Title | Semiconductor Physics and Devices 4th edition Neaman pdf |
---|---|
Author | 28-012-Md Ibrahim Shikder Mahin |
Course | Electric circuit |
Institution | Bangladesh University of Business and Technology |
Pages | 160 |
File Size | 3.4 MB |
File Type | |
Total Downloads | 16 |
Total Views | 206 |
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 1By D. A. Neamen Problem Solutions Chapter 1Problem Solutions1.(a) fcc: 8 corner atoms 8/1 1 atom6 face atoms 2/1 3 atomsTotal of 4 atoms per unit cell(b) bcc: 8 corner atoms 8/1 1 atom1 enclosed atom =1 atom Total of 2...
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
Chapter 1 Problem Solutions 1.1 (a) fcc: 8 corner atoms1 / 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 atoms1 / 2 3 atoms 4 enclosed atoms = 4 atoms Total of 8 atoms per unit ce ll _______________________________________ 1.2 (a) Simple cubic lattice: a 2r Unit cell vol a 3 2r 3 8r 3
4 r 3 1 atom per cell, so atom vol 1 3 Then 4 r 3 3 Ratio 100 % 52.4% 3 8r (b) Face-centered cubic lattice d d 4r a 2 a 2 2 r 2
Unit cell vol a 3 2 2 r
3
4 r Unit cell vol a 3
Ratio
4r 3 (d) Diamond lattice
3
100% 68%
8
Body diagonal d 8 r a 3 a 8r 3 Unit cell vol a 3
r
3
3
4 r 3 8 atoms per cell, so atom vol 8 3 Then 4 r 3 8 3 Ratio 100% 34% 3 8r 3 _______________________________________
o
(a) a 5.43 A ; From Problem 1.2d, a
8
r
3
o a 3 5.43 3 1.176 A 8 8 Center of one silicon atom to center of
Then r
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
4 r 3 3
2
1.3
16 2 r 3
4 r 3 4 atoms per cell, so atom vol 4 3 Then 4 3 4 r 3 100% 74% Ratio 16 2 r 3 (c) Body-centered cubic lattice 4 d 4r a 3 a r 3
Then
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 8 5.65 10
o
(b) a 21.035 2.07 A (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 8 3 5.65 10
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 o
(a) Simple cubic: a 2r 3.9 A
3
o
5.515 A 4.503 A 2 4r
o
9.007 A
3 _______________________________________
1.8 (a)
21.035 2 21.035 2rB o
rB 0.4287 A
4.510
8 3
1. 0974 10 12.5 23 6.02 10 0.228 gm/cm3
(b) a
4r 3
22
o
5.196 A
1 # of atoms 8 1 2 8
Number density
2
5.196 10
8 3
1.4257 1022 cm 3 22 1. 4257 10 12.5 Mass density 23 6.02 10 3 0.296 gm/cm _______________________________________
o
(d) diamond: a
1
1.097 10 22 cm 3 N At.Wt . Mass density NA
1.6
(c) bcc: a
1 1 8
Number density
0.70715.65 d 3.995 A _______________________________________
2
23
o
a (b) d 2 0.7071a 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 8 3 2.07 10
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
2.810
8 3
2.28 10 22 cm 3 3 22 Cl: Density 2.28 10 cm (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 23 4.85 10 2.21 grams/cm 3 3 2.8 10 8 _______________________________________
1.12 o
(a) a 3 22.2 21.8 8 A o
Then a 4.62 A Density of A: 1 1.0110 22 cm 3 8 3 4.62 10 Density of B: 1 1.0110 22 cm 3 3 4.62 10 8 (b) Same as (a) (c) Same material _______________________________________ 1.13
22.2 21.8
2
4.687 10 14 cm 2 o
For 1.12(b), B-atoms: a 4.619 A 1 4.687 1014 cm 2 2 a For 1.12(a) and (b), Same material
Surface density
o
For 1.12(b), A-atoms; a 4.619 A Surface density 1 2 3.315 10 14 cm 2 a 2 B-atoms; Surface density 1 2 3.315 1014 cm 2 a 2 For 1.12(a) and (b), Same material _______________________________________ 1.14 (a) Vol. Density
1 ao3
Surface Density
1 2 o
a 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 a
(b) For 1.12(a), A-atoms; a 4.619 A Surface density 1 2 3.315 10 14 cm 2 a 2 B-atoms; Surface density 1 3.315 1014 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 (c) d 3.048 A 3 _______________________________________
(b) d
1.19 (a) Simple cubic (i) (100) plane: Surface density
1 1 2 a 4.73 10 8
2
4.47 10 14 cm 2 1 2
a 2 3.16 1014 cm 2 (iii) (111) plane: 1 Area of plane bh 2 o
where b a 2 6.689 A Now a 2 h a 2 2 2
2
2
3 a 2 4
o 6 So h 4.73 5.793 A 2
6.32 1014 cm 2 (iii) (111) plane: 1 3 6 Surface density 19.3755 10 16 2.58 1014 cm 2 (c) fcc (i) (100) plane: 2 14 Surface density 2 8.94 10 cm 2 a (ii) (110) plane: 2 Surface density 2 a 2 6.32 1014 cm 2 (iii) (111) plane: 1 1 3 3 6 2 Surface density 19.3755 10 16 1.03 10 15 cm 2 _______________________________________
(ii) (110) plane: Surface density
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 1014 cm 2 (b) bcc (i) (100) plane: 1 Surface density 2 4.47 1014 cm 2 a (ii) (110) plane: 2 Surface density a2 2
2
1.20 (a) (100) plane: - similar to a fcc: 2 Surface density 8 2 5.43 10
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
o 4r 4 2.37 6.703 A 2 2 1 1 8 6 4 8 2 (a) #/cm 3 3 a 6.703 10 8
a
1.328 10 cm 1 1 4 2 2 4 (b) #/cm 2 a2 2 22
8 2
5 10 17 100% 10 3 % 5 10 22 2 10 15 100 % 4 10 6 % (b) 5 10 22 _______________________________________
3
1.25 (a) Fraction by weight 2 1016 10.82 1.542 10 7 5 1022 28.06 (b) Fraction by weight 18 10 30.98 2.208 10 5 22 5 10 28.06 _______________________________________
2
3.148 10 cm o a 2 6.703 2 4.74 A (c) d 2 2 1 1 (d) # of atoms 3 3 2 6 2 Area of plane: (see Problem 1.19)
o
#/cm 2
14
1 2 10 16 cm 3 3 d o
6 So d 3.684 10 cm d 368.4 A
2 3.8909 10 15
= 5.14 10
Volume density
o 6a 8.2099 A 2
Area 1 1 8 8 bh 9.4786 10 8.2099 10 2 2 15 2 3.8909 10 cm
1.26
b a 2 9.4786 A h
(a)
3
2
14
1.24
2
6.703 10
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 1023 cm 3 _______________________________________
cm 2
o a 3 6.703 3 3.87 A 3 3 _______________________________________
d
1.22 Density of silicon atoms 510 22 cm 3 and 4 valence electrons per atom, so 23 Density of valence electrons 2 10 cm 3 _______________________________________
o
We have ao 5.43 A 368.4 d 67.85 5.43 ao _______________________________________
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 630 d 116 ao 5.43 _______________________________________
Then
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________
Chapter 2 2.1 Sketch _______________________________________ 2.2 Sketch _______________________________________ 2.3 Sketch _______________________________________ 2.4 From Problem 2.2, phase
2 x
t
= constant Then dx 2 dx 0, p dt dt 2 2 x From Problem 2.3, phase t = constant Then dx 2 dx 0, p dt dt 2 _______________________________________
2.6
6.625 1034 550 10 9 1.205 10 27 kg-m/s p 1.2045 1027 1.32 103 m/s m 9.11 10 31 5 or 1.32 10 cm/s h 6.625 10 34 (b) p 440 10 9 1.506 10 27 kg-m/s p 1.5057 10 27 1.65 103 m/s m 9.11 10 31 5 or 1.65 10 cm/s (c) Yes _______________________________________ (a) p
h
2.7 (a) (i)
p 2mE 2 9.11 10 31 1.2 1.6 10 19 25
5.915 10 kg-m/s h 6.625 10 34 1.12 109 m p 5.915 1025 o
or 11.2 A
(ii) p 2 9.1110 31 12 1.6 10 19
2.5 hc
hc E h E Gold: E 4.90 eV 4.90 1.6 10 19 J So, 34 10 6.625 10 3 10 2.54 10 5 cm 19 4.90 1.6 10 or 0.254 m
Cesium: E 1.90 eV 1.90 1.6 10 19 J So, 6.625 10 34 3 1010 6.54 10 5 cm 1.90 1.6 10 19 or 0.654 m _______________________________________
24
1.87 10 kg-m/s 34 6.625 10 3.54 10 10 m 24 1.8704 10 o
or 3.54 A
(iii) p 2 9.1110 31 120 1.6 10 19 24
5.915 10 kg-m/s 34 6.625 10 1.12 1010 m 5.915 10 24 o
or 1.12 A
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)
p 2 1.67 10
27
2.10
1.2 1.6 10 19
2.532 10 23 kg-m/s 6.625 10 34 2.62 10 11 m 2.532 10 23 o
or 0.262 A _______________________________________
or
3 3 kT 0. 0259 0.03885 eV 2 2
or
Now
pavg 2mEavg
(b)
2 9.1110 31 0.03885 1.6 10 19
or
or
pavg 1.064 10 25 kg-m/s
h
2.8
E avg
6.625 10 34 10 85 10 26 7.794 10 kg-m/s p 7.794 10 26 8.56 104 m/s m 9.11 10 31 8.56 106 cm/s 1 1 E m 2 9.11 10 31 8.56 10 4 2 2 3.33 10 21 J 3.334 1021 E 2.08 10 2 eV 19 1.6 10 1 2 E 9.11 10 31 8 10 3 2 2.915 10 23 J 2.915 1023 E 1.82 10 4 eV 19 1.6 10 31 8 10 3 p m 9.1110
(a) p
Now
h 6.625 10 34 6.225 10 9 m p 1.064 10 25
2
7.288 10 27 kg-m/s h 6.625 10 35 9.09 108 m p 7.288 10 27
or o
62.25 A
o
_______________________________________
or 909 A _______________________________________
2.9
E p h
p
2.11
hc
p
(a) E h
Now pe2 1 h h Ee and pe 2m e 2m e Set E p Ee and p 10 e Then Ee
1 h p 2m e hc
2
1 10h 2m p
which yields 100 h p 2 mc
Ep E
hc
p
2 hc 2 mc 2 mc 100h 100
2
31 8 2 9.11 10 3 10 100 1.64 10 15 J 10.25 keV _______________________________________
1.99 10
2
2
hc
6.625 10 3 10 34
8
1 10 10
15
J
Now
E 1.99 10 15 19 e 1.6 10 4 V 1.24 10 V 12.4 kV E e V V
31 1.99 10 15 (b) p 2mE 2 9.1110
23
6.02 10 kg-m/s Then h 6.625 10 34 1.10 1011 m p 6.02 10 23 or o
0.11 A _______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.12
1.054 10 34 p 6 x 10 28 1.054 10 kg-m/s _______________________________________ 2.13 (a) (i) px
1.054 10 34 8.783 10 26 kg-m/s 10 12 10 2 dE d p (ii) E p p dp dp 2m 2p pp p 2m m p
Now p 2mE
2 9 10 31 16 1.6 10 19 24
2.147 10 kg-m/s 24 26 2.1466 10 8.783 10 so E 31 9 10 19 2.095 10 J 19 2.095 10 or E 1.31 eV 19 1.6 10 (b) (i) p 8.783 10 26 kg-m/s
(ii) p 2 5 10 28 16 1.6 10 19
23
5.06 10 kg-m/s 23 26 5.06 10 8.783 10 E 28 5 10 21 8.888 10 J 8.888 10 21 or E 5.55 10 2 eV 19 1.6 10 _______________________________________
2.14
1.054 10 34 1.054 10 32 kg-m/s x 10 2 p 1.054 10 32 p m m 1500 7 10 36 m/s _______________________________________ p
2.15 (a) Et 1.054 10 34 8.23 10 16 s t 0.8 1.6 10 19
1.054 10 34 x 1.5 1010 7.03 10 25 kg-m/s _______________________________________ (b) p
2.16 (a) If 1x, t and 2 x, t are solutions to Schrodinger's wave equation, then
x, t 2 2 1 x, t V x1 x , t j 1 2m t x 2 and 2 2 x , t 2 2 x, t V x 2 x, t j 2 t 2m x Adding the two equations, we obtain 2 2 1 x ,t 2 x ,t 2 m x 2 V x 1 x , t 2 x , t 1 x, t 2 x, t t which is Schrodinger's wave equation. So 1 x, t 2 x, t is also a solution. j
(b) If 1 x , t 2 x, t were a solution to Schrodinger's wave equation, then we could write 2 2 1 2 V x1 2 2 m x 2 j 1 2 t which can be written as 2 2 2 1 2 2 2 1 1 2 2 2 x x 2m x x 1 2 2 V x 1 2 j 1 t t Dividing by 1 2 , we find
2 2m
2 1 2 2 1 1 2 ...