Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron $= 1.1 M e V$ and for helium $= 7.0 M e V$ )

30.2 M e V

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23.6 M e V

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25.8 M e V

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32.4 M e V

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Solution

The correct option is B $23.6 M e V$
$_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e + E n e r g y$

Mass of deuterium $= 2.013550$

Mass of He $= 4.013490$

The energy released in the process is :

$Δ E = 4.002602 \times 7 M e V - 2 \times 2.01355 \times 1.1 M e V$

$Δ E = 23.588 M e V$

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