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

Step-by-step solution

Radius of each nucleus R=R₀(A)^ 1 / 3 =1.2(64)^ 1 / 3 =4.8 ~m Distance between two nuclei (r)=2 R So, potential energy U= k q^2 r = 9 10^9 (1.6 10⁻¹⁹ 29 )^2 2 4.8 10⁻¹⁵ 1.6 10⁻¹⁹ =126.15 MeV