If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _____$ imes 10^{-2} \mathrm{MeV}$. (Given $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$, atomic mass of helium $=4.002603 \mathrm{u}$ )

Question

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _____$	imes 10^{-2} \mathrm{MeV}$. (Given $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$, atomic mass of helium $=4.002603 \mathrm{u}$ )

Aicharya · Accepted Answer

Reaction : $\begin{aligned} & 3{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+\gamma 	ext { rays } \ & 	ext { Mass defect }=\Delta \mathrm{m}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}ight) \ & =(3 	imes 4.002603-12)=0.007809 \mathrm{u} \end{aligned}$ Energy released $=931 \Delta \mathrm{m} \mathrm{MeV}$ $=7.27 \mathrm{MeV}=727 	imes 10^{-2} \mathrm{MeV}$

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is…

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