Semiconductor Physics and Devices: Basic Principles (4th Edition) by Donald A. Neamen – Solutions Manual Overview.|NEW UPDATES|2026/2027

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SOLUTION MANUAL

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


Chapter 1
Problem Solutions F 4 r I
3




1.1
4 atoms per cell, so atom vol. = 4G
H 3 JK
(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
Ratio =
H3 K  100%  Ratio = 74%
(b) bcc: 8 corner atoms  1/8 = 1 atom
3
16 2 r
1 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  1/8 = 1 atom
6 face atoms  ½ = 3 atoms F I
4
3 3




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




1.2 2 atoms per cell, so atom vol. = 2G
H 3 K
J
F 4r I
(a) 4 Ga atoms per unit cell
4 Then 3
Density = 
2G
b g H 3 JK
−8 3
5.65x10
−3
Density of Ga = 2.22 x10 cm Ratio = 68%
22




4 As atoms per unit cell, so that
Ratio =
F4r I  100% 
3



−3
Density of As = 2.22 x10 cm
22

(d) Diamond lattice
(b) 8

F I
8 Ge atoms per unit cell Body diagonal = d = 8r = a3 3  a = r
Density = 8 8r 3

b5.65x10 g −8 3

Unit cell vol. = a =
H 3 K F 4r I
3
−3
Density of Ge = 4.44 x10 cm
22 3




1.3
8 atoms per cell, so atom vol. 8G
H 3 JK
a = (2ra) ==2r8r 8G 4r J
Then

HF 3 KI 100% Ratio 34%
(a) Unit
Simple
cell cubic
vol =lattice;
3 3 3 3



F 4r I 3



1 atom per cell, so atom vol. = (1)G J
Ratio

H3 K
=
F 8r I   =
3



Then H 3K
FG 4r IJ 3




Ratio =
H 3 K  100%  Ratio = 52.4% 1.4
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
d = 4r = a 2  a = =2 2r
2
Unit cell vol = a =
3
c2 2 rh = 16 2 r 3
3




3

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

Then mass density is
−23
1.5 4.85x10
=
b g
8 
From 1.3d, a = −8 3
(a)  r
a = 5.43 A 2.8x10
3
 = 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 density 
a = 4.62 A
b g
= 8  −3
Density = 5x10 cm
3 22
−8
5.43x10 1 22 −3
Density of A =  1.01x10 cm
(c) Mass density
b (28.09) g b4.62 x10 g −8 3



N ( At.Wt.)
22
5x10 1
== =  22
1.01x10 cm
−3




b g
23
NA 6.02 x10 Density of B = −8
4.62 x10
 = 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 density
Now 1 1
= 2 = 
2r + 2r = a 3  2r = 2.04 3 − 2.04 a 2
A B B

so that rB = 0.747 A 14
3.31x10 cm
−2


(b) A-type; 1 atom per unit cell Same for A atoms and B atoms
1
Density = (b) Same as (a)
b g 
−8 3 (c) Same material
2.04 x10
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 3
ao
1
1.7 Surface density = 2
ao 2
(b)
 (b) Same as (a)
a = 1.8 + 1.0  a = 2.8 A
(c) 1.11
12 −3 Sketch
Na: Density = = 2.28x10 cm
22



1.12
−3
Cl: Density (same as Na) = 2.28x10 cm
22
(a)
(d) FH1 , 1 ,1IK  (313)
Na: At.Wt. = 22.99 1 3 1

F 1 1 1 I (121)
Cl: At. Wt. = 35.45 (b)
So, mass per unit cell
1 1
(22.99) + (35.45) H 4 , 2 , 4K 
= 2 2 −23
= 4.85x10
23
6.02 x10


4

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


b g
1.13 = ph
2 atoms
ph
2 14 p h −2
(a) Distance between nearest (100) planes 4.50x10
−8 ph 9.88x10 cm

ph ph ph ph

is:
ph p h


d = a = 5.63 A ph ph ph ph ph


(b) Distance between nearest (110) planes
ph ph ph ph
(ii) (110) plane, surface density,
ph ph ph

is: ph


1 a 5.63 2 −2
= 
14 p h
6.99 x10 cm
d = 2= = atoms p h ph
ph
ph ph

a ph 2 2
2
or (iii) (111) plane, surface density,
FH3 1 IK
ph ph ph



d = 3.98 ph ph ph ph ph + 31
ph ph ph

A ph


(c) Distance between nearest (111) planes 6 2 4
= =
p h ph
ph ph ph ph
ph 3
is: ph

1 a 5.63 2
d = ph 3= = ph a
a ph 3 3 2
3
15 −2
or or 1.14 x10 ph
p h
cm
d = 3.25 ph ph

A ph
1.15
1.14 (a)
(a) (100) plane of silicon – similar to a fcc,
ph ph ph ph ph ph ph ph

 2 atoms
Simple cubic: a = 4.50 A
ph



b g
ph ph ph ph ph
surface = 2
ph
−8 ph
(i) (100) plane, surface ph ph density
ph
5.43x10
density, ph  p h




1 atom −2


b4.50x10 g
=
14 14 p h
ph
2
4.94 x10 ph 6.78x10 cm
−8 ph p h −2
cm (b)
p h


(ii) (110) plane, surface density,
ph ph ph (110) plane, surface density,
ph ph ph
−2
cm−23.49 x10 = 
14 14
= 1 atom ph p h
p h ph 4 9.59 x10 ph ph ph
p h
cm
atoms ph




b4.50x10 g
2 ph
−8 ph
2



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

3 F Iatoms
ph ph ph




HK
1 1 (111) plane, surface
ph ph

density,
ph 14 −2
h
p
6 ph
2 1 4 atoms ph

= p h = = =  p h p h 7.83x10 p h p h cm

c(x)a h
1 2
2

2 1 a 3 3a
p h p h
a 2 
p h 2 2
2 1.16
1
=  2.85x10
14

−2 p h d = 4r 2 ph ph
cm
= a
4(2.25)
ph ph

(b) then
4r
Body-centered cubic ph
(ii) (110) plane,
(i) (100) plane, surface density, ph ph ph
2 2 surface density,
ph ph ph



14
Same as (a),(i); surface density 4.94x10
ph ph ph ph ph
p h −2
cm
5