(Neamen)solution manual for semiconductor physics and devices 3ed

Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions
Chapter 1
3
Problem Solutions
1.1
(a) fcc: 8 corner atoms × 1/8 = 1 atom
6 face atoms × ½ = 3 atoms
Total of 4 atoms per unit cell
(b) bcc: 8 corner atoms × 1/8 = 1 atom
1 enclosed atom = 1 atom
(c) Diamond: 8 corner atoms × 1/8 = 1 atom
6 face atoms × ½ = 3 atoms
4 enclosed atoms = 4 atoms
1.2
(a) 4 Ga atoms per unit cell
Density = ⇒
−
4
5 65 10 8 3
b . x g
Density of Ga = − 2.22 1022 3 x cm
4 As atoms per unit cell, so that
Density of As = − 2.22 1022 3 x cm
(b)
8 Ge atoms per unit cell
Density = ⇒
−
8
565 10 8 3 b . x g
Density of Ge = − 4.44 1022 3 x cm
J
IK GFH 1.3
(a) Simple cubic lattice; a = 2r
Unit cell vol = a 3 = ( 2 r) 3 = 8
r 3 4
πr
3 1 atom per cell, so atom vol. = ( 1
)3
Then
Ratio
r
π
= × ⇒
r
FH G
IK J
4
3
8
100%
3
3
Ratio = 52.4%
(b) Face-centered cubic lattice
d
d = 4r = a 2 ⇒ a
= = r
2
2 2
Unit cell vol = a = r = r 3 3 3 c2 2 h 16 2
4 atoms per cell, so atom vol. = FH G
IK J
4
3 πr
4
3
Then
Ratio
r
4
3
16 2
= × ⇒
r
FH G
IK J
4
100%
3
3
π
Ratio = 74%
(c) Body-centered cubic lattice
d = 4r = a 3 ⇒ a = r
4
3
Unit cell vol. = =FH
IK
3 4
3
a r 3
2 atoms per cell, so atom vol. = FH G
IK J2
3 πr
4
3
Then
Ratio
r
= × ⇒
r
FH G
IK J
FH
IK
2
4
3
4
3
100%
3
3
π
Ratio = 68%
(d) Diamond lattice
Body diagonal = d = 8r = a 3 ⇒ a = r
8
3
Unit cell vol. = =FH
IK
a
3 8
3
r 3
3 πr FH G
8 atoms per cell, so atom vol. 8
4
3
IK J
Then
Ratio
r
= × ⇒
r
FH G
IK J
FH
IK
8
4
3
8
3
100%
3
3
π
Ratio = 34%
1.4
From Problem 1.3, percent volume of fcc atoms
is 74%; Therefore after coffee is ground,
Volume = 0 74 3 . cm

4

1.5
(a) a = A° 5.43 From 1.3d, a = r
8
3
so that r
( ) ° 3
8
a
= = = A
543 3
8
118
.
.
Center of one silicon atom to center of nearest
neighbor = 2r ⇒ 2.36 A°
(b) Number density
= ⇒
−
8
543 10 8 3 b . x g Density = − 5 1022 3 x cm
(c) Mass density
N ( AtWt
. . ) ( .
)
= ρ
= = ⇒
N
x
5 10 2809
6 02 10
.
x A
22
23
b g
ρ = 2.33 3 grams / cm
1.6
(a) a r A A = = ( ) = ° 2 2 1.02 2.04
Now
2r 2r a 3 2r 2.04 3 2.04 A B B + = ⇒ = −
so that r A B = ° 0.747
(b) A-type; 1 atom per unit cell
Density = ⇒
−
1
2.04 10 8 3 b x g
Density(A) =118 1023 3 . x cm−
B-type: 1 atom per unit cell, so
Density(B) =118 1023 3 . x cm−
1.7
(b)
a = 1.8 + 1.0⇒ a = A° 2.8
(c)
Na: Density =
1 2
b g = 2.28 x 1022 cm
− 3 2.8 x 10 −
8 3 Cl: Density (same as Na) = − 2.28 1022 3 x cm
(d)
Na: At.Wt. = 22.99
Cl: At. Wt. = 35.45
So, mass per unit cell
=
+
=
( ) ( )
−
1
2
22.99
1
2
35 45
6 02 10
4.85 10 23
23
.
. x
x
Then mass density is
−
ρ= ⇒
4.85 x
10
b 2.8 x 10
−
g
ρ = 2.21 gm / cm
3 23
8 3
1.8
(a) a 3 = 2(2.2) + 2(18) = 8 A° .
so that
a = A° 4.62
Density of A = ⇒
−
1
4.62 10 8 3 b x g 101 1022 3 . x cm−
1
4.62 10 8 b x g 101 1022 3 . x cm−
Density of B = ⇒ −
(b) Same as (a)
(c) Same material
1.9
(a) Surface density
= = ⇒
−
1
2
1
4.62 10 2 2 8 2 a b x g
331 1014 2 . x cm−
Same for A atoms and B atoms
(b) Same as (a)
(c) Same material
1.10
(a) Vol density =
1
3 ao
Surface density =
1
2 2 ao
(b) Same as (a)
1.11
Sketch
1.12
(a)
1
1
1
3
1
1
, , FH
IK
⇒ (313)
(b)
1
4
1
2
1
4
, , FH
IK
⇒ (121)

5
1.13
(a) Distance between nearest (100) planes is:
d = a = A° 5.63
(b)Distance between nearest (110) planes is:
a
d a
1
2
= = =
2
2
563
.
2
or
d = A° 3.98
(c) Distance between nearest (111) planes is:
a
d a
1
3
= = =
3
3
563
.
3
or
d = A° 3.25
1.14
(a)
Simple cubic: a = A° 4.50
(i) (100) plane, surface density,
= ⇒
1
atom
b g 4.94 x 1014 cm−
2 4.50 x 10 −
8 2
(ii) (110) plane, surface density,
=
atom
x −
1
2 4.50 10 8 2
⇒ b g 3 49 1014 2 . x cm−
(iii) (111) plane, surface density,
1
6
1
2
= =
⋅ ⋅
=
FH
IK
( )
3
2
1
2
1
2
2
3
2
1
3 2
atoms
a x a
c h a a
= ⇒
−
1
3 4.50 10 8 2 b x g 2.85 1014 2 x cm−
(b)
Body-centered cubic
(i) (100) plane, surface density,
Same as (a),(i); surface density 4.94 1014 2 x cm−
= ⇒
atoms
b x g 6 99 1014 2 . x cm−
2
2 4.50 10 −
8 2
Same as (a),(iii), surface density 2.85 1014 2 x cm−
(c)
Face centered cubic
(i) (100) plane, surface density
2
atoms
b g 9.88 x 1014 cm−
2 4.50 x 10 8 2
= ⇒
−
= ⇒
atoms
b x g 6 99 1014 2 . x cm−
2
2 4.50 10 −
8 2
=
⋅ + ⋅
=
FH
IK
b a x −
g
3
1
6
3
1
2
3
2
4
2 3 4.50 10
8 2
or 114 1015 2 . x cm−
1.15
(a)
(100) plane of silicon – similar to a fcc,
surface density = ⇒
atoms
b . x g
2
543 10 −
8 2
6 78 1014 2 . x cm−
(b)
(110) plane, surface density,
= ⇒
atoms
b . x g 9.59 1014 2 x cm−
4
2 5 43 10 −
8 2
(c)
(111) plane, surface density,
= ⇒
atoms
b . x g 7.83 1014 2 x cm−
4
3 5 43 10 −
8 2
1.16
d = 4r = a 2
then
a
4
r
4 ( 2.25
) ° 2
= = = A
2
6.364
(a)
Volume Density = ⇒
atoms
b . x g
4
6 364 10 −
8 3
155 1022 3 . x cm−
(b)
Distance between (110) planes,
= = = ⇒
1
2
2
2
6 364
2
a
a .
or

6
4.50 A°
(c)
Surface density
= =
atoms
a b . x −
g
2
2
2
2 6364 10 2 8 2
or
3 49 1014 2 . x cm−
1.17
Density of silicon atoms = − 5 1022 3 x cm and 4
valence electrons per atom, so
Density of valence electrons 2 1023 3 x cm−
1.18
Density of GaAs atoms
= =
22 3 atoms
x
8
4.44 x 10
cm
− 565 10
−
8 3
b . g
An average of 4 valence electrons per atom,
Density of valence electrons 1 77 1023 3 . x cm−
1.19
(a) Percentage = ⇒
2 10
5 10
100%
16
22
x
x
x
4 10 5 x − %
1 10
5 10
(b) Percentage = 100
⇒
15
22
x
x
x %
2 10 6 x − %

1.20
(a) Fraction by weight ≈ ⇒
b g
b g
( )
( )
16
5 10 3098
5 10 22
2806
x
x
.
.
110 10 6 . x −
(b) Fraction by weight
≈
b 18
g
10 10 .
82
b 16 g b 22
g
x . +
x .
7.71 x 10 −
6 ⇒
( )
( ) ( )
5 10 30 98 5 10 28 06
1.21
Volume density = = − 1
2 10 3
15 3
d
x cm
So
d = x cm = A − ° 7.94 10 794 6
We have a A O = ° 5.43
So
d
aO
794
5.43
= ⇒
d
aO
= 146

Chapter 2
9
Problem Solutions
2.1 Computer plot
2.2 Computer plot
2.3 Computer plot
2.4
For problem 2.2; Phase = − =
2π
λ
ω
x
t constant
Then
2
0
π
λ
dx
dt
⋅ −ω =
or
dx
dt
= v
=p
FH
+ IK
ω
λ
2π
2π
λ
For problem 2.3; Phase = + ω
=
x
t constant
Then
2
0
π
λ
dx
dt
⋅ +ω =
or
dx
dt
vp = =− FH
IK
ω
λ
2π
2.5
E h
hc hc
E
= ν = ⇒ =
λ
λ
Gold: E = 4 . 90 eV = ( 4 . 90 ) b 1 . 6 x 10 − 19 g
J So
b x −
34 gb x
10
g
λ= ⇒
6 625 10 3 10
4.90 1 6 10
b g 2.54 10 5 x cm −
( ) −
19
.
.
x
or
λ = 0.254 μm
Cesium: E = eV = ( ) x − J 190 190 16 10 19 . . b . g
So
λ= ⇒
b −
gb g
x cm − 6 625 10 3 10
190 16 10
5 .
. .
b g
( ) −
654 10
34 10
19
.
x x
x
or
λ = 0.654 μm
2.6
(a) Electron: (i) K.E. = = = − T 1 eV 1 6x10 J 19 .
p = mT = x x − − 2 2 9.11 10 1 6 10 b 31gb . 19 g
or
p = x kg − m s − 54 10 25 . /
−
λ= = ⇒
−
h
p
x
x
6 625 10
54 10
34
25
.
.
or
λ = ° 12.3 A
(ii) K.E. = = = − T 100 eV 1 6x10 J 17 .
p = 2mT ⇒ p = x kg − m s − 54 10 24 . /
h
p
λ= ⇒
λ = ° 1.23 A
(b) Proton: K.E. = = = − T 1 eV 1 6x10 J 19 .
p = mT = x x − − 2 216710 1610 27 19 b . gb . g
or
p = x kg − m s − 2.31 10 23 /
−
λ= = ⇒
−
h
p
x
x
6 625 10
2.31 10
34
23
.
or
λ = ° 0.287 A
(c) Tungsten Atom: At. Wt. = 183.92
For T = eV = x J − 1 161019 .
p = 2mT
= ( ) − − 2 18392 166 10 16 10 27 19 . b . x gb . x g
or
p = x kg = m s − 313 10 22 . /
−
λ= = ⇒
−
h
p
x
x
6 625 10
313 10
34
22
.
.
or
λ = ° 0.0212 A
(d) A 2000 kg traveling at 20 m/s:
p = mv = (2000)(20)⇒
or
p = 4x10 kg − m s 4 /
34
h
6 625 x
10
− p
4 x
10
λ= = ⇒
4
.
or
λ = − ° 166 10 28 . x A

10

2.7
Eavg = kT = ( )⇒ 3
2
3
2
0.0259
or
E eV avg = 0.01727
Now
p mE avg avg = 2
= 2 9.11 10− (0 01727) 1 6 10− 31 19 b x g . b . x g
or
p x kg ms avg= − − 7.1 10 26 /
Now
−
λ= = ⇒
−
h
p
x
x
6 625 10
7.1 10
34
26
.
or
λ = ° 93 3 . A
2.8
E h
hc
p p
p
= ν =
λ
Now
E
2
2
p
= e
e
m and p
h
e
= ⇒
λ
e
E
m
h
e
e
= FH G
IK J
1
2
2
λ
Set E E p e = and λ λ p e = 10
Then
hc
1
h
= FH = 2
G
m
10 2 2
1
2
m
h
IK J
FH G
IK J
p e p λ λ λ
which yields
λ p
h
mc
=
100
2
E E
hc hc
= = = ⋅ =
h
mc
mc
p
λ 100
p
2
2
2
100
31 8 2 b x gb x g
− 2 9.11 10 3 10
= ⇒
100
So
E = x J = keV − 164 10 10 3 15 . .
2.9
(a) E = mv = x x − 1
2
9.11 10 2 10 2 b 31gb 4 g2
1
2
or
E = x J ⇒ − 1822 10 22 . E = x eV − 114 10 3 .
Also
p = mv = x x ⇒ − 9.11 10 2 10 b 31gb 4 g
p = x kg − m s − 1822 10 26 . /
Now
−
λ= = ⇒
−
h
p
x
x
6 625 10
1822 10
34
26
.
.
λ = ° 364 A
(b)
p
h x
6 625 10
125 10
= = ⇒
x
−
− λ
34
10
.
p = x kg − m s − 53 10 26 . /
Also
v
p
m
x
x
−
4 .
= = = x m s
−
53 10
9.11 10
582 10
26
31
. /
or
v = 5 82x10 cm s 6 . /
Now
E = mv = x x − 1
2
9.11 10 582 10 2 b 31 gb . 4 g2
1
2
or
E = x J ⇒ − 154 10 21 . E = x eV − 9.64 10 3
2.10
(a) E h
b . 34 gb 8
g
hc x x
x
= = =
−
− ν
λ
6 625 10 3 10
1 10
10
or
E = x J − 199 10 15 .
Now
E = e ⋅V ⇒ x = x V − − 199 10 16 10 15 19 . b . g
so
V = 12.4x10 V = 12.4 kV 3
(b) p = mE = x x − − 2 2 9.11 10 199 10 b 31 gb . 15 g
= − − 6 02 10 23 . x kg m / s
Then
−
λ= = ⇒
−
h
p
x
x
6 625 10
6 02 10
34
23
.
.
λ = ° 0.11 A

11
2.11
(a) Δ
h 1054 10
Δ
p
x
x
−
= = ⇒
−
10
34
6
.
Δp = x kg − m s − 1054 10 28 . /
(b)
E
FH
hc
= = hc
= pc p
h
IK
λ
So
ΔE = c(Δp) = x x − ⇒ 3 10 1054 10 b 8 gb . 28 g
or
ΔE = x J ⇒ − 316 10 20 . ΔE = 0.198 eV
2.12
(a) Δ
h 1054 10
Δ
p
x
x
x
−
= = ⇒
−
12 10
34
10
.
Δp = x kg − m s − 8 78 10 26 . /
(b)
Δ
b . g
2 26 2
( Δ
) −
E
p
m
x
x
= ⋅ = ⋅ ⇒
−
1
2
1
2
8 78 10
5 10
29
ΔE = x J ⇒ − 7.71 10 23 ΔE = x eV − 4.82 10 4
2.13
(a) Same as 2.12 (a), Δp = x kg − m s − 8 78 10 26 . /
(b)
Δ
b . g
2 26 2
( Δ
) −
E
p
m
x
x
= ⋅ = ⋅ ⇒
−
1
2
1
2
8 78 10
5 10
26
ΔE = x J ⇒ − 7.71 10 26 ΔE = x eV − 4.82 10 7
2.14
Δ
− h 1054 10
Δ
p
x
x
−
32 .
= = = x
−
10
1054 10
34
2
.
p mv v
p
m
.
x
32 −
Δ
Δ 1054 10
= ⇒ = = ⇒
1500
or
Δv = x m s − 7 10 36 /
2.15
(a) Δ
h 1054 10
Δ
p
x
x
−
= = ⇒
−
10
34
10
.
Δp = x kg − m s − 1054 10 24 . /
(b) Δt
x
x
−
1054 10
1 16 10
34
.
b . g
= ⇒
( ) −
19
or
Δt = x s − 6 6 10 16 .
2.16
(a) If Ψ1 (x, t) and Ψ2 (x, t) are solutions to
Schrodinger’s wave equation, then
−
h ∂
( ) ⋅
( ) ( ) ( )
∂
+ =
∂Ψ
∂
h
2 2
1
2 1
1
2m
x t
x
V x x t j
x t
t
Ψ
Ψ
,
,
,
and
−
h ∂
( ) ⋅
( ) ( ) ( )
∂
+ =
∂Ψ
∂
h
2 2
2
2 2
2
2m
x t
x
V x x t j
x t
t
Ψ
Ψ
,
,
,
Adding the two equations, we obtain
−
h2 2
⋅
( ) + ( ) ∂
∂
Ψ x, t Ψ x, t
+V(x) Ψ (x t) + Ψ (x t) 1 2 , ,
2 1 2 2m x
=
∂
∂
j ( ) + ( )
h Ψ x t Ψ x t 1 2 , ,
t
which is Schrodinger’s wave equation. So
Ψ Ψ 1 2 (x, t) + (x, t) is also a solution.
(b)
If Ψ Ψ 1 2 ⋅ were a solution to Schrodinger’s wave
equation, then we could write
− ∂
⋅ + ( ) ⋅ h2 2
2m ∂
x
2 1 2 1 2 aΨ Ψ f V x aΨ Ψ f
=
h Ψ Ψ 1 2 a f
∂
∂
j ⋅
t
which can be written as
− ∂
∂
+
∂
∂
+
∂Ψ
∂
⋅
∂Ψ
∂
LN M
OQ P
h2
Ψ
1
2
Ψ
Ψ
2
2 2
2
Ψ
1
2
1 2
2
2
m x x x x
+ ⋅ =
∂Ψ
∂
+
∂Ψ
∂
( ) LNM
OQP
V x j
Ψ Ψ Ψ Ψ 1 2 1
h
2
1 2
t t
Dividing by Ψ Ψ 1 2 ⋅ we find
−
⋅
∂
∂
+ ⋅
∂
∂
+
∂Ψ
∂
∂Ψ
∂
LN M
OQ P
h2
2
2
Ψ
2
2
1
2
1
2
1 2
1 2
2
1 1 1
m Ψ x Ψ
x x x
Ψ
ΨΨ
+ =
∂Ψ
∂
+
∂Ψ
∂
( ) LN M
OQ P
V x j
t x
h
1 1
2
2
1
1
Ψ Ψ

12
Since Ψ1 is a solution, then
−
⋅ ⋅
∂
∂
+ = ⋅ ⋅
∂Ψ
∂
Ψ
( ) h
h
2
1
2
1
2
1
1
2
1 1
m x
V x j
Ψ Ψ
t
Subtracting these last two equations, we are left
with
− ∂
∂
+
∂Ψ
∂
∂Ψ
∂
LN M
OQ P
h2
2
2
Ψ
2
2
1 2
1 2
2
1 2
m Ψ x ΨΨ
x x
=
∂Ψ
∂
j
t
h
1
2
2
Ψ
Since Ψ2 is also a solution, we may write
− ∂
∂
+ =
∂Ψ
∂
Ψ
( ) h
h
2
2
2
2
2
2
2
2
1 1
m x
V x j
Ψ Ψ
t
Subtracting these last two equations, we obtain
−
h2
⋅ ⋅
− ( ) = ∂Ψ
∂
⋅
∂Ψ
∂
1 2
1 2
2
2
0
m x x
V x
ΨΨ
This equation is not necessarily valid, which
means that ΨΨ 1 2 is, in general, not a solution to
Schrodinger’s wave equation.

2.17
Ψ(x, t) = A sin(πx) exp(− jωt)
+ z = = z 2
1
1
Ψ(x, t) dx A sin ( x)dx
+
−
1 π
−
1
2 2
1
or
+
LNM
⋅ − (2 ) = 1 1 1
2
A x x 2
1
1
4
OQP
−
π
sin π
which yields
A2 = 1 or A = +1, − 1, + j , − j
2.18
Ψ(x, t) = A sin(nπx) exp(− jωt)
+ + z = = z 2
0
1
Ψ(x, t) dx A sin (n x)dx
1
2 2
1 π
0
or
⋅ − (2 ) = 1 LNM
1 1
2
n x 2
A x
n
0
1
4
OQP
+
π
sin π
which yields
A 2 = 2 or
A = + 2 , − 2 , + j 2 , − j 2
2.19
Note that Ψ⋅Ψ =
∞z *
0
dx 1
Function has been normalized
(a) Now
P
a
x
a
dx
o o
ao
=
− FH G
IK J
LN M
OQ P
z 2
2
0
4
exp
=
− FH G
IK J
2 ao
z 2
exp
0
4
a
x
a
dx
o o
=
− − FH
IK
FH G
IK J
2
2
2
0
4
a
a x
o
a o
o
ao
exp
or
P
a
o
− a
G
FH o
= −
−
IK J
LN M
OQ P
1
2
4
exp 1 = −
− FH
IK
1
1
2
exp
which yields
P = 0.393
(b)
P
IK Jz 2
IK JFH G
a
x
a
dx
a
o
=
2
a o o
o
− FH G
2
4
exp
=
− FH GIK J
2 z 2
o
2
exp
4
a
x
a
dx
a
o a o
o
=
FH G
− − FHIK
IK J
2
2
2
2
4
a
a x
o
a o
a
o a
o
o
exp
or
P = − − −
− ( ) FH
IK
LNM
OQP
1 1
1
2
exp exp
which yields
P = 0.239
(c)
P
a
x
a
dx
o o
ao
=
− FH G
IK J
FH G
IK J
z 2
2
0
exp
=
− FH G
IK J
2 z 2
ao
exp =
0 a
x
a
dx
o o
− − FH
IK
FH G
IK J
2
a x
2
2
o
a
a o
o
0 ao
exp
or
P = −1 exp(−2) − 1
which yields
P = 0.865

13
2.20
(a) kx −ωt = constant
Then
k
dx
dt
−ω = 0⇒
dx
dt
+ = v
=ω
p k or
v
x
x
15 10
15 10
4 .
.
= =
10
m s p13
9
/
v cms p = 106 /
(b)
k
2 2 2
= ⇒ = =
15 109
k x
π
λ
λ
π π
.
or
λ = ° 41 9 . A
Also
p
h x
6 625 10
419 10
= = ⇒
x
−
− λ
34
10
.
.
or
p = x kg − m s − 158 10 25 . /
Now
E h
b 34 gb 8
g
hc x x
x
= = =
−
− ν
λ
6 625 10 3 10
419 10
10
.
.
or
E = x J ⇒ − 4.74 10 17 E = 2.96x10 eV 2
2.21
ψ (x) = Aexp − jbkx +ωtg
where
k
mE
=
2
h
=
b g . b . g
2 9.11 10 (0 015) 1 6 10
− −
31 19
x x
−
1054 10
34
x
.
or
k = x m− 6 27 108 1 .
Now
ω= =
b g
( ) −
−
E x
h x
0 015 16 10
1054 10
19
34
. .
.
or
ω = 2.28 1013 x rad / s

2.22
E
π b . g π
n
ma
x n
x x
= =
−
1054 10
b − gb −
g
h2 2 2
2
34 2 2 2
31 10 2 2
2 9.11 10 100 10
so
E = x n J 6 018 10− ( ) 22 2 .
or
E = x n eV 3 76 10− ( ) 3 2 .
Then
3 = 3 76 10− .
n = 1⇒ E x eV 1
2 = 150 10− .
n = 2 ⇒ E x eV 2
2 = 3 38 10− .
n = 3⇒ E x eV 3
2.23
(a) E
n
ma
=
h2 2 2
π
2 2
=
b 1054 . x 10
−
34 g
2 π
2 n
2
b x gb x
g
31 10 2
− −
2 9.11 10 12 10
= 4.81 10− ( ) 20 2 x n J
So
E x J 1
= 4.18 10 − 20 ⇒ E = 0.261
eV 1 E = 1 . 67 x 10− 19 J ⇒ E = 1.04
eV 2
2 (b)
E E h
hc hc
− = ν = ⇒ λ
=
2 1 λ
Δ
E or
λ =
b 34 gb 8
g
6 625 10 3 10
167 10 −
4.18 10
⇒
−
− −
19 20
.
.
x x
x x
λ = − 159 10 6 . x m
or
λ = 1.59 μm
2.24
(a) For the infinite potential well
E
n
ma
π 2
n
ma E
h
= ⇒ =
h
2 2 2
2
2
2
2
2 π
2 so
n
b − 5 gb − 2 g 2 b −
2
g
b . g
56 2 10 10 10
1054 10
x 2
= = 182 10
x
34 2 2
−
.
π
or

14
n = 1 35x1028 .
(b)
ΔE
= (n + ) − n h2 2
ma
2
2 2
2
1
π
= ( + ) h2 2
2 2
2 1
π
ma
n
or
ΔE
b 1 . 054 x 10 34 g 2 2 (2) b 1 . 35 x
10
28
g
=
−
π
b gb g
5 2 2
− −
2 10 10
ΔE = x J − 148 10 30 .
Energy in the (n+1) state is 1 48 10 30 . x − Joules
larger than 10 mJ.
(c)
Quantum effects would not be observable.
2.25
For a neutron and n = 1:
E
b g
b gb g
x
x 1
h 2 π 2
. π
ma
2
34 2
1054 10
27 14 2 2
2 1 66 10 10
= =
−
− −
.
or
6 = 2.06 10
E x eV 1
For an electron in the same potential well:
E
b . g
b gb g
x
x 1
34 2 2
π
1054 10
2 9.11 10 10
31 14 2
=
−
− −
or
9 = 3.76 10
E x eV 1
2.26
Schrodinger’s wave equation
∂
∂
( ) 2
( ( )) ( )
+ − =
2
2 2
0
ψ
ψ
x
x
m
E V x x
h
We know that
ψ (x) = 0 for x
a
≥
2
and x
a
≤
−
2
V(x) = 0 for
−
a +
x
≤ ≤
a
2 2
so in this region
∂
∂
( ) 2
( )
+ =
2
2 2
0
ψ
ψ
x
x
mE
x
h
Solution is of the form
ψ (x) = Acos Kx + B sin Kx
where K
mE
=
2
2 h
Boundary conditions:
ψ (x) = 0 at x
a
x
a
=
+
=
−
,
2 2
So, first mode:
ψ 1 (x) = Acos Kx
where K
a
=
π
so E
2 2
ma 1
2 2
=
π h
Second mode:
ψ 2 (x) = B sin Kx
where K
a
=
2π
so E
2 2
ma 2
2
4
2
=
π h
Third mode:
ψ 3 (x) = Acos Kx
where K
a
=
3π
so E
2 2
ma 3
2
9
2
=
π h
Fourth mode:
ψ 4 (x) = B sin Kx
where K
a
=
4π
so E
2 2
ma 4
2
16
2
=
π h
2.27
The 3-D wave equation in cartesian coordinates,
for V(x,y,z) = 0
∂
2 ( ) ( ) ( )
ψ x y z ψ ψ
∂
+
∂
∂
+
∂
∂
2
2
2
2
2
x
x y z
y
x y z
z
, , , , , ,
mE
2
+ ( x y z
) = 0 2
h
ψ , ,
Use separation of variables, so let
ψ (x, y, z) = X(x)Y(y)Z(z)
Substituting into the wave equation, we get
YZ
X
x
XZ
Y
y
XY
Z
z
mE
XYZ
∂
∂
+
∂
∂
+
∂
∂
2
+ =
2
2
2
2
2
2 2
0
h
Dividing by XYZ and letting k
mE 2
2
2
=
h
, we
obtain
(1)
∂
∂
1 1 1
0
2
2
2
2
2
2
2
X
X
x Y
Y
y Z
Z
z
⋅ + ⋅
k
∂
∂
+ ⋅
∂
∂
+ =
We may set

15
1 2
2
2
X
X
x
∂
∂
kx ⋅
= − so
∂
∂
+ =
2
2
2 0
X
x
k X x
Solution is of the form
X(x) = Asinakx xf+ B cosakx xf
Boundary conditions: X(0) = 0⇒ B = 0
and X x a k
n
a x
π
( = ) = 0⇒ = x
where nx = 1, 2 , 3,...
Similarly, let
1 2
2
2
Y
Y
y
∂
∂
ky ⋅
= − and
1 2
2
2
Z
Z
z
∂
∂
⋅
= −
kz Applying the boundary conditions, we find
k
n
a
π
, 1, 2 , 3,...
y
n y
y = =
k
n
a
π
, 1, 2 , 3,...
n z
= z
=
z From Equation (1) above, we have
−k 2 − k 2 − k 2 + k 2 = 0
x y z
or
k k k k
mE
2 2 2 2
x y z
2
2
+ + = =h
so that
E E
h2 2
nxnynz nx ny nz ⇒ = + +
ma
2
2 2 2
2
πb g
2.28
For the 2-dimensional infinite potential well:
( ) ( ) ( )
∂
∂
+
∂
∂
2
+ =
2
2
2
2 2
0
ψ ψ
ψ
x y
x
x y
y
mE
x y
, ,
,
h
Let ψ (x, y) = X(x)Y(y)
Then substituting,
Y
X
x
X
Y
y
mE
XY
∂
∂
+
∂
∂
2
+ =
2
2
2
2 2
0
h
Divide by XY
So
X
x Y
Y
y
∂
∂
∂
∂
1 1 2
0
2
2
2
mE
⋅
+ ⋅
+ =
X
2 h
2 Let
1 2
2
2
X
X
x
∂
∂
⋅
= −
kx or
∂
∂
+ =
2
2
2 0
X
x
k X x
Solution is of the form:
X A kx B kx x x = sina f+ cosa f
But X(x = 0) = 0⇒ B = 0
So
X A kxx = sina f
Also, X x a k a n x x ( = ) = 0⇒ = π
Where nx = 1, 2 , 3, . . .
So that k
n
a x
π
= x
We can also define
1 2
2
2
Y
Y
y
∂
∂
⋅
= −
ky Solution is of the form
Y C k y D k y y y = sinb g+ cosb g
But
Y(y = 0) = 0⇒ D = 0
and
Y y b k b n y y ( = ) = 0⇒ = π
so that
k
n
b y
π
y =
Now
mE
−k 2 − k 2
+ =
x y
2
2
0
h
which yields
x y ⇒ = + FH G
E E
π
n
n
nxny
m
a
b IK J
h2 2 2
2
2
2 2
Similarities: energy is quantized
Difference: now a function of 2 integers
2.29
(a) Derivation of energy levels exactly the same
as in the text.
(b) ΔE
h2 2
2
= n − n
ma
2 2
2
1
2
π b g
For n = 2 , n = 1
2 1 Then
h π
ΔE
ma
=
3
2
2 2
2

16
(i) a = A° 4
ΔE
b x
−
34 g
2 2
b gb g
3 1054 10
2 1 67 10 4 10
= ⇒
27 10 2
− −
x x
.
.
π
ΔE = x eV − 385 10 3 .
(ii) a = 0.5 cm
ΔE
b x
−
34 g
2 2
b gb g
3 1054 .
10
2 167 . 10 0 .
5 10
= ⇒
27 2 2
− −
x x
π
ΔE = x eV − 2.46 10 17
2.30
(a) For region II, x > 0
∂
∂
( ) 2
( )
+ − =
2
2
2 2 2
0
ψ
ψ
x
x
m
E V x O
h
a f
General form of the solution is
ψ(x) = A expajK xf+ B expa− jK xf
2 2 2 2 2 where
2
m
K
= E −
V2 2 O
h
a f
Term with B2 represents incident wave, and
term with A2 represents the reflected wave.
Region I, x < 0
∂
∂
( ) 2
( )
+ =
2
1
2 2 1
0
ψ
ψ
x
x
mE
x
h
The general solution is of the form
ψ(x) = A expa jK xf+ B expa− jK xf
1 1 1 1 1 where
2
mE
K
=
1 2
h
Term involving B1 represents the transmitted
wave, and the term involving A1 represents the
reflected wave; but if a particle is transmitted
into region I, it will not be reflected so that
A1 = 0 .
Then
ψ1 1 1 (x) = B expa− jK xf
ψ2 2 2 2 2 (x) = A expajK xf+ B expa− jK xf
(b)
(1) ψ ψ 1 2 (x = 0) = (x = 0)
(2)
∂
∂
( ) ( )
ψ ψ 1
=
∂
∂
= =
0
2
0
x
x
x
x x x
Applying the boundary conditions to the
solutions, we find
B A B 1 2 2 = +
K A K B K B 2 2 2 2 1 1 − =−
Combining these two equations, we find
A
K −
K
K +
K
B 2
=
2 1
2 2 1
FH G
IK J
and B
K
K K
2
B 1
2 1
2
2
=
+
FH G
IK J
The reflection coefficient is
R
A A
B B
*
* R
= 2 2 ⇒
2 2
K K
K K
=
−
+
FH G
IK J
2 1
2 1
2
The transmission coefficient is
T = 1− R ⇒ T
K K
K K
=
4 1 2
a +
f2
1 2
2.31
In region II, x > 0 , we have
ψ2 2 2 (x) = A expa−K xf
where
K
a − f
h
m V E O
2
=
2 2
For V eV O = 2.4 and E = 2.1 eV
K
b x |W|
UV 31 g ( RS |T| ) b x
19
g
b g
x 2
34 2
1 2
2 9.11 10 2.4 2.1 16 10
1054 10
=
− − −
−
.
.
/
or
K x m 2
9 1 = 2.81 10 −
Probability at x compared to x = 0, given by
P
( x
)
( )
ψ
ψ
expa 2 f
= 2
= − K x
2
2
2 0
(a) For x = A° 12
P = − x x ⇒ − exp 2 2.81 10 12 10 b 9 gb 10 g
P = x = − 118 10 0118 3 . . %
(b) For x = 48 A°
P = − x x ⇒ − exp 2 2.81 10 48 10 b 9 gb 10 g
P = x − 19 10 10 . %
2.32
For V eV E eV O= 6 , = 2.2
We have that

17
T
E
E
= − − V
V
G
FH K a
IK J
FH G
O O
IK J
16 1 2 2 expa f
where
K
a − f
h
m V E O
2
=
2 2
=
( ) RS |T|
b − 31 g − b −
19
g
b −
g
UV |W|
2 9.11 10 6 2.2 16 10
1054 10
34 2
1 2
x x
x
.
.
/
or
9 1 = 9.98 10 −
K x m 2
For a = m − 10 10
T = − − x FH
IK
FH
IK
2.2
6
16
1
− 2.2
6
2 9.98 10 10 9 10 exp b gb g
or
T = 0.50
For a = m − 10 9
T = x − 7.97 10 9
2.33
Assume that Equation [2.62] is valid:
T
E
E
= − − V
V
G
FH K a
IK J
FH G
O O
IK J
16 1 2 2 expa f
(a) For m mo = (0.067)
K
a − f
h
m V E O
2
=
2 2
=
− ( ) ( ) RS |T|
UV |W|
b − 31 g b −
19
g
b g
2 0067 9.11 10 08 02 16 10
−
1054 10
34 2
1 2
. . . .
.
/
x x
x
or
9 1 = 1 027 10 − .
K x m 2
Then
T = − − x x FH
IK
FH
IK
0 2
08
16
1
− 0 2
08
.
.
2 1027 10 9 15 10 10 .
.
exp b . gb g
or
T = 0.138
(b) For m mo = (1.08)
K
b / x 31 g b x
19
g
b g
. . . .
x 2
34 2
1 2
2 1 08 9.11 10 0 8 0 2 1 6 10
1054 10
=
− ( ) ( ) RS |T|
UV |W|
− −
−
.
or
9 1 = 4.124 10 −
K x m 2
Then
T = − x x − 3 2 4.124 10 15 10 9 10 exp b gb g
or
T = x − 127 10 5 .
2.34
V x eV E x eV a m O= = = − 10 10 3 10 10 6 6 14 , ,
and m = x kg − 167 10 27 .
Now
K
a − f
h
m V E O
2
=
2 2
=
b − 27 g − b 6 gb −
19
g
b −
g
( ) RS |T|
UV |W|
2 167 10 10 3 10 16 10
1054 10
34 2
1 2
. .
.
/
x x
x
or
14 1 = 5 80 10 − .
K x m 2
So
T = − − x FH
IK
FH
IK
3
10
16
1
− 3
10
2 580 10 10 14 14 exp b . gb g
or
T = x − 306 10 5 .
2.35
Region I, V = 0 (x < 0) ; Region II,
V V x a O = (0 < < ) ; Region III, V = 0 (x > a) .
(a) Region I;
ψ1 1 1 1 1 (x) = A expa jK xf+ B expa− jK xf
(incident) (reflected)
Region II;
ψ2 2 2 2 2 (x) = A expaK xf+ B expa−K xf
Region III;
ψ3 3 1 3 1 (x) = A expajK xf+ B expa− jK xf
(b)
In region III, the B3 term represents a reflected
wave. However, once a particle is transmitted
into region III, there will not be a reflected wave
which means that B3 = 0 .
(c)
For x = 0 = ⇒ A + B = A + B 1 2 1 1 2 2 :ψ ψ
d
ψ ψ 1 2
dx
d
dx
jK A jK B K A K B
1 1 1 2 2 2 2 2 = ⇒ − = −
For x = a:ψ =ψ ⇒ 2 3
A Ka B Ka A jKa 2 2 2 2 3 1 expa f+ expa− f = expa f
And also

18
d
dx
d
dx
ψ ψ 2 = 3 ⇒
K A K a K B K a 2 2 2 2 2 2 expa f− expa− f
= jK A jK a 1 3 1 expa f
Transmission coefficient is defined as
T
A A
A A
*
= 3 3
*
1 1
so from the boundary conditions, we want to
solve for A3 in terms of A1 . Solving for A1 in
terms of A3 , we find
A
jA
K K
2
2
3
K K Ka Ka 1
2 2 4
1 2
2
1
=
+
lb − g expa f− expa− f
−2 + − 1 2 2 2 2 jK K expaK af expa K af sexpa jK af
We then find that
A A
A A
= a f lb 2
− 2
g expa f
K K
−exp a −K a f 2
2
3 3
K K Ka 1 1
2 4
1 2
2 2
1
*
*
K 2
K 2
expaK af expa K af 2 r
+4 + − 1
2
2 2
We have
K
a − f
h
m V E O
2
2
2 =
and since V E O >> , then K a 2 will be large so
that
exp K a exp K a 2 2 a f >> a− f
Then we can write
A A
A A
= a f lb 2
− 2
g expa f
K K
3 3
K K Ka 1 1
1 2
2 2
1
2
2
4
*
*
K 2
K expaK af 2 r
+4 1
2
2
2
which becomes
A A
A A
= a f b + gexpa f
K K
*
2
2
2 3 3
K K Ka 1 1
2 4
1 2
2 2
1
*
Substituting the expressions for Kand K, we
1 2 find
mVK 2
K
O
1
2
2
2
2
+ =
h
and
K K
m V E mE O
2
1
2
2
2 2
2 2
=
− LNM
OQP
LNM
OQP
a f
h h
= = FH
IK
m
( ) 2
2
V E E O
h
a f
or
K K
2
m
2 FH
= V
1− E
V
E O
O
2
1
2
2
2
IK
FH G
IK J
( )
h
Then
A A
A A
mV
K a
m
E
V
( E
) LN M
V
O
O
O
1 1
3 3 2
2
2
2
2
2
2
16
2
1
*
* exp
=
−
FH
IK
FH
IK
FH G
IK J
OQ P
h
h
a f
=
IK J
FH G
A A
IK J
3 3
E
E
− − V
FH G
V
K a
2 16 1 2
O O
*
expa f
or finally
T
A A
= = − − A A
G
FH * expa f
E
V
E
V
K a
IK J
FH G
O O
IK J
3 3
1 1
2 16 1 2
*
2.36
Region I: V = 0
∂
∂
2
+ = ⇒
2
1
2 2 1
0
ψ
ψ
x
mE
h
ψ1 1 1 1 1 = A expa jK xf+ B expa− jK xf
(incident wave) (reflected wave)
where K
mE
2
=
1 2
h
Region II: V = V1
∂
∂
+
−
=
2
2
2
1
2 2
2
0
ψ
ψ
x
maE V f
h
⇒
ψ2 2 2 2 2 = A expa jK xf+ B expa− jK xf
(transmitted (reflected
wave) wave)
where K
m E V
2
1
2
2
=
a − f
h
Region III: V = V2
∂
∂
+
−
=
2
3
2
2
2 3
2
0
ψ
ψ
x
maE V f
h
⇒
ψ3 3 3 = A expajK xf
(transmitted wave)
where K
m E V
3
2
2
2
=
a − f
h
There is no reflected wave in region III.

19
The transmission coefficient is defined as
T
v
v
A A
A A
K
K
A A
A A
= 3 ⋅ = ⋅
1
3 3
1 1
3
1
3 3
1 1
*
*
*
*
From boundary conditions, solve for A3 in terms
of A1 . The boundary conditions are:
x = 0: ψ ψ 1 2 = ⇒ A B A B 1 1 2 2 + = +
∂
∂
ψ ∂
ψ 1 =
2
⇒
x ∂
x
K A K B K A K B 1 1 1 1 2 2 2 2 − = −
x = a: ψ ψ 2 3 = ⇒
A jK a B jK a 2 2 2 2 expa f+ expa− f
=A jKa 3 3 expa f
∂
∂
ψ ∂
ψ 2 =
3
⇒
x ∂
x
K A jK a K B jK a 2 2 2 2 2 2 expa f− expa− f
= K A jK a 3 3 3 expa f
But K a n 2 = 2 π ⇒
exp jK a exp jK a 2 2 a f = a− f = 1
Then, eliminating B A B 1 2 2 , , from the above
equations, we have
T
K
K
K
2
4
a f T
3 ⇒
K K
1
= ⋅
+
1
1 3
2
K K
K K
=
4 1 3
a +
f2
1 3
2.37
(a) Region I: Since V E O > , we can write
∂
∂
−
−
=
2
2
1
2 2 1
0
ψ
ψ
x
m V E O a f
h
Region II: V = 0 , so
∂
∂
2
+ =
2
2
2 2 2
0
ψ
ψ
x
mE
h
Region III: V → ∞ ⇒ψ = 3 0
The general solutions can be written, keeping in
mind that ψ 1 must remain finite for x < 0 , as
ψ1 1 1 = B expa+K xf
ψ2 2 2 2 2 = A sinaK xf+ B cosaK xf
ψ 3 = 0
where
K
a − f
h
m V E O
2
=
1 2
and K
mE
2
=
2 2
h
(b)
x = 0: ψ ψ 1 2 = ⇒ B B 1 2 =
∂
∂
ψ ∂
ψ 1 =
2
⇒
x ∂
x
K B K A 1 1 2 2 =
x = a: ψ ψ 2 3 = ⇒
A Ka B Ka 2 2 2 2 sin + cos = 0
or
B A Ka 2 2 2 = − tan
(c)
K B K A 1 1 2 2 = ⇒ A
FH = G
K
K
1 1
B 2
2
IK J
and since B B 1 2 = , then
A
FH = G
K
K
2 1
B 2
2
IK J
From B A Ka 2 2 2 = − tan , we can write
B
= −G
FH K
K
2 2 1
B Ka 2
2
IK J
tan
which gives
1 1
2 = −FH G
2
IK J
K
K
tan K a
In turn, this equation can be written as
1
LN M
2
⋅ −
2 = −
OQ P
V E
E
mE
O tan a
h
or
E
V E
mE
a
O −
= − ⋅ LN M
OQ P
tan
2
2 h
This last equation is valid only for specific
values of the total energy E . The energy levels
are quantized.
2.38
m e
E
J n
n
o
o
=
−
∈
( )
4
2 2 2 a4π f 2h
=
( ) m e
o eV
o
∈
n
3
2 2 2 a4π f 2h
=
−
b gb g
b g b g
⇒
31 19 3
− −
x x
x x n
9.11 10 16 .
10
12 2 34 2 2
− −
π . .
E
4 8 85 10 2 1054 10
− 13 .
58
( ) eV n =
n
2
Then

20
n = 1⇒ E eV 1 = −13.58
n = 2 ⇒ E eV 2 = −3.395
n = 3⇒ E eV 3 = −1.51
n = 4 ⇒ E eV 4 = −0.849

2.39
We have
ψ
3 2 1 1
= ⋅
π 100
− FH G
IK J
FH G
IK J
a
/
exp
r
a o o
and
P r r
1 1 2 2
* exp
a
r
a o o
= = ⋅ ⋅
− FH G
IK J
FH G
IK J
4 4
100 100
2
3
π ψ ψ π
π
or
P
a f 2 exp
a
r
r
a o o
= ⋅
− FH G
IK J
4 2
3
To find the maximum probability
dP r
( )
dr
= 0
=
− FH G
− −
r
a f exp +
exp
a o o o o IK J
FH G
IK J
FH G
IK J
RST
UVW
4 2 22
2
3
2
a
r
a
r
a
r
which gives
−
0 = 1
+ ⇒
r
ao
r ao =
or r ao = is the radius that gives the greatest
probability.
2.40
ψ 100 is independent of θ and φ , so the wave
equation in spherical coordinates reduces to
1 2
FH
r
2
+ − = ∂
∂
∂
∂
IK
( ( )) ψ
0 2
r r
r
2 m
⋅ o E V r
ψ
h
where
V r
− 2 2
e
r mar o o o
( ) =
−
∈
=
4π
h
For
ψ
⇒ FH G
3 2 1 1
= ⋅
π 100
−
IK J FH G
IK J
a
/
exp
r
a o o
d
dr a a
r
a o o o
ψ
π
100
3 2 1 1 1
= ⋅
− − FH G
IK J
FH G
IK J
FH G
IK J
/
exp
Then
r
d
dr a
r
r
a o o
2 100
5 2
2 ψ 1 1
π
=
−
⋅
− FH G
IK J
FH G
IK J
/
exp
so that
d
dr
ψ r
FH
2 100 d
dr
IK
=
−
⋅
5 2 2
FH G
− −
r
−
a o o o o
IK J
FH G
IK J
FH G
IK J
FH G
IK J
LN M
OQ P
1 1
2
π a
r
r
a
r
a
/
exp exp
Substituting into the wave equation, we have
−
r
⋅
π
a o o o o 5 2 −
2
− FH G
−
IK J
FH G
IK J
FH G
IK J
LN M
OQ P
1 1
2 2
r a
r
r
a
r
a
/
exp exp
= LN M
2 3 2 m
E
h
+ + ⋅ ⋅
−
OQ P
FH
IK
FH G
IK J
FH G
IK J
2 1 1
/
exp
0 2
m a r a
r
a
o
π
o o o o h
where
E E
m e o
a4π f 2 2h
2 o
= =
4
−
∈ ⋅
⇒ 1
E
m a o o
1
2
2 2
=
−h
Then the above equation becomes
FH G
3 2
− r
a o o o o
1 1 12
2
2
π
⋅
− −
IK J
FH G
IK JLN M
OQ PRSTLN M
OQ P
a
r
a ra r
/
exp
+
+ = FH G
m
−
2 2 IK J
UVW
2
h h
0 2
2
m a m a r
o
o o o o h
or
1 1 3 2
⋅
π
− FH G
IK J
FH G
IK J
LN M
OQ P
a
/
exp
r
a o o
×
−
+ +
−
+ = FH G
IK J
RST
UVW
2 1 1 2
0 2 2 a r a a a r o o o o
which gives 0 = 0, and shows that ψ 100 is indeed
a solution of the wave equation.
2.41
All elements from Group I column of the
periodic table. All have one valence electron in
the outer shell.

Chapter 3

23
Problem Solutions
3.1 If ao were to increase, the bandgap energy
would decrease and the material would
begin to behave less like a semiconductor
and more like a metal. If ao were to
decrease, the bandgap energy would
increase and the material would begin to
behave more like an insulator.
3.2
Schrodinger’s wave equation
−
h ( ) ( ) ( ) ( )
⋅
x t
x
∂
∂
+ ⋅ =
∂Ψ
∂
h
2 2
2 2m
V x x t j
x t
t
Ψ
Ψ
,
,
,
Let the solution be of the form
Ψ x t u x j kx
FH
( , ) ( )E
exp t IK
FH
IK
LNM
OQP
= −
h
Region I, V(x) = 0 , so substituting the proposed
solution into the wave equation, we obtain
−
⋅
∂
∂
( ) − FH
IK
FH
IK
LNM
OQP
h
h
2
2m x
jku x j kx
E
{ exp t
+
( ) FH
∂
∂
−
IK
FH
IK
LNM
OQP
UVW
u x
x
j kx
E
exp t
h
=
FH
−
⋅ − IK
( ) FH
IK
FH
IK
LNM
OQP
j
jE
u x j kx
E
h exp
t
h h
which becomes
−
( ) ( ) − FH
IK
FH
IK
LNM
OQP
h
h
2
2
2m
jk u x j kx
E
{ exp t
+
( ) FH
∂
∂
−
IK
FH
IK
LNM
OQP
2 jk
u x
x
j kx
E
exp t
h
+
u x
x
∂
∂
( ) FH
−
IK
FH
IK
LNM
OQPUVW
2
2
j kx
E
exp t
h
= + ( ) − FHIK
FH
IK
LNM
OQP
Eu x j kx
E
exp t
h
This equation can then be written as
− +
( ) + ⋅ = ( ) ( ) ( ) k u x jk
∂
∂
+
2
∂
∂
u x
x
u x
x
mE
2
u x 2
2 2 2
0
h
Setting u(x) = u (x) 1 for region I, this equation
becomes
( ) ( ) ( ) + − b −α g =
d u x
dx
jk
du x
dx
k ux
2
1
2
1 2 2
1 2 0
where
α 2
2
2
=
mE
h
Q.E.D.
( ) = . Assume the same form
In region II, V x VO
of the solution
Ψ x t u x j kx
FH
( , ) ( )E
exp t IK
FH
IK
LNM
OQP
= −
h
Substituting into Schrodinger’s wave equation,
we obtain
−
( ) ( ) − FH
IK
FH
IK
LNM
OQP
h
h
2
2
2m
jk u x j kx
E
{ exp t
+
( ) FH
∂
∂
IKIKFH
LNM
−
OQP
2 jk
u x
x
j kx
E
exp t
h
+
u x
x
∂
∂
( ) FH
FHIK
−
IK
LNM
OQP
UVW
2
2
j kx
E
exp t
h
+ ( ) − FH
IK
FH
IK
LNM
OQP
V u x j kx
E
t O exp
h
= ( ) − FH
IK
FH
IK
LNM
OQP
Eu x j kx
E
exp t
h
This equation can be written as
− +
( ) ( ) ( )
∂
∂
+
∂
∂
k u x jk
u x
x
u x
x
2
2
2 2
mV
mE
2 2
− O u ( x
) + u ( x
) = 0 2 2
h h
Setting u(x) = u (x) 2 for region II, this equation
becomes
d u x
dx
( ) ( )
jk
du x
dx
2
2
2
2 2
+
− − + = FH
IK
mV
k 2 2 O u ( x
)
2 2
2
α 0
h
where
α 2
2
2
=
mE
h
Q.E.D.

2 ( ) ( ) ( ) = α − exp α −
24
3.3
We have
d u x
( ) ( ) + − b −α g ( ) =
dx
jk
du x
dx
k ux
2
1
2
1 2 2
1 2 0
The proposed solution is
u x A j k x B j k x 1( ) = exp (α − ) + exp − (α + )
The first derivative is
du x
1 j k A j k x ( ) ( ) ( ) = α − exp α −
dx
− j(α + k)B exp − j(α + k)x
and the second derivative becomes
d u x
dx
j k A j kx
2
1
2
+ j(α + k ) B − j(α + k )x 2 exp
Substituting these equations into the differential
equation, we find
−(α − k ) A j(α − k )x 2 exp
−(α + k ) B − j(α + k)x 2 exp
+2 jk{ j(α − k )Aexp j(α − k )x
− j(α + k )B exp − j(α + k )x }
− k − {A j( − k)x 2 2 b α g exp α
+B exp − j(α + k )x } = 0
Combining terms, we have
− α − α + − (α − ) 2 2 l b 2 k k g 2k k
− k − A j( − k)x 2 2 b α gq exp α
+m−cα 2 + 2αk + k 2 h+ 2kaα + kf
− k − B − j( + k)x = 2 2 b α gq exp α 0
We find that
0 = 0 Q.E.D.
For the differential equation in u x 2( ) and the
proposed solution, the procedure is exactly the
same as above.
3.4
We have the solutions
u x A j k x B j k x 1( ) = exp (α − ) + exp − (α + )
for 0 < x < a
u x C j k x D j k x 2( ) = exp (β − ) + exp − (β + )
for −b < x < 0
The boundary conditions:
u u 1 2 (0) = (0)
which yields
A + B − C − D = 0
Also
du
=
1 2
dx
du
x dx x
=0 =0
which yields
(α − k)A − (α + k )B − (β − k)C + (β + k)D = 0
The third boundary condition is
u a u b 1 2 ( ) = (− )
which gives
Aexp j(α − k )a + B exp − j(α + k)a
= C exp j(β − k )(−b) + Dexp − j(β + k )(−b)
This becomes
Aexp j(α − k )a + B exp − j(α + k)a
−C exp − j(β − k)b − Dexp j(β + k )b = 0
The last boundary condition is
du
=
1 2
dx
du
x = a dx x =−
b
which gives
j(α − k )Aexp j(α − k)a
− j(α + k)B exp − j(α + k)a
= j(β − k)C exp j(β − k)(−b)
− j(β + k)Dexp − j(β + k)(−b)
This becomes
(α − k)Aexp j(α − k )a
−(α + k)B exp − j(α + k )a
−(β − k )C exp − j(β − k )b
+(β + k)Dexp j(β + k )b = 0
3.5 Computer plot
3.6 Computer plot
3.7
P′ + =
a
a
a ka
sin
cos cos
α
α
α
Let ka = y ,αa = x
Then
P′ + =
x
x
x y
sin
cos cos
Consider
d
dy
of this function

25
d
dy
P′ ⋅ x ⋅ x + x y − { b g 1 } = − sin cos sin
We obtain
′ (− )( ) + ( ) RST
UVW
dx
dy
− − P x x
x x
dx
dy
1 2 1 sin cos
dx
dy
− sin x = − sin
y
Then
dx
dy
P
−
+ LNM
− =− x
x
x
x
OQP
′ x y
RST
UVW
1
2 sin
cos
sin sin
For y = ka = nπ , n = 0 , 1, 2 , . . .
⇒ sin y = 0
So that, in general, then
dx
dy
( )
( )
d a
d ka
d
dk
= = =
0
α α
And
α
α
FH
= ⇒ = IK
FH
IK
− 2 1
2
/
2 2
2 2
1 2
2
mE d
dk
mE m dE
h h h dk
This implies that
d
dk
dE
dk
α
= 0 = for k
n
a
=
π
3.8
sin
( ) = 9 + α =
f a
a
a
α
α
α cos a cos
ka
(a) ka = π ⇒ cos ka = −1
1st point: αa = π : 2nd point: αa = 1.66π
(2nd point by trial and error)
Now
α
α
FH
= ⇒ = ⋅ a a
mE
E
IK
2 2
a
a m
2
2 2
h
h
So
E
( )
a
x
b 1054 .
x
10
g
b x
g
= ⋅ ⇒
b −
g 2 9.11 10
−
−
α 2
10 2
34 2
31 5 10
E = ( a) x − (J) α 2 20 2.439 10
or
E = (αa) ( ) (eV ) 2 0.1524
So
αa = π ⇒ E = eV 1 1.504
αa = 1 66π ⇒ E = 4.145 eV 2 .
Then
ΔE = 2.64 eV
(b)
ka = 2π ⇒ cos ka = +1
1st point: αa = 2π
2nd point: αa = 2.54π
Then
E eV 3 = 6.0165
E eV 4 = 9.704
so
ΔE = 3.69 eV
(c)
ka = 3π ⇒ cos ka = −1
Then
E eV 5 = 13.537
E eV 6 = 17.799
so
ΔE = 4.26 eV
(d)
ka = 4π ⇒ cos ka = +1
Then
E eV 7 = 24.066
E eV 8 = 28.724
so
ΔE = 4.66 eV
3.9
(a) 0 < ka < π
For ka = 0⇒ cos ka = +1
By trial and error: 1st point: αa = 0.822π
2nd point: αa = π
From Problem 3.8, E = (αa) ( ) (eV ) 2 0.1524
Then
E eV 1 = 1.0163
E eV 2 = 1.5041
so
ΔE = 0.488 eV
(b)
π < ka < 2π
Using results of Problem 3.8
1st point: αa = 1.66π
2nd point: αa = 2π
Then

26
E3 = 4.145 eV
E eV 4 = 6.0165
so
ΔE = 1.87 eV
(c)
2π < ka < 3π
Then
E eV 5 = 9.704
E eV 6 = 13.537
so
ΔE = 3.83 eV
(d)
3π < ka < 4π
Then
E7 = 17.799 eV
E8 = 24.066 eV
so
ΔE = 6.27 eV
3.10
6
sin
cos cos
α
α
α
a
a
+ a = ka
Forbidden energy bands
(a) ka = π ⇒ cos ka = −1
1st point: αa = π
2nd point: αa = 1.56π (By trial and error)
From Problem 3.8, E = (αa) ( ) eV 2 0.1524
Then
E eV 1 = 1.504
E eV 2 = 3.660
so
ΔE = 2.16 eV
(b)
ka = 2π ⇒ cos ka = +1
Then
E eV 3 = 6.0165
E eV 4 = 8.809
so
ΔE = 2.79 eV
(c)
ka = 3π ⇒ cos ka = −1
Then
E eV 5 = 13.537
E eV 6 = 16.679
so
ΔE = 3.14 eV
(d)
ka = 4π ⇒ cos ka = +1
Then
E eV 7 = 24.066
E eV 8 = 27.296
so
ΔE = 3.23 eV
3.11
Allowed energy bands
Use results from Problem 3.10.
(a)
0 < ka < π
1st point: αa = 0.759π (By trial and error)
2nd point: αa = π
We have
E = (αa) ( ) eV 2 0.1524
Then
E eV 1 = 0.8665
E eV 2 = 1.504
so
ΔE = 0.638 eV
(b)
π < ka < 2π
Then
E eV 3 = 3.660
E eV 4 = 6.0165
so
ΔE = 2.36 eV
(c)
2π < ka < 3π

27
Then
E5 = 8.809 eV
E eV 6 = 13.537
so
ΔE = 4.73 eV
(d)
3π < ka < 4π
Then
E eV 7 = 16.679
E eV 8 = 24.066
so
ΔE = 7.39 eV
3.12
T = 100K ; E
x
g= −
4 2
+
⇒
− ( )
1170
4.73 10 100
636 100
.
b g
E eV g = 1.164
T = 200K ⇒ E eV g = 1.147
T = 300K ⇒ E eV g = 1.125
T = 400K ⇒ E eV g = 1.097
T = 500K ⇒ E eV g = 1.066
T = 600K ⇒ E eV g = 1.032
3.13
The effective mass is given by
m
d E
dk
*= ⋅ FH G
IK J
1
2
− 2
2
1
h
We have that
d E
dk
a curve A
f > a f
2 d E
dk
curve B
2
2
2
so that
m curve A m curveB *a f < *a f
3.14
The effective mass for a hole is given by
m
*= ⋅ FH G
d E
dk p
IK J
1
2
− 2
2
1
h
We have that
d E
dk
a curve A
f > a f
2 d E
dk
curve B
2
2
2
so that
* a f < * a f
m curve A m curveB p p
3.15
Points A, B:
∂
∂
< ⇒
E
k
0 velocity in –x direction;
Points C, D:
∂
∂
>
E
x
0 ⇒ velocity in +x direction;
Points A, D;
E
k
∂
∂
< ⇒
2
2 0
negative effective
mass;
Points B, C;
E
k
∂
∂
> ⇒
2
2 0
positive effective
mass;
3.16
E E
2 2
2
k
h
− =
C m At k = A ⇒ ° − 01 1 . b g
1
10 10 9
k
= A = m ° −
So
k = m + − 109 1
For A:
b g b g b g
0 07 16 10
10 1054 10
2
19
9 2 34 2
. .
.
( ) −
−
x =
x
m
which yields
m = x kg − 4.96 10 31
so
curve A
m
mo
; = 0.544
For B:
b g b gb g
0 7 16 10
10 1054 10
2
19
9 34 2
. .
.
( ) −
−
x =
x
m
which yields
m = x kg − 4.96 10 32
so
Curve B
m
mo
: = 0.0544
3.17
E E
2 2
2
k
h
− =
Vm

28
k = A ⇒ m − − 01 10 1 9 1 . b * g
For Curve A:
b g b g b g
0 08 16 10
10 1054 10
2
19
9 2 34 2
. .
.
( ) −
−
x =
x
m
which yields
m = x kg ⇒ − 4.34 10 31
m
mo
= 0.476
For Curve B:
b g b g b g
0 4 16 10
10 1054 10
2
19
9 2 34 2
. .
.
( ) −
−
x =
x
m
which yields
m = x kg ⇒ − 8 68 10 32 .
m
mo
= 0.0953
3.18
(a) E = hν
Then
ν= = ⇒
b g
b g
( ) −
−
E
h
x
x
142 16 10
6 625 10
19
34
. .
.
ν = 343 1014 . x Hz
(b)
λ
= = = − c x
ν
x
x m
3 10
343 10
8 75 10
8
14
7
.
.
or
λ = 0.875 μm
3.19
(c) Curve A: Effective mass is a constant
Curve B: Effective mass is positive around
k = 0 , and is negative around k = ±
π
2
.
3.20
E E E k k O O = − − 1 cos αa f
dE
dk
E k kO = − (− ) − 1 a f α sin αa f
= + − E k kO
1α sin αa f
So
d E
dk
= E α 2 cos αa k − kf
O
2
2 1
Then
d E
dk
E
k kO
2
2 1
2
=
= α
We have
1 1
*= ⋅ =
2
2
d E
dk
2
α
1
2
E
m
h h
2 or
m
E
* =
h2
2 α
1
3.21
For the 3-dimensional infinite potential well,
V(x) = 0 when 0 < x < a , 0 < y < a , and
0 < z < a . In this region, the wave equation is
∂
2 ( ) ( ) ( )
ψ x y z ψ ψ
∂
+
∂
∂
+
∂
∂
2
2
2
2
2
x
x y z
y
x y z
z
, , , , , ,
mE
2
+ ( x y z
) = 0 2
h
ψ , ,
Use separation of variables technique, so let
ψ (x, y, z) = X(x)Y(y)Z(z)
Substituting into the wave equation, we have
YZ
X
x
XZ
Y
y
XY
Z
z
∂
∂
+
∂
∂
+
∂
∂
2
2
2
2
2
2
2
mE
XYZ
+ ⋅ =
0 2
h
Dividing by XYZ , we obtain
X
x Y
Y
y Z
Z
z
∂
∂
∂
∂
∂
∂
1 1 1 2
0
2
2
2
2
2
mE
⋅
+ ⋅
+ ⋅
+ =
X
2 h
2 Let
1
0
2
2
2
2
2
2
X
X
x
k
X
x
∂
∂
⋅
= − ⇒
k X x x ∂
∂
+ =
The solution is of the form
X x A k x B k x x x ( ) = sin + cos
Since ψ (x, y, z) = 0 at x = 0 , then X(0) = 0 so
that B ≡ 0 .
Also, ψ (x, y, z) = 0 at x = a , then X(a) = 0 so
we must have k a n x x = π , where
nx = 1, 2 , 3, . .
Similarly, we have
1 2
2
2
Y
Y
y
∂
∂
ky ⋅
= − and
1 2
2
2
Z
Z
z
∂
∂
⋅
= −
kz From the boundary conditions, we find

29
k a = n π and k a = n π
y y z z where n= 1, 2 , 3, . . . and n= 1, 2 , 3, . . .
y z From the wave equation, we have
mE
−k 2 − k 2 − k 2
+ =
x y z
2
2
0
h
The energy can then be written as
E
FH
= n + n + n
m
x y z a IK
h2
2 2 2
2
2
b g π
3.22
The total number of quantum states in the 3-
dimensional potential well is given (in k-space)
by
k 2
dk
g k dk
π
π
a T( ) = ⋅
3
3
where
k
mE 2
2
2
=
h
We can then write
1
k = ⋅ 2
mE
h
Taking the differential, we obtain
dk m
E
dE
m
E
1
= ⋅ 2
⋅ ⋅ ⋅ = ⋅ ⋅ dE
1
2
1 1
h h 2
Substituting these expressions into the density of
states function, we obtain
g E dE
3
( ) a FH
mE m
dE TE
IK
π
π
2 1
h h 2
= ⋅ ⋅ ⋅
3 2
Noting that
h =
h
2m
this density of states function can be simplified
and written as
g E dE
a
h
3 2 π /
4
( ) = ( 2
m ) ⋅ EdE ⋅
T3
3
Dividing by a3 will yield the density of states,
so that
( ) E ( )
g E
4 π 2 3 /
2
m
h
= ⋅
3
3.23
g E
π b * g /
4 2 3 2
m
h
( )= n
−
C E E C
3
Now
g
π b * g /
+ 4 2 z 3 2
m
h
n
E kT
E E dE T
C
E
C
C
= − ⋅
3
= − FH
IK
* / E kT
4 π 2 m
2 +
3 2 3
3 2
3
h
E E n
C
C
C
E
/ b g a f
/ b g
= FH
IK
* /
3 2 π m
( )
4 2 2
3
3 2
3
h
kT n
Then
g
b g /
b g
( ) FH
x
4 2 0 067 9.11 10
x T =
IK
−
−
6 625 10
2
3
31 3 2
34 3
π .
.
×( ) − 0 0259 16 10 19 3 2 . . / b x g
or
g x m x cm T= =− − 3 28 10 3 28 10 23 3 17 3 . .
3.24
g E
π b * g /
4 2 3 2
m
h
( )= p
−
V E E V
3
Now
g
π b * g /
4 2 3 2
m
h
p
− z
E
E E dE T
V
E kT
V
V
= − ⋅
3
=
− FH
* / E
π m
−
3 2 IK
−
4 2 2
3
3 2
3
h
E E p
V
V
V
E kT
/ b g a f
/ b g
= FH
IK
* /
3 2 π m
( )
4 2 2
3
3 2
3
h
kT p
g
b g /
b g
( ) FH
x
4 2 048 9.11 10
x T =
IK
−
−
6 625 10
2
3
31 3 2
34 3
π .
.
×( ) − 0 0259 16 10 19 3 2 . . / b x g
or
g x m x cm T= = − − 6 29 10 6 29 10 24 3 18 3 . .
3.25
(a) g E
π b * g /
4 2 3 2
m
h
( )= n
−
C E E C
3
b g
b g b g
( ) −
π . 19 1 2
− 4 2108 9.11 10
= −
−
6 625 10
16 10
31 3 2
34 3
.
.
/
x /
x
x E EC
= − − − 4.77 1046 3 1 x E E mJ C

30
or
g E x E E cm eV C C ( )= − − − 7.63 1021 3 1
Then
E gC
EC + 0.05 eV
E eV C + 0.10
E eV C + 0.15
E eV C + 0.20
171 1021 3 1 . x cmeV − −
2.41 1021 x
2.96 1021 x
341 1021 . x
π b * g /
4 2 3 2
V ( )= −
(b) g E
m
h
p
E E V
3
( ) b −
g
b g b g
= 178 . x 1046 E − E mJ − 3 − 1 V
g ( E )= 2.85 x 1021 E − E cm − 3 eV − 1
V V π . 19 1 2
− 4 2 056 9.11 10
= −
−
6 625 10
16 10
31 3 2
34 3
.
.
/
x /
x
x E E V
E g E V( )
E eV V − 0.05
E eV V − 0.10
E eV V − 0.15
E eV V − 0.20
0 637 1021 3 1 . x cm eV − −
0 901 1021 . x
110 1021 . x
127 1021 . x

3.26
g
g
b m
g
b m
g
C
V
* /
n
= ⇒
* /
p
3 2
3 2
g
g
m
m
C
V
n
p
=
FH G
IK J
*
*
3/2
3.27
Computer Plot
3.28
g
10
a i
f ( − )
!
N g N
! !
i i i
!
=
− 8 ! 10 8
!
( )( )( )
( )( )
( )( )
( )( )
10 9 8
8 2
10 9
2 1
!
= = ⇒
! !
= 45
3.29
(a) f E
1
1 expa f
E kT E
C C
kT
( ) LNM
OQP
=
+
+ −
=
+
⇒
( )
1
1 exp 1
f (E) = 0.269
(b)
1 1
1
1
− = −
+
a − f
−
exp V V ( ) LNM
OQP
f E
E kT E
kT
= −
+ −
⇒
( )
1
1
1 exp 1
1− f (E) = 0.269
3.30
f E
E E
kT
F
( ) FH
IK
=
+
−
1
1 exp
(a) E E kT F − = , f (E)
( )
=
+
⇒
1
1 exp 1
f (E) = 0.269
(b) E E kT F − = 5 , f (E)
( )
=
+
⇒
1
1 exp 5
f (E) = x − 6 69 10 3 .
(c) E E kT F − = 10 , f (E)
( )
=
+
⇒
1
1 exp 10
f (E) = x − 4.54 10 5
3.31
1 1
1
1
− = −
+
−
( ) FH
IK
f E
E E
kT
exp F
or
1
1
1
− =
+
−
( ) FH
IK
f E
E E
kT
exp F
(a) E E kT F− = , 1− f (E) = 0.269
(b) E E kT F− =5 , 1 6 69 10 3 − ( ) = − f E . x
(c) E E kT F− =10 , 1 4.54 10 5 − ( ) = − f E x
3.32
(a) T = 300K ⇒ kT = 0.0259 eV
f E
LNM
1 − −
F ( ) FH
E E
kT
E E
IK
F kT
OQP
=
+
−
≈
1 exp
exp a f

31
E f bE g
EC
EC + (1 2)kT
E kT C +
EC + (3 2)kT
E kT C +2
6 43 10 5 . x −
390 10 5 . x −
2.36 10 5 x −
143 10 5 . x −
087 10 5 . x −
(b) T = 400K ⇒ kT = 0.03453
E f (E)
EC
EC + (1 2)kT
E kT C +
EC + (3 2)kT
E kT C +2
7.17 10 4 x −
4.35 10 4 x −
2.64 10 4 x −
160 10 4 . x −
0 971 10 4 . x −
3.33
E
h2 2 2
n
ma n =
π
2 2
=
b 1054 . x 10
−
34 g
2 n
2 π
2
b x gb x
g
31 10 2
− −
2 9.11 10 10 10
or
E = 6 . 018 x 10 − 20 nJ 2 = 0 .
376 neV 2 nFor n = 4 ⇒ E = 6 . 02 eV ,
4 For n = 5⇒ E = 9.40 eV .
5 As a 1st approximation for T > 0 , assume the
probability of n = 5 state being occupied is the
same as the probability of n = 4 state being
empty. Then
1
1
1
1
IKexp exp
− 4 1 5
FH
−
+
−
=
+
IK
FH
E E
kT
E E
kT
F F
⇒
+
− FH
E −
E
kT
=
+
IK
FHIK
1
1
1
E E
kT
exp 4 1 exp 5
F F
or
E E E E E
E E
F F F − = − ⇒ =
+
4 5
4 5
2
Then
EF =
+
⇒
6 02 9.40
2
.
E eV F = 7.71

3.34
(a) For 3-Dimensional infinite potential well,
h2 2
E
n n n x y z = + +
ma
2
2 2 2
2
πb g
b . x
−
34 g
2 π
2
b gb g b 2 2 2 g x
1054 10
2 9.11 10 10
= ++
31 9 2
− −
n n n x y z
= 0 376 + + 2 2 2 . n n n eV x y z b g
For 5 electrons, energy state corresponding to
n n n x y z= 221 = 122 contains both an electron
and an empty state, so
EF= (0 376) 2 + 2 + 1 ⇒ 2 2 2 . b g
E eV F = 3.384
(b) For 13 electrons, energy state corresponding
to n n n x y z= 323 = 233 contains both an
electron and an empty state, so
EF= (0 376) 2 + 3 + 3 ⇒ 2 2 2 . b g
E eV F = 8.272
3.35
The probability of a state at E E E 1 F = +Δ
being occupied is
f E
+ FH
E E
kT
E
kT
F
1 1
1
1
1
1 1
a f =
+
−
=
IK
FH
IK
exp exp
Δ
The probability of a state at E = E −Δ
E 2 F being empty is
1
1 1
1
2 2
2
− =−
+
− FH
IK
f E
E E
kT
F
a f
exp
= −
+
E
kT
=
−
FH
− −
+
IK
FH
IK
FH
IK
1
1
1 exp 1
exp
exp
Δ
Δ
E Δ
kT
E
kT
or
1
1
1
2 2 − =
+
+ FH
IK
f E
E
kT
a f
exp
Δ
Hence, we have that
f E f E 1 1 2 2 a f = 1− a f Q.E.D.

32
3.36
(a) At energy E1 , we want
1 1
exp exp
1
1
1
0 01
F F
1 1
1
exp
.
E E
kT
E E
kT
E E
kT
F
−
−
+
−
+
−
=
FH
IK
FH
IK
FH
IK
This expression can be written as
1
1 001
1
1
+
−
−
− =
FH
IK
FH
IK
exp
exp
.
E E
kT
E E
kT
F
F
⇒ =
− ( ) FH
IK
E E
kT
1 0.01 exp 1
F
or
E E kT 1 F = + ln(100)
Then
E E kT 1 F = + 4.6
(b)
At E E kT 1 F = + 4.6 ,
f E
4.6 a f =
+ FH
E E
kT
kT
kT
F
1
1
1
1
1
1
+
−
=
IK
FH
IK
exp exp
which yields
f E1 a f = 0.00990 ≈ 0.01
3.37
(a) E eVT K F= 6.25 , = 300 , At E = 6.50 eV
= − 1
f E x ( ) FH
IK
=
+
−
1
650 6 25
0 0259
6 43 10 5
exp
. .
.
.
or
6 43 10 3 . x % −
(b)
T = K ⇒ kT = ( )FH
IK
950 0 0259
950
300
.
or
kT = 0.0820 eV
Then
f E ( ) FH
IK
=
+
−
=
1
1
650 6 25
0 0820
0 0453
exp
. .
.
.
or 4.53%
= FH
− = 0 99
(c) 1 001
1
1
0 30
+
−
IK
.
exp
.
.
kT
Then
1
0 30 1
= = FH
+ 1 0101
0 99
−
IK
exp
.
.
.
kT
which can be written as
exp
FH
.
= = .
+
IK
0 30 1
0 0101
99
kT
Then
0 30
99
0 30
99
0 06529
.
ln
.
ln
.
kT
= ( )⇒ kT = =
( )
So
T = 756K
3.38
(a)
f E ( ) FH
IK
=
+
−
=
1
1
7.15 7.0
0 0259
0 00304
exp
.
.
or
0.304%
(b)
At T = 1000K ⇒ kT = 0.08633 eV
Then
f E ( ) FH
IK
=
+
−
=
1
1
7.15 7.0
0 08633
01496
exp
.
.
or14.96%
(c)
f E ( ) FH
IK
=
+
−
=
1
1
685 7.0
0 0259
0 997
exp
.
.
.
or
99.7%
(d)
At E EF = , f (E) =
1
2
for all temperatures.
3.39
For E = E1 ,
f E
LNM
1 − −
F ( ) FH
E E
kT
E E
IK
F kT
OQP
=
+
−
≈
1 1
1
exp
exp a f
Then

33
f E1
⇒ FH
0 30
0 0259
a f=
−
IK
exp
.
.
f E x 1
a f = 9.3 10 6 −
For E = E, E − E = 1.12 − 0.3 = 0.82
eV 2 F2 1
1 1
1
082
0 0259
− = −
+
−
( ) FH
IK
f E
exp
.
.
or
1 1 1
− ( ) FH
082
0 0259
− ≈ − −
IK
LNM
OQP
f E exp
.
.
=
FH
−
⇒ IK
exp
082
.
0 .
0259
1 1 78 10 14 − ( ) = − f E . x
(b)
For E E E E eV F F − = ⇒ − = 2 1 0.4 0.72
At E = E1 ,
( ) F LNM
f E
E E
kT
OQP
FH
IK
=
− −
=
−
exp exp
.
.
1 0 72
0 0259
a f
so
f (E) = x − 8 45 10 13 .
At E = E2 ,
1
− ( ) LNM
0 4
0 0259
− −
− = 2
=
OQP
FH
IK
f E
E E
kT
exp F exp
.
.
a f
so
1 1 96 10 7 − ( ) = − f E . x
3.40
(a) At E = E1 ,
a f 0 30
( ) F LNM
f E
E E
kT
OQP
FH
IK
=
− −
=
−
exp exp
.
.
0 0259
or
f (E) = x − 9.3 10 6
At E = E2 , then
E E eV F− = − = 2 1.42 0.3 1.12 ,
So
1
OQPFH
− ( ) LNM
112
0 0259
− −
− = 2
=
IK
f E
E E
kT
exp F exp
.
.
a f
or
1 1 66 10 19 − ( ) = − f E . x
(b)
For E E E E eV F F − = ⇒ − = 2 1 0.4 1.02 ,
At E = E1 ,
( ) F LNM
f E
E E
kT
OQP
FH
IK
=
− −
=
−
exp exp
.
.
1 102
0 0259
a f
or
f (E) = x − 7.88 10 18
At E = E2 ,
1
− ( ) LNM
0 4
0 0259
− −
− = 2
=
OQP
FH
IK
f E
E E
kT
exp F exp
.
.
a f
or
1 1 96 10 7 − ( ) = − f E . x
3.41
( ) F FH
f E
E E
kT
IK
LNM
OQP
= +
− −
1
1
exp
so
df E
F ( ) ( ) FH
dE
E E
kT
IK
LNM
OQP
= − +
− −
1 1
2
exp
×
− FH
IK
FH
IK
1
kT
E E
kT
exp F
or
df E
dE
kT
E E
kT
E E
kT
F
F
( )
FH
IK
FHIK
LNM
OQP
=
− −
+
−
1
1
2
exp
exp
(a) T = 0 , For
df
E < E
⇒ exp(−∞) = 0⇒ = 0
F dE E E
df
dE F > ⇒ exp(+∞) = +∞ ⇒ = 0
At E E
df
dE F = ⇒ →−∞
3.42
(a) At E Emidgap = ,
f E
E E
kT
E
kT
F g
( ) FH
IK
FH G
IK J
=
+
−
=
+
1
1
1
1
2
exp exp
Si: E eV g = 1.12 ,
f (E)
( )
LN M
OQ P
=
+
1
1
112
2 0 0259
exp
.
.
or
f (E) = x − 4.07 10 10
Ge: E eV g = 0.66 ,

34
f (E)
( )
LN M
OQ P
=
+
1
1
0 66
2 0 0259
exp
.
.
or
f (E) = x − 2.93 10 6
GaAs: E eV g = 1.42 ,
f (E)
( )
LN M
OQ P
=
+
1
1
142
2 0 0259
exp
.
.
or
f (E) = x − 124 10 12 .
(b)
Using results of Problem 3.35, the answers to
part (b) are exactly the same as those given in
part (a).
3.43
f E
kT
( ) FH
IK
= =
+
− 10
1
1
055
6
exp
.
Then
1
055 1
FH
+ exp
= = + 6 ⇒ 10
10 6
IK
−
.
kT
exp
. .
ln
055
10
055
10 6 6
kT kT
FH
IK
FH
IK
≈ ⇒ = + b g
or
kT= ⇒
055
106
.
lnb g T K = 461

3.44
At E = E, f a Ef = 0.05
2 2 So
1
0 05
1 2
.
exp
=
+
− FH
IK
E E
kT
F
Then
E −
E
kT
= ln( )
2 F 19
By symmetry, at E = E − f E = 1 1 , 1 a f 0.05 ,
So
E E
kT
F −
1 = ln(19)
Then
E −
E
kT
= ln( )
2 1 2 19
(a)
At T = 300K , kT = 0.0259 eV
E E E 2 1 − = Δ = (0.0259)(2)ln(19)⇒
ΔE = 0.1525 eV
(b)
At T = 500K , kT = 0.04317 eV
E E E 2 1 − = Δ = (0.04317)(2)ln(19)⇒
ΔE = 0.254 eV


Chapter 4

Problem Solutions

4.1
n N N
− FH G
E
kT i C V
2 g =
IK J
exp
(a) Silicon
T(° K) kT (eV ) n cm i
b −3 g
200
400
600
0.01727
0.03453
0.0518
7.68 104 x
2.38 1012 x
9.74 1014 x

(b) Germanium (c) GaAs
T(° K) n cm i
b −3 g n cm i
b −3 g
200
400
600
2.16 1010 x
8 60 1014 . x
382 1016 . x
1.38
328 109 . x
572 1012 . x

4.2
n N N
− FH G
E
kT i C V
2 g =
IK J
exp
− FH
IK
3 b g = b gb g
FH
T
112 12 2 19 19
x x
10 2.8 10 104 10
. exp
300
IK
.
kT
Then
exp
FH
1.12
T =b 2.912 x
10
g
300
14
3
kT
IK
FH
IK
By trial and error
T = 381K

4.3
Computer Plot

4.4
n N N T
( ) FH G IK J
E
kT i CO VO
2 3 g = ⋅ ⋅ −
exp
So

2
a f 1 1
a f= − − FH GIK JFH G
n T
n T
T
T
E
exp
kT kT
i
2
i
g
2
1
2
1
3
2 1
IK J
LN M
OQ P
At T K kT eV 2 = 300 ⇒ = 0.0259
At T K kT eV 1 = 200 ⇒ = 0.01727
Then
37

2 3 .
.
583 10
182 10
300
200
1
0 0259
1
0 01727
7
2
exp
. .
x
x
Eg
FH G
IK J
FH
IK
FH
IK
LNM
OQP
= − −
or
1026 10 3 375 19.29 11 . x . exp Eg = ( )
which yields
E eV g = 1.25
For T = 300K ,
583 10 300
− ( ) FH
125
0 0259
7 2 3 . exp
.
.
x NNCO VO b g = a f
IK

or
N N x CO VO = 115 1029 .

4.5
(a) g f E E
exp a f
− − LNM
E E
kT C F C
∝ − F
OQP
∝ −
exp C exp C F a f a f
− − − − LNM
OQP
LNM
OQP
E E
E E
E E
C
kT
kT Let E E x C − ≡
Then
g f x
− FH
x
kT C F ∝
IK
exp
Now, to find the maximum value

d g f
dx
x
x
kT
C F a f ∝
− − FH
IK
1
2
1/2 exp
− ⋅
FH
−
= IK
1
0 1 2
kT
x
x
kT
/ exp
This yields

1 2
1
2 x
1 2
2 x
kT
x
kT
/
/
= ⇒ =
Then the maximum value occurs at
kT
E E
C = +
2

(b)
g f E E
a f a f exp
− − LNM
E E
1− ∝ − F
V F V
kT OQP
exp F V exp V a f a f
E E
kT V
∝ −
− − − − LNM
OQP
LNM
OQP
E E
E E
kT
Let E E x V− ≡


Then
g f x
− FH
a f exp
x
kT V F 1− ∝
IK
To find the maximum value

a f exp
d g −
f
V F 1
dx
d
dx
x
= FH
x
kT
0
∝
−
IK
LNM
OQP
Same as part (a). Maximum occurs at
kT
x
E E V = = −
2

or
E E
kT
V = −
2

4.6

a f
a f
n E
n E
E E
a f
E E
kT
1
−
− −
LNM
exp
= a f
E E
C
1
E E
kT
C
C
C
1
2
2
− −
2
−
OQP
LNM
OQP
exp

where
E E kT 1 C = + 4 and E E
kT
= +
2 C 2
Then

a f
a f
n E
n E
a f =
kT
kT
− − LNM
E E
kT
1
2
1 2 4
2
OQP
exp
= − − = − FH
IK
LNM
OQP
1
2
2 2 4 ( )
exp 2 2 exp 3.5
or

0 0854 a f
a f = .
n E
n E
1
2

4.7
Computer Plot

4.8

2
( )
( )
n A
n B
E
kT
E
kT
exp b g
E E
kT
i
2
i
−
FH GIK JLN M
gA
gB
− −
gA gB
FH G
IK J
OQ P
=
−
=
exp
exp
or

exp b g
LN M
gA gB ( )
( )
n A
n B
E E
kT
i
i
OQ P
=
− −
2

38

=
( ) +
( )
− −
=
LN M
OQ P
( )
LN M
OQ P
exp
.
.
exp
.
.
1 12
2 0 0259
0 20
2 0 0259

or

n A
i
n B
i
( )
( )
= 47.5

4.9
Computer Plot

4.10
E E kT
m
m Fi midgap
p
n
− = FH G
IK J
3
4
ln
*
*
* * = 0.56 , = 1.08
Silicon: m m m m p O n O
E E eV Fi midgap − = −0.0128
Germanium: m m m m p O n O
* * = 0.37 , = 0.55
Gallium Arsenide: m m m m p O n O
* * = 0.48 , = 0.067
E E eV Fi midgap − = +0.038

4.11
(a) E E kT
m
m Fi midgap
p
n
− = FH G
IK J
3
4
ln
*
*
= ( ) ⇒ FH
IK
3
4
. ln .
0 0259
14
0 .
62

(b)
E E Fi midgap − = ( ) ⇒ FH
IK
3
4
. ln .
0 0259
0 25
110
.


4.12
E E kT
N
N Fi midgap
V
C
− = ( ) FH G
IK J
1
2
ln
= ( ) = − FH G
IK J
x
x
.
1
104 10
ln 0 495
( kT
) 2
2.8 10
19
19 kT
.


T(° K) kT (eV ) E E eV Fi midgap − ( )

200
400
600
0.01727
0.03453
0.0518
−0.0085
−0.017
−0.0256

4.13
Computer Plot

4.14
Let g E K C( )= =constant
Then,
n g Ef EdE O C
∞z ( ) ( )
F =
EC
=
+
1
− FH
IK
∞K z
E E
kT
dE
1 exp
E F C

≈
exp a f
− − LNM
OQP
∞K z
E E
kT
F dE
EC
Let
η =
E − E
kT
F so that dE = kT ⋅ dη
We can write
E E E E E E F C F C − = a − f− a − f
so that
exp exp exp
F C F a f a f η
⋅ − LNM
( ) E E
kT
− −
=
E E
kT
− −
OQP
LNM
OQP
The integral can then be written as
n K kT
∞exp zexp a f η η
− LNM
E E
kT
− −
OQP
( )
= ⋅ ⋅ C F
d O
0

which becomes
n K kT
exp a f
− − LNM
E E
kT O
= ⋅ ⋅ C F
OQP

4.15
Let g E C E E C C ( )= − 1a f for E EC ≥
n g Ef EdE O C F
∞z
= ( ) ( )
EC

39

=
a −
f
exp
+
− FH
IK
∞C z
E E
C dE
E E
kT
1
1
E F C

or
n C E E
1 a f a f exp
∞ − − z LNM
E E
kT
≈ −
F
dE O C
EC
OQP
Let
η =
E − E
kT
C so that dE = kT ⋅ dη
We can write
E E E E E E F C C F a − f = a − f+ a − f
Then
n C
1 exp a f
− − LNM
E E
kT O
= C F
OQP
− − ∞z LNM
× −
OQP
E E
E E
kT
C
dE C
EC
a f a f exp
or
=
1 exp a f
− − LNM
OQP
C
E E
kT
C F
∞z kT kT d
0
× ( ) (− ) ( )
η exp η η
We find that
η η η η
∞ e
−
η
∞ z0
exp(− ) = (− − ) = +
d 1 1
0 1

So
n C kT
− − ( ) LNM
E E
kT O
= C F
OQP
1
2 exp a f

4.16
We have
1 =∈ O FH
r
a
m
r
m O
IK
*
For Germanium, ∈ = = r O 16 , m 0.55m *
Then
r a1 O 16
= ( ) = 29 0 53 FH
1
055
IK
( )
.
.
so
r A 1 = 15 4 ° .
The ionization energy can be written as
E
m
m
2
13 6 eV

O
O
S
=
∈
∈
FH G
IK J
FH G
IK J
( )
*
.


055
.
( ) 16
. E = 0.029 eV
= ⇒
( )
13 6 2

4.17
We have
1 =∈ O FH
r
a
m
r
m O
IK
*
For GaAs, ∈ = = r O 13.1, m 0.067m *
Then
r1 131
= ( ) 0 53 FH
1
0 067
IK
. ( )
.
.
or
r A 1 = 104 °
The ionization energy is
E
FH G
*
= m
mO
∈
∈
O
S
=
IK J
FH G
IK J
( )
0 .
067
131
.
2 13 6
( )
( )
.
13 .
6
2
or
E = 0.0053 eV

4.18
(a) p
b . g
2 10 2
n
x
O
n
x 15 10
5 10
= i
= ⇒
O
4
p x cm O = − 4.5 1015 3 , p n O O > ⇒ p-type
(b)
E E kT
− = FH G
p
n Fi F
ln O

i
IK J
=( ) FH G
IK J
0 0259
x
x
4.5 10
15 10
15
10 . ln.

or
E − E = 0.3266 eV
Fi F
4.19
a f p N
exp
− − LNM
E E
kT O V
= F V
OQP
=
− FHIK
19 . exp .
x
104 10
0 22
0 .
0259
so
p x cm O = − 2.13 1015 3
Assuming
E E eV C F − = 1.12 − 0.22 = 0.90
Then
40

exp a f
− − LNM
E E
kT O C
n N
= C F
OQP
=
− FH
IK
2.8 10
0 90
0 0259
18 x exp
.
.

or
n x cm O = − 2.27 104 3

4.20
(a) T = 400K ⇒ kT = 0.03453 eV
N x x cm C= = FH
IK
400
300
3 2
4.7 10
17
7.24 10 − 17 3
/

Then
n N
exp a f
− − LNM
E E
kT O C
= C F
OQP
=
− FH
IK
7.24 10
0 25
0 03453
17 x exp
.
.

or
n x cm O = − 519 1014 3 .
Also
N x x cm V= = FHIK
400
300
3 2
7 10
18
108 10 − 19 3
/
.
and
E E eV F V − = 1.42 − 0.25 = 1.17
Then
p x O =
− FH
IK
19 . exp .
108 10
117
0 .
03453

or
p x cm O = − 2.08 104 3
(b)
E E kT
N
n C F
C
O
− = FH G
IK J
ln
=( ) FH G
IK J
0 0259
17
x
x
4.7 10
519 10
14 . ln.

or E E eV C F − = 0.176
Then
E E eV F V − = 1.42 − 0.176 = 1.244
and
p x O =
− FH
IK
7 10
1244
0 0259
b 18 gexp
.
.

or p x cm O = − − 9.67 103 3


4.21
p N
exp a f
− − LNM
E E
kT O V
= F V
OQP
or
E E kT
− = FH G
N
p F V
ln V

O
IK J
= ( ) = FH G
IK J
x
15 . ln .
0 0259
104 10
10
0 24
19
.
eV
Then
E − E = 1.12 − 0.24 = 0.88 eV
C F So
a f n N
exp
− − LNM
E E
kT O C
= C F
OQP
=
− FH
IK
19 x exp
2.8 10
088
.
0 .
0259

or
n x cm O = − 4.9 104 3

4.22
(a) p n
− FH
E E
kT O i
= Fi F
IK
exp
x FH
10 . exp .
15 10
0 35
0 .
0259
IK

or
p x cm O = − 111 1016 3 .
(b)
From Problem 4.1, n K x cm i 400 2.38 10( ) = 12 3 −
kT = ( ) = eV FH
IK
0 0259
400
300
. 0.03453
Then
E E kT
− = FH G
p
n Fi F
ln O

i
IK J
=( ) FH G
IK J
x
x
12 . ln .
0 03453
16
111 10
2.38 10

or
E E eV Fi F − = 0.292
(c)
From (a)
n
b g
2 10 2
n
x
= i
=
O
p
x O
15 10
111 10
16
.
.
or
41

n x cm O = − 2.03 104 3
From (b)
n
b g
.
2 12 2
n
x
= i
=
O
p
x O
16
2.38 10
111 10

or
n x cm O = − 510 108 3 .

4.23
(a) p n
− FH
E E
kT O i
= Fi F
IK
exp
= FH
IK
p b 18 . x 10
6 g ex
.
0 35
0 .
0259
or
p x cm O = − 133 1012 3 .
(b) From Problem 4.1,
n K x cm i 400 3 28 10( ) = 9 3 − . , kT = 0.03453 eV
Then
E E kT
p
n Fi F
O
i
− = FH G
IK Jln
=( ) FH G
IK J
9 . ln .
0 03453
12
133 10
328 .
10
x
x

or
E E eV Fi F − = 0.207
(c) From (a)
n
b g
2 6 2
n
x
= i
=
O
p
x O
18 10
133 10
12
.
.
or
n cm O = − 2.44 3
From (b)
n
b g
x
x O =
9 2
328 10
133 10
12
.
.
or
n x cm O = − 8 09 106 3 .

4.24
For silicon, T KE E F V = 300 , =
′ =
E −
E
kT
η = ⇒ (η′) =
V F 0 F 0 60 1/ 2 .
We can write


π / b . g .
2 pO = NV F ( η
′) 2
= 104 x 10 19
( 0 60 ) 1 2
π
or
p = 7.04 x 1018 cm − 3
O
4.25
Silicon, T = 300 Kn , = 5 x 1019 cm − 3
O We have
2
n =
NF O C F η / a f
1 2 π
or
5 10
2
b g aη f /
2.8 10 19 19
1 2 x x F F =
π
which gives
F 1 2 F 158 / aη f = .
Then
η F
F C E E
kT
= =
−
13 .
or E E F C − = (1.3)(0.0259) ⇒
E E eV C F − = −0.034

4.26
For the electron concentration
n E g E f E C F ( ) = ( ) ( )
The Boltzmann approximation applies so
b g a f
( ) F LNM
n E
π * /
4 2 − − 3 2
m
h
E E
E E
kT
n
C
OQP
= −
3
exp
or
b g a f
( ) n C F LNM
n E
π * /
4 2 − − 3 2
m
h
E E
kT
OQP
=
3
exp
×
C exp C a f
− − − LNM
OQP
kT
E E
kT
E E
kT
Define
x
E −
E
kT
= C

Then
n(E)→ n(x) = K x exp(−x)
To find maximum n(E)→ n(x) , set
dn x
dx
( ) ( ) ( ) ( ) LNM
1
2
= 0
= − 1/ 2 exp − + 1/ 2 − 1 exp
− K x x x x
OQP
or
42
0
1 2 = − − − ( )LNM
1
2
OQP
Kx x x / exp
which yields
x
E −
E
kT
= = C
⇒
1
2
1
2
E E kT C = +

For the hole concentration
p E g E f E V F ( ) = ( ) 1− ( )
From the text, using the Maxwell-Boltzmann
approximation, we can write
b g a f
( ) F LNM
p E
π * /
4 2 − − 3 2
m
h
E E
E E
kT
p
V
OQP
= −
3
exp
or
b g a f
( ) p F V LNM
p E
π * /
m
h
E E
kT
OQP
=
4 2 − − 3 2
3
exp
×
V exp V a f
− − − LNM
OQP
kT
E E
kT
E E
kT
Define ′ =
−
x
E E
kT
V
Then
p(x′) = K′ x′ exp(−x′)
To find the maximum of p(E)→ p(x′) , set
dp ( x
′
)
=
dx
′
0 . Using the results from above, we
find the maximum at
E E kT V = −
1
2

4.27
(a) Silicon: We have
n N
exp a f
− − LNM
E E
kT O C
= C F
OQP
We can write
E E E E E E C F C d d F − = a − f+ a − f
For
E E eV E E kT C d d F − = 0.045 , − = 3
n x O =
− LNM
b −
19 gexp
3 OQP
2.8 10
0 .
045
0 .
0259

= 2.8 10 (−4.737) 19 b x gexp
or
n x cm O = − 2.45 1017 3
We also have


exp a f
− − LNM
E E
kT O V
p N
= F V
OQP
Again, we can write
E E E E E E F V F a a V − = a − f+ a − f
For
E E kT E E eV F a a V − = 3 , − = 0.045
Then
p x O= −− LNM
OQP
b 104 p . 10 19 g ex3

.
0 045
0 .
0259
= 104 10 (−4.737) 19 b . x gexp
or
p x cm O = − 9.12 1016 3
(b)
GaAs: Assume E E eV C d − = 0.0058
Then
n x O =
− LNM
b −
17 gexp
3 OQP
4.7 10
0 .
0058
0 .
0259

= 4.7 10 (−3 224) 17 b x gexp .
or
n x cm O = − 187 1016 3 .
Assume E E eV a V − = 0.0345
Then
p x O =
− LNM
b −
18 gexp
3 OQP
7 10
0 .
0345
0 .
0259

= 7 10 (−4.332) 18 b x gexp
or
p x cm O = − 9.20 1016 3

4.28
Computer Plot

4.29
(a) Ge:
p
+ FH
N −
N N −
N
IK
2
n O
2
=
a d +
a d
i 2 2
Then

10
13 10
13 2
p = + + x O2
2
IK J
FH Gb 2.4 10
13 g2
or
p x cm O = − 2.95 1013 3
and
43
n
b g
2 13 2
n
x
O
p
x = i
= ⇒
O
13
2.4 10
2.95 10
n x cm O = − 195 1013 3 .
(b)
n
+ FH
N −
N N −
N
n O
=
d a +
d a
i IK
2 2
2
2
Then
n
x x
= + + x OG
FH IK J
5 10
2
5 10
2
2.4 10
15 15 2
b 13 g2
or
n x cm O ≈ − 5 1015 3
and
p
b g
2 13 2
n
x
O
n
x 2.4 10
5 10
= i
= ⇒
O
15
p x cm O = − 115 1011 3 .

4.30
For the donor level

n
1
N E E
− FHIK
kT
d
=
+
1
1
2
exp
d d F

=
+ FH
IK
1
1
1
2
0 20
0 0259
exp
.
.

or

n
N
= − 885 10 4 .
d x
d
And
f E
E E
kT
F
F
( ) FH
IK
=
+
−
1
1 exp

Now
E E E E E E F C C F − = a − f+ a − f
or
E E kT F − = + 0.245
Then
f E F( ) FH
IK
=
+ +
⇒
1
1 1
0 245
0 0259
exp
.
.

f E x F( )= − 2.87 10 5


4.31
(a) n N x cm O d = = − 2 1015 3
p
b . g
2 10 2
n
x
O
n
x 15 10
2 10
= i
= ⇒
O
15
p x cm O = − 1125 105 3 .
(b)
p N cm O a = = − 1016 3
n
n
p
b . g
2 10 2
x
O
15 10
10
= i
= ⇒
O
16
n x cm O = − 2.25 104 3
(c)
n p n x cm O O i = = = − 15 1010 3 .
(d)
T = 400K ⇒ kT = 0.03453 eV
n x x i
b gb . g exp
2 19 19
FH
3
− 2.8 10 1 04 10
400
300
112
0 03453
=
IK
FH
IK
.
.
or
n x cm i = − 2.38 1012 3
p
N N
= + + i FH
IK
2
a a
n O
2 2
2
= 5 10 + 5 10 + 2.38 10 13 13 2 12 2 x b x g b x g
or
p x cm O = − 10 1014 3 .
Also
n
n
p
b g
2 12 2
x
O
2.38 10
10
= i
= ⇒
O
14
n x cm O = − 566 1010 3 .
(e)
T = 500K ⇒ kT = 0.04317 eV
n x x i
b gb . g exp
2 19 19
IK
3
− FH2.8 10 1 04 10
500
300
112
0 04317
=
FH
IK
.
.
or

n x cm i = − 854 1013 3 .
Now
n
N N
= + + i FH
IK
2
d d
n O
2 2
2
44
= 5 10 + 5 10 + 8 54 10 13 13 2 13 2 x b x g b . x g
or
n x cm O = − 149 1014 3 .
Also
p
b g
2 13 2
n
x
O
n
x = i
= ⇒
O
14
854 .
10
149 .
10
p x cm O = − 4.89 1013 3

4.32
(a) n N x cm O d = = − 2 1015 3
p
b . g
2 6 2
n
x
O
n
x 18 10
2 10
= i
= ⇒
O
15
p x cm O = − − 162 103 3 .
(b)
p N cm O a = = − 1016 3
n
n
p
b . g
2 6 2
x
O
18 10
10
= i
= ⇒
O
16

n = 324 . x 10− 4 cm − 3
O (c)
n = p = n = 18 . x 106 cm − 3
O O i (d)
kT = 0.03453 eV
3
n 2 4.7 x 10 17 7 x 10
18
i
400
300
142
0 03453
=
− FH
IK
FH
IK
b gb g exp
.
.

or
n x cm i = − 328 109 3 .
Now
p N cm O a = = − 1014 3
and
n
n
p
b . g
2 9 2
x
O
328 10
10
= i
= ⇒
O
14

n = 108 . x 105 cm − 3
O (e)
kT = 0.04317 eV
n 2 x 17 x 18
i
3
4.7 10 7 10
500
300
142
0 04317
=
− FH
IK
FH
IK
b gb g exp
.
.

or


n x cm i = − 2.81 1011 3
Now
n N cm O d = = − 1014 3
Also
p
n
n
b g
2 11 2
x
O
2.81 10
10
= i
= ⇒
O
14
p x cm O = − 7.90 108 3

4.33
(a) N N a d > ⇒ p-type
(b) Si:
pO = Na − Nd = 2.5x10 − 1x10 13 13
or
p x cm O = − 15 1013 3 .
Then
n
b g
2 10 2
n
x
O
p
x 15 .
10
15 .
10
= i
= ⇒
O
13
n x cm O = − 15 107 3 .
Ge:
p
+ FH
N −
N N −
N
IK
2
n O
2
=
a d +
a d
i 2 2
= + + FH G
IK J
FH G
IK J
13 13 2
b x g
. x . x 13 2
15 10
2
15 10
2
2.4 10
or
p x cm O = − 326 1013 3 .
Then
n
b g
.
2 13 2
n
x
O
p
x 2.4 10
326 10
= i
= ⇒
O
13

n x cm O = − 177 1013 3 .
GaAs:
p x cm O = − 15 1013 3 .
And
n
b g
2 6 2
n
x
O
p
x 18 .
10
15 .
10
= i
= ⇒
O
13
n cm O = − 0 216 3 .

45
4.34
For T = 450K
n x x i
2 19 19
3
2.8 10 104 10
450
300
= FH
IK
b gb . g
×
−
( )
LN M
OQ P
exp
.
.
112
0 0259 a450 300f
or
n x cm i = − 172 1013 3 .
(a)
N N a d > ⇒ p-type
(b)
p
+ FH
N −
N N −
N
n O
=
a d +
a d
i IK
2 2
2
2
=
15 14 .x x
15 10 − 8 10
2

+
FH G
−
+ IK J
15 10 8 10
2
172 10
15 14 2
. 13 2
.
x x
b x g
or
p N N x cm O a d ≈ − = − 7 1014 3
Then
n
b . g
2 13 2
n
x
O
p
x 172 10
7 10
= i
= ⇒
O
14
n = 4.23 x 1011 cm − 3
O (c)
Total ionized impurity concentration
N = N + N = 15 . x 10 15 + 8 x 10 14
I a d or
N = 2.3 x 1015 cm − 3
I
4.35
b . 10 g
2
2 n
n
x
O
p
x = i
= ⇒
O
5
15 10
2 10

n x cm O = − 1125 1015 3 .
n p O O > ⇒ n-type


4.36
kT = ( ) = eV FH
IK
0 0259
200
300
. 0.01727
= FH
b gb g
2 17 18
n x x i
3
4.7 10 7 10
200
300
IK
×
− LNM
OQP
exp
142
.
0 .
01727

or
n = 138 . cm − 3
i Now
n p = n 2 ⇒ 5 p 2 = n 2
O O i O i or
n
p
O
= i ⇒
5
p cm O = − 0 617 3 .
And
n p O O = 5 ⇒ n cm O = − 309 3 .

4.37
Computer Plot

4.38
Computer Plot

4.39
Computer Plot

4.40
n-type, so majority carrier = electrons
n
N N
= + + i FH
IK
2
d d
n O
2 2
2
= 10 + 10 + 2 10 13 b 13 g2 b x 13 g2
or
n x cm O = − 324 1013 3 .
Then
p
b g
.
2 13 2
n
x
O
n
x 2 10
324 10
= i
= ⇒
O
13

p x cm O = − 123 1013 3 .

4.41
(a) N N d a > ⇒ n-type
n N N x x O d a = − = 2 10 − 1 10 16 16
or
46
n x cm O = − 1 1016 3
Then
p
n
n
b . g
2 10 2
x
O
15 10
10
= i
= ⇒
O
16
p x cm O = − 2.25 104 3
(b)
N N a d > ⇒ p-type
p N N x x O a d = − = 3 10 − 2 10 16 15
or
p x cm O = − 2.8 1016 3
Then
n
b . g
2 10 2
n
x
O
p
x = i
= ⇒
O
16
15 10
2.8 10
n x cm O = − 8 04 103 3 .

4.42
(a) n n O i < ⇒ p-type
(b) n x cm O= ⇒− 4.5 104 3 electrons: minority
carrier
p
b . g
2 10 2
n
x
O
n
x = i
= ⇒
O
4
15 10
4.5 10
p x cm O= ⇒ − 5 1015 3 holes: majority carrier
(c)
p N N O a d = −
so
5 10 5 10 15 15 x N x a = − ⇒ N cm a = − 1016 3
Acceptor impurity concentration,
N x cm d = − 5 1015 3 Donor impurity
concentration

4.43
E E kT
p
n Fi F
O
i
− = FH G
IK J
ln
For Germanium:
T(° K) kT(eV ) n cm i
b −3 g
200
400
600
0.01727
0.03454
0.0518
2.16 1010 x
8 6 1014 . x
382 1016 . x


p
N N
= + + i FH
IK
2
a a
n O
2 2
2 and N cm a = − 1015 3

T(° K) p cm O
b −3 g E E eV Fi F − ( )
200
400
600
10 1015 . x
149 1015 . x
387 1016 . x
0.1855
0.01898
0.000674

4.44
E E kT
− = FH G
n
n F Fi
ln O

i
IK J
For Germanium,
T K n x cm i = ⇒ = − 300 2.4 1013 3
n
N N
= + + i FH
IK
2
d d
n O
2 2
2

N b cm −3 g n b cm −3 g E − E ( eV )
d
O
F Fi 1014
105 . x 1014
0.0382
1016
1016
0.156
1018
1018
0.2755

4.45
n
N N
= + + i FH
IK
2
d d
n O
2 2
2
Now
n n i O = 0.05
so
n x x n O O = 15 10 + 15 10 + (0 05) 15 15 2 2 . b . g .
which yields

n x cm O = − 30075 1015 3 .
Then
n x cm i = − 1504 1014 3 .
We have
n N N
− FH G
E
kT i C V
2 g =
IK J
exp
so

47
1504 10 4.7 10 7 10
300
14 2 17 18
3
. x x x
b g b gb g T = FH
IK

×
−
( )
LN M
OQ P
exp
.
.
142
0 0259 aT 300f
By trial and error
T ≈ 762K

4.46
Computer Plot

4.47
Computer Plot

4.48
(a) E E kT
m
m Fi midgap
p
n
− = FH G
IK J
3
4
ln
*
*
= ( ) ( )⇒ 3
4
0.0259 ln 10
(b)
Impurity atoms to be added so
E E eV midgap F − = 0.45
(i) p-type, so add acceptor impurities
(ii) E E eV Fi F − = 0.0447 + 0.45 = 0.4947
p n
= FH
E −
E
O i
kT = Fi F
IK
FH
IK
exp exp
.
.
10
0 4947
0 0259
5
or
p N x cm O a = = − 197 1013 3 .

4.49
n N N N
exp a f
− − LNM
E E
kT O d a C
= − = C F
OQP
so
N x x d= +
− FH
IK
5 10 2.8 10
0 215
0 0259
15 19 exp
.
.

= 5 10 + 6 95 10 15 15 x . x
so
N x cm d = − 12 1016 3 .

4.50
(a) p N N
exp a f
− − LNM
E E
kT O a V
= = F V
OQP
or


exp
a f 104 10
OQP
= = LNM
E E
kT
N
N
+ − .
x
x
F V V
a
7 10
19
15
= 149 103 . x
Then
E E x F V − = (0 0259) 149 103 . lnb . g
or
E E eV F V − = 0.189
(b)
If E E eV F V − = 0.1892 − 0.0259 = 0.1633
Then
N x a =
− FH
IK
19 . exp .
104 10
01633
0 .
0259

= − 190 1016 3 . x cm
so that
ΔN x x a= 1 90 10 − 7 10 ⇒ 16 15 .
ΔN x cm a = − 12 1016 3 .
Acceptor impurities to be added

4.51
(a) E E kT
− = = FH G
IK J
( ) FH G
N
ln d
. ln
F Fi
n x i
IK J
15
10
15 .
10
0 0259
10

or
E E eV F Fi − = 0.2877
(b)
E E kT
N
− = = n
G
FH IK J
eV Fi F
ln a
0.2877
i
(c)
For (a), n N cm O d = = − 1015 3
For (b)
n
n
p
b . g
2 10 2
x
O
15 10
10
= i
= ⇒
O
15
n x cm O = − 2.25 105 3

4.52
E E kT
− = FH G IK J
p
n Fi F
ln O

i
= ( ) = FH G
IK J
p
n
O eV
i
0.0259 ln 0.45

Then
48
p x O= ⇒ FH
IK
p b 18 . 10
6 .
g ex
0 45
0 .
0259
p x cm O = − 6 32 1013 3 .
Now
p N O a < , Donors must be added
p N N N N p O a d d a O = − ⇒ = −
so
N x d= 10 − 6 32 10 ⇒ 15 13 .
N x cm d = − 9.368 1014 3

4.53
(a) E E kT
N
n F Fi
d
i
− = FH G
IK J
ln
= ( ) ⇒ FH G
IK J
0 0259
15
x
x
2 10
15 10
10 . ln.

E E eV F Fi − = 0.3056
(b)
E E kT
N
n Fi F
a
i
− = FH G
IK Jln
= ( ) ⇒ FH G
IK J
0 0259
16
10
15 10
10 . ln.
x

E E eV Fi F − = 0.3473
(c)
E E F Fi =
(d)
kT eV n x cm i = = − 0 03453 2.38 1012 3 . ,
E E kT
p
n Fi F
O
i
− = FH G
IK J
ln
= ( ) ⇒ FH G
IK J
0 03453
14
10
2.38 10
12 . lnx

E E eV Fi F − = 0.1291
(e)
kT eV n x cm i = = − 0 04317 8 54 1013 3 . , .
E E kT
n
n F Fi
O
i
− = FH G
IK J
ln
= ( ) ⇒ FH G
IK J
13 . ln .
0 04317
14
149 10
854 .
10
x
x

E E eV F Fi − = 0.0024


4.54
(a) E E kT
− = FH G
N
n F Fi
ln d

i
IK J
= ( ) ⇒ FH G
IK J
0 0259
15
x
x
2 10
18 10
6 . ln.

E E eV F Fi − = 0.5395
(b)
E E kT
− = FH G
N
n Fi F
ln a

i
IK J
= ( ) ⇒ FH G
IK J
0 0259
16
10
18 10
6 . ln.
x

E E eV Fi F − = 0.5811
(c)
E E F Fi = EFi −
(d)
kT eV n x cm i = = − 0 03453 3 28 109 3 . , .
E E
x Fi F − = ( ) ⇒ FH G
IK J
0 03453
14
10
328 10
9 . ln.

E E eV Fi F − = 0.3565

49
(e)
kT eV n x cm i = = − 0 04317 2.81 1011 3 . ,
E E kT
n
n F Fi
O
i
− = FH G
IK J
ln
= ( ) ⇒ FH G
IK J
0 04317
14
10
2.81 10
11 . lnx

E E eV F Fi − = 0.2536

4.55
p-type
E E kT
p
n Fi F
O
i
− = FH G
IK J
ln
= ( ) ⇒ FH G
IK J
0 0259
15
x
x
5 10
15 10
10 . ln.

E E eV Fi F − = 0.3294


(page left blank)
50


Chapter 5

Problem Solutions

5.1
(a) n cm O = − 1016 3
and
p
n
n
b . g
2 6 2
x
O
18 10
10
= i
= ⇒
O
16
p x cm O = − − 324 104 3 .
(b)
J e nn O = μ Ε
For GaAs doped at N cm d = − 1016 3 ,
μ n ≈ 7500 cm V − s 2/
Then
J = x 16 10− (7500) 10 (10) 19 16 b . g b g
or
J = 120 A cm2 /
(b) (i) p cm O = − 1016 3 , n x cm O = − − 324 104 3 .
(ii) For GaAs doped at N cm a = − 1016 3 ,
μ p ≈ 310 cm V − s 2 /
J e p p O = μ Ε
= ⇒ 16 10− (310) 10 (10) 19 16 b . x g b g
J = 4.96 A cm2 /

5.2
(a) V = IR ⇒10 = (0.1R)⇒
R = 100 Ω
(b)
R
L
A
L
RA
σ
= ⇒ = ⇒
σ
−
3
b 3 g
10
100 10
σ= ⇒
( ) −
σ = ( − )− 0 01 1 . Ω cm
(c)
σ ≈ eμ N n d
or
0 01 16 10 1350 19 . = . x − ( )Nd b g
or
53
N x cm d = − 4.63 1013 3
(d)
σ ≈ eμ p ⇒ p O
0 01 16 10 480 19 . = . x − ( )pO b g
or
p x cm N N N O a d a = = − = − − 130 10 10 14 3 15 .
or
N x cm a = − 113 1015 3 .
Note: For the doping concentrations obtained,
the assumed mobility values are valid.

5.3
(a) R
L
A
L
A
ρ
= =
σ
and σ ≈ eμ N n d
For N = 5 x 1016 cm − 3 , μ ≈ 1100 cm 2/ V − s
d n Then
01
.
R
b
16 . x 10 19 g 1100 b 5 x
10 16 g 100 b 10 4 g 2
=
− ( ) ( ) −
or
R = 1136x104 . Ω
Then
I
V
5
R x
= = ⇒
1136 104 .
I = 0.44 mA
(b)
In this case
R = 1136x103 . Ω
Then
I
V
5
R x
= = ⇒
1136 103 .
I = 4.4 mA
(c)
Ε =
V
L

5
010
For (a), Ε = =
50
.
V / cm
And
vd n = μ Ε = (1100)(50) or v x cm s d = 55 104 . /
For (b), Ε = = =
V
L
V cm
5
0 01
500
.
/
And
vd= (1100)(500)⇒ v x cm s d = 55 105 . /


5.4
(a) GaAs:
R
L
A
V
I
k
L
A
ρ
10
20
05 . Ω
= = = = =
σ
Now
σ ≈ eμ N p a
For N cm a = − 1017 3 , μ p ≈ 210 cm V − s 2 /
Then
σ= = − − − 1 6 10 (210) 10 3 36 ( ) 19 17 1 b . x g b g . Ω cm
So
L = R A = ( )( ) x − σ 500 3 36 85 10 8 . b g
or
L = 14.3 μm
(b) Silicon
For N cm a = − 1017 3 , μ p ≈ 310 cm V − s 2 /
Then
σ= = − − − 1 6 10 (310) 10 4.96 ( ) 19 17 1 b . x g b g Ω cm
So
L = R A = ( )( ) x − σ 500 4.96 85 10 b 8 g
or
L = 21.1 μm

5.5
(a) Ε = = =
V
L
V cm
3
1
3 /
v
v
= μ Ε ⇒ μ = d =
d n n
Ε
10
3
4

or
μ n = 3333 cm V − s 2 /
(b)
vd n = μ Ε = (800)(3)
or
v x cms d = 2.4 103 /

5.6
(a) Silicon: For Ε = 1 kV / cm ,
v x cms d = 12 106 . /
Then
t
− 10
12 10
d
v x t
4
= = ⇒
d
6 .
t x s t = − 8 33 10 11 .
For GaAs, v x cms d = 7.5 106 /
54

Then
t
d
v x t
= = ⇒
d
10
− 4
7.5 10
6 t x s t = − 133 10 11 .
(b)
Silicon: For Ε = 50 kV / cm ,
v x cms d = 9.5 106 /
Then
t
d
v x t
= = ⇒
d
10
− 4
9.5 10
6 t x s t = − 105 10 11 .
GaAs, v x cm s d = 7 106 /
Then
t
d
v x t
= = ⇒
d
10
− 4
7 10
6 t x s t = − 143 10 11 .

5.7
For an intrinsic semiconductor,
σ μ μ i i n p = en b + g
(a)
For N N cm d a = = − 1014 3 ,
μ μ n p = 1350 cm V − s = 480 cm V − s 2 2 / , /
Then
σ i = x x + 16 10− 15 10 (1350 480) 19 10 b . gb . g
or
σ i = x − cm − − 4.39 10 ( ) 6 1 Ω
(b)
For N N cm d a = = − 1018 3 ,
μ μ n p ≈ 300 cm V − s ≈ 130 cm V − s 2 2 / , /
Then
σ i = x x + 16 10− 15 10 (300 130) 19 10 b . gb . g
or
σ i = x − cm − − 1 03 10 ( ) 6 1 . Ω

5.8
(a) GaAs
σ ≈ μ ⇒ = μ − e p x p p O p O 5 16 1019 b . g
From Figure 5.3, and using trial and error, we
find
p x cm cm V s O p ≈ ≈ − − 13 10 240 17 3 2 . , μ /
Then


n
b g or n x cm O = − − 2.49 105 3
2 6 2
n
x
= i
=
O
p
x O
17
18 .
10
13 .
10

(b) Silicon:
1
σ
e nn O
= ≈ μ
ρ
or
n
1 1
8 16 10 1350 19 ρ μ b . g
= =
( ) − ( )
e x O
n
or
n x cm O = − 579 1014 3 .
and
p
b g p x cm O = − 389 105 3 .
2 10 2
n
x
O
n
x 15 .
10
579 .
10
= i
= ⇒
O
14
Note: For the doping concentrations obtained in
part (b), the assumed mobility values are valid.

5.9
σ μ μ i i n p = en b + g
Then
10 16 10 1000 600 −6 −19 = . x ( + )ni b g
or
n K x cm i 300 3 91 10( ) = 9 3 − .
Now
n N N
− FH G
E
kT i C V
2 g =
IK J
exp
or
= = FH G
IK J
L
b 19 g
2
b g
Q PPln . ln
N MM
10
391 10
N N
n x g
E kT
C V
9 2 0 0259
. 2
i
( )
O
or
E eV g = 1.122
Now
n K i
=
b g − exp a f
2 19 2 500 10
1122
.
( )
0 0259 500 300
( )
LN M
OQ P
.

= 515 1026 . x
or
n K x cm i 500 2.27 10( ) = 13 3 −
Then
σ i = x x + 16 10− 2.27 10 (1000 600) 19 13 b . gb g
so
σ i 500K 5 81x10 cm 3 1 ( ) = ( − ) − − . Ω
55

5.10
(a) (i) Silicon: σ μ μ i i n p = en b + g
σ i = x x + 16 10− 15 10 (1350 480) 19 10 b . gb . g
or
σ i = x − cm − − 4.39 10 ( ) 6 1 Ω
(ii) Ge:
σ i = x x + 16 10− 2.4 10 (3900 1900) 19 13 b . gb g
or
σ i = x − cm − − 2.23 10 ( ) 2 1 Ω
(iii) GaAs:
σ i = x x + 1 6 10− 18 10 (8500 400) 19 6 b . gb . g
or
σ i = x − cm − − 2.56 10 ( ) 9 1 Ω
(b) R
L
A
=
σ

(i) R
x
−
200 10
4
b 6 gb 8 g
= ⇒
− −
x x
4.39 10 85 10
R = 5 36x109 . Ω
(ii) R
x
−
200 10
4
b 2 gb 8 g
= ⇒
− −
x x
2.23 10 85 10
R = 1 06x106 . Ω
(iii) R
x
−
200 10
4
b 9 gb 8 g
= ⇒
− −
x x
2.56 10 85 10
R = 9.19x1012 Ω

5.11
(a) ρ
μ
= 5 =
1
e Nn d

Assume μ = 1350 cm 2/ V − s
n Then
1
N
16 10 1350 5 19 b . g
x d= ⇒ − ( )()
N x cm d = − 9.26 1014 3
(b)
T = 200K → T = −75C
T = 400K → T = 125C
From Figure 5.2,
T CN cm d = − = ⇒ − 75 1015 3 ,


μ ≈ 2500 cm 2/ V − s
n T = 125 CN , = 1015 cm − 3 ⇒
d μ ≈ 700 cm 2/ V − s
n Assuming n = N = 9.26 x 1014 cm − 3 over the
O d temperature range,
For T = 200K ,
1
ρ= ⇒ − ( )
1 6 10 2500 9.26 10 19 14 b . x g b x g
ρ = 2.7 Ω − cm
For T = 400K ,
1
ρ= ⇒ − ( )
1 6 10 700 9.26 10 19 14 b . x g b x g
ρ = 9.64 Ω − cm

5.12
Computer plot

5.13
(a) Ε = 10V cm⇒ v = Ε d n / μ
vd= (1350)(10)⇒ v x cm s d = 135 104 . /
so
T mv x x n d = = ( ) − 1
2
1 08 9.11 10 135 10 * 2 31 2 2 . b gb . g
1
2
or
T = x J ⇒ x eV − − 8 97 10 56 10 27 8 . .
(b)
Ε = 1 kV / cm ,
v x cm s d= (1350)(1000) = 1 35 106 . /
Then
T = ( ) x − x 1
2
108 9.11 10 135 10 31 4 2 . b gb . g
or
T = x J ⇒ x eV − − 8 97 10 56 10 23 4 . .

5.14
(a) n N N
− FH G
E
kT i C V
2 g =
IK J
exp
19 19 b x gb x gexp
= 2 10 1 10
− FH
110
.
0 .
0259
IK

= ⇒ = − 7.18 10 8 47 10 19 9 3 x n x cm i .
For N cm n n cm d i O = >> ⇒ = − − 10 10 14 3 14 3
Then
56
J enn O = σΕ = μ Ε
= 16 10− (1000) 10 (100) 19 14 b . x g b g
or
J = 1 60 A cm2 . /
(b)
A 5% increase is due to a 5% increase in electron
concentration. So
i = = + + FH
n x
N N
d d
n O
IK
105 10
2 2
14
2
2 .
We can write
b 1 . 05 x 10 14 − 5 x 10 13 g 2 = b 5 x 10 13 g 2 + 2
so
n 2 = 5.25 x 10 26
i
ni
=
− FH
IK J2 10 1 10
IK
FH G
b gb g exp g
300
19 19
3
x x
T E
kT
which yields
2.625 10
110 12
300
3
x
T
kT
− =
− FH
IK
FH
IK
exp
.

By trial and error, we find
T = 456K

5.15
(a) σ = eμ n + eμ p n O p O and n
n
p O
i
O
=
2

Then
σ
e μ
n
p
n i e p
O
= + μ
p O
2

To find the minimum conductivity,

d
dp
e n
p
σ μ
e
O
= = n i
μ
O
p
−
+ ⇒
( )
0
1 2
2
which yields
p n O i
n
p
=
FH G
IK J
μ
μ
1/ 2
(Answer to part (b))
Substituting into the conductivity expression
σ σ
e μ
n /
b g b g
μ μ
= = + μ μ μ min /
n
n i e n
i n p
p i n p
2
1 2
1 2
which simplifies to
σ μμ min = 2eni n p
The intrinsic conductivity is defined as

(Neamen)solution manual for semiconductor physics and devices 3ed

(Neamen)solution manual for semiconductor physics and devices 3ed

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