Title | Semiconductor Physics and Devices - Donald A. Neamen 3rd Edition Solutions |
---|---|
Author | Musical Mayhem |
Course | Electronic Devices |
Institution | Birla Institute of Technology and Science, Pilani |
Pages | 188 |
File Size | 4.3 MB |
File Type | |
Total Downloads | 11 |
Total Views | 157 |
Solutions to the book Semiconductor Physics and Devices - Donald A. Neamen 3rd Edition...
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 1 Problem Solutions
Chapter 1 Problem Solutions
GHF 4π3r JKI 3
4 atoms per cell, so atom vol. = 4
1.1 (a) fcc: 8 corner atoms × 1/8 = 1 atom 6 face atoms × ½ = 3 atoms Total of 4 atoms per unit cell
Then
FG 4πr IJ H 3 K × 100% ⇒ 3
4 Ratio =
(b) bcc: 8 corner atoms × 1/8 = 1 atom 1 enclosed atom = 1 atom Total of 2 atoms per unit cell
Ratio = 74%
3
16 2 r (c) Body-centered cubic lattice 4 d = 4r = a 3 ⇒ a = r 3
(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
3
Unit cell vol. = a =
F 4 rI H3K
3
GHF 4π3r JKI 3
2 atoms per cell, so atom vol. = 2
1.2 (a) 4 Ga atoms per unit cell Density =
Then
4
b
5.65 x10
−8
g
3
FG 4πr IJ H 3 K × 100% ⇒ Ratio = F 4r I H 3K
⇒
3
2
Density of Ga = 2.22 x1022 cm
−3
4 As atoms per unit cell, so that 22 −3 Density of As = 2.22 x10 cm
(d) Diamond lattice
(b)
Body diagonal = d = 8r = a 3 ⇒ a =
8 Ge atoms per unit cell 8 Density = ⇒ −8 3 5.65x10
b
Ratio = 68%
3
g
Unit cell vol. = a 3 =
22
− Density of Ge = 4.44 x10 cm
3
F 8r I H 3K
8
r
3 3
FG 4πr JI H 3K 3
8 atoms per cell, so atom vol. 8 1.3 (a) Simple cubic lattice; a = 2 r
Then
Unit cell vol = a = ( 2r) = 8r 3
3
FG 4πr IJ H 3 K × 100% ⇒ Ratio = F 8r I H 3K 3
3
8
FG 4πr IJ H3K 3
1 atom per cell, so atom vol. = (1)
3
Then
FG 4πr IJ H 3 K × 100% ⇒ Ratio =
Ratio = 34%
3
1.4 From Problem 1.3, percent volume of fcc atoms is 74%; Therefore after coffee is ground, 3 Volume = 0.74 cm
Ratio = 52.4% 3 8r (b) Face-centered cubic lattice d d = 4r = a 2 ⇒ a = =2 2r 2
c
3 Unit cell vol = a = 2 2 r
h
3
3 = 16 2 r
3
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 1 Problem Solutions
Then mass density is 1.5 (a) a = 5.43 A
°
8
From 1.3d, a =
ρ= r
3 a 3
neighbor = 2r ⇒ 2.36 A
= ρ=
N ( At. Wt. )
a = 4.62 A
1.9 (a) Surface density 1 1 = 2 = a 2 4.62 x10 −8
°
b
2 rA + 2 rB = a 3 ⇒ 2 rB = 2.04 3 − 2.04 °
so that r B = 0.747 A (b) A-type; 1 atom per unit cell 1 Density = ⇒ −8 3 2.04 x10
14
23
. x 10 cm Density(B) = 118
1.10 (a) Vol density =
−3
1 3
ao Surface density =
1.7 (b)
1 2
ao
b2.8 x10 g −8
°
22
3
2
(b) Same as (a)
(c) Na: Density =
g
2
⇒ 2
−2
−3
B-type: 1 atom per unit cell, so
12
−3 22 1.01x10 cm
Same for A atoms and B atoms (b) Same as (a) (c) Same material
Density(A) = 118 . x 10 cm
a = 18 . + 1.0 ⇒ a = 2.8 A
−3 22 ⇒ 1.01x10 cm
3.31x10 cm
g
23
3
(b) Same as (a) (c) Same material
1.6
b
−8
−8
3
(a) a = 2r A = 2(1.02) = 2.04 A Now
1
b 4.62x10 g 1 Density of B = b 4.62x 10 g ⇒
6. 02 x1023
ρ = 2.33 grams / cm
3
°
Density of A =
b 5x10 g (28.09) ⇒
NA
⇒
(a) a 3 = 2 (2.2 ) + 2(1.8) = 8 A° so that
22
=
3
−8
1.8
°
g
(c) Mass density
b2.8x10 g
°
(b) Number density 8 −3 22 = ⇒ Density = 5 x10 cm −8 3 5.43x10
b
23
ρ = 2.21 gm / cm
( 5. 43) 3
= = 118 . A 8 8 Center of one silicon atom to center of nearest so that r =
4.85x 10−
= 2.28 x10 cm
1.11 Sketch
−3
1.12 (a)
−
Cl: Density (same as Na) = 2.28 x1022 cm 3 (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
(b)
4
F 1 , 1 , 1I ⇒ (313) H 1 3 1K FH 1 , 1 , 1 IK ⇒ (121) 4 2 4
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 1.13 (a) Distance between nearest (100) planes is: d = a = 5.63 A
=
°
d = 3.98 A
(iii)
or
(iii)
F KI H c h b
3 4.50 x10 −8
1 3a
2
g
2
14
cm −2
14
(iii)
g
. x1015 cm−2 114
g
−2
b
g
b
g
d = 4r = a 2 then 4r 4(2.25) a= = = 6.364 A ° 2 2 (a) 4 atoms Volume Density = −8 6 .364 x10
−2
b
(110) plane, surface density, 2 atoms −2 14 ⇒ 6 .99 x10 cm −8 2 2 4.50 x10
b
b
1.16 ⇒ 2.85 x10
Same as (a),(i); surface density 4.94x 10 cm
=
KI
(b) (110) plane, surface density, 4 atoms −2 14 = ⇒ 9.59 x10 cm −8 2 2 5.43x 10 (c) (111) plane, surface density, 4 atoms −2 14 = ⇒ 7.83 x10 cm −8 2 3 5.43x 10
(b) Body-centered cubic (i) (100) plane, surface density, (ii)
F H
6.78x 10 cm
g
1
g
b
(110) plane, surface density, 1 atom −2 14 ⇒ 3.49 x10 cm −8 2 2 4.50x 10
(111) plane, surface density, 1 1 atoms 3 6 2 = = = 1 1 a 3 a 2 ( x) ⋅a 2 ⋅ 2 2 2
=
b
14
g
b
(110) plane, surface density, 2 atoms −2 14 ⇒ 6.99 x10 cm 2 −8 2 4.50x 10
1.15 (a) (100) plane of silicon – similar to a fcc, 2 atoms surface density = ⇒ −8 2 5.43 x10
°
=
−2 14 ⇒ 9.88 x10 cm
°
Simple cubic: a = 4.50 A (i) (100) plane, surface density, 1 atom −2 14 = ⇒ 4.94 x10 cm −8 2 4.50 x10 (ii)
2
(111) plane, surface density, 1 1 3⋅ + 3 ⋅ 4 6 2 = = −8 2 3 2 3 4.50x 10 a 2
°
1.14 (a)
b
b4.50 x10 g −8
=
(c) Distance between nearest (111) planes is: 5.63 1 a = d = a 3 = 3 3 3 or d = 3.25 A
2 atoms
(ii)
(b)Distance between nearest (110) planes is: 5.63 1 a = d = a 2 = 2 2 2 or
Chapter 1 Problem Solutions
g
. x10 22 cm− 3 155 (b) Distance between (110) planes, a 6 .364 1 = a 2 = = ⇒ 2 2 2 or
(111) plane, surface density, 14
−2
Same as (a),(iii), surface density 2.85 x10 cm (c) Face centered cubic (i) (100) plane, surface density
5
g
3
⇒
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 4.50 A
Chapter 1 Problem Solutions
°
1.20
(c) Surface density 2 atoms = = 2 2a
b5 x10 g(30.98 ) ⇒ b5 x10 g(28.06) 16
(a) Fraction by weight ≈ 2
b
2 6. 364 x10
−8
g
2
−
. x10 6 110 (b) Fraction by weight
or 14
3.49 x10 cm
22
b10 g(10. 82) ≈ b5x10 g( 30.98) + b5x10 g( 28.06) ⇒ 18
−2
16
1.17
22
−
22
6 7.71x 10
−3
Density of silicon atoms = 5x 10 cm and 4 valence electrons per atom, so −3 23 Density of valence electrons 2x 10 cm
1.21 Volume density =
1 d
1.18 Density of GaAs atoms 8 atoms −3 22 = = 4.44x 10 cm −8 3 5.65x 10
b
−6 ° d = 7.94x 10 cm = 794 A °
We have aO = 5.43 A So d d 794 = 146 = ⇒ aO a O 5.43
An average of 4 valence electrons per atom, 23
Density of valence electrons 1. 77x 10 cm
−3
1.19 16
2 x10 5x 10
22
1x10
15
5x10
22
x100% ⇒
−5
4 x10 % (b) Percentage =
= 2 x10 cm
So
g
(a) Percentage =
15
3
x100 % ⇒
−
6 2 x10 %
6
−3
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 2 Problem Solutions
Chapter 2 Problem Solutions
p = 5.4 x10
2.1 Computer plot
λ=
2.2 Computer plot
h
−25
kg − m / s
− 34
=
p
6 .625 x10
− 25
⇒
5. 4 x 10
or
λ = 12.3 A
2.3 Computer plot
°
(ii) K.E. = T = 100 eV = 1.6x 10 2.4 For problem 2.2; Phase =
p=
2 πx
− ωt = constant
λ
λ=
For problem 2.3; Phase =
2 πx
λ
h
p = 5.4 x10
− 24
J
kg − m / s
⇒ λ = 1.23 A°
p
Then dx 2 π dx λ = v = +ω ⋅ − ω = 0 or dt 2π λ dt
FH KI
p
2mT ⇒
−17
− (b) Proton: K.E. = T = 1 eV = 1. 6 x10 19 J
p=
2 mT =
b
gb
2 1.67 x10 −27 1.6 x10 −19
or
+ ωt = constant
p = 2.31 x10 −23 kg − m/ s
Then dx 2 π dx λ ⋅ + ω = 0 or = vp = −ω dt 2π λ dt
F I H K
λ=
h p
=
6 .625 x10−34 2.31 x10
− 23
⇒
or
λ = 0.287 A
2.5 E = hν =
hc
λ
⇒λ =
hc E
b
. x10 Gold: E = 4 .90 eV = ( 4 .90) 16
So
− 19
g
(c) Tungsten Atom: At. Wt. = 183.92 − 19 For T = 1 eV = 1.6x 10 J J
p=
b 6.625x 10 gb 3x 10 g ⇒ 2.54x10 λ= (4.90)b1.6x10 g −34
2mT
b
2(183. 92) 1. 66x 10
=
10
−5
− 19
cm
b
Cesium: E = 1. 90 eV = (1. 90) 1. 6 x10 So
− 19
g
b 6.625x 10 gb 3x 10 g ⇒ 6.54x10 λ= (1.90)b 1.6x10 g −34
λ= J
h p
−5
6 .625 x10 3.13x10
−22
⇒
λ = 0.0212 A ° cm
(d) A 2000 kg traveling at 20 m/s: p = mv = ( 2000)( 20) ⇒ or
λ = 0.654 µm
p = 4 x 104 kg − m / s
2.6 (a) Electron: (i) K.E. = T = 1 eV = 1 .6 x10
b
− 34
=
or
10
or
2 mT =
gb1. 6x 10 g
. x10 −22 kg = m / s p = 313
− 19
p=
−27
or
or
λ = 0.254 µm
°
gb
2 9.11x10 −31 1.6 x10 −19
−19
λ=
J
g
h p
− 34
=
6 .625 x10 4
⇒
4 x10
or
λ = 1.66 x10
or
9
−28
A
°
−19
g
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 2 Problem Solutions
or
2.7 3
Eavg =
kT =
2
3
. x10 −3 eV E = 1.822 x10 −22 J ⇒ E = 114
(0.0259 ) ⇒
2
Also − 31 4 p = mv = 9.11x 10 2 x10 ⇒
b
or E avg = 0.01727 eV
p = 1822 x10 .
Now p avg =
b
g
b
2 9.11 x10 −31 (0 .01727 ) 1 .6 x10 −19
g
=
p
6 .625 x10 x10 1822 .
λ = 364 A pavg = 7.1 x10
−26
kg − m / s
− 34
=
p
=
λ
⇒
− 26
7.1x 10
⇒
°
− 34
h
p=
6. 625x 10
− 26
(b)
Now
λ=
kg − m / s
− 34
h
λ=
or
h
g
Now
2 mE avg
=
gb
−26
6 .625 x10 125x 10
⇒
−10
p = 5.3 x10 −26 kg − m / s
or
Also °
λ = 93.3 A
p
v= 2.8
=
m
5.3 x10 −26
4
9.11x 10− 31
= 5.82x 10 m / s
or
Ep = hν p =
hc
6
v = 5.82 x 10 cm / s
λp
Now
Now 2
Ee =
pe
2m
and pe =
h
λe
⇒ Ee =
FG h JI 2m H λ K 1
E=
2
hc
λp
FG h IJ = 2mH λ K
2
1
e
e
F 10hIJ = G 2mH λ K
=
b
2
b 9.11x 10 gb5.82x10 g 2 1
−31
2.10
2
1
(a) E = h ν =
p
hc
λ
b 6.625x10 gb 3x10 g 1x10 −15
8
− 10
J
Now hc
λp
hc
=
2
⋅ 2 mc =
100h −31
E = e ⋅ V ⇒ 1.99 x10
2 mc 100
gb3 x10 g 8
⇒
(b) p =
b
2mE =
2 9.11x10
= 6. 02x 10
J = 10. 3 keV
b9.11x10 gb2 x10 g 2 −31
−31
gb1.99 x10 g
kg − m / s
Then
2.9 1
g
3
λ= mv2 =
b
− 19 = 1.6 x10 V
V = 12.4 x 10 V = 12.4 kV
2
− 23
− 15
E = 1. 64 x 10
− 15
so
So
2
E = 9.64 x10 eV
− 34
=
E = 1. 99 x10
100
1
2
or
2 9.11 x10
(a) E =
4
−3
E = 1.54 x10− 21 J ⇒
which yields 100h λp = 2mc Ep = E =
2
mv =
or
Set E p = E e and λ p = 10λ e Then
1
4
2
10
h p
=
6. 625x 10
−34
− 23
6. 02x 10
⇒
° λ = 0.11 A
−15
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 2 Problem Solutions
2.11 h
(a) ∆p =
∆x
− 34
=
1. 054x 10
(b) ∆t =
⇒
−6
10
∆p = 1.054 x10 −28 kg − m / s
= hc
−20
2.12 1 .054 x10
(a) ∆p =
(b) ∆E =
1
⋅
(∆p )
2
=
m
2
∆E = 7.71 x10
1
5 x10
2
2
(∆p )2
⋅
m
∆E = 7.71x10
−26
=
1 2
b 8.78 x10 g ⋅ −26
5x 10
−26
kg − m / s
∂t
−26
2
∂
⋅
2
Ψ1( x , t) + Ψ2( x , t)
2
equation, then we could write
2
−h
⇒
2
2
∂
2m ∂x 2
aΨ ⋅ Ψ f +V (x )aΨ ⋅ Ψ f 1
2
1
2
−7
J ⇒ ∆E = 4.82 x10 eV
= jh
2.14 ∆p =
∂Ψ1(x , t )
∂ Ψ1 ( x, t ) + Ψ2 ( x, t ) ∂t which is Schrodinger’s wave equation. So Ψ1 ( x , t ) + Ψ2 ( x , t ) is also a solution. (b) If Ψ1 ⋅ Ψ2 were a solution to Schrodinger’s wave
2.13
1
+V (x )Ψ1 (x , t ) = j h
= jh
J ⇒ ∆E = 4.82 x10 −4 eV
(b)
∂x
2
+V ( x ) Ψ1 ( x, t ) + Ψ2 ( x, t )
− 29
(a) Same as 2.12 (a), ∆p = 8.78 x10
∂ Ψ1 ( x, t ) 2
⋅
2m ∂x
b8 .78x10 g ⇒
2 −23
−h
−26
⋅
s
∂Ψ2 ( x, t ) − h 2 ∂ 2Ψ2 (x , t ) + V ( x )Ψ2 ( x, t ) = jh ⋅ 2 ∂x 2m ∂t Adding the two equations, we obtain
−34
= ⇒ − 10 12 x10 ∆x ∆p = 8. 78 x10 −26 kg − m / s
∆E =
2
−h
J ⇒ ∆E = 0.198 eV
2m and h
−16
Schrodinger’s wave equation, then
or ∆E = 3.16 x10
g⇒
2.16 (a) If Ψ1 ( x , t ) and Ψ2 ( x , t ) are solutions to
− 28
8
b
(1) 1.6 x10 −19
∆t = 6.6 x10
F pI = pc H hK λ So ∆E = c( ∆p) = b3 x10 gb1.054 x10 g ⇒ hc
−34
or
(b)
E=
1.054 x10
∂ ∂t
aΨ ⋅Ψ f 1
2
which can be written as
h ∆x
=
1. 054x 10−34 10 −2
p = mv ⇒ ∆v =
∆p
=
m
= 1.054 x10 1.054 x10 −32 1500
LMΨ ∂ Ψ + Ψ ∂ Ψ + 2 ∂Ψ ⋅ ∂Ψ PO 2m N ∂x ∂x Q ∂x ∂x LM ∂Ψ + Ψ ∂Ψ PO +V ( x)Ψ ⋅ Ψ = j h Ψ N ∂ t ∂t Q Dividing by Ψ ⋅ Ψ we find 1 ∂ Ψ 1 ∂Ψ ∂Ψ O −h L 1 ∂ Ψ + ⋅ ⋅ + M P Ψ Ψ ∂ x ∂x Q 2 m N Ψ ∂x Ψ ∂x L 1 ∂Ψ + 1 ∂Ψ PO +V (x ) = jh M NΨ ∂t Ψ ∂x Q −h
− 32
⇒
−36
1
2
1
2
2
1
2
1
m/s
2
2
2
...