1000 Solved Problems in Modern Physics

2.3 Solutions 123

EJ=[J(J+1)^2 c^2 ]/(2)(0.5)(Mpc^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 ΔEJ=

J^2

Io

=

J^2

μr^2

ΔEJ=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 =

16 x

16 +x

(2)

Using (2) in (1) and solving forx, we getx = 13 .004. Hence the mass

number is 13.

2.66 En=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 forn=8.

2.3.7 Commutators .....................................

2.67 (a) Writingx=i

∂

∂p

(

eipα/

(

i

∂

∂p

)

e−ipα/

)

ψ|p|

=ieipα/

[

−

iα



e

ipα

ψ(p)+e−ipα/

∂ψ(p)

∂p

]

=α+

i∂ψ(p)

∂p

=α+x

(b) IfAandBare Hermitian

(AB)†=B†A†=BA

If the product is to be Hermitian then (AB)†=ABi.e.AB=BA. Thus,

AandBmust commute with each other.