Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 1
Problem Solutions
Then
1.1
(a) fcc: 8 corner atoms  1 / 8  1 atom

6 face atoms 1 / 2  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 1 / 2  3 atoms
4 enclosed atoms = 4 atoms
Total of 8 atoms per unit cell
_______________________________________
1.2
(a) Simple cubic lattice: a  2r
Unit cell vol  a 3  2r   8r 3
3
 4 r 3
1 atom per cell, so atom vol  1
 3
Then
 4 r 3 


 3 


Ratio 
100%  52.4%
8r 3
(b) Face-centered cubic lattice
d
d  4r  a 2  a 
 2 2 r
2

Unit cell vol  a 3  2 2  r

3

3
3
 4r 




 3
(d) Diamond lattice
3




100%  68%
8
Body diagonal  d  8r  a 3  a 
r
3
 8r 

Unit cell vol  a 3  

 3
3
 4 r 3 

8 atoms per cell, so atom vol  8

 3 
Then
3


8 4 r 
3 
Ratio  
 100%  34%
3
 8r 




 3
_______________________________________
1.3
o
(a) a  5.43 A ; From Problem 1.2d,
a
8
r
3
 16 2  r 3
 4 r
4 atoms per cell, so atom vol  4
 3
Then
3


4 4 r 
 3 
Ratio 
100%  74%
16 2  r 3
(c) Body-centered cubic lattice
4
d  4r  a 3  a 
r
3
 4

 r 
Unit cell vol  a  
 3 




Ratio 
2 4 r
3




o
nearest neighbor  2r  2.35 A
(b) Number density
8

 5 10 22 cm 3
3
5.43 10 8
(c) Mass density
N  At.Wt . 5 10 22 28.09


NA
6.02 10 23




   2.33 grams/cm 3
_______________________________________
3
3
 4 r 3
2 atoms per cell, so atom vol  2
 3
o
a 3 5.43 3

 1.176 A
8
8
Center of one silicon atom to center of
Then r 




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.4
(a) 4 Ga atoms per unit cell
4
Number density 
5.65 10 8

(b) a  21.035  2.07 A
o
(c) A-atoms: # of atoms  8 

3
Density 
 Density of Ga atoms  2.22 10 22 cm 3
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 
3
5.65 10 8

1.5
From Figure 1.15
 a  3 
 0.4330 a
(a) d   
 2  2 

 3.38 10 cm 3
_______________________________________
# of atoms  8 
o
a
2
2

  2
sin   

  54.74
a
2
3
2
 
3
2
   109.5
_______________________________________
1.7
(a) Simple cubic: a  2r  3.9 A
o
 5.515 A
 4.503 A
24r 
o
 9.007 A
3
_______________________________________
1.8
(a) 21.035 2  21.035  2rB
o
rB  0.4287 A
4.5 10 
8 3
1.0974 10 12.5
22
6.02 10 23
 0.228 gm/cm 3
(b) a 
4r
o
 5.196 A
3
1
# of atoms 8   1  2
8
2
5.196 10 
8 3
 1.4257 10 22 cm 3
1.4257 10 22 12.5
Mass density   
6.02 10 23
 0.296 gm/cm 3
_______________________________________

o
(d) diamond: a 

Number density 
o
1
 1.097 10 22 cm 3
N  At.Wt .
Mass density   
NA
1.6
3
1
1
8
Number density 
 0.70715.65  d  3.995 A
_______________________________________
(c) bcc: a 

23
o
a
(b) d    2  0.7071a
2
2
4r
8 3
(a) a  2r  4.5 A
 0.43305.65  d  2.447 A
4r
2.07 10 
1.9
o
(b) fcc: a 
1
 1.13 10 23 cm 3
1
B-atoms: # of atoms  6   3
2
3
Density 
3
2.07  10 8

 Density of Ge atoms  4.44 10 22 cm 3
_______________________________________
1
1
8

1.10
From Problem 1.2, percent volume of fcc
atoms is 74%; Therefore after coffee is
ground,
Volume = 0.74 cm 3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.11
o
o
(b) a  1.8  1.0  2.8 A
(c) Na: Density 
1 / 2
2.8 10 
8 3
 2.28 10 22 cm 3
Cl: Density  2.28 10 22 cm 3
(d) Na: At. Wt. = 22.99
Cl: At. Wt. = 35.45
So, mass per unit cell
1
1
 22.99   35.45
2
 
2

 4.85 10  23
6.02 10 23
Then mass density
4.85 10 23

 2.21 grams/cm 3
8 3
2.8 10
_______________________________________


1.12
(a) a 3  22.2  21.8  8 A
o
o
Then a  4.62 A
Density of A:
1

 1.0110 22 cm 3
3
4.62 10 8
Density of B:
1

 1.0110 22 cm 3
8 3
4.62 10
(b) Same as (a)
(c) Same material
_______________________________________


1.13
a


22.2  21.8


2
 4.687 1014 cm 2
o
For 1.12(b), B-atoms: a  4.619 A
1
 4.687  10 14 cm 2
a2
For 1.12(a) and (b), Same material
Surface density 
o
For 1.12(b), A-atoms; a  4.619 A
Surface density
1

 3.315 1014 cm 2
2
a 2
B-atoms;
Surface density
1

 3.315  10 14 cm 2
2
a 2
For 1.12(a) and (b), Same material
_______________________________________
1.14
(a) Vol. Density 
1
a o3
1
Surface Density 
a
2
o
2
(b) Same as (a)
_______________________________________
1.15
(i) (110) plane
(see Figure 1.10(b))
(ii) (111) plane
(see Figure 1.10(c))
o
 4.619 A
3
(a) For 1.12(a), A-atoms
1
1
Surface density  2 
a
4.619 10 8
(b) For 1.12(a), A-atoms; a  4.619 A
Surface density
1

 3.315 1014 cm 2
2
a 2
B-atoms;
Surface density
1

 3.315  10 14 cm 2
2
a 2
1 1 
(iii) (220) plane   , ,    1, 1, 0
2 2 
Same as (110) plane and [110] direction
 1 1 1
(iv) (321) plane   , ,   2, 3, 6
 3 2 1
Intercepts of plane at
p  2, q  3, s  6
[321] direction is perpendicular to
(321) plane
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.16
(a)

1 1 1
 , ,   313
1 3 1
(b)
1 1 1
 , ,   121
4 2 4
_______________________________________
1.17
1 1 1
Intercepts: 2, 4, 3   , ,  
 2 4 3
(634) plane
_______________________________________
1.18
o
(a) d  a  5.28 A
o
a 2
 3.734 A
2
o
a 3
 3.048 A
(c) d 
3
_______________________________________
(b) d 
1.19
(a) Simple cubic
(i) (100) plane:
Surface density 
1
1

2
a
4.73 10 8


2
(ii) (110) plane:
1
Surface density 
a
2
2
 3.16 1014 cm 2
(iii) (111) plane:
1
bh
2
o
where b  a 2  6.689 A
Now
 
h2  a 2
2
2
 
a 2
 3 a 2

 2 
4


o
6
4.73  5.793 A
So h 
2


 6.32 10 14 cm 2
(iii) (111) plane:
1
3
6
Surface density 
19.3755  10 16
 2.58 10 14 cm 2
(c) fcc
(i) (100) plane:
2
Surface density  2  8.94  10 14 cm 2
a
(ii) (110) plane:
2
Surface density  2
a 2
 6.32 10 14 cm 2
(iii) (111) plane:
1
1
3  3
6
2
Surface density 
19.3755  10 16
 1.03 1015 cm 2
_______________________________________
 4.47 1014 cm 2
Area of plane 
Area of plane
1
 6.68923 10 8 5.79304 10 8
2
 19.3755 10 16 cm 2
1
3
6
Surface density 
19.3755  10 16
 2.58 10 14 cm 2
(b) bcc
(i) (100) plane:
1
Surface density  2  4.47  10 14 cm 2
a
(ii) (110) plane:
2
Surface density  2
a 2
2
1.20
(a) (100) plane: - similar to a fcc:
2
Surface density 
2
5.43 10 8


 6.78 10 cm 2
14
(b) (110) plane:
Surface density 

4
2 5.43  10 8

2
 9.59 1014 cm 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) (111) plane:
Surface density 
2
 3 25.43 10 
8 2
 7.83 1014 cm 2
_______________________________________
1.21
a
4r

42.37 
o
 6.703 A
2
2
1
1
8  6
4
8
2 
3
(a) #/cm 
a3
6.703 10 8

5 1017
100%  10 3 %
22
5 10
2 1015
(b)
100%  4 10 6 %
5 10 22
_______________________________________

1.25
(a) Fraction by weight
2 1016 10.82

 1.542 10 7
22
5 10 28.06
(b) Fraction by weight
1018 30.98

 2.208 10 5
5 10 22 28.06
_______________________________________



 3.148 10 cm 2
14
o
a 2 6.703 2

 4.74 A
2
2
1
1
(d) # of atoms  3   3   2
6
2
Area of plane: (see Problem 1.19)
(c) d 


 

Volume density 
o
6a
h
 8.2099 A
2
Area
1
1
 bh  9.4786 10 8 8.2099 10 8
2
2
 3.8909 10 15 cm 2


1.26
o
b  a 2  9.4786 A


(a)
3
 1.328 10 cm
1
1
4  2
4
2
(b) #/cm 2 
a2 2
2

2
6.703  10 8
2


1.24
3
22
1.23
Density of GaAs atoms
8

 4.44  10 22 cm 3
8 3
5.65 10
An average of 4 valence electrons per atom,
So
Density of valence electrons
 1.77 10 23 cm 3
_______________________________________
1
 2  10 16 cm 3
d3
o
So d  3.684 10 6 cm  d  368.4 A

2
3.8909  10 15
= 5.14 1014 cm 2
#/cm 2 
o
a 3 6.703 3

 3.87 A
3
3
_______________________________________
d
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
_______________________________________
o
We have ao  5.43 A
d
368.4

 67.85
ao
5.43
_______________________________________
Then
1.27
Volume density 
1
 4  10 15 cm 3
d3
o
So d  6.30 10 6 cm  d  630 A
o
We have ao  5.43 A
d
630

 116
a o 5.43
_______________________________________
Then