Question

In the reaction ${ }_{1}^{2} H +{ }_{1}^{3} H \longrightarrow{ }_{2}^{4} He +{ }_{0}^{1} n$, if the binding energies of ${ }_{1}^{2} H ,{ }_{1}^{3} H$ and ${ }_{2}^{4} He$ are respectively $a, b$ and $c$ (in $MeV$, 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

Answer 1

The Correct Option is B

The energy released per nuclear reaction is the resultant binding energy. Binding energy of $\left({ }_{1}^{2} H +{ }_{1}^{3} H \right)=a+b$ 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 $\left({ }_{1}^{2} H +{ }_{1}^{3} H \right)$ Thus, energy released per nucleon $=$ resultant binding energy $=c-(a+b)=c-a-b$