_84Rn^219 is member of actinium series. Another member of the same series is — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Decay Series of Rn-219 The four natural radioactive decay series are classified by mass number modulo 4: | Series | Formula | Members satisfy | |--------|---------|----------------| | Thorium | $4n$ | $A \equiv 0 \pmod{4}$ | | Neptunium | $4n+1$ | $A \equiv 1 \pmod{4}$ | | Uranium | $4n+2$ | $A \equiv 2 \pmod{4}$ | | Actinium | $4n+3$ | $A \equiv 3 \pmod{4}$ | For $\ce{^{219}_{86}Rn}$: $219 = 4 \times 54 + 3 \implies A \equiv 3 \pmod{4}$ → **Actinium series** ($4n+3$). Step 2 - Check Each Option **(A) $\ce{^{235}_{92}U}$**: $235 = 4 \times 58 + 3 \implies A \equiv 3 \pmod{4}$ ✓ → Member of Actinium series. **Correct.** **(B) $\ce{^{232}_{90}Th}$**: $232 = 4 \times 58 \implies A \equiv 0 \pmod{4}$ → Thorium series. Not the Actinium series. Incorrect. **(C) $\ce{^{225}_{88}Ra}$**: $225 = 4 \times 56 + 1 \implies A \equiv 1 \pmod{4}$ → Neptunium series. Incorrect. **(D) $\ce{^{35}_{15}P}$**: $35 = 4 \times 8 + 3 \implies A \equiv 3 \pmod{4}$, but P-35 is a light artificially produced nuclide, not a member of the natural heavy-element Actinium series. Incorrect. $$\boxed{\text{Answer: A — }\ \ce{^{235}_{92}U}\ \text{is also a member of the Actinium series}}$$