Question

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

Answer 1

Let there be $x\alpha$ and $y\beta$ particles be emitted.

${}_{92}^{238}{\to }_{82}^{214}Pb+x{}_2^{4}He+y{}_{-1}^{0}e$

Equating atomic numbers and atomic masses we get,

$92=82+2x-y$

$⟹2x-y=10$

$238=214+4x$

$⟹x=6$

$y=2$

So 6 alpha particles and 2 beta particles.