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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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1.1 MeV

7 MeV

23.6 MeV

6 MeV

answer is C.

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Detailed Solution

${}_{1}H^{2} + {}_{1}H^{2} \to {}_{2}{He}^{4} For each {}_{1}H^{2} \to {BE}_{1} = 2 (1.1) = 2.2 MeV For {}_{2}{He}^{4} \to {BE}_{2} = 4 (7) = 28 MeV E = {BE}_{2} - 2 {BE}_{1} = 28 - 2 (2.2) = 23.6 MeV$

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