3.3 Solutions 181
Using (11) in (9) and notingk^21 +k^22 =^2 mV 2 bk^22 =2 mVb
^2andk^21 k^22 =(
2 m
^2) 2
E(Vb−E)we findT=
1
1 +V
b^2 sin^2 k^2 a
4 E(E−Vb)(13)
andR=
1
1 +^4 VE 2 (E−V^0 )
0 sin^2 k^2 a(14)
A typical graph forTversusVEbis shown in Fig. 3.12Fig. 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.