174
3 Quantum Mechanics – II
The ground state corresponds to n = 1 and the first excited state to n =
2, m = 8m e and L = 1 nm = 10
6 fm. Putting n = 1 in (3)
hv = E 2 − E 1 =
3h
2
8m L 2 = 3π
2
2 c
2
/16m e c
2 L
2
= 3π
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 +
2m
2
[E − V (r )] ψ = 0
Region 1 (E < V 0 )
d
2
ψ
dx 2 −
2m
2
(V 0 − E)ψ = 0
( 1 )
d
2
ψ
dx 2 − β
2
ψ = 0
( 2 )
where β
2
=
2m
2
(V 0 − E)
( 3 )
ψ 1 = Ae
βx
+ Be
−βx
(4)
where A and B are constants of integration.
Since x is negative in region 1, and ψ 1 has to remain finite we must set B = 0,
otherwise the wave function grows exponentially. The physically accepted
solution is
ψ 1 = Ae
βx
(5)
Region 2; (V = 0)
Fig. 3.8 Square potential
well of finite depth
3 Quantum Mechanics – II
The ground state corresponds to n = 1 and the first excited state to n =
2, m = 8m e and L = 1 nm = 10
6 fm. Putting n = 1 in (3)
hv = E 2 − E 1 =
3h
2
8m L 2 = 3π
2
2 c
2
/16m e c
2 L
2
= 3π
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 +
2m
2
[E − V (r )] ψ = 0
Region 1 (E < V 0 )
d
2
ψ
dx 2 −
2m
2
(V 0 − E)ψ = 0
( 1 )
d
2
ψ
dx 2 − β
2
ψ = 0
( 2 )
where β
2
=
2m
2
(V 0 − E)
( 3 )
ψ 1 = Ae
βx
+ Be
−βx
(4)
where A and B are constants of integration.
Since x is negative in region 1, and ψ 1 has to remain finite we must set B = 0,
otherwise the wave function grows exponentially. The physically accepted
solution is
ψ 1 = Ae
βx
(5)
Region 2; (V = 0)
Fig. 3.8 Square potential
well of finite depth