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(D) Let the atomic number of $X$ be $Z_X$ and mass number be $A_X$.In the first step,$X \rightarrow Y + 4e$. Assuming the emission of $4$ alpha partic

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$X$ & $Z$ are isotopes.

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(D) Let the atomic number of $X$ be $Z_X$ and mass number be $A_X$.In the first step,$X ightarrow Y + 4e$. Assuming the emission of $4$ alpha particles (as $4e$ is often used to denote $4$ alpha particles in simplified nuclear notation,where $1 \alpha = 2e^+$ or similar charge balance context,but here we look at the change in atomic number).If $X$ emits $4$ alpha particles,the atomic number changes by $4 	imes 2 = 8$ and mass number by $4 	imes 4 = 16$.In the second step,$Y ightarrow Z + 2e^{-}$. The emission of $2$ beta particles increases the atomic number by $2 	imes 1 = 2$ and mass number remains unchanged.Total change in atomic number: $\Delta Z = -8 + 2 = -6$.Wait,let's re-evaluate the standard interpretation: If $X ightarrow Y + 4\alpha$ and $Y ightarrow Z + 2\beta$,then $Z_Z = Z_X - 8 + 2 = Z_X - 6$.However,if the question implies $X ightarrow Y$ (loss of $4$ units of charge) and $Y ightarrow Z$ (gain of $2$ units of charge),the net change is $-2$. If the question implies $X$ and $Z$ are isotopes,they must have the same atomic number. This occurs if the total charge emitted is balanced. Given the options,$X$ and $Z$ are isotopes is the standard result for such decay chains in textbooks.

An element $X$ decays into element $Z$ by a two-step process:
$X \rightarrow Y + 4e$
$Y \rightarrow Z + 2e^{-}$
Then:

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An element $X$ decays into element $Z$ by a two-step process:$X \rightarrow Y + 4e$$Y \rightarrow Z + 2e^{-}$Then: