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

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$4 (x_{2} - x_{1})$

$4 (x_{1} + x_{2})$

$2 (x_{2} - x_{1})$

$2 (x_{1} + x_{2})$

answer is B.

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Detailed Solution

$Q = BE product - {BE}_{Reactant}$

$= 4 (x_{2} - x_{1})$

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