1000 Solved Problems in Modern Physics

450 8 Nuclear Physics – II

U=

∫

dU=

4 πρ^2

3 ε 0

∫R

0

r^4 dr=

4 πρ^2 R^5

1. 5 ε 0

=

3 z^2 e^2

20 πε 0 R

where we have substituted the value ofρ.

8.8 Choose a point at distancer(<R) from the centre of the nucleus. Letq′be the
charge within the sphere of radiusr.

Thenq′=q

(r

R

) 3

The electric field will beE=

q′

4 πε 0 r^2

=

qr

4 πε 0 R^3

The potentialV(r)=−

∫

Edr=−

∫ qr

4 πε 0 R^3

+C

=−

qr^2

8 πε 0 R^3

+C (1)

whereCis a constant.

Atr=R, the point is just on the surface and the potential will be given by

Coulomb’s law.

V(R)=

q

4 πε 0 R

(2)

Using (2) in (1), the value ofCis determined asC=

3

2

q

4 πε 0 R

and (1) becomes

V(r)=

q

8 πε 0 R

(

3 −

r^2

R^2

)

8.3.3 Nuclear Spin and Magnetic Moment ..............

8.9 The rotational kinetic energy is given by

ER=

1

2

Iω^2 (1)

whereERis the rotational energy,Ithe rotational inertia andωis the angular

velocity

The angular momentum is given by

J=Iω (2)

Combining (1) and (2)

ER=

1

2

Jω