Actinium series starts with A and ends at Z. A and Z are — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - The Four Natural Radioactive Decay Series | Series | $A$ 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 - Verify the Actinium (4n+3) Series **Starting nuclide:** $\ce{^{235}_{92}U}$ $$235 = 4 \times 58 + 3 \implies A = 4n+3\ (n=58)\ ✓$$ **Terminal nuclide:** $\ce{^{207}_{82}Pb}$ $$207 = 4 \times 51 + 3 \implies A = 4n+3\ (n=51)\ ✓$$ Note: Although the series starts with U-235, it is historically called the "Actinium series" because $\ce{^{227}_{89}Ac}$ is a chemically prominent intermediate member. Step 3 - Evaluate Options - **(A) $\ce{_{90}Th^{232}}, \ce{_{82}Pb^{206}}$**: Th-232 starts the Thorium ($4n$) series; Pb-206 ends the Uranium ($4n+2$) series. Neither is correct for Actinium. Incorrect. - **(B) $\ce{_{92}U^{235}}, \ce{_{82}Pb^{207}}$**: U-235 and Pb-207 both satisfy $4n+3$ ✓. **Correct.** - **(C) $\ce{_{92}U^{238}}, \ce{_{82}Pb^{207}}$**: U-238 belongs to $4n+2$ (Uranium series), not $4n+3$. Incorrect. - **(D) $\ce{_{90}Ac^{227}}, \ce{_{82}Bi^{209}}$**: Ac-227 is an intermediate; Bi-209 ends the Neptunium ($4n+1$) series. Incorrect. $$\boxed{\text{Answer: B} — \ce{^{235}_{92}U}\ \text{to}\ \ce{^{207}_{82}Pb}}$$