Question

The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction ${}_{92}^{238}U{\to }_{82}^{214}Pb$ is:

Answer 1

Number of $\alpha$ particles emitted $=\frac{massofU-massofPb}{massof\alpha particles}=\frac{238-214}{4}=6$

Number of $\beta$ particles emitted $=2\times no.of\alpha particlesemitted-(92-82)=2$

Reaction:
${}_{92}{U}^{238}{\stackrel{-6\alpha }{\to }}_{80}{X}^{214}{\stackrel{-2\beta }{\to }}_{82}P{b}^{214}$

$(6\alpha ,2\beta ),$ i.e., total 8 paritcles.