1000 Solved Problems in Modern Physics

(Grace) #1
3.3 Solutions 193

3.44 (a)
Fig. 3.18Penetration of a
rectangular barrier


(b) Region 1,x< 0
d^2 ψ
dx^2

+k^2 ψ= 0
withk^2 =^2 mE 2
ψ 1 =Aeikx+Be−ikx
Incident reflected atx= 0
Region 2, 0<x<L
d^2 ψ
dx^2

−α^2 ψ= 0

withα^2 =^2 m(W 2 −E)
ψ 2 =Ce−αx+Deαx
Region 3,x>L
d^2 ψ
dx^2

+k^2 ψ= 0
withk^2 = 2 mE/^2
ψ 3 =Feikx
The second term is absent as there is no reflected wave coming from
right to left
The transmission coefficientT=|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 −WE

)

e−^2 αL

α^2 = 2 m

(

W−E

^2

)

→α=


2 mc^2 (W−E)
c

=


(2× 0. 511 ×(5−2)× 10 −^6

197. 3 × 10 −^15

= 8. 8748 × 109 m−^1
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