In the reaction 1H2 + 1H3 →2He4 + 0n1 if binding energies of 1H2, 1H2 and 2H3 are respectively a, b, and c (in MeV), then the energy released in this reaction is:

Mass decay = mass of 2He4+ mass of 0n1− mass of 1He2+ mass of 1He3 ) ∴Δm=E×u2= B.E. of 2He4+0−B.E.of1H2−B.E. of 1H3u2 Mass decay =c−a−bu2 Now, E= mass decay ×u2=c−a−bu2×u2=c−a−b