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NEET 2010 (AIPMT, Mains) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from NEET 2010 (AIPMT, Mains)

Q29NEET 2010 · AIPMT · MAINSEasy

Nuclear Fission & Fusion

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

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Answer: (a)

$Q = \text{BE(products)} - \text{BE(reactants)}$ .
BE of ${}^4\text{He} = 4\times7.0 = 28.0\text{ MeV}$ ; BE of two deuterons $= 2\times(2\times1.1) = 4.4\text{ MeV}$ .
$Q = 28.0 - 4.4 = 23.6\text{ MeV}$ .
So the answer is (a) $23.6\text{ MeV}$ .

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