Question

A radioactive element ${}_{90}{X}^{238}$ decays to ${}_{83}{Y}^{222}.$ How many $\beta$-particles are emitted during this decay?

Answer 1

Number of $\alpha$ particles emitted$=\frac{238-222}{4}$
$=4$
$\therefore$number of $\beta$ particle emiited$=1$ as the atomic number becomes 83.