The radioactive series to which _88Ra^224 belongs is — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - The Four Natural Radioactive Decay Series Each natural decay series is classified by the remainder of its members' mass numbers when divided by 4 (only $\alpha$-decay changes $A$, always by 4): | Series | Formula | Start | End | |--------|---------|-------|-----| | Thorium | $A = 4n$ | $\ce{^{232}_{90}Th}$ | $\ce{^{208}_{82}Pb}$ | | Neptunium | $A = 4n+1$ | $\ce{^{237}_{93}Np}$ | $\ce{^{209}_{83}Bi}$ | | Uranium | $A = 4n+2$ | $\ce{^{238}_{92}U}$ | $\ce{^{206}_{82}Pb}$ | | Actinium | $A = 4n+3$ | $\ce{^{235}_{92}U}$ | $\ce{^{207}_{82}Pb}$ | Step 2 - Check the Mass Number of $\ce{^{224}_{88}Ra}$ $$A = 224$$ $$\frac{224}{4} = 56 \quad \text{(remainder = 0)}$$ $$224 = 4 \times 56 \implies A = 4n \text{ with } n = 56$$ Step 3 - Identify the Series $A = 4n$ → **Thorium series**. Step 4 - Evaluate Options - **(A) Actinium series** ($4n+3$): $224 \mod 4 = 0 \neq 3$. Incorrect. - **(B) Thorium series** ($4n$): $224 \mod 4 = 0$ ✓. **Correct.** - **(C) Uranium series** ($4n+2$): $224 \mod 4 = 0 \neq 2$. Incorrect. - **(D) Neptunium series** ($4n+1$): $224 \mod 4 = 0 \neq 1$. Incorrect. $$\boxed{\text{Answer: B — Thorium series}}$$