半导体物理与器件第四版课后习题答案1
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Chapter 1
Problem Solutions
1.1
1/ 8 1 atom
(a)fcc: 8 corner atoms
6 face atoms
1 /
2
3 atoms
Total of 4 atoms per unit cell
(b)bcc: 8 corner atoms 1 / 81atom
1 enclosed atom
=1 atom
Total of 2 atoms per unit cell
(c)Diamond: 8 corner atoms 1 / 8 1atom
6 face
atoms 1 / 2 3 atoms
atoms= 4 atoms
4 enclosed
Total of 8 atoms per unit cell
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1.2
a2r
(a)Simple cubic lattice:
Unit cell vol a 32r 38r 3
1 atom per cell, so atom vol
4 r 3 1
3
Then
4r 3
Ratio
3
100% 52.4% 8r3
(b)Face-centered cubic lattice
d 4r a 2a
d
2 2 r 2
3
Unit cell vol a 3 2 2 r 16 2 r 34 atoms per cell, so atom vol
4r 3
4
3
Then
4r 3
43
Ratio100% 74%
16 2 r 3
(c)Body-centered cubic lattice
4
d 4r a 3a r
3
3
Unit cell vol a 3
4
r
3
2 atoms per cell, so atom vol
4r 3
2
3
Then
4r 3
23
Ratio3100% 68%
4r
3
(d)Diamond lattice
Body diagonal
d 8r a 3a
8
r
3
3
Unit cell vol a 38r
3
8 atoms per cell, so atom vol
4r 3
8
3 Then
4r 3
8
3
Ratio100 %34%
3
8r
3
_______________________________________ 1.3
o
(a) a 5.43 A ; From Problem 1.2d,
8
r
a
3
a 3 5.433o
Then r 1.176 A
88
Center of one silicon atom to center of
nearest neighbor
o
2r 2.35 A
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(b) Number density
8
8
35
10 22 cm
3
a 2
2
5.43 10
sin
2 54.74
3
(c)
Mass density
2
a 2
3
N At.W t.
5 10 22 28.09
2
109.5
N A
6.02 10 23
grams/cm 3
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2.33
1.7
_______________________________________
o
(a) Simple cubic:
a 2r 3.9 A
4r o
(b)
fcc:
5.515 A
a
2
1.4 (a)
4 Ga atoms per unit cell
4
Number density
5.65 10 8 3
Density of Ga atoms
22
3
2.22 10 cm
4 As atoms per unit cell Density of As atoms
2. 22 10 22 cm 3
(b) 8 Ge atoms per unit cell
8
Number density
8
3
5.65 10
Density of Ge atoms
4. 44 10 22 cm 3
_______________________________________ 1.5
From Figure 1.15
(a)
a 3
d
0.4330 a
2
2
o
0.4330 5.65
d
2.447 A
(b)
a 2 0.7071 a
d
2
o
0.7071 5.65 d 3.995 A
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1.6
4r
o
(c) bcc:
a
4.503 A
3
2 4r
o
(d) diamond:
a
9.007 A
3
_______________________________________ 1.8
(a)
2 1.035
2 2 1.035 2r B
o
r B
0.4287 A
o
(b)
a
2 1.035 2.07 A
(c)
A-atoms:
# of atoms
8 1 1
8
Density
1
2.07
8 3
10
1.13 10 23 cm
3
B-atoms: # of atoms 6 1
3
2
Density
3
2.07 10 8 3
3.38 10 23 cm 3
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1.9
o
a 2r
(a)
4.5 A
# of atoms
8 1 1
8
Number density
1
4.5 10 8 3
1.097 10 22 cm 3
By D. A. Neamen Problem Solutions
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Mass density N At .W t. 4.85 10 23
N A 2.8 8
3 2.21
10
3
22
grams/cm
1.0974 10 1
2.5
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6.02 10 23
1.12
0.228gm/cm
3
4r o
(b)
5.196 A a
3
# of atoms
8
1
1 2
8
Number density
2
5.196 10 8 3
1.4257 10 22
cm 3
1022
Mass density
1.4257 1
2.5
6.02 10
23
0.296 gm/cm 3
_______________________________________ 1.10
From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is
o
(a)
a 3
2 2.2 2 1.8
8 A
o
Then
a 4.62 A
Density of A:
4.62 1 8 3 1.01 10 22 cm 3
10
Density of B:
1
1.01 10 22 cm
3
4.62 10
8 3
(b) Same as (a) (c)
Same material _______________________________________ 1.13
2 2.2 2 1.8
o 4.619 A
a 3
(a) For 1.12(a), A-atoms
Surface density
1 1 ground, Volume = 0.74 cm
3
a 2 4.619 10
8 2
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1.11
o
(b) a 1.8 1.0 2.8 A
1 / 2
(c)
Na:
Density
8
3
2.8 10
4.687 10 14 cm 2
o
For 1.12(b), B-atoms:
a 4.619 A
Surface density
1
4.687 1014
cm
2
a 2
For 1.12(a) and (b), Same material
2.28 10 22 cm 3
Cl: Density
3
(d)
Na: At. Wt. = 22.99 Cl: At. Wt. = 35.45
So, mass per unit cell
o
(b) For 1.12(a), A-atoms; a 4.619 A
Surface density
1 22.99 1 35.45 1
2
2
4.85 10
23
6.02 3.315 10 14
cm
2
1023
a 2
2
Then mass density
B-atoms;
Surface density
2.28 10 22 cm
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(a)
1 3.31510 14 cm2
a 22
o For 1.12(b), A-atoms; a 4.619 A
Surface density
1 3.31510 14 cm2
a 22
B-atoms;
Surface density
1 3.31510 14 cm2
a 22
For 1.12(a) and (b), Same material _______________________________________
1,1,1313
131
(b)
111
121
4
,,
24
_______________________________________ 1.17
Intercepts: 2, 4, 3 1 , 1,
1
243
(634)plane
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1.18
o
(a) d a 5.28 A
1.14
(a)Vol. Density 1
a2o
(b) 3.734 A
d2
1.16
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Surface density
2
8 2
5.43 10
Area of plane
1 6.68923 10 8 5.79304 10 8
2 10 16 cm 2
19.3755
3 1
6
Surface density
16
19.3755 10
2.58 1014 cm 2
(b) bcc
(i) (100) plane:
Surface density
1 4.47 1014 cm 2
(ii) (110) plane: a 2
Surface density
2
a 2
2
6.32 1014
cm 2
(iii) (111) plane:
3
1
6
Surface density
16
19.3755 10
2.58 1014
cm
2 (c) fcc
(i) (100) plane:
Surface density
2 8.94 1014
cm 2
(ii) (110) plane:
a 2
Surface density
2
a 2
2
6.32 1014 cm 2 (iii) (111) plane:
3
1 3 1 Surface density
6 2 16
19.3755 10
6.78 10 14 cm 2
(b) (110) plane:
4
Surface density
8
2
2 5.4
3 10
9.59 10 14 cm 2
(c)
(111) plane:
Surface density
2
2 5.4
3 10 8 2
3
7.83 10 14 cm 2
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1.21
4r 4 2.37
o
6.703 A
a
2 2
8
1 6
1
8 2
4
(a)
#/cm
3
3
3
a
6.703 10
8
1.328 10 22 cm 3
4
1 2 1
4 2
(b)
#/cm 2
a
2
2
2
6.703 10
8
2
2
3.148 10 14 cm
2
(c)
d
a 2
6.703 2
o
2
2 4.74 A
(d)
# of atoms
3
1 3
1
2
6
2
Area of plane:
(see Problem 1.19)
o
1.03 1015
cm
2
_______________________________________ 1.20
(a)
(100) plane: - similar to a fcc:
b a 2
9.4786 A
6a
o
h
8.2099 A
2
Area
1 bh 1 9.4786 10 8 8.2099 10 8 2
2
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3.8909 10 15 cm 2 Volume density 1 2 16 3
d 3 10 cm
o
10 6
#/cm
2
2
So d
3.684
cm
d 368.4 A
10 15
o
3.8909
We have
a o 5.43 A
= 5.14 10 14 cm 2
d
368.4 67.85
a 3
6.703 3
o
Then
5.43
a o
d
3
3.87 A
3
_______________________________________ _______________________________________ 1.27
1.22
Density of silicon atoms
5 10 22 cm
3
and
4 valence electrons per atom,
so Density of valence electrons 2 10 23 cm 3
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Volume density
1 4 1015 cm 3
d 3
o
6 cm
So d
6.30 10
d 630 A
o
We have a o 5.43 A
Then
d
630 116
5.43
a o
1.23
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Density of GaAs atoms
8
8 3 4.44 10 22 cm 3 5.65 10
An average of 4 valence electrons per atom,
So
Density of valence electrons 1.77 10 23 cm 3
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1.24 1017
(a)
5 100% 10 3%
5 1022
15
(b) 5 1022 100% 4 10 6
%
210
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1.25 (a)
Fraction by weight
2 10 16 10.82 1.542 10 7
5 10 22
28.06
(b)
Fraction by weight
10 18 30.98
2.208 10
5
5 10 22
28.06
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1.26