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Grade: 12

                        

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

one year ago

Answers : (1)

Arun
24742 Points
							
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
one year ago
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