A rock, recovered from far underground, is found to contain 0.86 mg of , 0.15 mg of and 1.6 mg of . — Nuclear Chemistry and Radioactivity Chemistry Question
Question
A rock, recovered from far underground, is found to contain 0.86 mg of $\text{U}^{238}$, 0.15 mg of $\text{Pb}^{206}$ and 1.6 mg of $\text{Ar}^{40}$. How much $\text{K}^{40}$ will it likely contain? Half-lives of $\text{U}^{238}$ and $\text{K}^{40}$ are $4.47 \times 10^9 \text{ years}$ and $1.25 \times 10^9 \text{ years}$, respectively.
💡 Solution & Explanation
Determine age using U-Pb: $N_U = 0.86/238 = 0.00361 \text{ mmol}$, $N_{Pb} = 0.15/206 = 0.000728 \text{ mmol}$. $N_0/N_U = 1 + 0.000728/0.00361 = 1.20$. Age $t = \frac{4.47 \times 10^9}{0.693} \ln(1.20) = 1.18 \times 10^9 \text{ years}$. Now use age for K-Ar: $\ln(1 + \frac{N_{Ar}}{N_K}) = \frac{t \ln 2}{t_{1/2}} = \frac{1.18}{1.25} \times 0.693 = 0.654$. $1 + N_{Ar}/N_K = e^{0.654} = 1.92 \implies N_{Ar}/N_K = 0.92$. Moles Ar = $1.6/40 = 0.04 \text{ mmol}$. Moles K = $0.04/0.92 \approx 0.043 \text{ mmol}$. Mass of K = $0.043 \times 40 \approx 1.7 \text{ mg}$. Therefore, correct answer is A.