JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 8)
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 :
62.5 MeV
6.25 MeV
4.95 MeV
49.5 MeV
Explanation
$$\begin{aligned}
& { }_6 \mathrm{C}^{13}+\text { Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \\
& \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \\
& =-0.00531 \mathrm{u} \\
& \therefore \text { Energy required }=0.00531 \times 931.5 \mathrm{~MeV} \\
& =4.95 \mathrm{~MeV}
\end{aligned}$$
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