1000 Solved Problems in Modern Physics

(Grace) #1

3.3 Solutions 181


Using (11) in (9) and notingk^21 +k^22 =^2 mV 2 b

k^22 =

2 mVb
^2

andk^21 k^22 =

(

2 m
^2

) 2

E(Vb−E)

we find

T=

1

1 +V

b^2 sin^2 k^2 a
4 E(E−Vb)

(13)

and

R=

1

1 +^4 VE 2 (E−V^0 )

0 sin^2 k^2 a

(14)

A typical graph forTversusVEbis shown in Fig. 3.12

Fig. 3.11Transmission
through a rectangular
potential barrier


Fig. 3.12Transmission as a
function ofE/Vb


3.32 The form of potential corresponds to that of a linear Simple harmonic Oscil-
lator. The energy of the oscillator will beE 1 = 2 ωandE 2 =^3  2 ω.

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