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 \(=4.002603 \mathrm{u}\) )
Pick an option to check your answer.
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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