Bismuth is the end product of radioactive disintegration series known as — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - The Four Natural Radioactive Decay Series | Series | Formula | End Product | |--------|---------|-------------| | Thorium | $4n$ | $\ce{^{208}_{82}Pb}$ | | Neptunium | $4n+1$ | $\ce{^{209}_{83}Bi}$ | | Uranium | $4n+2$ | $\ce{^{206}_{82}Pb}$ | | Actinium | $4n+3$ | $\ce{^{207}_{82}Pb}$ | Step 2 - Identify the Series for Bismuth-209 The only stable isotope of Bismuth is $\ce{^{209}_{83}Bi}$ ($A = 209$). $$\frac{209}{4} = 52\ \text{remainder}\ 1 \implies 209 = 4 \times 52 + 1$$ Since $A = 4n+1$ with $n = 52$, Bi-209 belongs to the **4n+1 (Neptunium) series**. Step 3 - Evaluate Options - **(A) $4n$**: End product is Pb-208. Bismuth is not the end of this series. Incorrect. - **(B) $4n+1$**: End product is Bi-209 ✓. **Correct.** - **(C) $4n+2$**: End product is Pb-206. Not Bismuth. Incorrect. - **(D) $4n+3$**: End product is Pb-207. Not Bismuth. Incorrect. $$\boxed{\text{Answer: B — }4n+1\ \text{series (Neptunium series)}}$$