The binding energy of deuteron _ 1 ^ 2 H is 1.112 MeV per nucleon and an α–particle _ 2 ^…
Physics · JEE Advanced · NTA Exams — Atoms and Nuclei
The binding energy of deuteron \({ }_{1}^{2} \mathrm{H}\) is 1.112 MeV per nucleon and an α–particle \({ }_{2}^{4} \mathrm{H}\) has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{H}+\mathrm{Q},\)the energy Q released is
- 1 MeV
- 11.9 MeV
- 23.8 MeV
- 931 MeV
Answer
(C) 23.8 MeV
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