Download Exercises 5

EMF
Exercise Class 5
Where does the energy of a fission
bomb come from?
An unstable atomic nucleus can be regarded as a uniform
sphere of charge with radius R and total charge Q.
It breaks into two identical spherical fragments.
Assume that the fragments have the same charge density as
the original nucleus.
The electric potential energy (“self-energy”) of a uniform
sphere of total charge Q and radius a is given by
Ui 
INITIAL
STATE
JUST AFTER
BREAK-UP
3Q 2
20  o a
Radius = a
Q
Q/2
(derived in lectures).
Initial energy
3Q 2
Ui 
20  o a
Radius of each fragment = ?
Q/2
Self-energy of each fragment = ?
Energy of each fragment due to
the proximity of the other one = ?
AFTER FRAGMENTS HAVE FLOWN APART BY A LARGE
DISTANCE (r >> a):
r
Q/2
Q/2
Total energy = ?
Just after break-up:
(a)
Radius of each fragment = ?
Hint: - The charge density is the same
as for the original nucleus
- The total charge is unchanged.
(b)
“Self-energy” of each fragment = ?
This is the PE it has due to the fact
that it is a sphere of charge.
(c)
“External energy” of each fragment = ?
This is the PE each fragment has due
to the proximity of the other one.
Hint: - It’s equivalent to a system of two point charges.
(d)
From (b) and (c), total energy of the system = ?
What fraction of Ui has been lost?
After the fragments have flown
apart by a large distance:
(a)
Self energy of each fragment = ?
(b)
External energy of each fragment = ?
(c)
From (a) and (b), total energy of the system = ?
Now what fraction of Ui has been lost?
What has happened to this energy?
Ui 
The initial energy of the system is
3Q 2
.
20 0 R
When the two fragments are just separated, there are three contributions to the total energy
U1:
U1a
Each of the two fragments has self energy
3(Q / 2)2

20 0 R f
where Rf is the radius of the fragment.
To find Rf:
The charge density of the fragments is the same as that of the original
nucleus, and the total charge is the same. Therefore the total volume has
not changed. Volume is proportional to (radius)3, so
R3  2Rf 3
R
.
21/ 3

Rf 
U1a
3(2)1/ 3 Q 2
.

80 0 R
Therefore, the self energy of each fragment is
Each of the fragments also lies in the electric field of the other. As they are spherically
symmetrical, we can regard them as point charges for the purpose of working out the energy:
So, we have two charges Q/2 a distance 2Rf apart. The potential at the position of each
charge (due to the other one) is
V 

(Q / 2 )
.
4 0 (2 R f )
The electrostatic energy of each fragment is
1 Q (Q / 2 )

2 2 4  0 (2 R f )
U 1b 
Q2
64  0 R f

(21/ 3 )Q 2
.
64  0 R
The total energy of the system is
U1  2(U1a  U1b ) 
Therefore
(ii)
Q 2  3(2)1/ 3
21/ 3 
3Q 2  (2)1/ 3
5(2)1/ 3 






.
 0 R  40
32 
20 0 R  2
24 
U1
Ui

17(2)1/ 3
24
 0.892 .
When the fragments are far apart, then U1b is small because it decreases as 1/distance.
We are then left with just the self energy of the two fragments, so the final potential
energy of the system is
U tot  2 U1a
(iii)
(2)1/ 3
 Ui
 (0. 630)U i .
2
The missing potential energy has been converted to kinetic energy of the fragments,
which have flown apart at very high speed. It is this generation of kinetic energy
through electrostatic repulsion that is responsible for the explosion of a nuclear
fission bomb, and for the generation of heat in a nuclear reactor.