3.3 Solutions
193
3.44 (a)
Fig. 3.18 Penetration of a
rectangular barrier
(b) Region 1, x < 0
d
2
ψ
dx 2 + k
2
ψ = 0
with k
2
=
2mE
2
ψ 1 = Ae
ikx
+ Be
−ikx
Incident reflected at x = 0
Region 2, 0 < x < L
d
2
ψ
dx 2 − α
2
ψ = 0
with α
2
=
2m(W −E)
2
ψ 2 = Ce
−αx
+ De
αx
Region 3, x > L
d
2
ψ
dx 2 + k
2
ψ = 0
with k
2
= 2m E/
2
ψ 3 = Fe
ikx
The second term is absent as there is no reflected wave coming from
right to left
The transmission coefficient T =
|F|
2
| A| 2
(c) Boundary conditions
ψ 1 (0) = ψ 2 (0)
dψ 1
dx
x=0
=
dψ 2
dx
x=0
ψ 2 (L) = ψ 3 (L)
dψ 2
dx
x=L
=
dψ 3
dx
x=L
(d) T = 16
E
W
1 −
E
W
e
−2αL
α
2
= 2m
W − E
2
→ α =
2mc 2 (W − E)
c
=
(2 × 0.511 × (5 − 2) × 10 −6
197.3 × 10 −15
= 8.8748 × 10
9 m
−1