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 $4.44MeV.{}^{222}Rn$ 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 the light]

Answer 1

Answer: 135

${}_{88}^{226}Ra\to {}_{86}^{222}{R}_n{}^{\ast }+{}_{2}H{e}^4+Q_1$
Given: (K.E.) ${}_{\infty }=4.44MeV$
$Q_1=\Delta m\times 931MeV$
$\Delta m=m_{Ra}-m_{Rn}+m_{He}$
$\Rightarrow 226.005-(222+4)$
$\Rightarrow 226.005-226$
$\Rightarrow 0.005$
So, $Q=0.005\times 931$
$\Rightarrow 4.655MeV$
Now, the energy of $\gamma$-photon
$k_{\alpha }=(Q-E)\frac{A-4}{A}$
$\Rightarrow 4.44=(4.655-E)\frac{222}{226}$
$\Rightarrow E=0.135MeV$
$\Rightarrow E=135KeV$