Question

An element $\text{A}$ decays by a two-step process into element $\text{C}$.
$$A\to B+H{e}_2^4$$$$B\to C+2e^{-}$$Then-

• A. $\text{A and C}$ are isotopes.
• B. $\text{A and B}$ are isotopes.
• C. $\text{A and C}$ are isobars.
• D. $\text{A and B}$ are isobars.

Answer 1

The correct option is A $\text{A and C}$ are isotopes.

The first process is an $\alpha$-decay,

$A\to B{+}_{2}H{e}^4$

$\therefore \text{A}$ losses two protons.

The second process is ${\beta }^{-}$-decay,

$B\to C+2e^{-}$

$\therefore \text{C}$ gains two protons.

Hence, atomic number $Z$ is same for $\text{A and C}$ and thus they are isotopes.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.