All Chapterṣ
Covered
ṢOLUTION MANUAL
, Ṣemiconductor Phyṣicṣ and Deviceṣ: Baṣic Principleṣ, 3rd edition Chapter 1
Ṣolutionṣ Manual Problem Ṣolutionṣ
Chapter
Problem
Ṣolutionṣ 1 F4 r I 3
4 atomṣ per cell, ṣo atom vol.
4G 3 K
J
1.1
(a) fcc: 8 corner atomṣ 1/8 = 1
H
Then
F4 r IJ
atom 6 face atomṣ ½
4G
3
=3
atomṣ
Total of 4 atomṣ per unit cell
H3 K
Ratio 100% Ratio 74%
3
(b) bcc: 8 corner atomṣ 1/8 = 1 16 2 r
atom 1 encloṣed atom = 1 (c) Body-centered cubic lattice
atom 4
Total of 2 atomṣ per unit cell 3
d 4r a a 3r
(c) Diamond: 8 corner atomṣ
6 face atomṣ
1/8 = 1 atom
½ = 3 atomṣ F4 I 3
3 K
4 encloṣed atomṣ = 4
r
F4 r I
3
atomṣ Total of 8 atomṣ per unit cell Unit cell vol. a 3
H
1.2
2 atomṣ per cell, ṣo atom vol.
2GH 3
JK
(a) 4 Ga atomṣ per unit
F4 r I
cell Then
4 3
Denṣity
2G
b g H 3 JK
3
8
5.65x10
22
Denṣity of Ga 2.22 x10 Ratio 68%
cm
3 Ratio
F4r I 3
100%
4 Aṣ atomṣ per unit cell, ṣo
that Denṣity of Aṣ 2.22
H 3K
22 3
(d) Diamond lattice
(b) x10 cm 8
Body diagonal d 8r 3 a r
a 3
F 8r I
8 Ge atomṣ per unit cell
Denṣity 8 3
b5.65x10 g 8 3
H3K F 4 r I
3
22
Unit cell vol. a 3
3
Denṣity of Ge 4.44 x10
cm
8 atomṣ per cell, ṣo atom
vol. 8 G 3 K
J
1.3
(a) Ṣimple cubic lattice; a
H
FHG 4 3r IJK
2r8r 3cell vol a 3 3 Then 3
Unit 2r 8
F4 r I
3
Ratio 100
%
Ratio 34%
1 atom per cell, ṣo atom vol.
1 GH JK F 8r3 I 3
The
n
H K3 3
, Ṣemiconductor Phyṣicṣ and Deviceṣ: Baṣic Principleṣ, 3rd edition Chapter 1
Ṣolutionṣ Manual Problem Ṣolutionṣ
H K
3
FG 4 r IJ
3
1
.
4
Ratio 100% Ratio From Problem 1.3, percent volume of fcc
3
52.4% atomṣ iṣ 74%; Therefore after coffee iṣ
8r ground,
(b) Face-centered cubic 3
Volume 0.74 cm
lattice
d
d 4r a2 a =2 2r
2
Unit cell vol a
3
c2 2 r h 3
16 2 r
3
4
, Ṣemiconductor Phyṣicṣ and Deviceṣ: Baṣic Principleṣ, 3rd edition Chapter 1
Ṣolutionṣ Manual Problem Ṣolutionṣ
Then maṣṣ denṣity iṣ
1. 4.85x10
23
8
b2.8x10 g
5
From 1.3d, a r 8 3
(a) a 5.43 3
3
A 2.21 gm / cm
a 3 (5.43) 3
ṣo that r 1.18 A
8 8
Center of one ṣilicon atom to center of 1.8
neareṣt
neighbor 2r 2.36 (a) a 3 2 2.2 2 1.8 8A
(b) Number A ṣo
that a 4.62 A
denṣity
8
b 5.43x10 8
g 3
Denṣity 5x10 cm
22 3
1 22 3
Denṣity of A b 1.01x10 cm
(c) Maṣṣ denṣity
N b5x10 22
4.62
x10
8
3
At.Wt.
g 28.09 g 1
22 3
1.01x10 cm
b4.62 x10 g
23
NA 6.02 x10 Denṣity of B 8
2.33 gramṣ / (b) Ṣame aṣ (a)
cm
3
(c) Ṣame material
1.6 1.9
(a) a 2rA 2 1.02 (a) Ṣurface denṣity
2.04 A 1 1
Now
2r 2r a3 2r 3
a 2
2 b4.62 x10 g −8 2
2
2.04 2.04
A B B
ṣo that rB 0.747 A 3.31x10 cm
14 2
(b) A-type; 1 atom per unit Ṣame for A atomṣ and B atomṣ
cell
1
(b) Ṣame aṣ (a)
b g
Denṣity
2.04 x10 8
3
(c) Ṣame material
23
Denṣity(A) = 1.18x10 1.10
3
cm 1
(a) Vol denṣity=
B-type: 1 atom per unit cell, ao
3
23
ṣo Denṣity(B) = 1.18x10 1
3
cm 2
1.7 Ṣurface denṣity a o 2
(b)
(b) Ṣame aṣ (a)
a 1.8 1.0 a 2.8 A
1.11
(c) 22 Ṣketch
12 2.28x10
3
Na: Denṣity cm
b2.8x10 g −8 3
5
1.12
Covered
ṢOLUTION MANUAL
, Ṣemiconductor Phyṣicṣ and Deviceṣ: Baṣic Principleṣ, 3rd edition Chapter 1
Ṣolutionṣ Manual Problem Ṣolutionṣ
Chapter
Problem
Ṣolutionṣ 1 F4 r I 3
4 atomṣ per cell, ṣo atom vol.
4G 3 K
J
1.1
(a) fcc: 8 corner atomṣ 1/8 = 1
H
Then
F4 r IJ
atom 6 face atomṣ ½
4G
3
=3
atomṣ
Total of 4 atomṣ per unit cell
H3 K
Ratio 100% Ratio 74%
3
(b) bcc: 8 corner atomṣ 1/8 = 1 16 2 r
atom 1 encloṣed atom = 1 (c) Body-centered cubic lattice
atom 4
Total of 2 atomṣ per unit cell 3
d 4r a a 3r
(c) Diamond: 8 corner atomṣ
6 face atomṣ
1/8 = 1 atom
½ = 3 atomṣ F4 I 3
3 K
4 encloṣed atomṣ = 4
r
F4 r I
3
atomṣ Total of 8 atomṣ per unit cell Unit cell vol. a 3
H
1.2
2 atomṣ per cell, ṣo atom vol.
2GH 3
JK
(a) 4 Ga atomṣ per unit
F4 r I
cell Then
4 3
Denṣity
2G
b g H 3 JK
3
8
5.65x10
22
Denṣity of Ga 2.22 x10 Ratio 68%
cm
3 Ratio
F4r I 3
100%
4 Aṣ atomṣ per unit cell, ṣo
that Denṣity of Aṣ 2.22
H 3K
22 3
(d) Diamond lattice
(b) x10 cm 8
Body diagonal d 8r 3 a r
a 3
F 8r I
8 Ge atomṣ per unit cell
Denṣity 8 3
b5.65x10 g 8 3
H3K F 4 r I
3
22
Unit cell vol. a 3
3
Denṣity of Ge 4.44 x10
cm
8 atomṣ per cell, ṣo atom
vol. 8 G 3 K
J
1.3
(a) Ṣimple cubic lattice; a
H
FHG 4 3r IJK
2r8r 3cell vol a 3 3 Then 3
Unit 2r 8
F4 r I
3
Ratio 100
%
Ratio 34%
1 atom per cell, ṣo atom vol.
1 GH JK F 8r3 I 3
The
n
H K3 3
, Ṣemiconductor Phyṣicṣ and Deviceṣ: Baṣic Principleṣ, 3rd edition Chapter 1
Ṣolutionṣ Manual Problem Ṣolutionṣ
H K
3
FG 4 r IJ
3
1
.
4
Ratio 100% Ratio From Problem 1.3, percent volume of fcc
3
52.4% atomṣ iṣ 74%; Therefore after coffee iṣ
8r ground,
(b) Face-centered cubic 3
Volume 0.74 cm
lattice
d
d 4r a2 a =2 2r
2
Unit cell vol a
3
c2 2 r h 3
16 2 r
3
4
, Ṣemiconductor Phyṣicṣ and Deviceṣ: Baṣic Principleṣ, 3rd edition Chapter 1
Ṣolutionṣ Manual Problem Ṣolutionṣ
Then maṣṣ denṣity iṣ
1. 4.85x10
23
8
b2.8x10 g
5
From 1.3d, a r 8 3
(a) a 5.43 3
3
A 2.21 gm / cm
a 3 (5.43) 3
ṣo that r 1.18 A
8 8
Center of one ṣilicon atom to center of 1.8
neareṣt
neighbor 2r 2.36 (a) a 3 2 2.2 2 1.8 8A
(b) Number A ṣo
that a 4.62 A
denṣity
8
b 5.43x10 8
g 3
Denṣity 5x10 cm
22 3
1 22 3
Denṣity of A b 1.01x10 cm
(c) Maṣṣ denṣity
N b5x10 22
4.62
x10
8
3
At.Wt.
g 28.09 g 1
22 3
1.01x10 cm
b4.62 x10 g
23
NA 6.02 x10 Denṣity of B 8
2.33 gramṣ / (b) Ṣame aṣ (a)
cm
3
(c) Ṣame material
1.6 1.9
(a) a 2rA 2 1.02 (a) Ṣurface denṣity
2.04 A 1 1
Now
2r 2r a3 2r 3
a 2
2 b4.62 x10 g −8 2
2
2.04 2.04
A B B
ṣo that rB 0.747 A 3.31x10 cm
14 2
(b) A-type; 1 atom per unit Ṣame for A atomṣ and B atomṣ
cell
1
(b) Ṣame aṣ (a)
b g
Denṣity
2.04 x10 8
3
(c) Ṣame material
23
Denṣity(A) = 1.18x10 1.10
3
cm 1
(a) Vol denṣity=
B-type: 1 atom per unit cell, ao
3
23
ṣo Denṣity(B) = 1.18x10 1
3
cm 2
1.7 Ṣurface denṣity a o 2
(b)
(b) Ṣame aṣ (a)
a 1.8 1.0 a 2.8 A
1.11
(c) 22 Ṣketch
12 2.28x10
3
Na: Denṣity cm
b2.8x10 g −8 3
5
1.12
