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Calculate the electric potential energy of interaction due to the electric repulsion between two nuclei of ${}^{12}C$ when they touch each other at the surface. Assume that the potential energy of interaction between two nuclei is given by $U = \frac{q_{1} q_{2}}{4 {πε}_{0} r}$

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$10.2 M e V$

$102 M e V$

$1.02 M e V$

$0.102 M e V$

answer is D.

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Detailed Solution

The radius of a ${}^{12}C$ nucleus is $R = R_{0} A^{1 / 3}$

$\Rightarrow R = (1.1 f m) {(12)}^{1 / 3} = 2.52 f m$

The separation between the centres of the nuclei is $2 R = 5.04 f m$ . The potential energy of the pair is

$U = \frac{q_{1} q_{2}}{4 {πε}_{0} r}$

$\Rightarrow U = (9 \times 10^{9}) \frac{{(6 \times 1.6 \times 10^{- 19})}^{2}}{5.04 \times 10^{- 15}} \Rightarrow U = 1.64 \times 10^{- 12} J = 10.2 M e V$

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