Question

Two ${Cu}^{64}$ nuclei touch each other. The electrostatics repulsive energy of the system will be

• A. $0.788MeV$
• B. $7.88MeV$
• C. $126.15MeV$
• D. $788MeV$

Answer 1

The correct option is C $126.15MeV$

Radius of each nucleus $R=R=R_0{\left(A\right)}^{1/3}=1.2{\left(64\right)}^{1/3}=4.8m$
distance between two nuclei $\left(r\right)=2R$ $U=\frac{k.q^2}{r}=\frac{9\times {10}^9\times {(1.6\times {10}^{-19}\times 29)}^2}{2\times 4.8\times {10}^{-15}\times 1.6\times {10}^{-19}}=126.15MeV$
So, potential energy