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The mass of 3Li7 nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of 3Li7 nucleus is nearly

1. 46 MeV
2. 5.6 MeV
3. 3.9 MeV
4. 23 MeV

Correct answer: 5.6 MeV

Solution

The mass defect is 0.042 u, which corresponds to an energy of 0.042 × 931 MeV = 39.1 MeV (using 1 u = 931 MeV). Dividing this by the number of nucleons (7) gives the binding energy per nucleon as approximately 5.6 MeV.

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