3.3 Solutions
207
c
1
λ j+1
−
1
λ j
= c × (20.556 cm
−1 ) =
2
I 0
Moment of inertia I 0 =
4π 2 c × 20.556
=
6.63 × 10
−27 erg − s
−1
4π 2 × 3 × 10 10 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
E N = E k + E l + E m =
3
2
+ n k + n l + n m
ω
where ω is the angular frequency
N = n k + n l + n m = 0, 1, 2 . . .
For a given value of N , various possible combinations of n k , n l and n m are
given in Table 3.5, and the degeneracy indicated.
Table 3.5 Possible combinations of n k , n l and n m and degeneracy of energy levels
N
n l
n m
n n
Degeneracy (D)
0
0
0
0
Non-degenerate
1
1
0
0
T h r e ef o l d( 1+ 2)
0
1
0
0
0
1
2
1
1
0
Sixfold (1 + 2 + 3)
1
0
1
0
1
1
2
0
0
0
2
0
0
0
2
3
1
1
1
Tenfold (1 + 2 + 3 + 4)
1
2
0
1
0
2
2
1
0
2
0
1
0
2
1
0
1
2
3
0
0
0
3
0
0
0
3