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Q.
$_{92}U^{235}$ undergoes successive disintegrations with the end product of $_{82}Pb^{203}$.The number of $\alpha$ and $\beta$ particles emitted are
Let number of a particles decayed be $x$ number of $p$ particles decayed be $y$.
Then equation for the decay is given by
${ }_{92} U^{235}----->x a_{2}^{4}+y \beta_{1}^{0}+P b_{8 z}^{203}$
Equating the mass number on both sides
$235=4 x+203\,\,\,\,\,\,\,...(i)$
Equating atomic number on both sides
$92=2 x-y+82\,\,\,\,\,\,\,\,\,\,\,...(ii)$
Solving Eqs. (i) and (ii), we get
$x=8, y=6$
$\therefore 8 \alpha$ particles and $6 \beta$ particles are emitted in disintegration.