The nuclear reaction 2H+2H→4He (mass of deuteron = 2.0141 a.m.u. and mass of He = 4.0024 a.m.u.) is

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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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Fission reaction releasing 0.0258 MeV energy

Fusion reaction releasing 24 MeV energy

Fusion reaction absorbing 24 MeV energy

Fission reaction absorbing 0.0258 MeV energy

answer is A.

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Detailed Solution

Total mass of reactants $= (2.0141) \times 2 = 4.0282 amu$
$Total mass of products = 4.0024 amu Mass defect = 4.0282 amu - 4.0024 amu = 0.0258 amu ∴ Energy released E = 931 \times 0.0258 = 24 MeV$

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