What are the respective numbers of $α$ and $β$ particles emitted respectively in the following radioactive decay?

${^{200}_{90}X}→{^{168}_{80}Y}$

8 and 6

The correct answer is Option (2) → 8 and 6

$ \textbf{Concept:} $

$ \text{Emission of an }\alpha\text{ particle reduces mass number by }4\text{ and atomic number by }2. $

$ \text{Emission of a }\beta\text{ particle increases atomic number by }1\text{ without changing mass number.} $

$ \textbf{Formula Used:} $

$ \text{After }n\alpha\text{ decays: }A=A_0-4n,\;Z=Z_0-2n $

$ \text{After }m\beta\text{ decays: }Z=Z+m $

$ \textbf{Where:} $

$ \text{A = mass number} $

$ \text{Z = atomic number} $

$ \text{n = number of }\alpha\text{ decays} $

$ \text{m = number of }\beta\text{ decays} $

$ \textbf{Given:} $

$ A_0=200,\;Z_0=90 $

$ A=168,\;Z=80 $

$ \textbf{Calculations:} $

$ 200-4n=168 $

$ 4n=32 $

$ n=8 $

$ \text{Atomic number after }8\alpha\text{ decays} $

$ Z=90-2\times8=74 $

$ \text{To reach final atomic number }80 $

$ m=80-74=6 $

$ \text{Hence }8\alpha\text{ and }6\beta\text{ particles are emitted} $