NEETPhysicsAtoms and NucleiMCQ+4 / −1
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
- A23.6 MeV
- B2.2 MeV
- C28.0 MeV
- D30.2 MeV
View written solutionFree
Correct answer: A
1H2
- 1H2 2He4
- E
The binding energy per nucleon of a deuteron = 1.1 MeV
Total binding energy = 2 × 1.1 = 2.2 MeV
The binding energy per nucleon of a helium nuclei = 7 MeV
Total binding energy = 4 × 7 = 28 MeV
Hence, energy released
E = (28 – 2 × 2.2) = 23.6 MeV
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