Question

The binding energy per nucleon for a deuteron and an $\alpha $ -particle are x₁ and x₂ respectively. The energy (Q) released in the reaction
${ }_1^2 H +{ }_1^2 H \rightarrow{ }_2^4 He + Q$ is :

• A. $2\left(\right.x2-x1\left.\right)$
• B. $2\left(\right.x_{1}+x_{2}\left.\right)$
• C. $4\left(\right.x_{1}+x_{2}\left.\right)$
• D. $4\left(\right.x_{2}-x_{1}\left.\right)$

Answer 1

Answer: (D)

$H_{1}^{2}+\_{1}^{2}H \rightarrow \_{2}^{}He_{}^{4}+Q$
$\frac{B \cdot E}{A}\text{ is }\_{1}^{}H_{}^{2} \rightarrow x_{1}\text{ and }\_{2}^{}He_{}^{4} \rightarrow x_{2}$
$Q=4x_{2}-4x_{1}=4\left(x_{2} - x_{1}\right)$