HW 2 Solutions - UTA HEP WWW Home Pagebrandta/teaching/sp2010/HW/HW2-sol.pdf1. Chapter 3 The...

1.

Chapter 3 The Experimental Basis of Quantum Theory

18. a)

T =2.898× 10−3 m · K

10−14 m= 2.898× 1011 K

b)

T =2.898× 10−3 m · K

10−9 m= 2.898× 106 K

c)

T =2.898× 10−3 m · K

670× 10−9 m= 4325 K

d)

T =2.898× 10−3 m · K

1 m= 2.898× 10−3 K

e)

T =2.898× 10−3 m · K

204 m= 1.42× 10−5 K

* 19.P1

P0=

σT 41

σT 40

so P1 = P0T 4

1

T 40

=�

1900 K900 K

�4

P0 = 19.9 P0

The power increases by a factor of 19.9.

* 20. a)

λmax =2.898× 10−3 m · K

310 K= 9.35 µm

b) At this temperature the power per unit area is

R = σT 4 =�5.67× 10−8 W · m−2 · K4

�(310 K)4 = 524W/m2

The total surface area of a cylinder is 2πr (r + h) = 2π (0.13 m) (1.78 m) = 1.45 m2 so thetotal power is

P =�524 W/m2

� �1.45 m2

�= 760W .

21.λmax =

2.898× 10−3 m · K200, 273 K

= 1.447× 10−8m

22. We know from example 3.8 that in the long-wavelength limit, the two expressions are thesame. By comparing the expressions, we can also note that the Rayleigh-Jeans spectraldistribution will be larger than the Planck expression for a given λ and T . (Compare graphsor calculations using Equations (3.22) and (3.23) one can obtain using Excel, Mathcad, orsimilar program.) So want

(0.95)�

2πckT

λ4

�=

�2πc2h

λ5

� �1�

ehc/λkT − 1��

Simplifying we find

(0.95)�

λkT

hc

�=

1�ehc/λkT − 1

� .

31

2.


18. a)

T =2.898× 10−3 m · K

10−14 m= 2.898× 1011 K

b)

T =2.898× 10−3 m · K

10−9 m= 2.898× 106 K

c)

T =2.898× 10−3 m · K

670× 10−9 m= 4325 K

d)

T =2.898× 10−3 m · K

1 m= 2.898× 10−3 K

e)

T =2.898× 10−3 m · K

204 m= 1.42× 10−5 K

* 19.P1

P0=

σT 41

σT 40

so P1 = P0T 4

1

T 40

=�

1900 K900 K

�4

P0 = 19.9 P0

The power increases by a factor of 19.9.

* 20. a)

λmax =2.898× 10−3 m · K

310 K= 9.35 µm

b) At this temperature the power per unit area is

R = σT 4 =�5.67× 10−8 W · m−2 · K4

�(310 K)4 = 524W/m2

The total surface area of a cylinder is 2πr (r + h) = 2π (0.13 m) (1.78 m) = 1.45 m2 so thetotal power is

P =�524 W/m2

� �1.45 m2

�= 760W .

21.λmax =

2.898× 10−3 m · K200, 273 K

= 1.447× 10−8m

22. We know from example 3.8 that in the long-wavelength limit, the two expressions are thesame. By comparing the expressions, we can also note that the Rayleigh-Jeans spectraldistribution will be larger than the Planck expression for a given λ and T . (Compare graphsor calculations using Equations (3.22) and (3.23) one can obtain using Excel, Mathcad, orsimilar program.) So want

(0.95)�

2πckT

λ4

�=

�2πc2h

λ5

� �1�

ehc/λkT − 1��

Simplifying we find

(0.95)�

λkT

hc

�=

1�ehc/λkT − 1

� .

31

3.


* 32.

energy per photon = hf =�6.626× 10

−34J · s

� �107.7× 10

6s−1

�= 7.14× 10

−26J

�5.0× 10

4J/s

� 1 photon

7.14× 10−26 J= 7.00× 10

29photons/s

33. a)


−34J · s

� �1100× 10

3s−1

�= 7.29× 10

−28J

(150 J/s)1 photon

7.29× 10−28 J= 2.06× 10

29photons/s

b)

energy per photon = hc

λ=

�6.626× 10

−34J · s

� �3.00× 10

8m/s

8× 10−9 m

�= 2.48× 10

−17J

(150 J/s)1 photon

2.48× 10−17 J= 6.05× 10

18photons/s

c)

(150 J/s)1 photon

4 MeV

1 MeV

1.60× 10−13 J= 2.34× 10

14photons/s

34.

ft =φ

h=

2.9 eV

4.136× 10−15 eV · s= 7.01× 10

14Hz

eV0 =hc

λ− φ so V0 =

1

e

�hc

λ− φ

�

V0 =1

e

�1240 eV · nm

400 nm− 2.9 eV

�= 0.20 eV

* 35.

λt =hc

φ=

1240 eV · nm

4.64 eV= 267.2 nm

If the wavelength is halved (to λ = 133.6 nm)

K =hc

λ− φ =

1240 eV · nm

133.6 nm− 4.64 eV = 4.64 eV

36. Notice thathc

λ= 2.34 eV > φ, so photoelectrons will be produced.

�2× 10

−3J/s

� �10−5

� �1 photoelectron

2.34 eV

� �1 eV

1.60× 10−19 J

� �1.60× 10

−19C

electron

�= 8.55 nA

37.

φ =hc

λt=

1240 eV · nm

270 nm= 4.59 eV

K = 2.0 eV = hf − φ

f =K + φ

h=

2.0 eV + 4.59 eV

4.136× 10−15 eV · s= 1.59× 10

15Hz

35

4.


* 32.


−34J · s

� �107.7× 10

6s−1

�= 7.14× 10

−26J

�5.0× 10

4J/s

� 1 photon

7.14× 10−26 J= 7.00× 10

29photons/s

33. a)


−34J · s

� �1100× 10

3s−1

�= 7.29× 10

−28J

(150 J/s)1 photon

7.29× 10−28 J= 2.06× 10

29photons/s

b)


λ=

�6.626× 10

−34J · s

� �3.00× 10

8m/s

8× 10−9 m

�= 2.48× 10

−17J

(150 J/s)1 photon

2.48× 10−17 J= 6.05× 10

18photons/s

c)

(150 J/s)1 photon

4 MeV

1 MeV

1.60× 10−13 J= 2.34× 10

14photons/s

34.

ft =φ

h=

2.9 eV

4.136× 10−15 eV · s= 7.01× 10

14Hz

eV0 =hc

λ− φ so V0 =

1

e

�hc

λ− φ

�

V0 =1

e

�1240 eV · nm

400 nm− 2.9 eV

�= 0.20 eV

* 35.

λt =hc

φ=

1240 eV · nm

4.64 eV= 267.2 nm


K =hc

λ− φ =

1240 eV · nm

133.6 nm− 4.64 eV = 4.64 eV

36. Notice thathc


�2× 10

−3J/s

� �10−5


2.34 eV

� �1 eV

1.60× 10−19 J

� �1.60× 10

−19C

electron

�= 8.55 nA

37.

φ =hc

λt=

1240 eV · nm

270 nm= 4.59 eV

K = 2.0 eV = hf − φ

f =K + φ

h=

2.0 eV + 4.59 eV

4.136× 10−15 eV · s= 1.59× 10

15Hz

35

5.


* 32.


−34J · s

� �107.7× 10

6s−1

�= 7.14× 10

−26J

�5.0× 10

4J/s

� 1 photon

7.14× 10−26 J= 7.00× 10

29photons/s

33. a)


−34J · s

� �1100× 10

3s−1

�= 7.29× 10

−28J

(150 J/s)1 photon

7.29× 10−28 J= 2.06× 10

29photons/s

b)


λ=

�6.626× 10

−34J · s

� �3.00× 10

8m/s

8× 10−9 m

�= 2.48× 10

−17J

(150 J/s)1 photon

2.48× 10−17 J= 6.05× 10

18photons/s

c)

(150 J/s)1 photon

4 MeV

1 MeV

1.60× 10−13 J= 2.34× 10

14photons/s

34.

ft =φ

h=

2.9 eV

4.136× 10−15 eV · s= 7.01× 10

14Hz

eV0 =hc

λ− φ so V0 =

1

e

�hc

λ− φ

�

V0 =1

e

�1240 eV · nm

400 nm− 2.9 eV

�= 0.20 eV

* 35.

λt =hc

φ=

1240 eV · nm

4.64 eV= 267.2 nm


K =hc

λ− φ =

1240 eV · nm

133.6 nm− 4.64 eV = 4.64 eV

36. Notice thathc


�2× 10

−3J/s

� �10−5


2.34 eV

� �1 eV

1.60× 10−19 J

� �1.60× 10

−19C

electron

�= 8.55 nA

37.

φ =hc

λt=

1240 eV · nm

270 nm= 4.59 eV

K = 2.0 eV = hf − φ

f =K + φ

h=

2.0 eV + 4.59 eV

4.136× 10−15 eV · s= 1.59× 10

15Hz

356.


38.

E = 100

�hc

λ

�= 100

1240 eV · nm

580 nm= 214 eV

39. eV01 = hc/λ1−φ and eV02 = hc/λ2−φ. Subtracting these equations and rearranging we find

h =e (V02 − V01)

c�

1λ2− 1

λ1

� =e (2.3 V − 1.0 V)

(3.00× 108 m/s)�

1207 nm −

1260 nm

� = 4.40× 10−15

eV · s

This is about 6% from the accepted value. For the work function we use the first set of data

(the second set should give the same result):

φ =hc

λ1− eV01 =

�4.40× 10

−15eV · s

� �3.00× 10

8m/s

�

260× 10−9 m− 1.0 eV = 4.1 eV

40. For 400 nm:

E =hc

λ=

1240 eV · nm

400 nm= 3.10 eV f =

c

λ=

3.00× 108

m/s

400 nm= 7.50× 10

14Hz

For 700 nm:

E =hc

λ=

1240 eV · nm

700 nm= 1.77 eV f =

c

λ=

3.00× 108

m/s

700 nm= 4.29× 10

14Hz

41.

λmin =1.24× 10

−6V · m

V=

1.24× 10−6

V · m

30 kV= 0.0413 nm

42.

λ =hc

K=

1240 eV · nm

5× 1010 eV= 2.48× 10

−17m

A photon produced by bremsstrahlung is still an x ray, even though this falls outside the

normal range for x rays.

* 43.

λ =1240 eV · nm

2× 104 eV= 0.0620 nm

44. Since the electron is accelerated through a potential difference, it has kinetic energy equal to

K = eV0. As described in the text, Equation (3.37) gives the minimum wavelength (assuming

the work function is small). So

λmin =

�hc

e

� �1

V0

�=

1.240× 10−6

V · m

3.5× 104 V= 3.54× 10

−11m = 3.54× 10

−2nm

The work function for tungsten from Table 3.3 is 4.63 eV. If we include the work function,

the energy available for the photon and the wavelength of the photon will change by 4.63

parts out of 35000 or about 0.013 % which is very small.

36

7.


38.

E = 100

�hc

λ

�= 100

1240 eV · nm

580 nm= 214 eV

39. eV01 = hc/λ1−φ and eV02 = hc/λ2−φ. Subtracting these equations and rearranging we find

h =e (V02 − V01)

c�

1λ2− 1

λ1

� =e (2.3 V − 1.0 V)

(3.00× 108 m/s)�

1207 nm −

1260 nm

� = 4.40× 10−15

eV · s

This is about 6% from the accepted value. For the work function we use the first set of data

(the second set should give the same result):

φ =hc

λ1− eV01 =

�4.40× 10

−15eV · s

� �3.00× 10

8m/s

�

260× 10−9 m− 1.0 eV = 4.1 eV

40. For 400 nm:

E =hc

λ=

1240 eV · nm

400 nm= 3.10 eV f =

c

λ=

3.00× 108

m/s

400 nm= 7.50× 10

14Hz

For 700 nm:

E =hc

λ=

1240 eV · nm

700 nm= 1.77 eV f =

c

λ=

3.00× 108

m/s

700 nm= 4.29× 10

14Hz

41.

λmin =1.24× 10

−6V · m

V=

1.24× 10−6

V · m

30 kV= 0.0413 nm

42.

λ =hc

K=

1240 eV · nm

5× 1010 eV= 2.48× 10

−17m

A photon produced by bremsstrahlung is still an x ray, even though this falls outside the

normal range for x rays.

* 43.

λ =1240 eV · nm

2× 104 eV= 0.0620 nm

44. Since the electron is accelerated through a potential difference, it has kinetic energy equal to

K = eV0. As described in the text, Equation (3.37) gives the minimum wavelength (assuming

the work function is small). So

λmin =

�hc

e

� �1

V0

�=

1.240× 10−6

V · m

3.5× 104 V= 3.54× 10

−11m = 3.54× 10

−2nm

The work function for tungsten from Table 3.3 is 4.63 eV. If we include the work function,

the energy available for the photon and the wavelength of the photon will change by 4.63

parts out of 35000 or about 0.013 % which is very small.

36

8.

Modern Physics 3313Spring 2010

HW 2


* 45. From Figure (3.19) we observe that the two characteristic spectral lines occur at wavelengthsof 6.4× 10−11 m and 7.2× 10−11 m. Rearrange Equation (3.37) to solve for the potential V0;

V0 =hc

e

1λmin

=1.24× 10−6 V · m

7.2× 10−11 m= 17.2 kV

and we have used the larger of the two wavelengths in order to determine the minimumpotential.

46.∆λ =

h

mc(1− cos θ)

so at maximum cos θ = −1 and∆λ

λ=

2h

λmc=

2hc

mc2λ=

2 (1240 eV · nm)(511.0 keV) (530 nm)

= 9.16× 10−6

This corresponds to ∆λ = 4.9× 10−12 m and therefore is not easily observed.

47. The maximum change in the photon’s energy is obtained in backscattering (θ = 180◦), so

1− cos θ = 2 and ∆λ =2h

mc= 4.853× 10−12 m. The photon’s original wavelength was

λ =hc

E=

1240 eV · nm40000 eV

= 0.0310 nm = 3.10× 10−11 m

and the new wavelength is λ� = λ+∆λ = 3.585×10−11 m. The electron’s recoil energy equalsthe change in the photon’s energy, or

K =hc

λ− hc

λ�=

1240 eV · nm3.10× 10−2 nm

− 1240 eV · nm3.585× 10−2 nm

= 5411 eV = 5.41 keV

48. Use the Compton scattering formula but with the proton’s mass and θ = 90◦:

∆λ =h

mc(1− cos θ) =

h

mc=

hc

mc2=

1240 eV · nm938.27 MeV

= 1.32 fm

49. We find the Compton wavelength using

λc =h

mc=

hc

mc2=

1240 eV · nm938.27 MeV

= 1.32 fm

The photon energy is

E =hc

λc= 938 MeV

In principle this could be observed, but the energy requirements are high.

* 50.∆λ

λ= 0.004 =

λc

λ(1− cos θ) so λ = 250λc (1− cos θ)

a)λ = 250

�2.43× 10−12 m

�(1− cos 30◦) = 8.14× 10−11 m

b)λ = 250

�2.43× 10−12 m

�(1− cos 90◦) = 6.08× 10−10 m

c)λ = 250

�2.43× 10−12 m

�(1− cos 170◦) = 1.21× 10−9 m

37

9.


51. By conservation of energy we know the electron’s recoil energy equals the energy lost by thephoton:

K =hc

λ− hc

λ�=

hc (λ� − λ)λ�λ

=hc ∆λ

λ�λ

Using λ� = λ + ∆λ

K =hc ∆λ

λ (λ + ∆λ)=

hf ∆λ

λ + ∆λ=

hf ∆λ

λ (1 + ∆λ/λ)=

(∆λ/λ)1 + ∆λ

λ

hf

Conservation of px:

pe cos φ +h

λ�cos θ =

h

λ

pe cos φ =h

λ− h

λ�cos θ (1)

Conservation of py:

pe sinφ− h

λ�sin θ = 0

pe sinφ =h

λ�sin θ (2)

Dividing equation (2) by equation (1)

tanφ =hλ� sin θ

hλ −

hλ� cos θ

Using λ� = λ + hmc (1− cos θ)

tanφ =h sin θ

λ+ hmc (1−cos θ)

hλ −

h cos θλ+ h

mc (1−cos θ)

Multiplying numerator and denominator by λ�λ + h

mc (1− cos θ)�,

tanφ =λh sin θ

λh + h2

mc (1− cos θ)− λh cos θ=

λ sin θ�λ + h

mc

�(1− cos θ)

Trig identity:sin θ

(1− cos θ)= cot

�θ

2

�

tanφ =λ

λ + hmc

cot�

θ

2

�=

11 + h

mcλ

cot�

θ

2

�=

11 + hf

mc2

cot�

θ

2

�

Inverting the equation gives

cot φ =�1 +

hf

mc2

�tan

�θ

2

�

38


HW 2


51. By conservation of energy we know the electron’s recoil energy equals the energy lost by thephoton:

K =hc

λ− hc

λ�=

hc (λ� − λ)λ�λ

=hc ∆λ

λ�λ

Using λ� = λ + ∆λ

K =hc ∆λ

λ (λ + ∆λ)=

hf ∆λ

λ + ∆λ=

hf ∆λ

λ (1 + ∆λ/λ)=

(∆λ/λ)1 + ∆λ

λ

hf

Conservation of px:

pe cos φ +h

λ�cos θ =

h

λ

pe cos φ =h

λ− h

λ�cos θ (1)

Conservation of py:

pe sinφ− h

λ�sin θ = 0

pe sinφ =h

λ�sin θ (2)

Dividing equation (2) by equation (1)

tanφ =hλ� sin θ

hλ −

hλ� cos θ

Using λ� = λ + hmc (1− cos θ)

tanφ =h sin θ

λ+ hmc (1−cos θ)

hλ −

h cos θλ+ h

mc (1−cos θ)

Multiplying numerator and denominator by λ�λ + h

mc (1− cos θ)�,

tanφ =λh sin θ

λh + h2

mc (1− cos θ)− λh cos θ=

λ sin θ�λ + h

mc

�(1− cos θ)

Trig identity:sin θ

(1− cos θ)= cot

�θ

2

�

tanφ =λ

λ + hmc

cot�

θ

2

�=

11 + h

mcλ

cot�

θ

2

�=

11 + hf

mc2

cot�

θ

2

�

Inverting the equation gives

cot φ =�1 +

hf

mc2

�tan

�θ

2

�

3810.


52.λ� = λ + λc (1− cos θ) =

hc

E+ λc (1− cos θ)

λ� =1240 eV · nm700× 103 eV

+�2.43× 10−3 nm

�(1− cos 110◦) = 5.03 pm

E� =hc

λ�=

1240 eV · nm5.03× 10−3 nm

= 2.46× 105 eV = 246 keV

By conservation of energy

Ke = E−E� = 700 keV −246 keV = 454 keV (agrees with K formula in the previous problem)

From Problem 51

cot φ =�1 +

hf

mc2

�tan

�θ

2

�=

�1 +

700 keV511 keV

�tan

�110◦

2

�= 3.3845

φ = 16.5◦

* 53. For θ = 90◦ we know λ� = λ + λc = 2.00243 nm

∆λ

λ=

λc

λ=

2.43× 10−3 nm2 nm

= 1.22× 10−3 = 0.122%

* 54.E = 2mc2 = 2 (938.3 MeV) = 1877 MeV

This energy could come from a particle accelerator.

55. a) To find the minimum energy consider the zero-momentum frame. Let Ee be the energy ofthe electron in that frame, and E0 is the rest energy of the electron. From conservation ofenergy and momentum:

momentum:hf

c= pe =

�E2

e − E20

cor hf =

�E2

e − E20

energy: hf + Ee = 3E0 or hf = 3E0 − Ee

Squaring and subtracting these two equations gives

0 = −10E20 + 6EeE0 or Ee =

53E0

This tells us that for the transformation form the lab frame to the zero-momentum frame,γ = 5/3 and v = 0.8c. Then from the momentum equation we have in the zero-momentumframe

hf =

�25E2

0

9− E2

0 =43E0

In the lab the photon’s energy is obtained using a Doppler shift:

hflab = hf

�1 + β

1− β=

43E0

�1 + 0.81− 0.8

= 4E0 = 2.04 MeV

39


HW 2

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