Question

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 $a$, $b$ and $c$ (in MeV) respectively, then energy (in MeV) released in this reaction is:

• A. $a+b-c$
• B. $c+a-b$
• C. $c-a-b$
• D. $a+b+c$

Answer 1

The correct option is C $c-a-b$

${}_1^{2}H{+}_1^{3}H\to {}_2^{4}He{+}_0^{1}n$

Nuclear reaction of ${}_1^{2}H$ & ${}_1^{3}H$ will absorb a & b Mev of energy to proceed reaction.

Formation of ${}_2^{4}He$ will release c Mev of energy.

Total energy released = $c-(a+b)$.