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 \(=4.002603 \mathrm\ {u}\))

Answer 1

Correct answer is :  727

Reaction :

\(3{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{6}^{12} \mathrm{C}+\gamma\) rays

Mass defect \(=\Delta \mathrm{m}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right)\)

\(=(3 \times 4.002603-12)=0.007809 \mathrm\ {u}\)

Energy released

\(=931 \ \Delta \mathrm{m} \ \mathrm{MeV}\)

= 7.27 MeV \(= 727 \times 10^{–2}\) MeV