Question

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

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

Answer 1

The correct option is C 126.15 MeV ​

Radius of each nucleus $R=R_0(A{)}^{\frac{1}{3}}=1.2(64{)}^{\frac{1}{3}}=4.8fm$
Distance between two nuclei (r) = 2R
So potential energy $U=\frac{k.q^2}{r}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}\times 29{)}^2}{2\times 4.8\times {10}^{-15}\times 1.6\times {10}^{-13}}=126.15MeV$.