Question:
\({}^{228}_{88}X \xrightarrow{-3\alpha, -\beta} Y\). The element Y is
\({}^{228}_{88}X \xrightarrow{-3\alpha, -\beta} Y\). The element Y is
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\(\alpha\) decay: \({}^{A}_{Z}X \rightarrow {}^{A-4}_{Z-2}Y + {}^{4}_{2}\mathrm{He}\). \(\beta^-\) decay: \({}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + e^- + \bar{\nu}\).
Updated On: Apr 20, 2026
- \({}^{216}_{82}\mathrm{Pb}\)
- \({}^{217}_{82}\mathrm{Pb}\)
- \({}^{218}_{83}\mathrm{Bi}\)
- \({}^{216}_{83}\mathrm{Bi}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
\(\alpha\) decay: mass number decreases by 4, atomic number by 2. \(\beta\) decay: mass number unchanged, atomic number increases by 1.
Step 2: Detailed Explanation:
Start: \({}^{228}_{88}X\)
After 3\(\alpha\) decays: Mass = 228 - 3×4 = 216, Atomic = 88 - 3×2 = 82.
After 1\(\beta\) decay: Mass = 216 (unchanged), Atomic = 82 + 1 = 83.
Element with Z=83 is Bi. So Y = \({}^{216}_{83}\mathrm{Bi}\).
Step 3: Final Answer:
\({}^{216}_{83}\mathrm{Bi}\)
\(\alpha\) decay: mass number decreases by 4, atomic number by 2. \(\beta\) decay: mass number unchanged, atomic number increases by 1.
Step 2: Detailed Explanation:
Start: \({}^{228}_{88}X\)
After 3\(\alpha\) decays: Mass = 228 - 3×4 = 216, Atomic = 88 - 3×2 = 82.
After 1\(\beta\) decay: Mass = 216 (unchanged), Atomic = 82 + 1 = 83.
Element with Z=83 is Bi. So Y = \({}^{216}_{83}\mathrm{Bi}\).
Step 3: Final Answer:
\({}^{216}_{83}\mathrm{Bi}\)
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