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Question

# Two 29Cu64 nuclei almost touch each other. The electronics repulsive energy of the system will be (R0=1 fm)

A
0.788 MeV
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B
7.88 MeV
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C
126.15 MeV
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D
788 MeV
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Solution

## The correct option is C 126.15 MeVRadius of each nucleus, R=R0(A)1/3=1.2(64)1/3=4.8 fm=4.8×10−15 m Distance between two nuclei (r)=2R=2×4.8×10−15 m So, potential energy U=kq2r=9×109×(1.6×10−19×29)22×4.8×10−15 Potential energy in (eV), U=kq2rqelectron=9×109×(1.6×10−19×29)22×4.8×10−15×1.6×10−19 =126.15×106 eV=126.15 MeV Hence, (C) is the correct answer.

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