Question 110: Atoms & Nuclei

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 ............ keV.[Given : atomic mass of $$_{86}^{226}Ra$$ = 226.005 u, atomic of $$_{86}^{222}Rn$$ = 222.000 u, atomic mass of $$\alpha $$ particle = 4.000 u, 1 u = 931 MeV/e2, c is speed of the light]

Accepted Answer

Given, atomic mass of ${ }_{88}^{226} \mathrm{R} a=226.005 u$,
atomic mass of ${ }_{86}^{222} \mathrm{R} n=222.000 u$.
atomic mass of $\alpha$ particle $=4.000 u$
$$
1 u=931 \frac{\mathrm{MeV}}{c^2}
$$
where, c = speed of the light
$$
{ }_{88}^{226} \mathrm{R} a \stackrel{\alpha-	ext { decay }}{\longrightarrow}{ }_{86}^{222} \mathrm{R} n+{ }_2^4 \mathrm{He}
$$
Total energy emitted $=(\Delta m) c^2$
$$
\begin{aligned}
& =[226.005-(222+14)] 	imes 931.5 \mathrm{Mev} \\
& =0.005 	imes 931.5 \mathrm{MeV}
\end{aligned}
$$
Kinetic energy of $\alpha=4.44 \mathrm{MeV}$
Kinetic energy of Rn = $$4.44 \mathrm{MeV} 	imes \frac{4}{222}$$ = 0.08 MeV
$$
\begin{aligned}
& 	ext { Energy of photon }(\gamma)=(\Delta m) c^2-\left(k . \mathrm{E}_\alpha+k \mathrm{E}_{\mathrm{R} n}ight) \\
& \mathrm{E}_r=[[0.05 	imes 931.5]-(4.44+0.08)] \mathrm{MeV} \\
& \mathrm{E}_r=[4.655-4.52] \mathrm{MeV} \\
& \mathrm{E}_r=0.135 \mathrm{MeV}=0.135 	imes 10^3 \mathrm{keV} \\
& \mathrm{E}_r=135 \mathrm{keV}
\end{aligned}
$$

Atoms & Nuclei

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