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

Answer 1

${ }_{88}^{226} Ra \rightarrow{ }_{86}^{222} R _{ n }{ }^{*}+{ }_{2} He ^{4}+ Q _{1}$
Given: (K.E.) ${ }_{\infty}=4.44\,MeV$
$Q _{1}=\Delta m \times 931 \,MeV$
$\Delta m=m_{R a}-m_{R n}+m_{H e}$
$\Rightarrow 226.005-(222+4)$
$\Rightarrow 226.005-226$
$\Rightarrow 0.005$
So, $Q=0.005 \times 931$
$\Rightarrow 4.655\, MeV$
Now, the energy of $\gamma$-photon
$k_{\alpha}=(Q-E) \frac{A-4}{A}$
$\Rightarrow 4.44=(4.655- E ) \frac{222}{226}$
$\Rightarrow E=0.135\, MeV$
$\Rightarrow E=135\, KeV$