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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 3 
By D. A. Neamen Problem Solutions 
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3.9 
(a) At ka ,  a1 
 a
Emo
2
12

 
 
   
  21031
2342
1
102.41011.92
10054.1






E 
 19104114.3  J 
 At 0ka , By trial and error, 
  859.0ao 
 
   
  21031
2342
102.41011.92
10054.1859.0






oE 
 19105172.2  J 
 oEEE  1 
 1919 105172.2104114.3   
 2010942.8  J 
 or 559.0
106.1
10942.8
19
20






E eV 
(b) At 2ka ,  23 a 
 2
2
2
3 a
Emo

 
 
   
  21031
2342
3
102.41011.92
10054.12






E 
 
18103646.1  J 
 At ka . 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 
 23 EEE  
 
1818 100198.1103646.1   
 
19104474.3  J 
 or 15.2
106.1
104474.3
19
19






E eV 
_______________________________________ 
 
 
 
 
 
 
 
 
3.10 
(a)  a1 
 a
Emo
2
12

 
 
   
  21031
2342
1
102.41011.92
10054.1






E 
 19104114.3  J 
 From Problem 3.6,  515.12 a 
 515.1
2
2
2 a
Emo

 
 
   
  21031
2342
2
102.41011.92
10054.1515.1






E 
 
1910830.7  J 
 12 EEE  
 
1919 104114.310830.7   
 
19104186.4  J 
 or 76.2
106.1
104186.4
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 3.6,  375.24 a 
 375.2
2
2
4 a
Emo

 
 
   
  21031
2342
4
102.41011.92
10054.1375.2






E 
 
18109242.1  J 
 34 EEE  
 
1818 103646.1109242.1   
 
1910597.5  J 
 or 50.3
106.1
10597.5
19
19






E eV 
_____________________________________