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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 2 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
Chapter 2 
 
 
2.1 
 Sketch 
_______________________________________ 
 
2.2 
 Sketch 
_______________________________________ 
 
2.3 
 Sketch 
_______________________________________ 
 
2.4 
 From Problem 2.2, phase t
x




2
 
 = constant 
 Then 
 











2
,0
2
p
dt
dx
dt
dx
 
 From Problem 2.3, phase t
x




2
 
 = constant 
 Then 
 











2
,0
2
p
dt
dx
dt
dx
 
_______________________________________ 
 
2.5 
 
E
hchc
hE  

 
 Gold: 90.4E eV   19106.190.4  J 
 So, 
 
  
  
5
19
1034
1054.2
106.190.4
10310625.6 





 cm 
 or 
  254.0 m 
 Cesium: 90.1E eV   19106.190.1  J 
 So, 
 
  
  
5
19
1034
1054.6
106.190.1
10310625.6 





 cm 
 or 
  654.0 m 
_______________________________________ 
 
 
 
2.6 
 (a) 
9
34
10550
10625.6






h
p 
 2710205.1  kg-m/s 
 3
31
27
1032.1
1011.9
102045.1






m
p
 m/s 
 or 51032.1  cm/s 
 (b) 
9
34
10440
10625.6






h
p 
 
2710506.1  kg-m/s 
 3
31
27
1065.1
1011.9
105057.1






m
p
 m/s 
 or 51065.1  cm/s 
 (c) Yes 
_______________________________________ 
 
2.7 
(a) (i) 
    1931 106.12.11011.922   mEp 
 2510915.5  kg-m/s 
 9
25
34
1012.1
10915.5
10625.6 






p
h
 m 
 or 
o
A2.11 
(ii)    1931 106.1121011.92  p 
 
241087.1  kg-m/s 
 10
24
34
1054.3
108704.1
10625.6 





 m 
 or 
o
A54.3 
(iii)    1931 106.11201011.92  p 
 
2410915.5  kg-m/s 
 10
24
34
1012.1
10915.5
10625.6 





 m 
 or 
o
A12.1