3
The Quantum Theory of Light
3-1
(a)
dB 
E2 r   r 2 

 dt 
r  dB 
E


2  dt 

(b)
If r remains constant, then: E  Eq 
 2r   dBdt  e so that Fdt   2r   dBdt  dt  m dv , or
e
 re 
dv  
 dB
 2 me 
B
E
v  v

v
(c)
 
r
E
v
eB
1T

 1.6  1019 C 
 9.1 1031 kg   8.8  1010 rad sec
r
2me
2
  2 f 
(d)
 er  B
dv  
  dB
 2 me  0
erB
v 
2 me
2 c

3.0  108 m s 

 2
 3.8  1015
 500  109 m 
rad sec ;


 2.3  105
For the 0 line the electrons’ plane is parallel to B, therefore, the magnetic flux,  B
is always zero. This means that F and E are zero and as a consequence, there is no
force on the electrons and there will be no  v for the electrons. The 0    is the
case calculated in parts (a)–(c). The 0    will have the same magnitude for F, B,
and  v as in (a)–(c) but the direction will be opposite.
e
B
29
30
CHAPTER 3
THE QUANTUM THEORY OF LIGHT
3-2
Assume that your skin can be considered a blackbody. One can then use Wien’s displacement
law, maxT  0.289 8  102 m  K with T  35 0 C  308 K to find
max 
3-3
(a)
0.289 8  102 m  K
 9.41  106 m  9 410 nm .
308 K
The total energy of a simple harmonic oscillator having an amplitude A is
kA 2
,
2
 0.4 m 2
kA2
  25 N m 
 2.0 J . The frequency of oscillation will be
2
2
1  k 1 2
1 25 1 2
f
 0.56 Hz .
  
2  m 
2 2
therefore, E 
 
(b)
  
If energy is quantized, it will be given by En  nhf and from the result of (a) there
follows En  nhf  n  6.63  10-34 J s   0.56 Hz   2.0 J . Upon solving for n one obtains
n  5.4  1033 .
3-4
(c)
The energy carried away by one quantum of charge in energy will be
E  hf   6.63  10 34 J s   0.56 Hz   3.7  10 34 J .
(a)
From Stefan’s law, one has
P
  T 4 . Therefore,
A
P
  5.7  108 W m2 K 4   3000 K 4  4.62  106 W m 2 .
A
3-5
P
75 W

 16.2 mm 2 .
6
2
4.62  10 W m
4.62  106 W m 2
(b)
A
(a)
Planck’s radiation energy density law as a function of wavelength and temperature is
8 hc
u
hc
given by u   , T   5 hc  T
. Using
, yields an
 0 and setting x 
B

max kBT
 e
1


extremum in u   , T  with respect to . The result is
hc

 hc max kBT
0  5  
 e
 max kBT 

(b)
max kBT
1

1
or x  5  1  e  x  .
Solving for x by successive approximations, gives x  4.965 or
 hc 
maxT     4.965   2.90  10 3 m  K .
 kB 
12
3-6
 e hc 
hG 
35
Planck length  
  4.05  10 m
 c 
hG 1 2
Planck time   5   1.35  1045 s
c 
MODERN PHYSICS
31
hG 1 2
8
Planck mass  
  5.46  10 kg
 c 
3-7
(a)
n
where n  1, 2, 3,
defines a mode or standing wave pattern
2
with a given wavelength. As we wish to find the number of possible values of n
2L
between 2.0 and 2.1 cm, we use n 

In general, L 
200
 200
2.0
200
n  2.1 cm    2 
 190
2.1
 n  10
n  2.0 cm    2 
As n changes by one for each allowed standing wave, there are 10 standing waves of
different wavelength between 2.0 and 2.1 cm.
2m
(b)
The number of modes per unit wavelength per unit length is
n
10
 200   0.5 cm2 .

L  0.1
(c)
For short wavelengths n is almost a continuous function of . Thus one may use
n
1  dn 
2L dn 2 L
1  dn  2
calculus to approximate
,
and
.





 . As n 

d  2
L  d   2
L 
L  d 


This gives approximately the same result as that found in (b):
1  dn  2
2
 0.5 cm2 .

 2 
L  d  
 2.0 cm 2

(d)
For short wavelengths n is almost a continuous function of , n 
function.
3-8
Using E  hf with h  4.136  1015 eV gives
(a)
for f  5  1014 Hz , E  2.07 eV
(b)
for f  10 GHz , E  4.14  105 eV
(c)
for f  30 MHz , E  1.24  107 eV
2L

is a discrete
32
CHAPTER 3
3-9
Use E 
hc
(a)
  600 nm
(b)
  0.03 m
(c)
  10 m
3-10

THE QUANTUM THEORY OF LIGHT
or  
hc
(where hc  1 240 eV nm ) and the results of Problem 3-7 to find
E
The energy per photon, E  hf and the total energy E transmitted in a time t is Pt where
power P  100 kW . Since E  nhf where n is the total number of photons transmitted in the
time t, and f  94 MHz , there results nhf   100 kW  t   10 5 W  t , or
105 J s
n 105 W


 94  106 s1   1.60  1030 photons s .
t
hf
6.63  1034 J s
3-11
Following the same reasoning as in Problem 3-9, one obtains
n P P
500  10 9 s 1


  3.74  10 26 J s 
 3  108 s1   9.45  1044 photons s .
t hf
hc
6.63  10 34 J s
3-12
As in Problems 3-9 and 3-10,
n P P
589  10 9 m


  10 W 
 3.0  1019 photons s .
t hf
hc
1.99  10 25 J m
K  hf   
hc
3-13
 
1 240 eV nm
hc


 2.92 eV  2.04 eV
 K
250 nm
3-14
(a)
K  hf   
(b)
At   c , K  0 and  
(a)
At the cut-off wavelength, K  0 so
3-15
cut-off 
hc

hc
 


1 240 eV nm
 2.24 eV  1.30 eV
350 nm
hc


1 240 eV nm
 554 nm
2.24 eV
hc

   0 , or
1 240 eV nm
 300 nm . The threshold frequency, f0 is given by
4.2 eV
f0 
c
cut-off

3.0  108 m s
3.0  102  109 m
 1.0  1015 Hz .
MODERN PHYSICS
3-16
3-17
eVs  K  hf   
hc
(b)


hc  1 240 eV nm
Vs 
 
 4.2 eV e  2.0 V
e e
200 nm e
(a)

(b)
Vs 
hc

K,  
1 240 eV nm
 2.23 eV  1.90 eV
300 nm
1 240 eV nm
 1.90 eV e  1.20 V
400 nm e
The energy of one photon of light of wavelength   300 nm is
hc
E
3-19
1 240 eV nm
 4.13 eV .
300 nm
As lithium and beryllium have work functions that are less than 4.13 eV, they will
exhibit the photoelectric effect for incident light with this energy. However, mercury
will not because its work function is greater than 4.13 eV.
(b)
The maximum kinetic energy is given by K 
hc
 
Kmax  eVs  s  0.45 V   0.45 eV
(b)

(c)
c 
1 240 eV nm
hc

 0.45 eV  2.03 eV
 K
500 nm
hc

1 240 eV nm
 612 nm
2.03 eV

  2.00 eV , Kmax  eV0  hf   
 hc   
e
Kmax  hf   
, so
1 240 eV nm
1 240 eV nm
 2.3 eV  1.83 eV , and K  Be  
 3.9 eV  0.23 eV .
300 nm
300 nm
(a)
 V0 
3-20


(a)
K  Li  
3-18
33
hc

 4.141015 eV s  3108
 2.00 eV
e
hc
First Source:  
m s
35010 9 m


 .
   
hc

Second Source:  
hc

 Kmax ;
 1.00 eV .
hc

2
 4.00 eV=
2hc

 4.00 eV .
 1.55 V .
34
CHAPTER 3
THE QUANTUM THEORY OF LIGHT
As the work function is the same for both sources (a property of the metal),
hc
2hc
hc
 100 eV 
 4.00 eV 
 3.00 eV and


3-21
3-22

hc


 1.00 eV  3.00 eV  1.00 eV  2.00 eV .
h f 
h

. Choosing two points on the graph, one has    4  1014 Hz    0 and
Vs   
e e
e
e
 h  8  1014 Hz  1.7 eV . Combining these two expressions one obtains:

 
e
(a)
  1.6 eV
(b)
h
 4.0  1015 Vs
e
(c)
For cut-off wavelength, c 
(d)
Accepted
hc


1 240 eV nm
 775 nm .
1.6eV
h
 4.14  1015 Vs , about a 3% difference.
e
The force acting on a charge moving perpendicular to a magnetic field has a magnitude given
by qvB. For constant B and v the charge moves in a circle of radius r, and from Newton’s
qBr
mv2
second law we have F  qvb 
, or v 
. Hence, one can express the kinetic energy of
m
r
hc
mv2  qBr 
. Using the photoelectric equation K 
, there results



2
2m
2
the charge q as K 

hc


 qBr 2
2m
. Substituting in the values hc  1 240 eV nm ,   450 nm , B  2  105 T ,
r  0.2 m , q  1.6  1019 C , and m  9.11 1031 kg , gives   2.76 eV  1.41 eV  1.35 eV .
3-23
6.626  1034 J s  3  108 m s 

E

 2.48 eV

 5  107 m 1.6  1019 J eV 
19
J eV 
h E  2.48 eV   1.6  10
p  
 1.32  10 27
hc

3-24
c
3  108 m s
kg m s
h
1  cos    0.002 43 nm  1  cos   . When   90 ,   0.002 43 nm .
me c
(a)
 
(b)
Conservation of energy requires that
hc
0

hc
   Ke
1
 1
 
or K e  hc 



 0
MODERN PHYSICS
  6.625  10 34 J s  3  108 m s  
10
Ke  
m   1   2.024 3  10 10 m   1
  2  10
19
1.6  10 J eV 


 74.4 eV
3-25
E  300 keV ,   30
(a)
     0 
h
 1  cos     0.002 43 nm  1  cos  30   3.25  10 13 m
me c
 3.25  10 4 nm
(b)
hc
E
0
 0 
15
eVs  3  108 m s 
hc  4.14  10

 4.14  10 12 m ; thus,
3
E0
300  10 eV
   0    4.14  1012 m  0.325  1012 m  4.465  1012 m , and
E 
(c)
hc
0
 4.14  1015 eV s  3  108 m s   2.78  105 eV .
hc
 E 

4.465  10 12 m

hc
 Ke , (conservation of energy)

15
eV s  3  108 m s 
 1 1   4.14  10
Ke  hc    
 22 keV
1
1

 0   
4.141012
4.46510 12
3-26
(a)
 
h
1  cos   2.426  1012 m 1  cos  
me c
For   30
  2.426  1012 m 1  cos 30  3.25  1011 m ;    0   ,
   0.04  109 m  3.25  1013 m  4.03  1011 m
(b)
1
hc
 1
 
, Ke hc 
0    Ke
 0   
For   30
 6.63  1034 J s  3  108 m s   3.70  1017 J
Ke 

 1.6  10119 J eV  231 eV .
1
1

0.04109
4.0310 11
hc

The remaining calculations are similar and the following table summarizes the
values to three significant figures
35
36
CHAPTER 3
(c)
3-27
THE QUANTUM THEORY OF LIGHT

  nm     nm  Ke  eV 
30
0.000 325 0.040 3
60
0.001 21
0.041 2
905
90
0.002 43
0.042 4
1 760
120 0.003 64
0.043 6
2 570
150 0.004 53
0.044 5
3 140
180 0.004 85
0.044 8
3 330
210 0.004 53
0.044 5
3 140
231
The electron which is backscattered corresponding to   180 has the greatest
energy.
Conservation of energy yields hf  Ke  hf  (Equation A). Conservation of momentum yields
E hf
there results
pe2  p2  p2  2pp cos . Using pphoton  
c
c
hf  2  hf 2
 hf   hf   cos  (Equation B). If the photon transfers all of its energy,
pe2  
     2 

 c  c 
 c   c 
2
hf
f   0 and Equations A and B become Ke  hf and pe2   respectively. Note that in
 c 
2
general, Ke  Ee  me c 2   pe2 c 2   me c 2  


12
 me c 2 . Finally, substituting Ke  hf and
2
hf
2 12
2 12
2
Pe2    into Ke   pe2 c 2   me c 2    me c 2 , yields hf   hf    me c 2    me c 2




 c 
(Equation C). As Equation C is true only if h, or f, or me , or c is zero and all are non-zero this
contradiction means that f  cannot equal zero and conserve both relativistic energy and
momentum.
3-28
(a)
From conservation of energy we have E0  E  Ke  120 keV  40 keV=160 keV . The
hc
photon energy can be written as E0 
. This gives
0
0 
(b)
hc 1 240 nm eV

 7.75  103 nm  0.007 75 nm .
E0 160  103 eV
Using the Compton scattering relation    0  c 1  cos  where
h
hc 1 240 nm eV

 10.3  103 nm  0.010 3 nm .
c 
 0.002 43 nm and   
E 120  103 eV
me c
Solving the Compton equation for cos , we find
c cos      0  c
0.010 3 nm  0.007 5 nm
   0
cos   1 
 1
 1  1.049  0.049
c
0.002 43 nm
The principle angle is 87.2 or   92.8 .
MODERN PHYSICS
(c)
37
Using the conservation of momentum Equations 3.30 and 3.31 one can solve for the
recoil angle of the electron.
p  p cos  pe cos 
pe sin   p sin  ; dividing these equations one can solve for the recoil angle of the
electron
p sin 
h
  
p  p cos     
tan  

h
0
sin 
hc
  
h

   cos
 

hc
0
sin 
hc
   cos

120 keV  0.998 8 
 0.723 2
160 keV  120 keV  0.049 
and   35.9 .
3-29
Symmetric Scattering,    . First, use the equations of conservations of momentum given by
Equations 3.30 and 3.31 for this two dimensional scattering process with    :
(a)
h
   cos  pe cos 
  
h
0
h

(1)
sin   pe sin  or pe 
h
(2)

Substituting (2) into (1) yields    20 cos
(3)
Next, express the Compton scattering formula as
   0  c 1  cos 
(4)
  0
h
. In this
 0.002 43 nm . Combining (3) and (4) yields cos   c
me c
c  20
hc
case, because E  1.02 MeV , and E 
there results
where c 
0
0 
Thus, cos 
hc 1 240 eV nm

 0.001 22 nm .
E 1.20  106 eV
0.002 43 nm  0.001 22 nm
 0.749 5 , and solving for the scattering
0.002 42 nm  0.002 44 nm
angle,   41.5 .
(b)
   0  c 1  cos 
   0.001 22 nm   0.002 43 nm  1  cos  41.5   0.001 83 nm
E
hc 1 240 eV  nm

 0.679 MeV
  0.001 83 nm
38
CHAPTER 3
THE QUANTUM THEORY OF LIGHT
3-30
Maximum energy transfer occurs when the scattering angle is 180 degrees. Assuming the
electron is initially at rest, conservation of momentum gives
hf  hf   pe c 
 me c2  K 
2
 m2 c 4   511  50 2  178 keV
while conservation of energy gives hf  hf   K  30 keV . Solving the two equations gives
E  hf  104 keV and hf  74 keV . (The wavelength of the incoming photon is
hc

 0.012 0 nm .
E
3-31
(a)
E 
hc
,    0  

0 
34
J  s  3  108 m s 
hc  6.63  10

 1.243  10 11 m
E0
0.1 MeV
 6.63  1034 J  s  1  cos 60  1.215  1012 m
 h 


  
1

cos



 9.11  1034 kg  3  108 m s 
 me c 
   0    1.364  10 11 m
E 
(b)
hc
0

34
J  s  3  108 m s 
hc  6.63  10

 9.11  10 4 eV

1.364  10 11 m
hc
 Ke

Ke  0.1 MeV  91.1 keV  8.90 keV
e–

0

photon
(c)
Conservation of momentum along x:
photon
h
   cos   me v cos  . Conservation of
0    
h
h
momentum along y:   sin    me v sin  . So that
  
 me v sin   h 
h
 h 

   sin       cos  


 me v cos    
 0    

0 sin 
tan  
    0  cos 
MODERN PHYSICS
39
Here,   60 , 0  1.243  1011 m , and    1.364  1011 m . Consequently,
1.24  1011 m   sin 60
tan  
1.36  1.24 cos 60   10 11 m
  55.4
3-32
Initially assume at rest; K  0 . Excitation energy, E  14.4 keV . Final energy, E , momentum,
E
E
. Conserve momentum and energy momentum: mv 
(m  mass of Fe57) energy:
c
c
E  E  mv2 (heavy Fe moves slowly so classical kinetic energy formula is used).
(a)
E
2E
for
 1 as we see here. Thus the kinetic
mc
mc 2
energy of the Fe57, which is the amount by which the photon energy is reduced, is
Solving simultaneously gives v 
K
1
1  E2 
E2
 14.4 keV 2
mv 2  m  2 2  

2
2  m c  2mc 2 2  57   931.5 MeV c 2  c 2
 1.95  10 3
(b)
3-33
 1.451
 keV 2
 1.95  10 3 eV
MeV
The energy of the emitted photon is 14.4 keV. The energy of the photon is given by
hc
hc 1 240 eV  nm
and 0 

 86.1  103  0.086 1 nm .
E
3
E
0
14.4  10
Substituting equations 3-33 and 3-34 of the text, Ee  h  f0  f   me c2 and
pe2 c 2  h 2  f 2  f02   2h 2 f f0 cos 
into the relativistic energy expression Ee2  pe2 c 2   me c 2  yields
2
h2  f 2  f02  2 f0 f    me2 c 4  2h  f0  f   me c 2  h2  f 2  f02   2h2 f0 f  cos  2   me c 2  .
2
Canceling and combining there results
 f 2  f02  2 f0 f  
2me c 2  f0  f  
 f 2  f02  2 f0 f  cos 
h
me c 2  f0  f  
 f0 f  1  cos   . Using  f  c one obtains
h
h 1  cos  
, which is the Compton scattering or Compton shift relation.
   0 
me c
which reduces to
40
CHAPTER 3
THE QUANTUM THEORY OF LIGHT
3-34
Maximum energy transfer occurs when the scattering angle is 180 degrees. Assuming the
electron is initially at rest, conservation of momentum gives
hf  hf   pe c 
 mec2  K 
2
 m2 c 4   511  50 2  232 keV
while conservation of energy gives hf  hf   K  50 keV . Solving the two equations gives
hc
E  hf  141 keV . (The wavelength of the incoming photon is  
 8.79 pm .)
E
3-35
(a)
The energy vs wavelength relation for a photon is E 
hc

. For a photon of
wavelength given by 0  0.071 1 nm the photon’s energy is
6.626  1034 J  s  3  108 m s 

E
 17.4 keV
 0.071 1 109 m 1.602  1019 J eV 
(b)
For the case of back scattering,    the Compton scattering relation becomes
 2hc 
   0  
. Setting 0  0.071 1 nm we obtain
2 
 me c 
   0.711 nm 
2hc
 7.60  1011
me c 2
or 0.076 0 nm.
3-36
 6.626  1034 J  s  3  108 m s   16.3 keV .
hc

   7.60  1011 m 1.602  1019 J eV 
(c)
E 
(d)
E  17.45 keV  16.33 keV  1.12 keV ~ 1.1 keV .
A scattered photon has an energy of 80 keV and the recoiled electron has an energy of 25 keV.
(a)
From conservation of energy we require that: Ephoton  80 keV  25 keV  105 keV .
As E0 
(b)
hc
0
, we have 0 
34
J  s  3  108 m s 
hc  6.626  10

 0.011 8 nm .
E0
105 keV  1.602  1019 J eV 
The incident photon energy is E0 
E 
hc
0
, and the energy of the scattered photon is
hc
. One can then take their ratio,

E0  
E
105 keV 

    0 0  0.011 8 nm  
  0.015 4 nm .
 80 keV 
E 0
E
MODERN PHYSICS
41
Using the Compton scattering formula we have:
 h 
 me c  
   0  

  1  cos     1  cos    
 h 
 me c 

(c)
1
 0.015 4 nm  0.011 8 nm   1.487;   119 .
0.002 43 nm
The relations between pe , p1 , p2 , f, and  are given by pe2  p12  p22  2p1 p2 cos and
p1  pe cos   p2 cos . (see the figure to solution 3-31). Substituting in the first of
these equations we have
12
2


1
1
2  0.487 
pe   6.626  10 34 J s  




 0.015 4 nm 0.011 8 nm  0.015 4 nm  0.011 8 nm  
 8.58  10 23 kg  m s.
Now rearrange terms and substitute in the second equation:
cos  
 p1  p2  cos
pe
  6.626  1034 J s 
1  0.015 4 nm   0.487 0.011 8 nm
8.58  1023
 0.819
and so   35.0 .
3-37
When waves are scattered between two adjacent planes of a single crystal, constructive wave
interference will occur when the path length difference between such reflected waves is an
integer multiple of wavelengths. This condition is expressed by the Bragg equation for
constructive interference, 2d sin   n where d is the distance between adjacent crystalline
planes,  is the angle of incidence of the x-ray beam of photons, n is an integer for
constructive interference, and  is the wavelength of the photon beam which is in this case,
0.062 6 nm. Ignoring the incident beam that is not scattered, the first three angles for which
maxima of x-ray intensities are found are 1  2d sin 1 or
0.626  1010 m
2d
8  10 10 m
1  0.078 3 radians  4.49
sin 1 


2  2d sin 2 or
sin 2 

d

0.626  1010 m
 0.156 5 ,   9.00
4.0  1010 m
3  2d sin 3 or
sin  3 
10
m
3 3  0.626  10

 0.234 75 , 3  13.6
10
2d
8  10
m
42
3-38
CHAPTER 3
THE QUANTUM THEORY OF LIGHT
d0
A
A
d0
d1 1
(a)
d2 2
For a single crystal of cubic structure having the principal interplanar separation d0 ,
the two successive subsidiary interplanar separations are identified as d1 and d2 .
1
Referring to the accompanying figure, one can see that cos 1  1 2  1  45 .
 2
d
d
1
Thus f1  d0 sin 45  01 2 . Also, sin 2  1 2 so d2  d0 sin 2  01 2 .
 2
 5
 5
(b)
To obtain the angles of the x-ray intensity maxima, the Bragg equation is used and
the angles with respect to the d1 planes are given by 2d1 sin 1  n or
sin 1 
 1 
0.626  10 10 m

 0.111  1  6.37
2d1
2  2.83  10 10 m 
2
 0.221   2  12.8
2d1
3
sin  3 
 0.331   3  19.3
2d1
sin  2 
Hence, the angles with respect to AA are,
1  6.37  45  51.4
2  12.8  45  57.8
3  19.3  45  64.3
3-39
The first x-ray intensity maximum in the diffraction pattern occurs at   6.41 . To determine
d use the Bragg diffraction condition n  2d sin  for n  1 .
d

2 sin 
From Figure P3.39 there are  4 

0.626 Å
 2.80 Å  2.80  108 cm .
2 sin 6.41
 81  Cl

and  4 
 81  Na

ions per primitive cell. This works
out to half a NaCl formula unit per primitive cell. The formula weight of NaCl is 58.4 g/mole.
Setting the mass per unit volume of the primitive cell equal to the density we have
 58.4 g mole   formula unit 
  where N A is the number of formula units per mole
2d 3 N A
(Avogadro’s number). So
MODERN PHYSICS
NA 

43
 58.4 g mol   formula unit 
2d 3
 58.4 g mol   formula unit 
2  2.8  10 8 cm   2.17 g cm 3 
3
N A  6.13  10 23 formula units mole
3-40
(a)
 G  hf c 2  MS 
hdf  FG dr  
 dr
r2


f


S 
 f    2   2
c
f
RS r
df
GM
dr

GM
GM
 f
1 
ln     2 S 
 2 S
c RS
 f   c  r RS
 GM 
or f   f exp   2 S 
 c RS 
(b)
3-41
For small
 GM 
 GM 
GM
MS
, exp   2 S   1  2 S so f   f 1  2 S  .
RS
c RS
 c RS 
 c RS 
The classical definition of a black hole is a star so massive that no object nor any
electromagnetic radiation can escape its gravitational attraction. This occurs when the star’s
 GM 
mass is such that the gravitational term  2 S  is greater than unity. Setting this term to be
 c RS 
unity and using MS to be the mass of the sun yields a value of
(a)
Rblackhole 
GMS
c2

 6.67  1011 N  m 2 kg 2 1.99  1030   1 470 m  1 500 m
2
 3.00  108 m s 
for a
black hole of one solar mass. The normal sun has a radius of 6.96  108 m .
(b)
The density of the black hole is  blackhole 
sun is S 
MS
4  RS 3 3
MS
4  Rblackhole 3 3
and the density of the
. Then
 6.96  108 m   1.1 1017
 blackhole 3 MS 4  Rblackhole 3
3 MS
RS3




S
 blackhole
4  RS 3  Rblackhole 3  1.47  103 m 3
3
or an increase in density over that of the sun by the factor of a wicked 1017 or
seventeen orders of magnitude.
44
CHAPTER 3
3-42
(a)
eVS 
(b)
eVS  

3-43
(a)
THE QUANTUM THEORY OF LIGHT
hc

     

1 240 eV  nm 
  1.70 eV  0.571 eV
546.1 nm 
1 240 eV  nm 
  0.571 eV  VS  1.54 V
587.5 nm 
A 4000 Å wavelength photon is backscattered,    by an electron. The energy
transferred to the electron is determined by using the Compton scattering formula
 hc 
   0     1  cos   where we take Ee  me c2 for the rest energy of the electron
 Ee 
and so Ee  0.511 MeV . Upon substitution, one obtains
  2  0.002 43 nm   0.004 86 nm .
The energy of a photon is related to its wavelength by the relation E 
hc

, so the
change in energy associated with a corresponding change in wavelength is given by
hc
E    2   . Upon making substitutions one obtains the magnitude
 
E  6.037 9  1024 J and using the conversion factor 1 Joule of energy is equivalent
to 1.602  1019 eV . The result is E  3.77  105 eV .
(b)
This may be compared to the energy that would be acquired by an electron in the
photoelectric effect process. Here again the energy of a photon of wavelength  is
hc
given by E 
. With   400 nm , one obtains

E
 6.626  1034 J  s  3.0  108
400  10
9
m s
m
and upon converting to electron volts, E  3.10 eV .
 4.97  10 19 J
E
 10 5 . The maximum
Ephoton
energy transfer is about five orders of magnitude smaller than the energy necessary
for the photoelectric effect.
(c)
3-44
Could “a violet photon” eject an electron from a metal by Compton scattering? The
answer is no, because the maximum energy transfer occurring at    is not
sufficient.
Each emitted electron requires an energy
hf 
 9.11  10 31 kg 
2
1
5
19
mv 2    
J eV 
  4.2  10 m s    3.44 eV  1.6  10
2
2


MODERN PHYSICS
45
E  6.3  10 19 J per emitted electron.
Therefore, with an incident intensity of
0.055 J m 2 5.5  10 6 J cm 2
, the number

s
s
of electrons emitted per cm 2 per second is
electron flux 
6.3  10
5.5108 J cm 2
s
19
J emitted electron

8.73  1012
.
cm 2 s
3-45
From a plot of Vs versus f one finds h  6.7  10 34 J  s , f0  7.1 1014 Hz , and   2.29 eV .
3-46
Using Compton shift formula
   0  c 1  cos   0.5 nm  0.002 43 nm 1  cos134  0.504 nm .
Conservation of py : p0y
 me ve sin 

 me ve cos 
3-47
h
h
h
h
.
  me ve cos  ,  me ve cos  

  cos
0   cos
h sin 
h sin 
, and so
  me ve sin  ,  me ve sin  
 py  pey , 0 


conservation of px : p0x  px  pex ,
0

  sin 
h
h
0
 h cos 
tan  
0 sin 
   0 cos 
tan  
0.5 nm sin 134
 0.424,   22.9
0.504 nm   0.5 nm  cos134
Head on collision means the point of contact is through the center of mass of the electron,
therefore, it is a case of back scattering. Again, using the Compton shift formula
1  cos180
2h

me c
me c
hc
  h
E0 
0
f  0   
Eelectron
hc 2 h

E0 me c
hc
 E0 
 E0 
0  
hc
E0
hc

 m2 hc
e
 m2hc  E0 
e
hc
E0
 m2 hc
e
2 E0 h
me c
 Ehc  1   m2 Ec
0
e
0
2

 2 E02    2 E0 1 
E
Eelectron  
1   2    2 hf   1  2 1 where   0 2 .
2 
me c
 me c    mc  
photon
electron
46
CHAPTER 3
THE QUANTUM THEORY OF LIGHT
3-48
The electron’s kinetic energy is K 
2
1
1
mv2  9.11 1031 kg  2.18  106 m s   2.16  1018 J .
2
2
hc hc
This is the energy lost by the photon, hf0  hf  ,

 2.16  1018 J . We also have
0  
   0 
h
6.63  10 34 Js s
 1  cos   
1  cos17.4 
me c
9.11  10 31 kg  3  108 m 
   0  1.11  10 13 m
(a)
Combining the equations by substitution,
1
0

1
2.16  10 18 J s
1.09  107


34
8
0  0.111 pm 6.63  10 J s  3  10 m 
m
0  0.111 pm  0 1.09  107

m
02  0  0.111 pm 
 1.09  107  2
6
0.111 pm  
 0  1.21  10 0
m


1.09  107 02  1.21  10 6 m0  1.11  10 13 m 2  0
0 
1.21  10 6 m 
1.21  106 m   4 1.09  107  1.11  1013 m 2 
2  1.09  107 
2
only the positive answer is physical: 0  1.01 1010 m .
(b)
Then    1.01 1010 m  1.11 1013 m  1.01 1010 m . Conservation of momentum
in the transverse direction:
0
h
sin    me v sin 

9.11  10 31 kg  2.18  106 m s  sin 
6.63  10 34 J  s
sin 17.4 
6 2
1.01  10 10 m
1  2.18108

1.96  10 24  1.99  10 24 sin 
  81.1
310
