1000 Solved Problems in Modern Physics

8.3 Solutions 451

ThereforeJ=

2 ER

ω

=

2 × 537. 5 × 1. 6 × 10 −^13

2. 6 × 1023

= 6. 61 × 10 −^34 Js=h

8.10 (a) IfJbe the electronic angular momentum andIthe nuclear spin, the mul-

tiplicity is given by (2J+1) or (2I+1), whichever is smaller.

Now, 2J+ 1 = 2 ×

5

2

+ 1 =6. But only four terms are found. Therefore,

the multiplicity is given by 2I+ 1 =4, whenceI= 3 /2.

(b) The magnetic field produced by the electron interacts with the nuclear

magnetic moment resulting in the energy shift in hyperfine structure.

ΔE≈ 2 I.J=F(F+1)−I(I+1)−J(J+1)

where,F=I+Jtakes on integral values from 4 to 1.

F= 4 ,ΔE= 20 − 25 / 2

F= 3 ,ΔE= 12 − 25 / 2

F= 2 ,ΔE= 6 − 25 / 2

F= 1 ,ΔE= 2 − 25 / 2

The intervals are 8, 6 and 4 in the ratio 4:3:2

8.11 The resonance frequencyνis given by

ν=

μβ

Ih

whereμis the magnetic moment, B the magnetic induction,Ithe nuclear spin

in units of. A nuclear magnetonμN= 5. 05 × 10 −^27 JT−^1.

B=

νIh

μ

=

60 × 106 ×(1/2)× 6. 625 × 10 −^34

2. 79 × 5. 05 × 10 −^27

= 1 .4T

8.12 f=
μB
Jh

=

3. 46 × 3. 15 × 10 −^14 × 1. 6 × 10 −^13 × 0. 8

(

7

2

)

× 6. 63 × 10 −^34

= 6. 012 × 106 Hz

= 6 .012 MHz

8.3.4 Electric Quadrupole Moment........................

8.13 Quadrupole Moment
Assume that the charge is uniformly distributed over an ellipsoid of revolution,
with the axis of symmetry along thez/-axis, the semi-major axis of lengtha,
and the other two semi-axes of lengthb. The charge densityρis

ρ=

Ze

volume

=

Ze

4 π

3 ab

2