1000 Solved Problems in Modern Physics

8.3 Solutions 453

8.15 (a) The electrical quadrupole moment is the expectation value of the operator

Qij=

∑z

k= 1

ek(3xixj−δijr^2 )k (1)

whereδijis the kronecker delta.

TheQzz– component,

Qzz=

∑z

k= 1

ek(3z^2 k−rk^2 )(2)

is zero for a spherically charge distribution. This feature also becomes obvi-

ous from the formula

Q=

2

5

ze(a^2 −b^2 ) for a homogenous ellipsoid of semi axesa,b

(Problem 8.54).

For a spherea=band thereforeQ=0.

(b) The quadrupole moment which is a tensor has the property that it is sym-

metric, that isQij=Qjiand that its trace (the sum of the diagonal ele-

ments),Qxx+Qyy+Qzz =0. Using these two proiperties,Qijcan be

expressed in terms of the spin vectorIwhich specifies the quantized state

of the nucleus.

Qij=C

(

IiIj+IjIi−

2

3

I^2 δij

)

(3)

whereCis a constant. SubstitutingI^2 =I(I+1) in (3)

Q=

2

3

CI(2I−1) (4)

which is zero forI=0orI= 1 / 2

8.16 From the results of Problem 8.15

Q=^25

(

a^2 −b^2

)

Z

We can write

Q=

4

5

Z

(

a+b

2

)

(a−b)

CallingR=

(

a+b

2

)

andΔR=a−b, we get

Q=

4

5

ZRΔR

Q= 4 .2 barn= 4. 2 × 10 −^24 cm^2 =420 fm^2

=

2

5

×71(a^2 −b^2 )