# Example 17-2 Electric Potential Energy and Nuclear

```Example 17-2 Electric Potential Energy and Nuclear Fission
When a nucleus of uranium-235 (92 protons and 143 neutrons) absorbs an additional neutron, it undergoes a process
called nuclear fission in which it breaks into two smaller nuclei. One possible fission is for the uranium nucleus to divide
into two palladium nuclei, each of which has 46 protons and is 5.9 * 10215 m in radius. The palladium nuclei then fly apart
due to their electric repulsion. If we assume that the two palladium nuclei begin at rest and are just touching each other,
what is their combined kinetic energy when they are very far apart?
Set Up
We can treat the two spherical nuclei as
though they were point charges of q = +46e
located at the centers of the two nuclei. (To
motivate this, recall from Section 16-7 that
the electric field outside a sphere of charge
Q is the same as that of a point charge Q
located at the sphere’s center.) Equation 17-3
then tells us the electric potential energy of
the two palladium nuclei when they begin at
rest. We’ll then use energy conservation to
find the combined kinetic energy when the
palladium nuclei are very far apart.
Solve
Use Equation 17-3 to solve for the initial
electric potential energy when the two nuclei
are just touching.
Electric potential energy of two
point charges:
kq1q2
(17-3)
Uelectric =
r
nucleus
q = +46e
nucleus
q = +46e
Conservation of mechanical energy:
Ki + Uelectric, i = Kf + Uelectric, f
(6-23)
2 &times; (5.9 &times; 10–15 m)
equivalent to:
q = +46e
q = +46e
2 &times; (5.9 &times; 10–15 m)
Each nucleus has charge +46e:
q1 = q2 = +46(1.60 * 10219 C) = +7.36 * 10218 C
The separation between the charges is twice the radius of either nucleus:
r = 2(5.9 * 10215 m)
The electric potential energy is
Uelectric, i =
18.99 * 109 N # m2 &gt;C 2 2 17.36 * 10-18 C2 2
kq1q2
=
r
215.9 * 10-15 m2
= 4.1 * 10211 N &middot; m = 4.1 * 10211 J
Use energy conservation to find the combined
final kinetic energy of the two palladium
nuclei.
The palladium nuclei begin at rest, so the initial kinetic energy is zero:
Ki = 0
The palladium nuclei end up very far apart, so their separation is
&shy;essentially infinite (r S  ). The final electric potential energy is
&shy;therefore zero:
Uelectric, f = 0
From the energy conservation equation, the final kinetic energy is
Kf = Ki + Uelectric,i 2 Uelectric,f
= 0 + 4.1 * 10211 J 2 0 = 4.1 * 10211 J
Reflect
All of the initial electric potential energy is converted into kinetic energy of the palladium nuclei. The energy released by
the fission of a single uranium-235 nucleus, 4.1 * 10211 J, is very small. But imagine that you could get 1.0 kg of uranium-235, which contains about 2.6 * 1024 uranium atoms, to undergo fission at once. (This is the principle of a fission
bomb.) The released energy would be (2.6 * 1024) * (4.1 * 10211 J) = 1.1 * 1014 J, equivalent to the energy given off by
exploding 26,000 tons of TNT. This gives an inkling of the terrifying amount of energy that can be released by a fission
weapon.
```