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

Two cu64 nuclei touch each other. The electrostatics repulsive energy of the system will be
 

Grade:12

1 Answers

Arun
25763 Points
one year ago
Dear student
 
Radius of each nuclei = R (A)^1/3 = 1.2 * (64)^1/3 = 1.2 * 4 = 4.8
distance between two nuclei = 2 * 4.8 = 9.6
 
Hence potential energy = k q^2 / r = 126.15 Mev

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