${ }_{92}{\mathrm{U}}^{235}$ undergoes successive disintegrations with the end product of ${ }_{82}\mathrm{P}{\mathrm{b}}^{203}$. The number of $\alpha \mathrm{a}\mathrm{n}\mathrm{d }\mathrm{\beta }$- particles emitted are

Let number of $\mathrm{\alpha }$particles decayed be $x$ and number of $\mathrm{\beta }$particles decayed be $y$.

Then equation for the decay is given by

${ }_{92}{\mathrm{U}}^{235}\to {x\alpha }_2^4+ {y\beta }_{-1}^0+{\mathrm{P}\mathrm{b}}_{82}^{203}$

Equating the mass number on both sides

$235=4x+203 \dots \left(i\right)$

Equating atomic number on both sides

$92=2x-y+82 \dots \left(ii\right)$

Solving Equ.(i) and(ii), we get

$x=8,y=6$

$\therefore 8\mathrm{\alpha } \mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d} 6\mathrm{\beta } \mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }$are emitted in disintegration.