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,  max T  0.289 8  10 2 m  K with T  35 0 C  308 K to find
 max 
3-4
(a)
0.289 8  10 2 m  K
308 K
 9.41  10 6 m  9 410 nm .
From Stefan’s law, one has
P
  T 4 . Therefore,
A


P
4
8
2 4
6
2
 5.7  10 W m K 3000 K   4.62  10 W m .
A
3-5
P
75 W
(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



 e
1
max kB T
4.62  10 W m
6
2

4.62  10 W m
6

2
2
 16.2 mm .

extremum in u  , T  with respect to . The result is
 hc  hc max kBT hc  max kBT
1
e
1
or x  5 1  e x .
0  5  
 e
 max k BT 



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
5
5


10 J s
n 10 W


94  106 s 1  1.60  10 30 photons s .
t
hf
6.63  10 34 J s
3-14

Solving for x by successive approximations, gives x  4.965 or
hc 
 max T   4.965  2.90  10 3 m  K .
k B 
(b)
3-10

1 240 eV nm
hc

 2.24 eV  1.30 eV
350 nm
 
(a)
K  hf   
(b)
At   c , K  0 and  
hc


1 240 eV nm
 554 nm
2.24 eV
3-16
3-17
hc
(a)

(b)
Vs 

 K,  
1 240 eV nm
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
3-18
3-20
 2.23 eV  1.90 eV


1 240 eV nm
300 nm
 4.13 eV .
(a)
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 
(a)
K max  eVs  s0.45 V  0.45 eV
(b)

(c)
c 
hc
, so
 
1 240 eV nm
1 240 eV nm
 2.3 eV  1.83 eV , and K Be 
 3.9 eV  0.23 eV .
K Li 
300 nm
300 nm
hc

 K
hc

K max  hf   

500 nm
1 240 eV nm
hc

First Source:  
1 240 eV nm
2.03 eV
   
hc

Second Source:  
 0.45 eV  2.03 eV
 612 nm
hc

 K max ;
 1.00 eV .
hc

 4.00 eV =
2 hc
2

 4.00 eV .
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


3-25

hc


 1.00 eV  3.00 eV  1.00 eV  2.00 eV .
E  300 keV ,   30
(a)
   0 
 3.25  10
h
1  cos    0.002 43 nm 1  cos 30 3.25  10 13 m
mec
4
nm
(b)



4.14  10 15 eVs 3  10 8 m s
hc
E
 0 

 4.14  10 12 m ; thus,
0
E0
300  10 3 eV
hc
 0    4.14  10 12 m  0.325 10 12 m  4.465  10 12 m , and



4.14  10 15 eV s 3  108 m s
hc
 E 
 2.78  10 5 eV .
E 
12

4.465  10
m
(c)
hc
0

hc

 K e , (conservation of energy)



15
eV s 3  10 8 m s
 1
1  4.14  10
 22 keV
K e  hc   
1
1
0 
12 
12
4.14 10
3-27
4.46510
Conservation of energy yields hf  K e  hf  (Equation A). Conservation of momentum yields
E hf
2
2
there results
pe  p2  p  2ppcos  . Using pphoton  
c
c
2
2
hf  hf 
 hf  hf 
pe2        2  cos  (Equation B). If the photon transfers all of its energy,
 c   c 
 c  c 
f  0 and Equations A and B become K e  hf and


general, K e  Ee  me c 2  p2e c2  m e c2

2
2

 respectively. Note that in
 c 
1 2
 
2
2  hf
pe 
 me c 2 . Finally, substituting K e  hf and
12
12
2 
2 
hf 

 2
Pe2    into K e  pe2 c 2  me c 2   m e c2 , yields hf  hf   me c 2   m e c2
 c 




(Equation C). As Equation C is true only if h, or f, or m e , 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  K e  120 keV  40 keV = 160 keV . The
hc
. This gives
photon energy can be written as E0 
0
0 
(b)
hc 1 240 nm eV
3

 7.75 10 nm  0.007 75 nm .
E0 160  10 3 eV
Using the Compton scattering relation  0   c 1  cos   where
h
hc 1 240 nm eV
c 
 0.002 43 nm and 

 10.3  103 nm  0.010 3 nm .
mec
E 120  10 3 eV
Solving the Compton equation for cos  , we find
 c cos     0  c
cos   1 
0.010 3 nm  0.007 5 nm
  0
1 
 1 1.049  0.049
0.002 43 nm
c
The principle angle is 87.2 or   92.8 .
(c)
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
tan  
psin 
 h  sin 
hc  sin 
   h
   hc
h
p  pcos     cos     hc
cos 
0

120 keV 0.998 8 
0
160 keV 120 keV 0.049
 0.723 2
and   35.9 .
3-30
Maximum energy transfer occurs when the scattering angle is 180 degrees. Assuming the
electron is initially at rest, conservation of momentum gives
m e c2  K   m 2 c 4 
2
hf  hf  pe c 
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 
0 
hc

,  0  



6.63  10 34 J  s 3  10 8 m s
hc

 1.243  10 11 m
E0
0.1 MeV




J s 1  cos 60
6.63  10
 h 
12
  
1  cos   
 1.215  10
m


34
8
m e c 
9.11  10
kg 3  10 m s
 0    1.364  10
E 
(b)

34
11
m

6.63  10 34 J s 3  10 8 m s
hc

 9.11  10 4 eV
11

1.364  10
m
hc
 Ke

K e  0.1 MeV  91.1 keV  8.90 keV
hc
0



e–
0


ph ot on
(c)
ph ot on
 h 
  cos    me v cos  . Conservation of
 0 
h
Conservation of momentum along x:
 h 
momentum along y:  sin    me v sin  . So that

 h   h 

 m e v sin   h 
  sin      cos  
 m e v cos  


 0  

tan  
0 sin 
 0 cos 
Here,   60 ,  0  1.243  10 11 m , and  1.364  10 11 m . Consequently,
tan  
1.24  10 11 m sin 60
 1.451
1.36  1.24 cos 60  10 11 m
  55.4
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, 2dsin   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  2 d sin 1 or

10
0.626  10
m
10
2d
m
8  10
1  0.078 3 radians  4.49
sin 1 

2  2dsin  2 or
sin  2 
3  2 d sin  3 or

d

0.626  10
4.0  10
10
10
m
m
 0.156 5 ,   9.00


10
m
3 3 0.626  10
sin  3 

 0.234 75 , 3  13.6
10
2d
8  10
m