Question

A radioactive element ${}_{90}{X}^{238}$ decays in to ${}_{82}{Y}^{222}$. The number of $\beta$-particles emitted is:

• A. $0$
• B. $4$
• C. $6$
• D. $1$

Answer 1

The correct option is A $0$

$\text{Given-}\ast {\text{Radioactive element}}_{90}{X}^{238}{⟶}_{82}{Y}^{222}+\text{particles}$

$\begin{array}{c}\text{we know that -}\\ \text{* For}1\alpha \text{particle emission}⟶\text{mass reduces by (4)}\\ \text{Atomic num. reduces by (2) .}\\ \text{* For 1}\beta \text{particle emission}⟶\text{No change in mass}\\ \text{Atomic number increases by (1).}\end{array}$

$\text{For}\to {}_{90}{X}^{238}{⟶}_{82}{Y}^{222}$

$\text{Change in mass num}=16$

$\text{we can say that (4)}\alpha \text{-particles are released.}$

$\text{since,}\beta \text{- barticles bring no change in mass.}$

$\text{Hence, 4 alpha particles reduces the atomic number}$

$\text{by a value of (8).}$

$\begin{array}{c}\text{And in the decay exactly atomic number reduces}\\ \text{by (}8\text{) .}\\ \text{Hence,}4\alpha \text{-particles are released and}0\beta \text{-particles}\\ \text{are emitted}\end{array}$