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QJEEM24Ph69ee: JEE Main 2024 Physics - Class 12 JEEM24Ph69ee Solution: The atomic mass of is and that of is . The required energy to…

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 :

Multiple Choice Options

A 62.5 MeV
B 6.25 MeV
C 4.95 MeV
D 49.5 MeV

Step-by-Step Verified Solution

$\begin{aligned}& { }_6 \mathrm{C}^{13}+\text { Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \\& \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \\& =-0.00531 \mathrm{u} \\& \therefore \text { Energy required }=0.00531 \times 931.5 \mathrm{~MeV} \\& =4.95 \mathrm{~MeV}\end{aligned}$

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