Question

Two Cu⁶⁴ 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

Answer: (C)

126.15 MeV

Answer 2

Radius of each nucleus $R = R_{0}(A)^{1/3} = 1.2(64)^{1/3} = 4.8fm$Distance between two nuclei (r) = 2R

So potential energy

$U = \frac{k \cdot q^{2}}{r} = \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19} \times 29)^{2}}{2 \times 4.8 \times 10^{- 15} \times 1.6 \times 10^{- 19}} = 126.15MeV.$