Question

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

• A. $\alpha =6,\beta =4$
• B. $\alpha =6,\beta =0$
• C. $\alpha =8,\beta =6$
• D. $\alpha =3,\beta =3$

Answer 1

Answer: (C)

Let number of a particles decayed be $x$ number of $p$ particles decayed be $y$.
Then equation for the decay is given by
${}_{92}{U}^{235}----->x{a}_2^4+y{\beta }_1^0+P{b}_{8z}^{203}$
Equating the mass number on both sides
$235=4x+203...(i)$
Equating atomic number on both sides
$92=2x-y+82...(ii)$
Solving Eqs. (i) and (ii), we get
$x=8,y=6$
$\therefore 8\alpha$ particles and $6\beta$ particles are emitted in disintegration.