Semiconductor Physics and Devices Exam (2026/2027) | Electronics Engineering | University (PDF)

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SOLUTION MANUAL

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


Chapter 1
Problem Solutions F4 r I 3




1.1
4 atoms per cell, so atom vol.
H 3 JK
4Ḡ

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


Total of 4 atoms per unit cell
Ratio
H3 K3
100% Ratio 74%
(b) bcc: 8 corner atoms 1/8 = 1 atom 1 16 2 r
enclosed atom = 1 atom (c) Body-centered cubic lattice
Total of 2 atoms per unit cell 4
d 4r a3 a r
(c) Diamond: 8 corner atoms
6 face atoms ½
1/8 = 1 atom
= 3 atoms F I4
3 3




H 3 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 2 atoms per cell, so atom vol. 2Ḡ
H J
3 K
F4 r I
(a) 4 Ḡa atoms per unit cell
4 Then 3
Density
2Ḡ
bx ḡ H 3 JK
3
8
5.65 10
22 3
Density of Ḡa 2.22 x10 cm Ratio 68%
Ratio
F 4r I 100% 3

4 As atoms per unit cell, so that
Density of As 2.22 x10 cm
22 3 H 3K
(d) Diamond lattice
(b) 8
Body diaḡonal d 8r a 3
F 8r I
8 Ḡe atoms per unit cell r a
Density 8 3
3




b5.65x10 ḡ 8 3



H 3K F 4 r I
3

22 3
Unit cell vol. a 3
Density of Ḡe 4.44 x10 cm


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


8Ḡ 4 r J
Simple
(a) Unit r cubic a a
vol lattice; 22rr Then
FH 3 KI
3 3 3
8cell 3



F4 r I 3

Ratio 100% Ratio 34%
1 atom per cell, so atom vol. 1 ḠH3 JK F 8r I 3



Then H 3K
FḠ 4 r IJ 3




Ratio
H K3 100% Ratio 52.4%
1.4
From Problem 1.3, percent volume of fcc atoms is 74%;
3
8r Therefore after coffee is ḡround,
(b) Face-centered cubic lattice Volume 0.74 cm
3

d
d 4r a2 a =2 2r
2
Unit cell vol a
3
c2 2 r h 3
16 2 r
3




3

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

Then mass density is
1.5 4.85x10
23

8
(a) a 5.43 A From 1.3d, a
3
r b2.8x10 ḡ 8 3



3
2.21 ḡm / cm
a 3 (5.43) 3
so that r 1.18 A
8 8
Center of one silicon atom to center of nearest 1.8
neiḡhbor 2r 2.36 A (a) a 3 2 2.2 2 1.8 8A
(b) Number density so that
8 a 4.62 A
5.43x10 b
8 3
ḡ Density 5x10 cm
22 3

1 22 3



b 4.62 x10 ḡ
Density of A 3
1.01x10 cm

b5x10 ḡ
8
(c) Mass density 22
N 28.09 1
At.Wt.
22 3
1.01x10 cm
b ḡ
23
NA 6.02 x10 Density of B
4.62 x10 8

2.33 ḡrams / cm
3
(b) Same as (a)
(c) Same material

1.6 1.9
(a) a 2rA 2 1.02 (a) Surface density
2.04 A 1 1
Now
2r 2r a3 2r 2.04 3 2.04
a
2
2 b4.62 x10 ḡ −8 2
2
A B B

so that r B 0.747 A 3.31x10 cm
14 2


(b) A-type; 1 atom per unit cell Same for A atoms and B atoms
1
(b) Same as (a)
b ḡ
Density
2.04 x10 8
3
(c) Same material
23 3
Density(A) = 1.18x10 cm 1.10
B-type: 1 atom per unit cell, so 1
23 3
(a) Vol density =
Density(B) = 1.18x10 cm ao
3


1
1.7 Surface density 2
ao 2
(b)
(b) Same as (a)
a 1.8 1.0 a 2.8 A
1.11
(c) 22 3 Sketch
12 2.28x10 cm

b2.8x10 ḡ2.28x10
Na: Density
−8 3
1.12
(a)
F 1 , 1 , 1I
22 3
Cl: Density (same as Na) cm
(d)
Na: At.Wt. = 22.99 Cl: At. H1 3 1 K (313)


F 1 , 1 ,1 I
Wt. = 35.45 (b)
So, mass per unit cell
1
2
22.99
1
2
35.45
23
H4 2 4K 121

4.85x10
23
6.02 x10
4

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

1.13 2 atoms
(a) Distance between nearest (100) planes is: b 4.50x10 8 ḡ 2
14
9.88x10 cm
2



d a 5.63
A (ii) (110) plane, surface density,
(b) Distance between nearest (110) planes is:
2 atoms
1 a 5.63 14 2


b4.50x10 ḡ 6.99 x10 cm
d a 2 2
−8 2

2 2 2
or (iii) (111) plane, surface density,
d 3.98 A FH 3 1 3 I 1 K 4
(c) Distance between nearest (111) planes is: 6 2
1 a 5.63 3b4.50x10 ḡ
−8 2
3 2
d a 3 a
3 3 3 2
or or
15
1.14 x10 cm
2


d 3.25 A
1.15
1.14 (a)
(a) (100) plane of silicon – similar to a fcc,
Simple cubic: a 4.50 A
2 atoms
(i) (100) plane, surface density,
1 atom
surface density
5.43x10 8
2
b ḡ
14 2 14 2
6.78x10 cm
ḡ
4.94 x10 cm
b
4.50x10
8 2

(b)
(ii) (110) plane, surface density, (110) plane, surface density,
= 1 atom 14
3.49 x10 cm
2
4 atoms 14
9.59 x10 cm
2




b ḡ
2 4.50x10
8 2
2 b5.43x10 ḡ −8 2


(iii) (111) plane, surface density, (c)

3 F I atoms
1 (111) plane, surface density,
HK
1
4 atoms 14 2
1 7.83x10 cm
b5.43x10 ḡ
6 2
1
c a h2
−8 2

1 a 3 2
3
3a
2 a 2  2
x
2 1.16
1
14 2


b4.50x10 ḡ 2.85x10 cm d 4r
−8 2 2
3 a
(b) then 4 2.25
4r
Body-centered cubic a 6.364 A
(i) (100) plane, surface density, 2 2
14 2 (a)
Same as (a),(i); surface density 4.94x10 cm
4 atoms

b
Volume Density
(ii) (110) plane, surface density,
ḡ
8
6.364 x10 3
2 atoms

b ḡ
14 2
6.99 x10 cm
8 2 22 3
2 4.50x10 1.55x10 cm
(iii) (111) plane, surface density, (b)
14
Same as (a),(iii), surface density 2.85x10 cm
2 Distance between (110) planes,
1 a 6.364
(c) a 2
Face centered cubic 2 2 2
(i) (100) plane, surface density or
5