Nuclear Physics | Physics JEE previous year questions with Solutions

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Q 21 :

The binding energy for the following nuclear reactions are expressed in MeV.

${}_{2}{He}^{3} + {{}_{0}n}^{1} \to {}_{2}{He}^{4} + 20 MeV$

${}_{2}{He}^{4} + {{}_{0}n}^{1} \to {}_{2}{He}^{5} - 0.9 MeV$

If $X_{3}, X_{4}, X_{5}$ denote the stability of ${}_{2}{He}^{3}, {}_{2}{He}^{4}$ and ${}_{2}{He}^{5}$ respectively, then the correct order is: [2026]

$X_{4} > X_{5} > X_{3}$
$X_{4} < X_{5} < X_{3}$
$X_{4} > X_{5} < X_{3}$
$X_{4} = X_{5} = X_{3}$

(1)

${BE}_{{He}^{4}} - {BE}_{{He}^{3}} = 20 MeV \dots (1)$

${BE}_{{He}^{5}} - {BE}_{{He}^{4}} = - 0.9 MeV \dots (2)$

$From eq (1) & (2)$

${BE}_{{He}^{4}} > {BE}_{{He}^{5}} > {BE}_{{He}^{3}}$

$X_{4} > X_{5} > X_{3}$

Q 22 :

The average energy released per fission for the nucleus of ${}_{92}^{235}U$ is 190 MeV. When all the atoms of 47 g pure ${}_{92}^{235}U$ undergo fission process, the energy released is $α \times 10^{23} MeV$ . The value of $α$ is ______.

(Avogadro Number $= 6 \times 10^{23}$ per mole) [2026]

0%

(228)

$Total number of U-235 atoms is$

$47 g = \frac{47}{235} moles = \frac{1}{5} moles$

$∴$ $Total energy released = \frac{1}{5} \times 6 \times 10^{23} \times 190 MeV$

$= 228 \times 10^{23} MeV$

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