• # question_answer The binding energies per nucleon of deuteron $({{\,}_{1}}{{H}^{2}})$ and helium atom $({{\,}_{2}}H{{e}^{4}})$ are 1.1 MeV and 7 MeV If two deuteron atoms react to form a single helium atom, then the energy released is: A) 13.9 MeV                    B) 26.9 MeV        C)      23.6 MeV        D)      19.2 MeV

Solution :

Total binding energy of helium atom $({{\,}_{2}}H{{e}^{4}})$ $=4\times 7=28\,\text{MeV}$ Total binding energy of deuteron ${{\,}_{1}}{{H}^{2}}(Ip+1n)$ $=2\times 1.1=2.2\,\text{MeV}$ $\therefore$Binding energy of 2 deuterons $=2\times 2.2=4.4\,\text{MeV}$ so, the energy released in forming helium nucleus from two deuterons$=(28-4.4)MeV$ $=23.6\,\text{MeV}$ Hence, the correction option is (c).

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