Question

In a radioactive decay series, three successive decays each result in a particle being emitted. The first decay results in the emission of a ${\beta }^{-}$ particle. The second decay results in the emission of an $\alpha$ particle. The third decay results in the emission of another ${\beta }^{-}$ particle.
Nuclides $P$ and $S$ are compared.
Which statements is correct?

• A. $P$ and $S$ are identical in all respects.
• B. $P$ and $S$ are isotopes of the same element.
• C. $S$ is a different element of lower atomic number
• D. $S$ is a different element of reduced mass.

Answer 1

The correct option is B $P$ and $S$ are isotopes of the same element.

$\text{Given -* Radioactive decay series-}$

$p\stackrel{{\beta }^{-}\text{emit}}{⟶}Q\stackrel{\alpha \text{emit}}{⟶}R\stackrel{\beta \text{emit}}{⟶}S$

$\text{we know that,}$

$\text{* one}\alpha \text{- particle}$

$\text{- decreases atomic mass by 4 units.}$

$\text{- decreases atomic num by 2 units.}$

$\begin{array}{c}\text{* one}\beta \text{-particle.}\\ \text{- bring no change in atomic mass.}\\ \text{- increases atomic no. by l unit.}\\ \Rightarrow \text{when one}\alpha \text{-particle is emitted}\\ \Rightarrow \text{(s)will have atomic mass less than}4\text{units}\\ \text{from(p).}\\ \Rightarrow \text{(s) will have atomic no. less by}2\text{units}\\ \text{from (p).}\end{array}$

$\begin{array}{c}\Rightarrow \text{When two}{\beta }^{-}\text{particles are emitted}\\ \text{Increases the atomic no. by}2\text{units.}\\ \text{Hence,}1\alpha \text{-particle,}2{\beta }^{-}\text{particle emission bring no}\\ \text{change in atomic Num.}\\ \text{Pand}S\text{differs only in atomic mass and are isotopes.}\end{array}$