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

Text Solution

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The radius of `.^(12)C` nucleus is
`R =R_(0) A^(1//3)`
` =(1.2 fm)(12)^(1//3) =2.75 fm`
The separation between the centers of the nuclei is `2R =5.50 fm`. The potential energy of the pair is
` U=(q_(1)q_(2))/(4piomega_(0)r)=(9xx10^(9) Nm^(2) c^(-2)) ((6xx1.6xx10^(19)C)^(2))/(5.50xx10^(-15)m)`
=1.50xx 10^(-12) J=9.39 MeV`.
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