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

Answer 1

Answer: 135

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