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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 $=4.002603 \mathrm{u}$ )
Solution:
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Verified Answer
The correct answer is:
727
Reaction :
$\begin{aligned}
& 3{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+\gamma \text { rays } \\
& \text { Mass defect }=\Delta \mathrm{m}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right) \\
& =(3 \times 4.002603-12)=0.007809 \mathrm{u}
\end{aligned}$
Energy released
$=931 \Delta \mathrm{m} \mathrm{MeV}$
$=7.27 \mathrm{MeV}=727 \times 10^{-2} \mathrm{MeV}$
$\begin{aligned}
& 3{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+\gamma \text { rays } \\
& \text { Mass defect }=\Delta \mathrm{m}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right) \\
& =(3 \times 4.002603-12)=0.007809 \mathrm{u}
\end{aligned}$
Energy released
$=931 \Delta \mathrm{m} \mathrm{MeV}$
$=7.27 \mathrm{MeV}=727 \times 10^{-2} \mathrm{MeV}$
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