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

Option: 1

Fusion reaction releasing 24 \, \mathrm{MeV} energy


Option: 2

Fusion reaction absorbing 24\, \mathrm{MeV}energy


Option: 3

Fission reaction releasing 0.0258 \, \mathrm{MeV} energy


Option: 4

Fission reaction absorbing 0.0258 \, \mathrm{MeV}energy


Answers (1)

best_answer

Total mass of reactants

\mathrm{=(2.0141) \times 2=4.0282 \, \mathrm{amu}}

Total mass of products \mathrm{=4.0024\, \mathrm{amu}}

Mass defect \mathrm{=4.0282 \, \mathrm{amu}-4.0024\, \mathrm{amu}}

\mathrm{=0.0258\, \mathrm{amu}}

\therefore \text{Energy released}\mathrm{\: E=931 \times 0.0258=24 \, \mathrm{MeV}}

Posted by

Divya Prakash Singh

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