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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4.95 \ \mathrm{MeV}\)
Step-by-step Solution
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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