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JEE Main 2004 (AIEEE) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from JEE Main 2004 (AIEEE)

Q34JEE Main 2004 · AIEEEEasy

Nuclear Fission & Fusion

The binding energy per nucleon of deuteron $\left({}^{2}_{1}H\right)$ and helium nucleus $\left({}^{4}_{2}He\right)$ is $1.1\text{ MeV}$ and $7\text{ MeV}$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

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Answer: (a)

$Q = 4\times7 - 2\times(2\times1.1)$ .
$Q = 28 - 4.4 = 23.6\text{ MeV}$ .
So the answer is (a) $23.6\text{ MeV}$ .

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