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An element \( X \) decays into element \( Z \) by two-step process.
\[
\begin{array}{l}
X \rightarrow Y+4 e \\
Y \rightarrow Z+2 e^{-} \text {then }:
\end{array}
\]

A \( X \& Z \) are isobars.
B \( X \& Y \) are isotopes
C \( X \& Z \) are isotones
D \( X \& Z \) are isotopes

Verified Solution

Answer & Solution

Correct Answer

(D) \( X \& Z \) are isotopes

Step-by-step Solution

Detailed explanation

The two processes are as follows:
\[
\mathrm{z}^{\mathrm{A}} X \rightarrow_{\mathrm{Z}-2}^{\mathrm{A}-{ }^{4} Y+{ }_{2}{ }^{4} \mathrm{He}}
\]
Since \( X \) and \( Z \) have same atomic number, that is, same number of protons but different atomic mass, that is, sum of
number of protons and neutrons.
We know isotopes are atoms with same number of protons but different number of neutrons. Therefore, \( X \) and \( Z \) are
isotopes.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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An element \( X \) decays into element \( Z \) by two-step process.\[\begin{array}{l}X \rightarrow Y+4 e \\Y \rightarrow Z+2 e^{-} \text {then }:\end{array}\]

Step-by-step Solution

An element \( X \) decays into element \( Z \) by two-step process.
\[
\begin{array}{l}
X \rightarrow Y+4 e \\
Y \rightarrow Z+2 e^{-} \text {then }:
\end{array}
\]