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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2024
  • Hard
  • Numerical

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

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Answer & Solution

Correct Answer

(C) \(727\)

Step-by-step Solution

Detailed explanation

Reaction:

\(3{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+\gamma \text { rays }\)

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

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

Energy released

\(=931 \Delta \mathrm{m} \mathrm{MeV}\)

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

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