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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 3 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
3.6 
 (b) (i) First point:  a 
 Second point: By trial and error, 
  515.1a 
 (ii) First point:  2a 
 Second point: By trial and error, 
  375.2a 
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3.7 
 kaa
a
a
P coscos
sin
 


 
 Let yka  , xa  
 Then 
 yx
x
x
P coscos
sin
 
 Consider 
dy
d
 of this function. 
     yxxxP
dy
d
sincossin
1


 
 We find 
     








dy
dx
xx
dy
dx
xxP cossin1
12
 
 y
dy
dx
x sinsin  
 Then 
 yx
x
x
x
x
P
dy
dx
sinsin
cos
sin
1
2















 
 For nkay  , ...,2,1,0n 0sin  y 
 So that, in general, 
 
 
  dk
d
kad
ad
dy
dx 
 0 
 And 
 
2
2

mE
 
 So 
 
dk
dEmmE
dk
d














2
2/1
2
22
2
1


 
 This implies that 
 
dk
dE
dk
d
 0

 for 
a
n
k

 
_______________________________________ 
 
 
 
 
 
 
3.8 
(a)  a1 
 a
Emo
2
12

 
 
   
  21031
2342
2
22
1
102.41011.92
10054.1
2 





am
E
o

 
 19104114.3  J 
 From Problem 3.5 
  729.12 a 
 729.1
2
2
2 a
Emo

 
 
   
  21031
2342
2
102.41011.92
10054.1729.1






E 
 
18100198.1  J 
 12 EEE  
 
1918 104114.3100198.1   
 
19107868.6  J 
or 24.4
106.1
107868.6
19
19






E eV 
(b)  23 a 
 2
2
2
3 a
Emo

 
 
   
  21031
2342
3
102.41011.92
10054.12






E 
 
18103646.1  J 
 From Problem 3.5, 
  617.24 a 
 617.2
2
2
4 a
Emo

 
 
   
  21031
2342
4
102.41011.92
10054.1617.2






E 
 
18103364.2  J 
 34 EEE  
 
1818 103646.1103364.2   
 
1910718.9  J 
 or 07.6
106.1
10718.9
19
19






E eV 
_______________________________________