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

The atomic mass of \({ }_6 \mathrm{C}^{12}\) is \(12.000000\ \mathrm{u}\) and that of \({ }_6 \mathrm{C}^{13}\) is \(13.003354 \ \mathrm{u}\). The required energy to remove a neutron from \({ }_6 \mathrm{C}^{13}\), if mass of neutron is \(1.008665 \ \mathrm{u}\), will be :

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Answer & Solution

Correct Answer

(C)  \(4.95 \ \mathrm{MeV}\)

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Detailed explanation

\( { }_6 \mathrm{C}^{13}+\text { Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \)

\( \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \)

\( =-0.00531\  \mathrm{u} \)

\( \therefore \text { Energy required }=0.00531 \times 931.5 \mathrm{MeV} \)

\( =4.95\  \mathrm{MeV}\)

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