Question:

A radioactive element A decays into radioactive element C by the following processes in succession. \( A \rightarrow B + _2\text{He}^4; B \rightarrow C + 2e^- \) Then elements}

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One $\alpha$ decay followed by two $\beta$ decays returns the atom to the same position in the periodic table (isotope).
Updated On: Apr 26, 2026
  • A and B are isobars.
  • A and C are isobars.
  • A and C are isotopes.
  • A and B are isotopes.
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The Correct Option is C

Solution and Explanation

Step 1: First Decay ($\alpha$ decay)
$_Z\text{A}^M \to _{Z-2}\text{B}^{M-4} + _2\text{He}^4$.
Step 2: Second Decay ($2\beta$ decay)
$_{Z-2}\text{B}^{M-4} \to _{Z-2+2}\text{C}^{M-4} + 2e^-$.
So, element C is $_Z\text{C}^{M-4}$.
Step 3: Compare A and C
Element A and C have the same atomic number $Z$ but different mass numbers. Therefore, they are isotopes.
Final Answer: (C)
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