A rock, recovered from far underground, is found to contain 0.86 mg of U^238, 0.15 mg of Pb^206 and — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Determine Rock Age from U-238/Pb-206 Ratio Using first-order decay: $N = N_0 e^{-\lambda t}$, and all Pb-206 came from U-238 decay. Moles of current $\ce{^{238}U}$: $n_U = \dfrac{0.86}{238} = 3.613 \times 10^{-3}$ Moles of $\ce{^{206}Pb}$: $n_{Pb} = \dfrac{0.15}{206} = 7.28 \times 10^{-4}$ Initial moles of U-238: $N_0 = n_U + n_{Pb} = 3.613 \times 10^{-3} + 7.28 \times 10^{-4} = 4.341 \times 10^{-3}$ $$\frac{N_0}{N} = \frac{4.341}{3.613} = 1.2015$$ $$t = \frac{t_{1/2}}{\ln 2} \ln\!\left(\frac{N_0}{N}\right) = \frac{4.47 \times 10^9}{0.6931} \times \ln(1.2015) = 6.449 \times 10^9 \times 0.1833 = 1.182 \times 10^9\ \text{yr}$$ Step 2 - Find Current K-40 from Ar-40 and Rock Age All $\ce{^{40}Ar}$ came from $\ce{^{40}K}$ decay. Let $N_K$ = current moles K-40. $$e^{\lambda_K t} - 1 = \frac{N_{Ar}}{N_K}$$ Moles Ar-40: $n_{Ar} = \dfrac{1.6}{40} = 0.040$ $$\lambda_K = \frac{\ln 2}{t_{1/2}(K)} = \frac{0.6931}{1.25 \times 10^9} = 5.545 \times 10^{-10}\ \text{yr}^{-1}$$ $$\lambda_K t = 5.545 \times 10^{-10} \times 1.182 \times 10^9 = 0.655$$ $$e^{0.655} - 1 = 1.925 - 1 = 0.925$$ $$N_K = \frac{n_{Ar}}{0.925} = \frac{0.040}{0.925} = 0.0432$$ $$m_{K} = 0.0432 \times 40 = \boxed{1.73\ \text{mg} \approx 1.7\ \text{mg}}$$ Step 3 - Evaluate Options - **(A) 1.7 mg**: Matches the calculation. **Correct.** - **(B) 3.4 mg**: Twice the answer. Incorrect. - **(C) 5.1 mg**: Three times the answer. Incorrect. - **(D) 0.85 mg**: Half the answer. Incorrect. $$\boxed{\text{Answer: A — approximately 1.7 mg of K-40}}$$