Calculate the electric potential energy due to the electric 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=5.0 fm`. The potential energy of the pair is
`U=(q_(1)q_(2))/(4piepsilon_(0)r)`
`=(9xx10^(9)N-m^(2)//C^(2))((6xx1.6xx10^(-19)C)^(2))/(5.04xx10^(-15)m)=(9xx10^(9)N-m^(2)//C^(2))((6xx1.6xx10^(-19)C)^(2))/(5.04xx10^(-15)m)`
`1.64xx10^(-12)J=10.2 MeV`.