Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-59 continued)
Applying the quadratic formula,


2
0.04541   0.04541  4 2.36  103 


1 
2
1/ 2
1  0.03092nm and 2  0.1259nm
hc
(b) E1 
1

3-60. Let x 
kT



1240eV nm
 40.1 keV
0.03092nm

Eelectron  9.90keV
hf
in Equation 3-15:
kT

   e 
f n  A e nx  A e0  e x  e x


n 0
n 0
2
3
x


  A 1  y  y 2  y3 

 1
Where y  e x . This sum is the series expansion of
1  y 
1
, i.e., 1  y   1  y  y 2  y3 
1
. Then
f
 A 1  y   1 gives A  1  y.
1
n
Writing Equation 3-16 in terms of x and y.



n 0
n 0
n 0
E   En Ae En / kT  A nhfe nhf / kT  Ahf  ne nx
Note that
 ne
 nx
 ne

 nx
   d / dx   e nx . But  e nx  1  y  , so we have
1
d
d
dy 
1
2 
2
e nx   1  y   1  y      y 1  y 

dx
dx
 dx 
 
x
dy d e
Since

 e x   y.
dx
dx
Multiplying this sum by hf and by A  1  y  , the average energy is

E  hfA ne nx  hf 1  y  y 1  y  
2
n 0
hfy
hfe x

1  y 1  e x
Multiplying the numerator and the denominator by e x and substituting for x, we obtain
E
hf
e
hf / kT
1
, which is Equation 3-17.
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