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 $keV$.

[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 light]

Answer 1

$R{a}^{226}\to R{n}^{222}+\alpha$
Mass defect $\Delta m=226.005-222.000-4.000$
$=0.005amu$

$\therefore Q$ value = $0.005\times 931.5=4.655MeV$

Also $\frac{K.E_{\alpha }}{K.E_{Rn}}=\frac{m_{Rn}}{m_{\alpha }}$

$\Rightarrow K.E_{Rn}=\frac{m_{\alpha }}{m_{Rn}}.K.E_{\alpha }=\frac{4}{222}\times 4.44=0.08MeV$

$\therefore$ Energy of $\gamma$ -Photon $=Q-(E_{Rn}+E_{\alpha })$

$4.655-(4.44+0.08)$

$=0.135MeV=135KeV$