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In the reaction 21H + 31H → 42He + 10n, if the binding energies of 21H, 31H and 42He are respectively, a, b and c (in MeV), then the energy (in MeV) released in this reaction is

1. a + b + c
2. a + b − c
3. c − a − b
4. c + a + b

Correct answer: c − a − b

Solution

The energy released in a nuclear reaction is the difference in binding energy between the products and the reactants. Here, the energy released is given by the binding energy of 42He (c) minus the sum of the binding energies of 21H (a) and 31H (b), i.e., c − a − b.

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