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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 14 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
Chapter 14 
 
14.1 
 
gE
24.1
max  m 
(a) Si:  11.1
12.1
24.1
max  m 
(b) Ge:  88.1
66.0
24.1
max  m 
(c) GaAs:  873.0
42.1
24.1
max  m 
(d) InP:  919.0
35.1
24.1
max  m 
_______________________________________ 
 
14.2 
 (a) For 480 nm, 
 58.2
480.0
24.124.1


E eV 
 For 725 nm, 
 71.1
725.0
24.1
E eV 
 (b) For 87.0E eV, 
  43.1
87.0
24.124.1

E
m 
 For 32.1E eV, 
  939.0
32.1
24.1
 m 
 For 90.1E eV, 
  653.0
90.1
24.1
 m 
_______________________________________ 
 
14.3 
(a)  752.0
65.1
24.1
 m 
(i) From Figure 14.4, 
3109 cm 1 
 (ii) 
 
 d
I
dI


  exp
0
 
    43 102.1109exp  
 340.0 
 Fraction absorbed 66.034.01  
 
 
 
 
 
 
(b)  653.0
90.1
24.1
 m 
 (i) From Figure 14.4, 4106.2  cm 1 
 (ii) 
 
 d
I
dI


  exp
0
 
    44 1080.0106.2exp  
 125.0 
 Fraction absorbed 875.0125.01  
_______________________________________ 
 
14.4 
 
 


h
xI
g  
 For 3.1h eV,  95.0
3.1
24.1
 m 
 For silicon: 
2103 cm 1 
 Then for   210xI W/cm 2 , we obtain 
 
  
  3.1106.1
10103
19
22




g 
 or 
 191044.1 g cm 3 s 1 
 The excess concentration is 
   619 101044.1   gn 
 or 
 
131044.1 n cm 3 
_______________________________________ 
 
14.5 
(a) 
0
0
p
p
p
ggp


  
 22
7
15
105.2
102
105





g cm 3 s 1 
 For 65.1h eV, 
  752.0
65.1
24.1
 m 
 From Figure 14.4, 
3109 cm 1 
 
  



hg
I

0 
 
   
3
1922
109
65.1106.1105.2




 
 733.0 W/cm 2