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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2024 · 5 April · Morning Shift · Physics · Nuclear Physics

If three helium nuclei combine to form a carbon nucleus then the...

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _____×10−2MeV\times 10^{-2} \mathrm{MeV}. (Given 1u=931MeV/c21 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2, atomic mass of helium =4.002603u=4.002603 \mathrm{u} )

Correct answer

0

Step-by-step solution

Reaction : 324He⟶612C+γ rays  Mass defect =Δm=(3mHe−mC)=(3×4.002603−12)=0.007809u\begin{aligned} & 3{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+\gamma \text{ rays } \\ & \text{ 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} \end{aligned} Energy released =931ΔmMeV=7.27MeV==931 \Delta \mathrm{m} \mathrm{MeV} =7.27 \mathrm{MeV}=727 \times 10^{-2} MeV\mathrm{MeV}
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