The end product of (4n+2) disintegration series the — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - The Four Radioactive Decay Series | Series | Formula | Start | End | |--------|---------|-------|-----| | Thorium | $4n$ | $\ce{^{232}_{90}Th}$ | $\ce{^{208}_{82}Pb}$ | | Neptunium | $4n+1$ | $\ce{^{237}_{93}Np}$ | $\ce{^{209}_{83}Bi}$ | | **Uranium** | **$4n+2$** | $\ce{^{238}_{92}U}$ | $\ce{^{206}_{82}Pb}$ | | Actinium | $4n+3$ | $\ce{^{235}_{92}U}$ | $\ce{^{207}_{82}Pb}$ | Step 2 - Identify the (4n+2) Series End Product The $(4n+2)$ series is the **Uranium series**. Verify: - Start: $238 = 4(59)+2$ ✓ - End: $206 = 4(51)+2$ ✓ → stable $\ce{^{206}_{82}Pb}$ Step 3 - Evaluate Options - **(A) $\ce{_{82}Pb^{204}}$**: $204 \div 4 = 51$ remainder 0 → belongs to $4n$ series. Not the end of $4n+2$. Incorrect. - **(B) $\ce{_{82}Pb^{208}}$**: $208 \div 4 = 52$ remainder 0 → belongs to $4n$ (Thorium) series. Incorrect. - **(C) $\ce{_{82}Pb^{209}}$**: $209 \div 4 = 52$ remainder 1 → belongs to $4n+1$ series; also radioactive (not the stable endpoint). Incorrect. - **(D) $\ce{_{82}Pb^{206}}$**: $206 \div 4 = 51$ remainder 2 → belongs to $4n+2$ ✓. Stable end product. **Correct.** $$\boxed{\text{Answer: D} — \ce{^{206}_{82}Pb}}$$