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Semiconductor Physics Devices Basic Principles 3rd Ed. by Neamen

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Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1 
Solutions Manual Problem Solutions 
 
 3 
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 
(a) 4 Ga atoms per unit cell 
Density = ⇒
−
4
5 65 10 8
3
. xb g 
Density of Ga = −2.22 1022 3x cm 
 4 As atoms per unit cell, so that 
 Density of As = −2.22 1022 3x cm 
(b) 
 8 Ge atoms per unit cell 
 Density = ⇒
−
8
5 65 10 8
3
. xb g 
 Density of Ge = −4.44 1022 3x cm 
 
1.3 
(a) Simple cubic lattice; a r= 2 
 Unit cell vol = = =( )a r r3 3 32 8 
 1 atom per cell, so atom vol. = ( )FHG
I
KJ1
4
3
3
πr
 
Then 
 Ratio
r
r
= × ⇒
F
HG
I
KJ
4
3
8
100%
3
3
π
 Ratio = 52.4% 
(b) Face-centered cubic lattice 
 d r a= = ⇒4 2 a
d
r= =
2
2 2 
 Unit cell vol = = =a r r3
3 32 2 16 2c h 
 
 
 4 atoms per cell, so atom vol. =
F
HG
I
KJ4
4
3
3
πr
 
Then 
 Ratio
r
r
= × ⇒
F
HG
I
KJ4
4
3
16 2
100%
3
3
π
 Ratio = 74% 
(c) Body-centered cubic lattice 
 d r a a r= = ⇒ =4 3
4
3
 
 Unit cell vol. = = FH
I
Ka r
3
34
3
 
 2 atoms per cell, so atom vol. =
F
HG
I
KJ2
4
3
3
πr
 
Then 
 Ratio
r
r
= × ⇒
F
HG
I
KJ
F
H
I
K
2
4
3
4
3
100%
3
3
π
 Ratio = 68% 
(d) Diamond lattice 
 Body diagonal = = = ⇒ =d r a a r8 3
8
3
 
 Unit cell vol. = = FH
I
Ka
r3
38
3
 
 8 atoms per cell, so atom vol. 8
4
3
3
πrF
HG
I
KJ 
Then 
 Ratio
r
r
= × ⇒
F
HG
I
KJ
F
H
I
K
8
4
3
8
3
100%
3
3
π
 Ratio = 34% 
 
1.4 
From Problem 1.3, percent volume of fcc atoms 
is 74%; Therefore after coffee is ground, 
 Volume = 0 74 3. cm 
 
 
 
 
 
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1 
Solutions Manual Problem Solutions 
 
 4 
 
1.5 
(a) a A= °5 43. From 1.3d, a r=
8
3
 
so that r
a
A= = =
( )
°
3
8
5 43 3
8
118
.
. 
Center of one silicon atom to center of nearest 
neighbor = ⇒2r 2.36 A° 
(b) Number density 
 = ⇒
−
8
5 43 10 8
3
. xb g Density =
−5 1022 3x cm 
(c) Mass density 
 = = = ⇒
( ) ( )
ρ
N At Wt
N
x
xA
. . .
.
5 10 28 09
6 02 10
22
23
b g
 
 ρ = 2.33 3grams cm/ 
 
1.6 
(a) a r AA= = =( ) °2 2 1 02 2.04. 
Now 
 2 2 3 2 2.04 3 2.04r r a rA B B+ = ⇒ = − 
so that r AB =
°0 747. 
(b) A-type; 1 atom per unit cell 
 Density = ⇒
−
1
2.04 10 8
3
xb g 
 Density(A) =118 1023 3. x cm− 
 B-type: 1 atom per unit cell, so 
 Density(B) =118 1023 3. x cm− 
 
1.7 
(b) 
 a = + ⇒18 1 0. . a A= °2.8 
(c) 
Na: Density =
−
1 2
2.8 10 8
3
xb g =
−2.28 1022 3x cm 
Cl: Density (same as Na) = −2.28 1022 3x cm 
(d) 
Na: At.Wt. = 22.99 
Cl: At. Wt. = 35.45 
So, mass per unit cell 
 =
+
=
( ) ( )
−
1
2
22.99
1
2
35 45
6 02 10
4.85 10
23
23
.
. x
x 
Then mass density is 
 ρ = ⇒
−
−
4.85 10
2.8 10
23
8 3
x
xb g 
 ρ = 2.21 3gm cm/ 
 
1.8 
(a) a A3 2 2.2 2 18 8= + =( ) ( ) °. 
so that 
 a A= °4.62 
Density of A = ⇒
−
1
4.62 10 8
3
xb g 1 01 10
22 3. x cm− 
Density of B = ⇒
−
1
4.62 10 8xb g 1 01 10
22 3. x cm− 
(b) Same as (a) 
(c) Same material 
 
1.9 
(a) Surface density 
= = ⇒
−
1
2
1
4.62 10 2
2 8 2a xb g 
3 31 1014 2. x cm− 
 Same for A atoms and B atoms 
(b) Same as (a) 
(c) Same material 
 
1.10 
(a) Vol density =
1
3ao
 
 Surface density =
1
22ao
 
(b) Same as (a) 
 
1.11 
Sketch 
 
1.12 
(a) 
 
1
1
1
3
1
1
, ,FH IK ⇒ ( )313 
(b) 
 
1
4
1
2
1
4
, ,FH IK ⇒ 121( ) 
 
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1 
Solutions Manual Problem Solutions 
 
 5 
1.13 
(a) Distance between nearest (100) planes is: 
 d a A= = °5 63. 
(b)Distance between nearest (110) planes is: 
 d a
a
= = =
1
2
2
2
5 63
2
.
 
or 
 d A= °3 98. 
(c) Distance between nearest (111) planes is: 
 d a
a
= = =
1
3
3
3
5 63
3
.
 
or 
 d A= °3 25. 
 
1.14 
(a) 
Simple cubic: a A= °4.50 
(i) (100) plane, surface density, 
 = ⇒
−
1
4.50 10 8
2
atom
xb g 4.94 10
14 2x cm− 
(ii) (110) plane, surface density, 
 =
1
2 4.50 10 8
2
atom
x −
⇒b g 3 49 10
14 2. x cm− 
(iii) (111) plane, surface density, 
 = =
⋅ ⋅
=
FH IK
( )
3
1
6
1
2
2
1
2
1
2
2
3
2
1
3 2
atoms
a x a
a ac h
 
 = ⇒
−
1
3 4.50 10 8
2
xb g 2.85 10
14 2x cm− 
(b) 
Body-centered cubic 
(i) (100) plane, surface density, 
Same as (a),(i); surface density 4.94 1014 2x cm− 
(ii) (110) plane, surface density, 
 = ⇒
−
2
2 4.50 10 8
2
atoms
xb g 6 99 10
14 2. x cm− 
(iii) (111) plane, surface density, 
Same as (a),(iii), surface density 2.85 1014 2x cm− 
(c) 
Face centered cubic 
(i) (100) plane, surface density 
 = ⇒
−
2
4.50 10 8
2
atoms
xb g 9.88 10
14 2x cm− 
(ii) (110) plane, surface density, 
 = ⇒
−
2
2 4.50 10 8
2
atoms
xb g 6 99 10
14 2. x cm− 
(iii) (111) plane, surface density, 
 =
⋅ + ⋅
=
FH IK
−
3
1
6
3
1
2
3
2
4
3 4.50 102 8
2
a xb g
 
or 114 1015 2. x cm− 
 
1.15 
(a) 
(100) plane of silicon – similar to a fcc, 
surface density = ⇒
−
2
5 43 10 8
2
atoms
x.b g 
 6 78 1014 2. x cm− 
(b) 
(110) plane, surface density, 
 = ⇒
−
4
2 5 43 10 8
2
atoms
x.b g 9.59 10
14 2x cm− 
(c) 
(111) plane, surface density, 
 = ⇒
−
4
3 5 43 10 8
2
atoms
x.b g 7.83 10
14 2x cm− 
 
1.16 
 d r a= =4 2 
then 
 a
r
A= = =
( )
°
4
2
4 2.25
2
6 364. 
(a) 
 Volume Density = ⇒
−
4
6 364 10 8
3
atoms
x.b g 
 155 1022 3. x cm− 
(b) 
Distance between (110) planes, 
 = = = ⇒
1
2
2
2
6 364
2
a
a .
 
or 
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1 
Solutions Manual Problem Solutions 
 
 6 
 4.50 A° 
(c) 
Surface density 
 = =
−
2

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