2.3 Solutions
123
E J = [J (J + 1)
2 c
2 ]/(2)(0.5)(M p c
2 )r
2
E 2 = (2)(3)(197.3)
2 (10
−15 )
2
/(940) × (10
−10 )
2
= 0.264 × 10
−10 MeV
= 2.64 × 10
−5 eV
2.65 ΔE J =
J
2
I o
=
J
2
μr 2
ΔE J = hc/λ
∴ λ ∝ μ
λ 1
λ 2
=
0.00260
0.00272
=
μ 1
μ 2
(1)
μ 1 =
16 × 12
16 + 12
; μ 2 =
16x
16 + x
(2)
Using (2) in (1) and solving for x, we get x = 13.004. Hence the mass
number is 13.
2.66 E n = ω
n +
1
2
=
k
μ
n +
1
2
4.5 =
1.054 × 10
−34
1.6 × 10 −19
573
0.5 × 1.67 × 10 −27
1/2
n +
1
2
whence n = 7.75
Therefore the molecule would dissociate for n = 8.
2.3.7 Commutators
2.67 (a) Writing x = i
∂
∂ p
e
i pα/
i
∂
∂ p
e
−i pα/
ψ| p|
= i e
i p α/
−
iα
e
ipα
ψ( p) + e
−i pα/ ∂ψ( p)
∂ p
= α +
i ∂ψ( p)
∂ p
= α + x
(b) If A and B are Hermitian
( AB)
†
= B
† A
†
= B A
If the product is to be Hermitian then (AB)
†
= AB i.e. AB = B A. Thus,
A and B must commute with each other.