AN INTRODUCTION TO SEMICONDUCTOR DEVICES SOLUTION
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An Introduction to Semiconductor Devices Chapter 1
Solutions Manual Problem Solutions
______________________________________________________________________________________
2
Chapter 1
Problem Solutions
1.1
(a) fcc: 8 corner atoms × 1/8 = 1 atom
6 face atoms
× ½ = 3 atoms
Total of 4 atoms per unit cell
(b) bcc: 8 corner atoms × 1/8 = 1 atom
1 enclosed atom = 1 atom
Total of 2 atoms per unit cell
(c) Diamond: 8 corner atoms
×
1/8 = 1 atom
6 face atoms × ½ = 3 atoms
4 enclosed atoms = 4 atoms
Total of 8 atoms per unit cell
_______________________________________
1.2
4 Ga atoms per unit cell
Density
=⇒
−
4
565 10
8
3
. x
b
g
Density of Ga
=
−
2.22 10
22 3
xcm
4 As atoms per unit cell, so that
Density of As
=
−
2.22 10
22 3
xcm
_______________________________________
1.3
8 Ge atoms per unit cell
Density
=⇒
−
8
565 10
8
3
. x
b
g
Density of Ge =
4.44 10
22 3
xcm
−
_______________________________________
1.4
(a) Simple cubic lattice; a
r
= 2
Unit cell vol
== =
()
arr
3
3
3
28
1 atom per cell, so atom vol.
=
()
F
H
G
I
K
J
1
4
3
3
π
r
Then
Ratio
r
r
=×⇒
F
H
G
I
K
J
4
3
8
100%
3
3
π
Ratio
=
52.4%
(b)
Face-centered cubic lattice
dra== ⇒42 a
d
r==
2
22
Unit cell vol
== =ar r
3
3
3
22 162
ch
4 atoms per cell, so atom vol.
=
F
H
G
I
K
J
4
4
3
3
π
r
Then
Ratio
r
r
=×⇒
F
H
G
I
K
J
4
4
3
16 2
100%
3
3
π
Ratio = 74%
(c)
Body-centered cubic lattice
dra a r== ⇒=43
4
3
Unit cell vol.
==
F
H
I
K
ar
3
3
4
3
2 atoms per cell, so atom vol.
=
F
H
G
I
K
J
2
4
3
3
π
r
Then
Ratio
r
r
=×⇒
F
H
G
I
K
J
F
H
I
K
2
4
3
4
3
100%
3
3
π
Ratio = 68%
(d)
Diamond lattice
Body diagonal
== = ⇒=dra a r83
8
3
Unit cell vol.
==
F
H
I
K
a
r
3
3
8
3
8 atoms per cell, so atom vol.
8
4
3
3
π
r
F
H
G
I
K
J
Then
Ratio
r
r
Ratio=×⇒=
F
H
G
I
K
J
F
H
I
K
8
4
3
8
3
100% 34%
3
3
π
_______________________________________
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