Explanation

(c) Let number of $\alpha\ $particles decayed be $x$ and number of $\beta\ $particles decayed be$y$.
Then equation for the decay is given by
$\ _{92}U^{235} \rightarrow \ {x\alpha}_{2}^{4} + \ {y\beta}_{- 1}^{0} + {Pb}_{82}^{203}$
Equating the mass number on both sides
$\ \ \ \ \ \ \ \ \ \ \ \ 235 = 4x + 203\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Equating atomic number on both sides
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 92 = 2x - y + 82\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Solving Equ.(i) and(ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 8,y = 6$
$\therefore 8\alpha\ particles\ and\ 6\beta\ particles\ $are emitted in disintegration.