An element A decays into an element C by a two-step process.
$A\to B+He_{2}^{4}$ and $B\to C+2e_{-1}^{0}$ . Then,
(A) A and C are isotopes
(B) A and C are isobars
(C) B and C are isotopes
(D) A and B are isobars

Hint: We will first understand the above radioactive process. In the first step, a Helium atom has been released. This means that the atomic number of the daughter nucleus has decreased by two and its mass has decreased by four. And, in the second step, the second daughter nucleus has gained atomic number by two but its mass remains the same. We will use this analogy to solve our problem.

Complete answer:
Let us first assign some terms that we are going to use in our solution later.
Let the atomic numbers of A,B and C be ${{Z}_{A}},{{Z}_{B}},{{Z}_{C}}$ respectively. And let the atomic mass of A,B and C be ${{M}_{A}},{{M}_{B}},{{M}_{C}}$ respectively. Then, in the first step of our radioactive decay process, we have:
$A\to B+He_{2}^{4}$
Here, we can see that due to the emission of 1 He atom, the atomic number of daughter nucleus B is affected. This can be written as follows:
$\Rightarrow {{Z}_{B}}={{Z}_{A}}-2$
$\therefore {{Z}_{B}}+2={{Z}_{A}}$ [Let this expression be termed as equation number (1)]
And,
$\Rightarrow {{M}_{B}}={{M}_{A}}-4$
Now, in the second step of the process, we have:
$B\to C+2e_{-1}^{0}$
In this case, the final atomic number of daughter nucleus in terms of its parent nucleus can be written as:
 $\Rightarrow {{Z}_{C}}={{Z}_{B}}+2$ [Let this expression be termed as equation number (1)]
And,
$\Rightarrow {{M}_{C}}={{M}_{B}}$
Thus, on equating equation number (1) and (2), we get:
$\Rightarrow {{Z}_{C}}={{Z}_{A}}$
Thus, A and C have the same atomic numbers but different masses.
Hence, we can say that A and C are isotopes of each other.

So, the correct answer is “Option A”.

Note: We should know the basic difference in definitions between these similar terms like, isotopes, isobars, isotones etc. Also, in the above problem, since B and C have the same atomic mass but different atomic numbers, they can be collectively termed as isobars.