1000 Solved Problems in Modern Physics

(Grace) #1

174 3 Quantum Mechanics – II


The ground state corresponds ton=1 and the first excited state ton=
2 ,m= 8 meandL=1nm= 106 fm. Puttingn=1in(3)

hv=E 2 −E 1 =

3 h^2
8 mL^2

= 3 π^2 ^2 c^2 / 16 mec^2 L^2

= 3 π^2 (197.3)^2 MeV^2 −fm^2 /(16× 0 .511 MeV)(10^6 )^2 fm^2
= 0. 14 × 10 −^6 MeV= 0 .14 eV

λ(nm)=

1 , 241

E(eV)

=

1 , 241

0. 14

=8864 nm

This corresponds to the microwave region of the electro-magnetic spectrum.

3.25 Consider a finite potential well. Take the origin at the centre of the well.


V(x)=V 0 ;|x|>a
=0;|x|<a

d^2 ψ
dx^2

+

(

2 m
^2

)

[E−V(r)]ψ= 0

Region 1 (E<V 0 )

d^2 ψ
dx^2


(

2 m
^2

)

(V 0 −E)ψ=0(1)

d^2 ψ
dx^2

−β^2 ψ=0(2)

whereβ^2 =

(

2 m
^2

)

(V 0 −E)(3)

ψ 1 =Aeβx+Be−βx (4)

whereAandBare constants of integration.
Sincexis negative in region 1, andψ 1 has to remain finite we must setB=0,
otherwise the wave function grows exponentially. The physically accepted
solution is
ψ 1 =Aeβx (5)

Region 2; (V=0)

Fig. 3.8Square potential
well of finite depth

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