Question

In the reaction21H+31H→42He+10n. If the binding energies of21H ,31H and 42He are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is

Answer 1

Applying principle of energy conservation, Energy of proton

=      total B.E. of 2a - energy ofLi7

=      8 × 7.06 - 7 × 5.6

=      56.48 - 39.2 = 17.28 MeV