state:

( ) 12

1 1 EnEn −=−

Substitute for En and En−1 and simplify to obtain:

( )

( )λhcEn

EnEnE

=−=

−−=Δ

1

12

12

12

1

Solving for n yields:

21

2 1

+=E

hcnλ

Chapter 34

50

Substitute numerical values and evaluate n: ( )( ) 5

21

eV1.2nm114.82nmeV1240

=+⋅

=n

67 •• [SSM] The Pauli exclusion principle states that no more than one electron may occupy a particular quantum state at a time. Electrons intrinsically occupy two spin states. Therefore, if we wish to model an atom as a collection of electrons trapped in a one-dimensional box, no more than two electrons in the box can have the same value of the quantum number n. Calculate the energy that the most energetic electron(s) would have for the uranium atom that has an atomic number 92. Assume the box has a length of 0.050 nm and the electrons are in the lowest possible energy states. How does this energy compare to the rest energy of the electron? Picture the Problem We can use the expression for the energy of a particle in a well to find the energy of the most energetic electron in the uranium atom. Relate the energy of an electron in the uranium atom to its quantum number n:

⎟⎟⎠

⎞⎜⎜⎝

⎛= 2

e

22

8 LmhnEn


( ) ( )( )( )

MeV3.1

MeV273.1J101.602

eV1nm050.0kg10109.98

sJ10626.692 19231

2342

92

=

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

××

×⋅×

= −−

−

E

The rest energy of an electron is:

( )( ) MeV512.0J101.602

eV1m/s10998.2kg10109.9 19

28312e =⎟⎟

⎠

⎞⎜⎜⎝

⎛×

××= −−cm

Express the ratio of E92 to mec2:

5.2MeV512.0MeV1.273

2e

92 ≈=cm

E

The energy of the most energetic electron is approximately 2.5 times the rest-energy of an electron. 68 •• A beam of electrons that each have the same kinetic energy illuminates a pair of slits separated by a distance 54 nm. The beam forms bright and dark fringes on a screen located a distance 1.5 m beyond the two slits. The arrangement is otherwise identical to that used in the optical two-slit interference


51

experiment described in Chapter 33 and in Figure 33-7 and the fringes have the appearance shown in Figure 34-18d. The bright fringes are found to be separated by a distance of 0.68 mm. What is the kinetic energy of the electrons in the beam? Picture the Problem We can express the kinetic energy of an electron in the beam in terms of its momentum. We can use the de Broglie relationship to relate the electron’s momentum to its wavelength and use the condition for constructive interference to find λ.

θ

θ

Δ θ

Δ

d

dr =

L

y

sin

n =

n =

1

0

Express the kinetic energy of an electron in terms of its momentum:

mpK2

2

= (1)

Using the de Broglie relationship, relate the momentum of an electron to its wavelength:

λhp =

Substitute for p in equation (1) to obtain:

2

2

2 λmhK = (2)

The condition for constructive interference is:

λθ nd =sin ⇒n

d θλ sin=

where d is the slit separation and n = 0, 1, 2, …

For θ << 1, sinθ is also given by: L

yΔ≈θsin

Substitute for sinθ to obtain:

nLydΔ

=λ

Substitute for λ in equation (2) to obtain:

( )22

222

2 ymdhLnKΔ

=

Chapter 34

52

Substitute numerical values (n = 1) and evaluate K:

( ) ( ) ( )( )( ) ( )

keV5.2J101.602

eV1mm68.0nm54kg10109.92

sJ106.626m5.11192231

23422

=××

⋅×= −−

−

K

69 •• When a surface is illuminated by light of wavelength λ, the maximum kinetic energy of the emitted electrons is 1.20 eV. If the wavelength λ′ = 0.800λ is used, the maximum kinetic energy increases to 1.76 eV. For wavelength λ′ = 0.600λ, the maximum kinetic energy of the emitted electrons is 2.676 eV. Determine the work function of the surface and the wavelength λ. Picture the Problem The maximum kinetic energy of the photoelectrons is related to the frequency of the incident photons and the work function of the illuminated surface through the Einstein equation. We can apply this equation to either set of data and solve the resulting equations simultaneously for the work function of the surface and the wavelength of the incident photons. Using the Einstein equation, relate the maximum kinetic energy of the emitted electrons to the frequency of the incident photons and the work function of the cathode material:

φλ

φ

−=

−=hchfKmax

Substitute numerical values for the light of wavelength λ:

φλ

−=hceV20.1

Substitute numerical values for the light of wavelength λ′:

φλ

φλ

−=−=800.0

eV76.1 hc'

hc

Solve these equations simultaneously for φ to obtain:

eV04.1=φ

Substitute in either of the equations and solve for λ:

nm554=λ

70 •• A simple pendulum has a length equal to 1.0 m and has a bob that has a mass equal to 0.30 kg. The energy of this oscillator is quantized, and the allowed values of the energy are given byEn = n + 1

2( )hf0 , where n is an integer and f0 is the frequency of the pendulum. (a) Find n if the angular amplitude is 1.0°. (b) Find n such that En+1 exceeds En by 0.010 percent.


53

Picture the Problem The diagram shows the pendulum when its angular displacement is θ. The energy of the oscillator is equal to its initial potential energy mgh = mgL(1 − cosθ). We can find n by equating this initial energy to

( ) 021 hfnEn += and solving for n.

θ

θ LL

h

U =

cos

g 0

m

(a) Express the nth-state energy as a function of the frequency of the pendulum:

( ) 021 hfnEn +=

Because the small-amplitude frequency of the pendulum is given

by Lgf

π21

0 = :

( )LghnEn π22

1+=

The energy of the pendulum is also equal to its initial gravitational potential energy:

( )θcos1−= mgLEn

Equating these expressions gives: ( ) ( )LghnmgL

πθ

2cos1 2

1+=−

Solving for n gives: ( )

21cos12 23

−−

=h

Lgmn

θπ

Substitute numerical values and evaluate n:

( ) ( ) ( )

3030

34

232

104.110357.1

21

sJ10626.60.1cos1m1.0m/s9.81kg30.02

×=×=

−⋅×

°−= −

πn

(b) Because En+1 exceeds En by 0.010 percent:

41 100.1 −+ ×=−

n

nn

EEE

Chapter 34

54

( ) ( )( )

4

021

021

021

100.11 −×=

++−++

hfnhfnhfn

Substituting for En+1 and En gives:

or ( ) ( ) ( )2

1421

21 100.11 +×=+−++ − nnn

Solving for n yields: 4

214 100.1100.1 ×=−×=n

71 •• [SSM] (a) Show that for large n, the fractional difference in energy between state n and state n + 1 for a particle in a one-dimensional box is given approximately by En +1 − En( )/ En ≈ 2 / n (b) What is the approximate percentage energy difference between the states n1 = 1000 and n2 = 1001? (c) Comment on how this result is related to Bohr’s correspondence principle. Picture the Problem We can use the fact that the energy of the nth state is related to the energy of the ground state according to to express the fractional change in energy in terms of n and then examine this ratio as n grows without bound.

12EnEn =

(a) Express the ratio (En + 1 − En)/En:

( )

nnn

nn

nnn

EEE

n

nn

212

121

2

22

221

≈+=

+=

−+=

−+

for n >> 1.

(b) Evaluate 1000

10001001

EEE − : %2.0

10002

1000

10001001 =≈−

EEE

(c) Classically, the energy is continuous. For very large values of n, the energy difference between adjacent levels is infinitesimal. 72 •• A mode-locked, titanium–sapphire laser has a wavelength of 850 nm and produces 100 million pulses of light each second. Each pulse has a duration of 125 femtoseconds (1 fs = 10–15 s) and consists of 5 × 109 photons. What is the average power produced by the laser? Picture the Problem We can apply the definition of power in conjunction with the de Broglie equation for the energy of a photon to derive an expression for the average power produced by the laser.


55

The average power produced by the laser is: t

EPΔΔ

=

Use the de Broglie equation to express the energy of the emitted photons:

λNhcNhfE ==Δ

where N is number of photons in each pulse.

Substitute for ΔE to obtain: t

NhcPΔ

=λ

Substitute numerical values and evaluate P:

( )( )( )( )( ) W1.0

s01nm850m/s10998.2sJ10626.6105

8

8349

≈×⋅××

= −

−

P

Remarks: Note that the pulse length has no bearing on the solution. 73 •• This problem estimates the time lag in the photoelectric effect that is expected classically but not observed. Let the intensity of the incident radiation falling on an atom be 0.010 W/m2. (a) If the area presented by the atom is 0.010 nm2, find the energy per second falling on an atom. (b) If the work function is 2.0 eV, how long would it take for this much energy to fall on the atom if the radiation energy was distributed uniformly rather than in compact packets (photons)? Picture the Problem We can find the rate at which energy is delivered to the atom using the definitions of power and intensity. We can also use the definition of power to determine how much time is required for an amount of energy equal to the work function to fall on one atom. (a) Relate the energy per second (power) falling on an atom to the intensity of the incident radiation:

IAtEP =

ΔΔ

=

Substitute numerical values and evaluate P:

( )( )

eV/s106.2

eV/s106.242J101.602

eV1sJ01

m100.010W/m0.010

4

4

1922

2182

−

−

−−

−

×=

×=

××=

×=P

Chapter 34

56

(b) Classically: PP

Et φ=

Δ=Δ

Substitute numerical values and evaluate Δt:

min53

s3204eV/s106.242

eV2.04

≈

=×

=Δ −t

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Ch34 ISM 070623