Explanation

(b) Let $x$ be the mass number of $A\ $and $y\ $the atomic number. Then , since atomic number and mass number remain conserved, we have
$\ _{y}A^{x} \rightarrow \ \ _{y - 2}B^{x - 4}\ + \ \ _{2}He^{4}$
$\ _{y - 2}B^{x - 4} \rightarrow \ _{y}C^{x - 4}\ + \ 2\ _{- 1}e^{0}$
Hence, we observe that $A$ and $C$ are isotopes as their atomic numbers are same but mass numbers are different.