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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

28.0MeV


Option: 4

30.2MeV


Answers (1)

best_answer

{ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4+\Delta \mathrm{E}

The binding energy per nucleon of a deuteron =1.1 \mathrm{MeV}

\therefore \text { Total binding energy }=2 \times 1.1=2.2 \mathrm{MeV}

The binding energy per nucleon of a helium nuclei =7\, \mathrm{MeV}

\therefore \text { Total binding energy }=4 \times 7=28 \mathrm{MeV}

Hence, energy released

\Delta \mathrm{E}=(28-2 \times 2.2)=23.6 \mathrm{MeV}

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