106
2 Quantum Mechanics – I
ΔE n,1 =
α
2 m e c
2
4
1
1 2 −
1
n 2
, n = 2, 3, 4, 5
Thus,
ΔE 21 =
1
137
2
0.511 × 10
6
4
1 −
1
4
= 5.1 eV
λ 21 =
1,241
5.1
= 243.3 nm = 2,433 ˚
A
The wavelengths of the other three lines can be similarly computed. They are
2,053, 1,946 and 1,901 ˚
A.
2.15 (a) Using Bohr’s theory of hydrogen atom
E n = −
μe
4
8ε
2
0 h 2 n 2
(1)
where μ is the reduced mass. But the fine structure constant
α =
e
2
4πε 0 c
(2)
Combining (1) and (2)
E n = −
α
2
μc
2
2n 2
(3)
For positronium, = m e /2. Therefore for positron
E n = −
α
2 m e c
2
4n 2
(4)
(b) r n = ε 0
n
2 h
2
πμc 2
(5)
r n ∝
1
μ
=
2
m e
Therefore the radii are doubled.
(c) E n ∝ μ =
m e
2
Therefore the transition energies are halved.
2.16 (a) U (r ) = −
f (r )dr =
kr dr + C
=
1
2
kr
2
+ C
U (0) = 0 → C = 0
U (r ) =
1
2
kr
2
(b) Bohr’s assumption of quantization of angular momentum gives
mvr = n
(1)
Equating the attracting force to the centripetal force.
2 Quantum Mechanics – I
ΔE n,1 =
α
2 m e c
2
4
1
1 2 −
1
n 2
, n = 2, 3, 4, 5
Thus,
ΔE 21 =
1
137
2
0.511 × 10
6
4
1 −
1
4
= 5.1 eV
λ 21 =
1,241
5.1
= 243.3 nm = 2,433 ˚
A
The wavelengths of the other three lines can be similarly computed. They are
2,053, 1,946 and 1,901 ˚
A.
2.15 (a) Using Bohr’s theory of hydrogen atom
E n = −
μe
4
8ε
2
0 h 2 n 2
(1)
where μ is the reduced mass. But the fine structure constant
α =
e
2
4πε 0 c
(2)
Combining (1) and (2)
E n = −
α
2
μc
2
2n 2
(3)
For positronium, = m e /2. Therefore for positron
E n = −
α
2 m e c
2
4n 2
(4)
(b) r n = ε 0
n
2 h
2
πμc 2
(5)
r n ∝
1
μ
=
2
m e
Therefore the radii are doubled.
(c) E n ∝ μ =
m e
2
Therefore the transition energies are halved.
2.16 (a) U (r ) = −
f (r )dr =
kr dr + C
=
1
2
kr
2
+ C
U (0) = 0 → C = 0
U (r ) =
1
2
kr
2
(b) Bohr’s assumption of quantization of angular momentum gives
mvr = n
(1)
Equating the attracting force to the centripetal force.