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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```