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Nuclear Force

If the bindin...

Question

If the binding energy per nucleon for $_{3} {L i}^{7}$ is $5.6 M e V$ , the total binding energy of a lithium nucleus is?

139.2 M e V

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39.2 M e V

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15.8 M e V

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115.8 M e V

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Solution

The correct option is B $39.2 M e V$
Binding energy per nucleon = 5.6 MeV
No. of nucleon = No. of proton + No. of neutron
=3 + 4 = 7
So, for 7 Nucleon $= 7 \times 5.6 = 39.2 M e V$

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_{3} {L i}^{7}

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