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The binding energy per nucleon for Deuteron and Helium $(_{1} H^{2} {and}_{2} {He}^{4})$ are 1.1 MeV and 7 MeV respectively. The energy released when two Deuterons fuse to form a Helium nucleus is

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12.1 MeV

14 MeV

23.6 MeV

6 MeV

answer is C.

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Detailed Solution

$_{1} H^{2} +_{1} H^{2} ⟶_{2} {He}^{4}$
For each $_{1} H^{2} \to {BE}_{1} = 2 (1 .1) = 2 .2 MeV$
For $_{2} {He}^{4} ⟶ {BE}_{2} = 4 (7) = 28 MeV$
$E = {BE}_{2} - 2 {BE}_{1} = 28 - 2 (2 .2) = 23 .6 MeV$

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