JEE Mains · Physics · STD 12 - 13. Nuclei
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 \(=4.002603 \mathrm{u}\) )
- A \(725\)
- B \(726\)
- C \(727\)
- D \(728\)
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…
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