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(D) Let the nucleus $X$ be represented as $_{Z}^{A}X$.$1$. When $X$ undergoes $\alpha$-decay,it emits an $\alpha$-particle $(_{2}^{4}He)$. The new nuc

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$X$ and $Z$ are isotopes

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(D) Let the nucleus $X$ be represented as $_{Z}^{A}X$.$1$. When $X$ undergoes $\alpha$-decay,it emits an $\alpha$-particle $(_{2}^{4}He)$. The new nucleus $Y$ will have an atomic number $Z-2$ and a mass number $A-4$:$_{Z}^{A}X \xrightarrow{\alpha} _{Z-2}^{A-4}Y + _{2}^{4}He$$2$. When $Y$ undergoes $2\beta$-decay,it emits two $\beta$-particles (electrons,$_{-1}^{0}e$). Each $\beta$-decay increases the atomic number by $1$ without changing the mass number:$_{Z-2}^{A-4}Y \xrightarrow{2\beta} _{Z-2+2}^{A-4}Z + 2_{-1}^{0}e$$3$. The resulting nucleus $Z$ is $_{Z}^{A-4}Z$.Comparing $X$ $(_{Z}^{A}X)$ and $Z$ $(_{Z}^{A-4}Z)$,we see they have the same atomic number $Z$ but different mass numbers. Therefore,$X$ and $Z$ are isotopes.

$A$ nucleus $X$ undergoes the following transformation:
$X \xrightarrow{\alpha} Y$
$Y \xrightarrow{2\beta} Z$
Then:

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$A$ nucleus $X$ undergoes the following transformation:$X \xrightarrow{\alpha} Y$$Y \xrightarrow{2\beta} Z$Then: