In the reaction 12 H + 13 H longrightarrow 24 He + 01 n, if the binding energies of 12 H , 13 H and 24 He are respectively a, b and c (in MeV, then the energy (in MeV ) released in this reaction is

Question

In the reaction 12 H + 13 H longrightarrow 24 He + 01 n, if the binding energies of 12 H , 13 H and 24 He are respectively a, b and c (in MeV, then the energy (in MeV ) released in this reaction is (A) c + a - b (B) c - a - b (C) a + b + c (D) a + b - c

Tardigrade Admin · Accepted Answer

The energy released per nuclear reaction is the resultant binding energy.
Binding energy of

(_{1}^{2} H +_{1}^{3} H) = a + b

Binding energy of

_{2}^{4} He = c

In a nuclear reaction the resultant nucleus is more stable than the reactants.
Hence, binding energy of

_{2}^{4} He

will be more than that of

(_{1}^{2} H +_{1}^{3} H)

Thus, energy released per nucleon

=

resultant binding energy

= c - (a + b) = c - a - b

Q. In the reaction 12​H+13​H⟶24​He+01​n, if the binding energies of 12​H,13​H and 24​He are respectively a,b and c (in MeV, then the energy (in MeV ) released in this reaction is

c + a - b

c - a - b

a + b + c

a + b - c

Q. In the reaction $_{1}^{2} H +_{1}^{3} H ⟶_{2}^{4} He +_{0}^{1} n$ , if the binding energies of $_{1}^{2} H,_{1}^{3} H$ and $_{2}^{4} He$ are respectively $a, b$ and $c$ (in $M e V$ , then the energy (in $M e V$ ) released in this reaction is