Report Issue

JEE MAIN 2024
05-04-24 S1
Question
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 $=4.002603 \mathrm{u}$ )
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Reaction : $$ \begin{aligned} & 3{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+\gamma \text { rays } \\ & \text { Mass defect }=\Delta \mathrm{m}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right) \\ & =(3 \times 4.002603-12)=0.007809 \mathrm{u} \end{aligned} $$ Energy released $$ \begin{aligned} & =931 \Delta \mathrm{mMeV} \\ & =7.27 \mathrm{MeV}=727 \times 10^{-2} \mathrm{MeV} \end{aligned} $$
Question Tags
JEE Main
Physics
Medium
Start Preparing for JEE with Competishun