1000 Solved Problems in Modern Physics

3.3 Solutions 207

c

(

1

λj+ 1

−

1

λj

)

=c×(20.556 cm−^1 )=

^2

I 0

Moment of inertia I 0 =



4 π^2 c× 20. 556

=

6. 63 × 10 −^27 erg−s−^1

4 π^2 × 3 × 1010 cm−s−^1 × 20 .556 cm−^1

= 2. 727 × 10 −^40 g−cm^2

I 0 =μr^2

μ=

m(H)m(Cl)

[m(H)+m(Cl)]

=

1 × 35 × 1. 67

1 + 35

× 10 −^24 g

= 1. 62 × 10 −^24 g

r=

(

I 0

μ

) 1 / 2

=

(

2. 727 × 10 −^40

1. 62 × 10 −^24

)

= 1. 3 × 10 −^8 cm= 1. 3 A ̊

3.60 For the 3-D isotropic oscillator the energy levels are given by

EN=Ek+El+Em=

(

3

2

+nk+nl+nm

)

ω

whereωis the angular frequency

N=nk+nl+nm= 0 , 1 , 2 ...

For a given value ofN, various possible combinations ofnk,nlandnmare

given in Table 3.5, and the degeneracy indicated.

Table 3.5Possible combinations ofnk,nlandnmand degeneracy of energy levels

Nnl nm nn Degeneracy (D)

0 0 0 0 Non-degenerate

1100Threefold(1+2)

010

001

2 1 1 0 Sixfold (1+ 2 +3)

101

011

200

020

002

3 1 1 1 Tenfold (1+ 2 + 3 +4)

120

102

210

201

021

012

300

030

003