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  • The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

Option: 1

23.6 MeV


Option: 2

2.2 MeV


Option: 3

28.0 MeV


Option: 4

30.2 MeV


Answers (1)

best_answer

\\ \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \rightarrow{ }_{2} \mathrm{He}^{4}+\Delta E$ \\ $$ \\ Total binding energy of a deuteron $=2 \times 1.1=2.2 \mathrm{MeV}$ \\ Total binding energy of two deuterium nuclei=2.2×2=4.4MeV \\ The binding energy per nucleon of a helium nuclei $=7 \mathrm{MeV}$ \\ Total binding energy $=4 \times 7=28 \mathrm{MeV}$ \\ Hence, energy released $\Delta E=(28-2 \times 2.2)=23.6 \mathrm{MeV}$

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Suraj Bhandari

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