Question

In successive emission of $\alpha$ - and $\beta$ - particles, the number of $\alpha$ - and $\beta$ - particles that should be emitted for the conversion of ${}_{92}{U}^{238}$ to ${}_{82}{Pb}^{206}$ are

• A. $7\alpha ,5\beta$
• B. $6\alpha ,4\beta$
• C. $4\alpha ,3\beta$
• D. $8\alpha ,6\beta$

Answer 1

The correct option is D $8\alpha ,6\beta$

${}_{92}{U}^{238}{⟶}_{82}{Pb}^{206}+m_2{He}^4+n_{-1}{e}^0$
On comparing both sides
$238=206+4m$
$\Rightarrow m=8$
$92=82+2m-n$
$2m-n=10$4
$\Rightarrow n=6$
$\therefore \alpha$ - particles emitted, $m=8$
$\beta$ - particles emitted, $n=6$