Question

Suppose a $_{88}^{226}Ra$ nucleus at rest and in ground state undergoes $\alpha $ -decay to a $\_{86}^{222}Rn$ nucleus in its excited state. The kinetic energy of the emitted $\alpha $ particle is found to be $4.44\,MeV$ . $\_{86}^{222}Rn$ nucleus then goes to its ground state by $\gamma -$ decay. The energy of the emitted $\gamma $ photon is (in $keV$ . [Given: atomic mass of $\_{88}^{226}Ra=226.005\,u$ , atomic mass of $\_{86}^{222}Rn=222.000\,u$ , atomic mass of $\alpha $ -particle $=4.000\,u,1\,u=931\,MeV\,c^{- 2},c$ is speed of the light ]

Answer 1

Mass defect $\Delta m=226.005-222.000-4.000=0.005\,u$
$Q$ value $=\Delta m\times 931.5=0.005\times 931.5=4.655\,MeV$
$\frac{\left(K E\right)_{\alpha }}{\left(K E\right)_{Rn}}=\frac{m_{Rn}}{m_{\alpha }}\Rightarrow \left(K E\right)_{Rn}=\frac{m_{\alpha }}{m_{Rn}}\times \left(K E\right)_{\alpha }=\frac{4}{222}\times 4.44=0.08\,MeV$
So, the energy of $\gamma ‐$ photon $=4.655-\left(4 . 44 + 0.08\right)=0.135\,MeV=135\,KeV$