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