Question

The binding energies per nucleon for a deuteron and an $\alpha$-particle are $x_{1}$ and $x _{2}$ respectively. The energy $Q$ released in the reaction ${ }^{2} H _{1}+{ }^{2} H _{1} \rightarrow{ }^{4} He _{2}+Q$, is

• A. $4(x_1+x_2)$
• B. $4(x_2-x_1)$
• C. $2(x_2-x_1)$
• D. $2(x_1 + x_2)$

Answer 1

Answer: (B)

Number of nucleon on reactant side $=4$
Binding energy for one nucleon $=x_{1}$
Binding energy for $4$ nucleons $=4 x_{1}$,
Similarly on product side binding energy $=4 x_{2}$
Now, $Q =$ change in binding energy $=4\left(x_{2}-x_{1}\right)$.