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  • The binding energy per nucleon of deuterium and helium atom is and <img alt="7.0 \, \mathrm{MeV}" src="/latex

The binding energy per nucleon of deuterium and helium atom is 1.1 \, \mathrm{MeV} and 7.0 \, \mathrm{MeV}. If two deuterium nuclei fuse to form helium atom, the energy released is

Option: 1

19.2\, \mathrm{MeV}


Option: 2

23.6 \, \mathrm{MeV}


Option: 3

26.9\, \mathrm{MeV}


Option: 4

13.9 \, \mathrm{MeV}


Answers (1)

best_answer

\mathrm{{ }_1 H^2+{ }_1 H^2\rightarrow_2 H e^4+ energy }

Binding energy of a \mathrm{\left({ }_1 H^2\right)}  deuterium nuclei

\mathrm{=2 \times 1.1=2.2 \, \mathrm{MeV}}

Total binding energy of two deuterium nuclei
\mathrm{=2.2 \times 2=4.4\, \mathrm{MeV}}

Binding energy of a \mathrm{\left({ }_2 \mathrm{He}^4\right)} nuclei \mathrm{=4 \times 7=28\, \mathrm{MeV}}

So, energy released in fusion \mathrm{=28-4.4=23.6\, \mathrm{MeV}}

Posted by

shivangi.bhatnagar

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