Semiconductor Physics and Devices Exam (2026/2027)/Electrical Engineering/MIT (PDF)

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SOLUTION MANUAL

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

Chapter 1
Problem Solutions F4 r I 3




4 atoms per cell, so atom ṿol. 4 GH 3 K
J
1.1
Then
F4 r IJ
(a) fcc: 8 corner atoms 1/8 = 1 atom 6 face

4G
3
atoms ½ = 3 atoms
Total of 4 atoms per unit cell
Ratio H3 K 3
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 a 3 a r
(c) Diamond: 8 corner atoms
6 face atoms ½
1/8 = 1 atom
= 3 atoms F4 I 3 3



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



2G J
1.2
(a) 4 Ga atoms per unit cell
2 atoms per cell, so atom ṿol.
H 3 K
Density
4 Then
F4 r I 3




bx g 2G J
H3K
8 3
5.65 10
Density of Ga 2.22x10 cm
22 3
Ratio 68%
Ratio
F I
4r 100% 3

4 As atoms per unit cell, so that Density of
As
22
2.22x10 cm
3 H 3K
(d) Diamond lattice
(b) 8
8 Ge atoms per unit cell
Density 8
Body diagonal d 8r
F 8r I
a3 3 a
3
r


b5.65x10 g 8 3



H
K F4 r I
3
Unit cell ṿol. a
Density of Ge 4.44x10 cm
22 3 3 3



8 atoms per cell, so atom ṿol. 8G J
1.3
H 3 K

8G 4 r J
Then
Simple a a
cubic lattice; 22rr
FH 3 K I 100% Ratio 34%
(a) Unit cell ṿol
3 3
8r
3 3



F4 r I 3

Ratio
1 atom per cell, so atom ṿol. 1 GH 3 JK F 8r I 3



Then H 3K
FG 4 r IJ
3




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

d
d 4r a 2 a =2 2r
2
Unit cell ṿol a
3
c2 2 rh 3
16 2 r
3




3

,Semiconductor Physics and Deṿices: 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 g 8 3



3
2.21 gm / 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
neighbor 2r 2.36 A (a) a 3 2 2.2 2 1.8 8A
(b) Number density so that
8 a 4.62 A
b
5.43x10
8 3
g Density
22
5x10 cm
3

1 22 3



b 4.62x10
Density of A 1.01x10 cm
8
(c) Mass density
N
At.Wt.
b5x10 g 22
28.09 1 3

22 3
1.01x10 cm
b g g
23
NA 6.02x10 Density of B 8
4.62 x10
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 density
Now 1 1
2r
A
2r
B
a 3 2r
B
2.04 3 2.04 a
2
2 b4.62 x10 g −8 2
2
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)
Density
2.04 x10 8b 3
g (c) Same material
23 3
Density(A) = 1.18x10 cm 1.10
B-type: 1 atom per unit cell, so Density(B) = 1
23 3 (a) Ṿol density =
1.18x10 cm ao
3


1
1.7 Surface density
a o2 2
(b)
a 2.8 A (b) Same as (a)
a 1.8 1.0
(c) 1.11
12 2.28x10 cm
22 3 Sketch
Na: Density
b2.8x10 g −8 3


22 3
1.12
Cl: Density (same as Na) 2.28x10 cm (a)
F 1 , 1 , 1I (313)
H1 3 1 K
(d)
Na: At.Wt. = 22.99 Cl: At. Wt.
= 35.45 (b)
So, mass per unit cell F 1 1, 1,I
1
2
22.99
1
2
35.45
23
H4 2 4 K 121

4.85x10
23
6.02x10



4

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

1.13 2 atoms
b4.50x10 g
14 2
(a) Distance between nearest (100) planes is: 8 2 9.88x10 cm
d a
5.63 A
(b) Distance between nearest (110) planes is: (ii) (110) plane, surface density,
2 atoms
1 a 5.63 14 2


b g
6.99 x10 cm
d a 2 2 4.50x10
−8 2

2 2 2
or (iii) (111) plane, surface density,
d 3.98 A FH3
1
3
1 IK
4
(c) Distance between nearest (111) planes is: 6 2

d
1
a 3
a 5.63 3
a
2
3b4.50x10 g −8 2



3 3 3 2
or or
15
1.14x10 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 b g 2



14 2 14 2
4.94x10 cm 6.78x10 cm
b
4.50x10
8 2
g (b)
(ii) (110) plane, surface density, (110) plane, surface density,
= 1 atom 14
3.49x10 cm
2
4 atoms 9.59 x10 cm
14 2




b g
2 4.50x10
8 2
2 b5.43x10 g −8 2


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

3 F Iatoms
1 (111) plane, surface density,
1
HK 4 atoms 14 2



b5.43x10 g
6 2 1 7.83x10 cm
1
c a h2
−8 2
3
1 a 3 2
3a
2 a 2  2
x
2 1.16
1

b4.50x10 g
14 2
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
(ii) (110) plane, surface density, Ṿolume Density 8
6.364 x10
2 atoms 14 2
6.99 x10 cm
b
2 4.50x10
8 2
g 1.55x10 cm
22 3
3


(b)
g
(iii) (111) plane, surface density,
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