Complete Solutions Manual for Semiconductor Physics and Devices: Basic Principles (4th Edition, 2012) (PDF)

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SOLUTION MANUAL

,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
(a) fcc: 8 corner atoms  1/8 = 1 atom Then
6 face atoms  ½ = 3 atoms F4r IJ
4G
3


Total of 4 atoms per unit cell
H3 K
Ratio =  100%  Ratio = 74%
(b) bcc: 8 corner atoms  1/8 = 1 atom
3
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




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




2G
b g H 3 JK
−8 3
5.65x10
−3
Densitỵ of Ga = 2.22 x10 cm Ratio = 68%
22
Ratio =
F4r I  100% 
3

4 As atoms per unit cell, so that
−3
Densitỵ of As = 2.22 x10 cm
22

(d) Diamond lattice
(b) 8
8 Ge atoms per unit cell Bodỵ diagonal = d = 8r = a a= r
Densitỵ =
8

3
F 8r I 3



−8

b5.65x10 g Unit cell vol. = a =
3

H K
Densitỵ of Ge = 4.44 x10 cm
22 −3
F 4r I 3




1.3 H 3 JK
8 atoms per cell, so atom vol. 8 G

a = (2ra) ==2r8r
(a) Unit
Simple Then
cell cubic
vol =lattice; 4 r
3 3 3 3


F I
3

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

F 4r I 3

Ratio 100% Ratio 34%

1 atom per cell, so atom vol. = (1)G J =   =
HK F 8r I
3



3
Then H K
FG 4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

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

Then mass densitỵ is
−23
1.5 4.85x10
8 = 
(a) a = 5.43 A

From 1.3d, a = r b
2.8x10
−8
3
g
 = 2.21 gm / cm
3

a 3 (5.43) 3 
so that r = = = 1.18 A
8 8
Center of one silicon atom to center of nearest 1.8

(a) a 3 = 2(2.2) + 2(1.8) = 8 A


neighbor = 2r  2.36 A
so that
(b) Number densitỵ 
8 a = 4.62 A
= 
b5.43x10 g
−8
3
Densitỵ = 5x10 cm
22 −3

1 22 −3



Densitỵ of A = b 4.62 x10 −8
 1.01x10 cm
(c) Mass densitỵ
N ( At.Wt.) b5x10 g(28.09)
22
1



== =  22
1.01x10 cm
−3
23
Densitỵ of B =
NA 6.02 x10
b4.62 x10 g  −8




 = 2.33 grams / cm (b) Same as (a)
3

(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
Densitỵ =  (b) Same as (a)
b 2.04 x10
−8
g 3
(c) Same material


23 −3
Densitỵ(A) = 1.18x10 cm 1.10
B-tỵpe: 1 atom per unit cell, so 1
23 −3
(a) Vol densitỵ =
3
Densitỵ(B) = 1.18x10 cm ao
1
2
1.7
Na: Densitỵ = o
(b)

a = 1.8 + 1.0  a = 2.8 A
(c)
12
5