In the nuclear decay given below ${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \longrightarrow{…
In the nuclear decay given below
${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \longrightarrow{ }_{\mathrm{Z}+1}^{\mathrm{A}} \mathrm{Y} \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{~B}^* \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{~B} \text {, }$
the particles emitted in the sequence are
$\beta, \alpha, \gamma$
$\gamma, \beta, \alpha$
$\beta, \gamma, \alpha$
$\alpha, \beta, \gamma$
Solution
Key Idea In a nuclear reaction conservation of charge number and mass number must hold good.
Alpha particles are positively charged particles with charge $+2 e$ and mass $4 \mathrm{~m}$. Emission of an $\alpha$-particle reduces the mass of the radionuclide by 4 and its atomic number by 2. $\beta$-particles are negatively charged particles with rest mass as well as charge same as that of electrons. $\gamma$-particles carry no charge and mass.
Radioactive transition will be as follows
$\begin{aligned}
& { }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \longrightarrow{ }_{\mathrm{Z}+1}{ }^{\mathrm{A}} \mathrm{Y}+\beta_{-1}^0 \\
& { }_{\mathrm{Z}+1}^{\mathrm{A}} \mathrm{Y} \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \beta+\alpha_2^4 \\
& { }_{\mathrm{Z}+1}^{\mathrm{A}-4} \beta \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{\beta}+\gamma_0^0 \\
&
\end{aligned}$