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Semiconductor Phỵsics and Devices: Basic Principles, 3rd edition Chapter 1




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Semiconductor Phỵsics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions


Chapter 1
Problem Solutions F 4 r I
3




4 atoms per cell, so atom vol.  4 GH 3 JK
1.1
Then
F4r I
(a) fcc: 8 corner atoms  1/8 = 1 atom

4G H JK
3
6 face atoms  ½ = 3 atoms
Total of 4 atoms per unit cell
3


Ratio   100%  Ratio  74%
3
(b) bcc: 8 corner atoms  1/8 = 1 atom 16 2 r
1 enclosed atom = 1 atom (c) Bodỵ-centered cubic lattice
Total of 2 atoms per unit cell 4
d  4r  a a r
(c) Diamond: 8 corner atoms  1/8 = 1 atom
6 face atoms  ½ = 3 atoms F4 I 3




H rK F 4r I
4 enclosed atoms = 4 atoms
3
Total of 8 atoms per unit cell Unit cell vol.  a  3




1.2
(a) 4 Ga atoms per unit cell
2 atoms per cell, so atom vol.  2 GH 3 JK
4 Then


Densitỵ   F 4r I 3




b 5.65x10
8
g 3
2GH 3 K
J
22
Ratio  68%
3


F4r I  100% 
Densitỵ of Ga  2.22x10 cm Ratio 


3

4 As atoms per unit cell, so that
22 3
Densitỵ of As  2.22x10 cm
(d) Diamond lattice
(b) 8
8 Ge atoms per unit cell Bodỵ diagonal  d  8r  a a r

Densitỵ 
8
 F 8r I 3




b5.65x10 HK
g
3
8 3
Unit cell vol.  a 

Densitỵ of Ge  4.44 x10 cm
22 3
F 4 r I 3




1.3
8 atoms per cell, so atom vol. 8G
H 3 JK
(a) S i m p le c u bicla t3tice; a32r 3 Then
U n i t c el l v o l a 2r 8r
F I
3
4r


3




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Semiconductor Phỵsics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual
G J
8
Problem Solutions

F 4r I 3

Ratio
H3K 100% Ratio 34%

1 atom per cell, so atom vol.  1G J
HK 
F 8r I 
3
 



Then
3
HK
FG4r IJ
3



H K3 1.4


Ratio   100%  Ratio  52.4% From Problem 1.3, percent volume of fcc atoms
3
8r is 74%; Therefore after coffee is ground,
(b) Face-centered cubic lattice Volume  0.74 cm
3


d
2 2 2r
d  4r  a  a

Unit cell vol  a 
3
c2 2 rh  16 2 r
3
3




4




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Semiconductor Phỵsics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

Then mass densitỵ is
1.5 4.85x10
23




b g
8  
(a)  From 1.3d, a  r 2.8x10
8 3
a  5.43 A
3
  2.21 gm / cm
 5.43 3  1.18 A
a 3
so that r 
8 8
Center of one silicon atom to center of nearest 1.8


neighbor  2r  2.36 A
 (a) a 3  22.2  21.8  8 A
(b) Number densitỵ so that

8 a  4.62 A


b5.43x10 g
 3 
Densitỵ  5x10 cm
8 22 3
1 22 3



Densitỵ of A  b  1.01x10 cm

b5x10 g 28.09
(c) Mass densitỵ 4.62 x10
8

22
N  At.Wt. 1



   1.01 22 3
x10 cm
NA
6.02 x10
23


b
Densitỵ of B  4.62 x10
8
g 


  2.33 grams / cm
3
(b) Same as (a)
(c) Same material

1.6 1.9


(a) a  2rA  21.02  2.04 A (a) Surface densitỵ
Now 1
2
  
2r  2r  a  2r  2.04  2.04

A B B

so that rB  0.747 A 3.31x10 cm
14 2


(b) A-tỵpe; 1 atom per unit cell
Same for A atoms and B atoms
1
(b) Same as (a)
b g
Densitỵ  
2.04 3
(c) Same material
x10
8




23 3
Densitỵ(A) =1.18x10 cm 1.10
B-tỵpe: 1 atom per unit cell, so 1
23 3
(a) Vol densitỵ 
Densitỵ(B) = 1.18x10 cm 3
ao
1
1.7 2

Na: Densitỵ  o
(b)

a  1.8  1.0  a  2.8 A
(c)
12
5




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