Two C u 64 nuclei touch each other. The electrostatics repulsive energy of the system will be Options: A . &nbsp0.788 MeV B . &nbsp7.88 MeV C . &nbsp126.15 MeV D . &nbsp788 MeV

Two Cu64 nuclei touch each other. The electrostatics repulsive energy of the sys

Question

Two

C u^{64}

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

Options:

A . 0.788 MeV

B . 7.88 MeV

C . 126.15 MeV

D . 788 MeV

Answer: Option C
:
C
Radius of each nucleus

R = R_{0} (A)^{\frac{1}{3}} = 1.2 (64)^{\frac{1}{3}} = 4.8 f m

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}} = 126.15 M e V

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