Chapter 3 – Quantization of Charge, Light, and Energy
3-3.
B
E
,
u
pc 
2
u pc

, and pc  E 2   mc 2 
c E
 0.561MeV    0.511MeV 
2
2
 0.2315MeV
u 0.2315MeV

 0.41
c 0.561MeV
2.0 105V / m
 B
 1.63 103 T  16.3G
0.41c
3-4.
F  quB and FG  mp g


19
6
5
FB quB 1.6 10 C  3.0 10 m / s   3.5 10 T 


 1.03 109
27
2
FG m p g
1.67 10 kg 9.80m / s 
3-5.
(a)
mu  2 Ek / e  e / m  
R

qB
 e / m  B 
1/ 2
4
1 2 Ek / e
1   2   4.5 10 eV / e  




B e/m
0.325T  1.76 1011 kg


(b)
frequency
f 
u
2 R

1/ 2
 2.2 103 m  2.2mm
 2 Ek / e  e / m 
2 R
 2   4.5 104 eV / e 1.76 1011 C / kg  


3
2  2.2 10 m 
1/ 2
 9.1109 Hz
period T  1/ f  1.11010 s
3-6.
(a) 1/ 2mu 2  Ek , so u 
 2Ek / e  e / m 
 u   2  2000eV / e  1.76 1011 C / kg 
(b)
t1 
1/ 2
x1
0.05m

 1.89 109 s  1.89ns
7
u 2.65 10 m / s
54
 2.65 107 m / s
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-6 continued)
(c) mu y  F t1  eEt1
 u y   e / m  Et1  1.76 1011 C / kg  3.33 103V / m 1.89 109 s   1.11106 m / s
3-7.
3-8.
NEk  W  CV T 
3
 4.186 J / cal   1.311014
CV T  5 10 cal / C   2C 
N


Ek
2000eV
1.60 1019 J / eV 
Q1  Q2   25.41  20.64 1019 C  4.47 1019 C   n1  n2  e
Q2  Q3   20.64  17.47  1019 C  3.17 1019 C   n2  n3  e
Q4  Q3  19.06  17.47  1019 C  1.59 1019 C   n4  n3  e
Q4  Q5  19.06  12.70 1019 C  6.36 1019 C   n4  n5  e
Q6  Q5  14.29  12.70 1019 C  1.59 1019 C   n6  n5  e
where the ni are integers. Assuming the smallest Δn = 1, then Δn12 = 3.0, Δn23 = 2.0,
Δn43 = 1.0, Δn45 = 4.0, and Δn65 = 1.0. The assumption is valid and the fundamental
charge implied is 1.59 1019 C.
3-9.
For the rise time to equal the field-free fall time, the net upward force must equal the
weight. qE  mg  mg  E  2mg / q.
3-10. (See Millikan’s Oil Drop Experiment on the home page at
www.whfreeman.com/tiplermodernphysics6e.) The net force in the y-direction is
mg  bv y  ma y . The net force in the x-direction is qE  bvx  max . At terminal speed
ax  ay  0 and vx / vt  sin  .
sin  
vx  qE / b  qE


vt
vt
bvt
55
Chapter 3 – Quantization of Charge, Light, and Energy
3-23. Equation 3-18: u    
8 hc 5
A 5
Letting
A


hc
,
B

hc
/
kT
,
and
U




ehc /  kT  1
eB /  1
  5  1 e B /    B 2 

du
d  A 5 
5 6 


 B/

  A
2
B/
d  d   eB /   1
e  1
e

1




6
6 B / 
A
 B B/
 A e
B


e  5  e B /   1  
 5 1  e  B /     0
2 
2 
B/
B
/

  e  1  

 e  1  
The maximum corresponds to the vanishing of the quantity in brackets.
Thus,
5 1  e B /    B . This equation is most efficiently solved by iteration; i.e., guess at a
value for B/λ in the expression 5 1  e B /   , solve for a better value of B/λ; substitute the
new value to get an even better value, and so on. Repeat the process until the calculated
value no longer changes. One succession of values is: 5, 4.966310, 4.965156, 4.965116,
4.965114, 4.965114. Further iterations repeat the same value (to seven digits), so we
have:
6.63 1034 J  s  3.00 108 m / s 

hc
hc
 4.965114 
 mT 

m
 m kT
 4.965114  k  4.965114  1.38 1023 J / K 
B
mT  2.898 103 m  K
(Equation 3-5)
3-24. Photon energy E  hf  hc / 
(a) For λ = 380nm: E  1240eV  nm / 380nm  3.26eV
For λ = 750nm: E  1240eV  nm / 750nm  1.65eV
(b) E  hf   4.14 1015 eV  s 100 106 s 1   4.14 107 eV
3-25. (a) hf  hc /   0.47eV .
max
 4.14 10
hc


4.87eV
15
eV  s  3.00 108 m / s 
4.87eV
60
 2.55 107 m  255nm
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-25 continued)
(b) It is the fraction of the total solar power with wavelengths less than 255nm, i.e., the
area under the Planck curve (Figure 3-6) up to 255nm divided by the total area. The
latter is: R   T 4   5.67 108W / m2  K 4   5800K   6.42 107W / m2 .
4
Approximating the former with u     with   127nm and   255nm :
 8 hc 127  109 m 5 
  255  109 m   1.23  104 J / m3
u 127nm   255nm   
9
 ehc / kT 12710   1 


R  0  255nm 
c
1.23  104 J / m 3  

4
R
8
4
3
 3.00  10 m / s 1.23  10 J / m  fraction = 1.4  104

 4   6.42  107W / m2 
R  0  255nm  
3-26. (a) t 
hc


1240eV  nm
 653nm,
1.9ev
ft 

h

1.9eV
 4.59 104 Hz
15
4.136 10 eV  s
1  hc
 1  1240eV  nm

(b) V0       
 1.9eV   2.23V
e 
 e  300nm

1  hc
 1  1240eV  nm

(c) V0       
 1.9eV   1.20V
e 
 e  400nm

3-27. (a) Choose λ =550nm for visible light. nhf  E 
dn
dE
hf 
P
dt
dt
 0.05  100W  550  109 m 
dn P P



 1.38  1019 / s
34
8
dt hf
hc  6.63  10 J  s  3.00  10 m / s 
(b) flux 
number radiated / unit time 1.38 1019 / s

 2.75 1017 / m2  s
2
area of the sphere
4  2m 
61
Chapter 3 – Quantization of Charge, Light, and Energy
3-28. (a) hf    ft 
(b) f  c /  

h

4.22eV
 1.02 1015 Hz
15
4.14 10 eV  s
3.00 108 m / s
 5.36 1014 Hz
9
560 10 m
No.
Available energy/photon hf   4.14 1015 eV  s  5.36 1014 Hz   2.22eV .
This is less than  .
3-29. (a) E  hf  hc /     hc / E
For E  4.26eV :   1240eV  nm /  4.26eV   291nm
and since f  c /  ,
f   3.00 108 m / s  /  291nm   1.03 1015 s 1
(b) This photon is in the ultraviolet region of the electromagnetic spectrum.
3-30. (a) First, add a row f 1014 Hz to the table in the problem, then plot a graph
Ek ,max versus f . The slope of the graph is h/e; the intercept on the  Ek ,max axis is
work function. The graph below is a least squares fit to the data.
λ nm
544
Ek ,max eV
0.360 0.199 0.156 0.117 0.062
f 1014 Hz
5.51
594
5.05
62
604
4.97
612
4.90
633
4.74
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-30 continued)
Slope  h / e  3.90 1015 eV  s
(b)
(3.90 1015  4.14 1015 ) eV  s
 5.6 percent
4.14 1015 eV  s
(c) The work function is the magnitude of the intercept on the Ek ,max axis, 1.78 eV.
(d) cesium
En
hc
3-32. (a)  
hc
3-31.



(b) Ek 

hc

 60   6.63 1034 J
s  3.00 108 m / s 
9
550 10 m
1240eV nm
 1.90eV
653nm
 
1240eV nm
 1.90eV  2.23eV
300nm
63
 2.17 1017 J
Chapter 3 – Quantization of Charge, Light, and Energy
3-33. Equation 3-25: 2  1   
 6.63 10
 
9.1110
34
31

1
100 
J s  1  cos135 
kg  3.00 10 m / s 
3-36.
p
h

8
 4.14 1012 m  4.14 103 nm
4.14 103 nm
100  5.8%
0.0711nm
3-34. Equation 3-24: m 
3-35.
h
1  cos 
mc
1.24 103
1.24 103
nm 
 0.016nm
V
80 103V
hc
c

(a) p 
1240eV nm
6.63 1034 J s
 3.10eV / c 
 1.66 1027 kg m / s
c  400nm 
400 109 m
(b) p 
1240eV nm
6.63 1034 J s
 1.24 104 eV / c 
 6.63 1024 kg m / s
c  0.1nm 
0.1109 m
(c) p 
1240eV nm
6.63 1034 J s
5

4.14

10
eV
/
c

 2.211032 kg m / s
3 102 m
c  3 107 nm 
(d) p 
1240eV nm
6.63  1034 J s
 620eV / c 
 3.32  1025 kg m / s
9
c  2nm 
2  10 m
2  1 
6.63 1034 J s  1  cos110 

h
1

cos


 3.26 1012 m


31
8
mc
 9.1110 kg  3.00 10 m / s 
6.63 1034 J s  3 108 m / s 

hc
1  
 2.43 1012 m
E1  0.511106 eV 1.60 1019 J / eV 
2  1  3.26 1012 m   2.43  3.26 1012 m  5.69 1012 m
64
Chapter 3 – Quantization of Charge, Light, and Energy
3-33. Equation 3-25: 2  1   
 6.63 10
 
9.1110
34
31

1
100 
J s  1  cos135 
kg  3.00 10 m / s 
3-36.
p
h

8
 4.14 1012 m  4.14 103 nm
4.14 103 nm
100  5.8%
0.0711nm
3-34. Equation 3-24: m 
3-35.
h
1  cos 
mc
1.24 103
1.24 103
nm 
 0.016nm
V
80 103V
hc
c

(a) p 
1240eV nm
6.63 1034 J s
 3.10eV / c 
 1.66 1027 kg m / s
c  400nm 
400 109 m
(b) p 
1240eV nm
6.63 1034 J s
 1.24 104 eV / c 
 6.63 1024 kg m / s
c  0.1nm 
0.1109 m
(c) p 
1240eV nm
6.63 1034 J s
5

4.14

10
eV
/
c

 2.211032 kg m / s
3 102 m
c  3 107 nm 
(d) p 
1240eV nm
6.63  1034 J s
 620eV / c 
 3.32  1025 kg m / s
9
c  2nm 
2  10 m
2  1 
6.63 1034 J s  1  cos110 

h
1

cos


 3.26 1012 m


31
8
mc
 9.1110 kg  3.00 10 m / s 
6.63 1034 J s  3 108 m / s 

hc
1  
 2.43 1012 m
E1  0.511106 eV 1.60 1019 J / eV 
2  1  3.26 1012 m   2.43  3.26 1012 m  5.69 1012 m
64
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-36 continued)
E2 
hc
2

1240eV nm
 2.18 105 eV  0.218MeV
3
5.69 10 nm
Electron recoil energy Ee  E1  E2 (Conservation of energy)
Ee  0.511MeV  0.218MeV  0.293MeV . The recoil electron momentum makes an
angle θ with the direction of the initial photon.
PE
θ
h/λ1
110°
h/λ2
20°
h
2
cos 20  pe sin   1/ c  E 2   mc 2  sin 
2
 3.00 10 m / s  6.63 10
8
sin  
 5.69 10
12
(Conservation of momentum)
34
J s  cos 20
2
2
m   0.804MeV    0.511MeV  


1/ 2
1.60 10
13
J / MeV 
 0.330 or   19.3
3-37. (a) First, add a row ( 1  cos  ) to the table in the problem, then plot a graph of  versus
( 1  cos  ). The slope of the graph is the Compton wavelength of the electron.
 pm
 degrees
1  cos 
0.647 1.67 2.45 3.98 4.80
45
75
90
135
170
0.293 0.741 1.000 1.707 1.985
65
Chapter 3 – Quantization of Charge, Light, and Energy
h
(4.5  1.0) nm
 slope 
 2.43nm
mc
(1.88  0.44)
(b) The graph above is a least squares fit to the data. The percent difference is
(2.43  2.426) nm
0.004 nm
100 
100  0.15percent
2.426 nm
2.426 nm
3-38.
  2  1   
1  100 
3-39. (a) E1 
h
1  cos   100 0.00243nm 1  cos90  0.243nm
mc
hc
1

(b) 2  1 
(c) E2 
h
1  cos   0.011 Equation 3-25
mc
hc
2
1240eV nm
 1.747 104 eV
0.0711nm
h
1  cos   0.0711nm   0.00243nm 1  cos180   0.0760nm
mc

1240eV nm
 1.634 104 eV
0.0760nm
(d) Ee  E1  E2  1.128 103 eV
66
Chapter 3 – Quantization of Charge, Light, and Energy
3-42. (a) Compton wavelength =
electron:
h
mc
h
6.63 1034 J s

 2.43 1012 m  0.00243nm
31
8
mc  9.1110 kg  3.00 10 m / s 
h
6.63 1034 J s
proton:

 1.32 1015 m  1.32 fm
27
8
mc 1.67 10 kg  3.00 10 m / s 
(b) E 
hc

(i) electron: E 
(ii) proton: E 
1240eV nm
 5.10 105 eV  0.510MeV
0.00243nm
1240eV nm
 9.39 108 eV  939MeV
6
1.32 10 nm
3-43. Photon energy E  hf  hc /  allows us to rewrite Equation 3-25 as
hc hc
h


(1  cos  )
E2 E1 mc
Rearranging the above,
hc
h
hc

(1  cos  ) 
E2 mc
E1
Dividing both sides of the equation by hc yields
1
1
1 ( E1 / mc 2 )(1  cos  )  1

(1  cos  )  
E2 mc 2
E1
E1
Or
E2 
E1
( E1 / mc )(1  cos  )  1
2
3-44. (a) eV0  hf    hc /   
e  0.52V    hc / 450nm   
(i)
e 1.90V    hc / 300nm   
(ii)
68