450
8 Nuclear Physics – II
U =
dU =
4πρ
2
3ε 0
R
0
r
4 dr =
4πρ
2 R
5
1.5ε 0
=
3z
2 e
2
20πε 0 R
where we have substituted the value of ρ.
8.8 Choose a point at distance r (
charge within the sphere of radius r .
Then q
= q
r
R
3
The electric field will be E =
q
4πε 0 r 2 =
qr
4πε 0 R 3
The potential V (r ) = −
Edr = −
qr
4πε 0 R 3 + C
= −
qr
2
8πε 0 R 3 + C
(1)
where C is a constant.
At r = R, the point is just on the surface and the potential will be given by
Coulomb’s law.
V (R) =
q
4πε 0 R
(2)
Using (2) in (1), the value of C is determined as C =
3
2
q
4πε 0 R
and (1) becomes
V (r ) =
q
8πε 0 R
3 −
r
2
R 2
8.3.3 Nuclear Spin and Magnetic Moment
8.9 The rotational kinetic energy is given by
E R =
1
2
I ω
2
(1)
where E R is the rotational energy, I the rotational inertia and ω is the angular
velocity
The angular momentum is given by
J = I ω
(2)
Combining (1) and (2)
E R =
1
2
J ω
8 Nuclear Physics – II
U =
dU =
4πρ
2
3ε 0
R
0
r
4 dr =
4πρ
2 R
5
1.5ε 0
=
3z
2 e
2
20πε 0 R
where we have substituted the value of ρ.
8.8 Choose a point at distance r (
charge within the sphere of radius r .
Then q
= q
r
R
3
The electric field will be E =
q
4πε 0 r 2 =
qr
4πε 0 R 3
The potential V (r ) = −
Edr = −
qr
4πε 0 R 3 + C
= −
qr
2
8πε 0 R 3 + C
(1)
where C is a constant.
At r = R, the point is just on the surface and the potential will be given by
Coulomb’s law.
V (R) =
q
4πε 0 R
(2)
Using (2) in (1), the value of C is determined as C =
3
2
q
4πε 0 R
and (1) becomes
V (r ) =
q
8πε 0 R
3 −
r
2
R 2
8.3.3 Nuclear Spin and Magnetic Moment
8.9 The rotational kinetic energy is given by
E R =
1
2
I ω
2
(1)
where E R is the rotational energy, I the rotational inertia and ω is the angular
velocity
The angular momentum is given by
J = I ω
(2)
Combining (1) and (2)
E R =
1
2
J ω