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The nuclear reaction $^{2} H +^{2} H \to^{4} He$ (mass of deuteron = 2.0141 a.m.u. and mass of He = 4.0024 a.m.u.) is

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Fusion reaction releasing 24 MeV energy

Fusion reaction absorbing 24 MeV energy

Fission reaction releasing 0.0258 MeV energy

Fission reaction absorbing 0.0258 MeV energy

answer is A.

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Detailed Solution

Total mass of reactants
$= (2 .0141) \times 2 = 4 .0282 amu$
$Total mass of products = 4 .0024 amu$
$Mass defect =4.0282amu−4.0024amu$
$= 0 .0258 cmu$
$∴ Energy released E = 931 \times 0 .0258 = 24 MeV$

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