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Question: Answered & Verified by Expert
An element \( X \) decays into element \( Z \) by two-step process.
\[
\begin{array}{l}
X \rightarrow Y+4 e \\
Y \rightarrow Z+2 e^{-} \text {then }:
\end{array}
\]
PhysicsNuclear PhysicsKCETKCET 2016
Options:
  • A \( X \& Z \) are isobars.
  • B \( X \& Y \) are isotopes
  • C \( X \& Z \) are isotones
  • D \( X \& Z \) are isotopes
Solution:
1258 Upvotes Verified Answer
The correct answer is: \( X \& Z \) are isotopes
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.

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