Question

In nuclear fission, a nucleus of uranium 238, which contains 92 protons, can divide into two smaller spheres, each having 46 protons and a radius of 5.90  10-15 m. Find the acceleration of each nucleus at the instant in which fission occur?

Answer 1

`~~1.77times 10^28 \ "m"\ "s"^-2`

Explanation:

The force on the two charges immediately at the instant of fission is

`1/(4 pi epsilon_0) (Ze)^2/(2r)^2`

where `Z=46` and `2r` is the distance between the two centers (we have made a rather unrealistic assumption that the charges are uniformly distributed over the two spheres)

We can estimate the mass of a nucleus to be `1/2` of a Uranium atom `M/2`

So, the acceleration is

`1/(4 pi epsilon_0) (Ze)^2/(2r)^2 div (M/2)= 1/(4 pi epsilon_0) (Ze)^2/(2r^2M)`

Substituting the actual values we get

`(9times 10^9) (46times 1.6times 10^-19)^2/(2 (5.90times 10^-15)^2( 238times 1.66times 10^-27)) \ "m"\ "s"^-2`

This works out to be

`1.77times 10^28 \ "m"\ "s"^-2`

a really gigantic value by macroscopic standards!