An Introduction to Semiconductor Devices
Chapter 1
Solutions Manual
Problem Solutions
______________________________________________________________________________________
Chapter 1
Problem Solutions
(b) Face-centered cubic lattice
d
d = 4r = a 2 ⇒ a =
=2 2r
2
1.1
(a) fcc: 8 corner atoms × 1/8 = 1 atom
6 face atoms × ½
= 3 atoms
Total of 4 atoms per unit cell
h = 16 2 r
F 4πr IJ
4 atoms per cell, so atom vol. = 4G
H3K
3
(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
_______________________________________
Ratio =
b
5.65 x10
−8
g
⇒
3
22
−3
Density of As = 2.22 x10 cm
_______________________________________
3
1.3
3
8 Ge atoms per unit cell
8
⇒
Density =
−8 3
5.65 x10
g
(d) Diamond lattice
Body diagonal = d = 8r = a 3 ⇒ a =
−3
22
Density of Ge = 4.44 x10 cm
_______________________________________
Unit cell vol. = a
1.4
(a) Simple cubic lattice; a = 2 r
Unit cell vol = a = (2r ) = 8r
3
FG 4πr IJ
H3K
3
3
FG 4πr IJ
H3K
Then
FG 4πr IJ
H 3 K × 100% ⇒ Ratio = 34%
Ratio =
F 8r I
H 3K
3
8
3
3
F 8r I
=
H 3K
r
3
8 atoms per cell, so atom vol. 8
Then
8r
3
8
3
3
1 atom per cell, so atom vol. = (1)
FG 4πr IJ
H 3 K × 100% ⇒
Ratio =
3
3
4 As atoms per unit cell, so that
3
F 4 rI
H 3K
F 4πr IJ
2 atoms per cell, so atom vol. = 2G
H3K
Then
F 4πr IJ
2G
H 3 K × 100% ⇒ Ratio = 68%
Ratio =
F 4r I
H 3K
−3
22
Ratio = 74%
3
Unit cell vol. = a =
Density of Ga = 2.22 x10 cm
b
FG 4πr IJ
H 3 K × 100% ⇒
16 2 r
(c) Body-centered cubic lattice
4
d = 4r = a 3 ⇒ a =
r
3
4 Ga atoms per unit cell
3
3
3
4
1.2
4
3
3
Then
Density =
c
Unit cell vol = a = 2 2 r
3
Ratio = 52.4%
_______________________________________
2
An Introduction to Semiconductor Devices
Chapter 1
Solutions Manual
Problem Solutions
______________________________________________________________________________________
B-type: 1 atom per unit cell, so
1.5
We have 8r = a o 3
so that
r=
5.65 3
= 1.223 A
°
1.9
8
Simple cubic; a o = 2r = 4.2 A
Then
°
d = 2 r = 2.45 A
_______________________________________
bcc: 4 r = a o 3 ⇒ a o =
fcc: 4 r = a o 2 ⇒ a o =
1.6
From Problem 1.4, percent volume of fcc atoms
is 74%. Therefore after coffee is ground,
3
8
°
so that r =
a = 18
. + 1.0 ⇒ a = 2.8 A
°
=ρ=
N ( At . Wt .)
Na: Density =
°
ρ = 2.33 grams / cm
23
_______________________________________
ρ=
1.8
°
°
22
−3
−3
4.85x10
−23
b2.8x10 g
−8
3
⇒
3
1.11
. )=8 A
(a) a 3 = 2(2.2) + 2(18
so that
g
23
= 2.28 x10 cm
3
_______________________________________
so that rB = 0.747 A
(b) A-type; 1 atom per unit cell
1
⇒
Density =
−8 3
2.04 x10
Density(A) = 118
. x10 cm
b2.8x10 g
−8
ρ = 2.21 gm / cm
2 rA + 2 rB = a 3 ⇒ 2 rB = 2.04 3 − 2.04
b
12
(d)
Na: At.Wt. = 22.99
Cl: At. Wt. = 35.45
So, mass per unit cell
1
1
(22.99) + (35.45)
−23
2
= 2
= 4.85x10
23
6.02 x10
Then mass density is
3
(a) a = 2 rA = 2(1.02) = 2.04 A
Now
°
22
22
6.02 x10
3
Cl: Density (same as Na) = 2.28 x10 cm
b5x10 g(28.09) ⇒
=
NA
8(2.1)
(c)
g
(c) Mass density
2
1.10
(b)
r
(b) Number density
8
22
−3
=
⇒ Density = 5x10 cm
−8 3
5.43 x10
b
°
°
=
= 118
. A
8
8
Center of one silicon atom to center of nearest
neighbor = 2r ⇒ 2.36 A
= 5.94 A
°
a o = 9.70 A
_______________________________________
3
(5.43) 3
3
4(2.1)
= 4.85 A
or
1.7
(a) a = 5.43 A From 1.3d, a =
°
4(2.1)
diamond: 8r = a o 3 ⇒ a o =
Volume = 0.74 cm
_______________________________________
a 3
−3
23
. x10 cm
Density (b) = 118
_______________________________________
a = 4.62 A
−3
Density of A =
3
°
°
1
b4.62 x10 g
−8
⇒ 1.01x10 cm
22
3
−3
An Introduction to Semiconductor Devices
Chapter 1
Solutions Manual
Problem Solutions
______________________________________________________________________________________
Density of B =
1
b4.62 x10 g
⇒ 1.01x10 cm
22
−8
1.15
(a) Planes intercept at:
(i) p = 1, q = ∞ , s = ∞
(ii) p = 3, q = 1, s = ∞
(iii) p = 3, q = ∞ , s = 2
−3
(b) Same as (a)
(c) Same material
_______________________________________
1.12
(a) Surface density
1
1
= 2
=
−8
a 2
4.62 x10
b
14
3.31x10 cm
g
(b) Vector components:
(i) p = 1, q = 1, s = 0
(ii) p = 3, q = 1, s = 1
(iii) p = 1, q = 2 , s = 3
_______________________________________
⇒
2
2
−2
1.16
(a)
Same for A atoms and B atoms
(b) Same as (a)
(c) Same material
_______________________________________
(b)
1.13
1
(a) Vol density =
3
ao
F 1 , 1 , 1I ⇒ (313)
H 1 3 1K
F 1 , 1 , 1 I ⇒ (121)
H 4 2 4K
_______________________________________
1
Surface density =
2
1.17
2
ao
(a) d = a o = 5.25 A
(b) Same as (a)
_______________________________________
(b) d =
ao 2
°
= 3.71 A
°
2
1.14
(100) plane: Density =
No. atoms
ao
=
2
b5.43x10 g
−8
= 6.78 x10 cm
14
2
(110) plane: Density =
b5.43x10 g
−8
−2
(a) d = a o = 5.20 A
No. atoms
2
4
2
(b) d =
= 9.59 x10 cm
14
2
(111) plane: Density =
°
(5.20) 2
= 3.68 A
°
2
ao 3
c 3 2h
=
(5.20) 3
°
4(2.25)
°
= 3.00 A
3
3
_______________________________________
No. atoms
2
=
2
(c) d =
a1 6f ⋅ 3 + a1 2f ⋅ 3 = 7.83x10
=
b5.43x10 g c 3 2h
2
ao 2
−2
2
ao
−8
°
1.18
ao
=
d=
ao 3
= 3.03 A
3
_______________________________________
(c)
2
1.19
14
cm
−2
ao =
4r
=
3
3
(a) Density =
2
2
=
3
−8
ao
. x10
5196
(a) (110) = highest density
(b) (100) = lowest density
_______________________________________
b
4
g
.
= 5196
A
= 1.43x10 cm
22
3
−3
An Introduction to Semiconductor Devices
Chapter 1
Solutions Manual
Problem Solutions
______________________________________________________________________________________
(b) d =
ao 2
=
5196
.
2
2
2
(d) Surface density
2
2
= 2
=
−8
ao 2
. x10
5196
b
1.23
= 3.67 A
FθI = a 4 =
H 2 K ca 3 h 4
°
We find
o
cos
o
g
1
3
so that
2
θ
2
= 54.74°
2
or
or
−2
= 5.24 x10 cm
_______________________________________
θ = 109.5°
_______________________________________
1.20
1.24
14
−3
Density of silicon atoms = 5 x10 cm and 4
valence electrons per atom, so
22
(a) Ratio =
5x10
−3
23
Density of valence electrons = 2 x10 cm
_______________________________________
(b) Ratio =
16
−5
22
2 x10
⇒ 8 x10 %
15
−6
⇒ 4 x10 %
22
5x10
_______________________________________
1.21
Density of GaAs atoms
8 atoms
22
−3
=
= 4.44 x10 cm
−8 3
5.65 x10
b
4 x10
1.25
g
b5x10 g(30.98) ⇒
b5x10 g(28.06)
16
(a) Fraction by weight ≈
22
An average of 4 valence electrons per atom,
−4
−3
23
110
. x10 %
Density of valence electrons = 1.77 x10 cm
_______________________________________
(b) Fraction by weight
b10 g(10.82)
b5x10 g(30.98) + b5x10 g(28.06) ⇒
18
≈
1.22
Silver: Group 1 = 1 valence electron per atom
At.Wt. = 107.88, Mass density = 10.50 gm / cm
Mass per unit cell,
107.88
−22
=
= 1.79 x10
23
6.02 x10
Now
ρ=
1.79 x10
−22
F 1.79 x10 IJ
⇒a =G
H 10.50 K
−22
7.71x10 %
_______________________________________
1.26
d
1/ 3
−6
15
−3
°
°
d 794
d
=
⇒
= 146
ao 5.43
ao
3
_______________________________________
or
22
= 2 x10 cm
d = 7.94 x10 cm = 794 A
We have a O = 5.43 A
So
g
3
So
°
b
1
Volume density =
or
a o = 2.57 A
Then, density of valence electrons
1
=
−8
2.57 x10
22
−4
3
o
3
ao
16
−3
Density = 5.89 x10 cm
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5