Calculate the electric potential energy due to the electri repulsion between two nuclei of `^12 C` when they `touch` each other at the surface.

Explanation

The radius of a `^12 C` nucleus is
`R = R_0 A^(1/3)`
`= (1.1 fm) (12)^(1/3) = 2.52 fm`.
The separation between the centres of the nuclei is `2R = 5.04 fm`. The potential energy of the pair is
`U = (q_1 q_2)/(4piepsilon_0) r`
`= (9 xx 10^9 N m^2 C^(-2) (6 xx 1.6 xx 10^(-19) C)^2/(5.04 xx 10^(-15) m)`
`= 1.64 xx 10^(-12) J = 10.2 MeV`.