1000 Solved Problems in Modern Physics

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