If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is × 10 -2 MeV . (Given 1 u =931 MeV / c 2 , atomic mass of helium =4.002603 u )

Physics · Nuclear Physics

JEE Main 2024 — 5 April, Shift 1 — Question 55

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is $\qquad$ $\times 10^{-2} \mathrm{MeV}$ . (Given $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2}$ , atomic mass of helium $=4.002603 \mathrm{u}$ )

Answer: 727

Numerical answer — enter this value.

Step-by-step solution

Reaction :

$3{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{6}^{12} \mathrm{C}+\gamma$ rays

Mass defect $=\Delta \mathrm{m}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right)$

$=(3 \times 4.002603-12)=0.007809 \mathrm{u}$

Energy released

$=931 \Delta \mathrm{~m} \mathrm{MeV}$

=7.27 \mathrm{MeV}=727 \times 10^{-2} \mathrm{MeV}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction