半导体物理与器件(尼曼)02章答案

2
Chapter 2 Problem Solutions
2.19 1 ⋅ ∂Ψ1 Note that
2 m Ψ1 ∂x Ψ1 ∂t Subtracting these last two equations, we are left with
2
⋅
1
⋅
∂ Ψ1
2
+ V ( x ) = jh ⋅
z
∞ 0
Ψ ⋅ Ψ dx = 1
−26 −34 −10 −20
J ⇒ ∆E = 0.198 eV
2m and −h
2
⋅
∂ Ψ1 ( x , t )
2
∂x
2
2
+ V ( x )Ψ1 ( x , t ) = jh
∂Ψ1 ( x , t ) ∂t ∂Ψ2 ( x , t ) ∂t
⇒
∆p = 8.78 x10 (b) ∆E = 1 2 ⋅
−26
kg − m / s
( ∆p)2
m
−26
=
1 2
b8.78x10 g ⋅
−26
2
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 equation, then we could write −h
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 2 Problem Solutions
Chapter 2
Problem Solutions
2.1 Computer plot 2.2 Computer plot 2.3 Computer plot 2.4 For problem 2.2; Phase = 2πx or p = 5.4 x10
aΨ ⋅ Ψ f
1 2 2
1.054 x10 10
−2
−34
= 1.054 x10 1.054 x10 1500
−32
−32
p = mv ⇒ ∆v = or ∆v = 7 x10 2.15 (a) ∆p = h ∆x =
∆p m
−36
=
⇒
m/s
1.054 x10 10
−10 −24
−34
⇒ kg − m / s
∆p = 1.054 x10
LMΨ ∂ Ψ + Ψ ∂ Ψ + 2 ∂Ψ ⋅ ∂Ψ OP 2m N ∂x ∂x Q ∂x ∂x LM ∂Ψ + Ψ ∂Ψ OP +V ( x )Ψ ⋅ Ψ = jh Ψ N ∂t ∂t Q Dividing by Ψ ⋅ Ψ we find −h L 1 ∂ Ψ 1 ∂ Ψ 1 ∂Ψ ∂Ψ O ⋅ + ⋅ + M P 2 m N Ψ ∂x Ψ ∂x Ψ Ψ ∂x ∂x Q L 1 ∂Ψ + 1 ∂Ψ OP +V ( x ) = jh M N Ψ ∂t Ψ ∂x Q
g
λ=
J
h p
=
6.625 x10 3.13x10
−22
b6.625x10 gb3x10 g ⇒ 6.54 x10 λ= (1.90)b1.6 x10 g
−19
or
−5
λ = 0.0212 A
cm
°
or
λ = 0.654 µm
(d) A 2000 kg traveling at 20 m/s: p = mv = (2000)(20) ⇒ or p = 4 x10 kg − m / s
−34 10
J
J
2 mT
2(183.92 ) 1.66 x10 p = 313 . x10
−22
−19
cm
b
−27
gb1.6x10 g
−19
or kg = m / s ⇒
−34
or
λ = 0.254 µm
So
Cesium: E = 1.90 eV = (1.90) 1.6 x10
−34 10
b
−19
−31 4
−23
⇒ λ = 0.11 A
2
10
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 2 Problem Solutions
2.11 (a) ∆p = h ∆x = 1.054 x10 10
J
Then λ 2π dx dx ⋅ + ω = 0 or = v p = −ω λ dt 2π dt 2.5 E = hν = hc
λ
+ ωt = constant
kg − m / s ⇒
λ=
or
h p
=
6.625 x10 2.31x10
−34
−23
λ = 0.287 A λ
⇒λ = hc E
−19
−31
−3
p = mv = 9.11x10
Now
b
gb2 x10 g ⇒
4
p = 1822 x10 . h p 6.625 x10 1822 . x10
°
−26
kg − m / s
b
−31
g(0.01727)b1.6x10 g
−19
−34 −26
λ=
=
⇒
pavg = 7.1x10
Now h p 6.625 x10 7.1x10
−6 −28 −34
⇒ kg − m / s
(b) ∆t = or
1.054 x10
−34
(1) 1.6 x10 −19
∆p = 1.054 x10 (b) E= hc = hc p
b
g⇒
s
F I = pc H hK λ So ∆E = c( ∆p) = b3x10 gb1.054 x10 g ⇒
8 −28
−34 8
1x10
−10
E = 1.99 x10 = hc 100h
−31
−15
J
= So
2 9.11x10
b
λp
⋅ 2 mc =
8 2
2 mc
2
E = e ⋅ V ⇒ 1.99 x10
so
3
−15
= 1.6 x10
b
−19
gV
−15
100
gb3x10 g
V = 12.4 x10 V = 12.4 kV ⇒ (b) p = Then 2 mE =
2
5 x10
−26
⇒
−7
∂
2 2
∆E = 7.71x10 2.14 ∆p = h ∆x =
J ⇒ ∆E = 4.82 x10 eV
2 m ∂x
aΨ ⋅ Ψ f + V ( x)aΨ ⋅ Ψ f
1 2 1 2
= jh which can be written as −h
2 2 2 1 2 2 2
∂ ∂t
Ψ1 ( x , t ) + Ψ2 ( x , t )
+V ( x ) Ψ1 ( x , t ) + Ψ2 ( x , t )
= jh ∂
∆E = 7.71x10 2.13
J ⇒ ∆E = 4.82 x10 eV
(a) Same as 2.12 (a), ∆p = 8.78 x10 (b) ∆E = 1 2 ⋅
4 −34
2.6 (a) Electron: (i) K.E. = T = 1 eV = 1.6 x10 p= or 2 mT = 2 9.11x10
−19
b
−31
gb1.6x10 g
−19
J or
λ=
h p
=
6.625 x10 4 x10
4
⇒
λ = 1.66 x10
−28
A
°
9
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
kg − m / s
−26 2
∂x 2m Adding the two equations, we obtain
2
⋅
∂ Ψ2 ( x , t )
+ V ( x )Ψ2 ( x , t ) = jh
−h ⇒
−4
2
( ∆p)2
m
−23
=
1 2
b8.78x10 g ⋅
5 x10
−29
2 m ∂x
⋅
∂
2 2
1 1 2 2 1 1 2 1 2 1 2 2 2 2 2 1 1 2 2 2 2 1 1 2 2 1 2 1
11
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Since Ψ1 is a solution, then −h
−26
λ = 364 A
kg − m / s
(b) p= ⇒ Also v= or p m = 5.3 x10 h
−34
λ=
or
=