Step 1: Understanding the Concept:
The problem involves radioactive decay series including alpha (\( \alpha \)) and beta (\( \beta \)) decays.
An alpha particle is a helium nucleus \( _2\text{He}^4 \), so its emission reduces the parent's mass number by 4 and atomic number by 2.
A beta particle is an electron \( _{-1}\text{e}^0 \), so its emission increases the parent's atomic number by 1 without changing the mass number.
Isotopes are atoms with the same atomic number but different mass numbers.
Isobars are atoms with the same mass number but different atomic numbers.
Step 2: Key Formula or Approach:
For a general nucleus \( _Z^M\text{X} \):
Alpha decay: \( _Z^M\text{X} \rightarrow _{Z-2}^{M-4}\text{Y} + _2^4\text{He} \)
Beta decay: \( _Z^M\text{X} \rightarrow _{Z+1}^M\text{Y} + _{-1}^0\text{e} \)
Step 3: Detailed Explanation:
Let the initial radioactive element A be represented as \( _Z^M\text{A} \), where \( M \) is the mass number and \( Z \) is the atomic number.
The first decay process is an alpha emission:
\[ _Z^M\text{A} \rightarrow \text{B} + _2^4\text{He} \]
From the conservation of mass and atomic numbers, element B must be \( _{Z-2}^{M-4}\text{B} \).
The second decay process is the emission of two beta particles:
\[ _{Z-2}^{M-4}\text{B} \rightarrow \text{C} + 2(_{-1}^0\text{e}) \]
Each beta decay increases the atomic number by 1. Since there are two beta decays, the atomic number increases by 2.
Therefore, element C must be \( _{Z-2+2}^{M-4}\text{C} \), which simplifies to \( _Z^{M-4}\text{C} \).
Now we compare element A (\( _Z^M\text{A} \)) and element C (\( _Z^{M-4}\text{C} \)).
They have the same atomic number \( Z \), but different mass numbers (\( M \) and \( M-4 \)).
By definition, elements with the same atomic number and different mass numbers are isotopes.
Thus, A and C are isotopes.
Step 4: Final Answer:
Elements A and C are isotopes.