In nature a decay chain starts with Th^232 and finally terminates at Pb^208. A thorium ore sample wa — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - He Atoms Produced Per Th-232 Decay The complete decay chain: $\ce{^{232}_{90}Th -> ^{208}_{82}Pb + 6^4_2He + 4 e^-}$ Each Th-232 that decays produces **6 helium atoms** (6 alpha particles). Step 2 - Find Moles of He and Th Decayed Volume of He at STP = $6.72 \times 10^{-5}$ mL: $$n_{He} = \frac{6.72 \times 10^{-5}}{22400} = 3.0 \times 10^{-9}\ \text{mol}$$ Moles of Th-232 that decayed: $$n_{Th,\ \text{decayed}} = \frac{n_{He}}{6} = \frac{3.0 \times 10^{-9}}{6} = 5.0 \times 10^{-10}\ \text{mol}$$ Step 3 - Find Current and Initial Moles of Th Current Th-232: $n_{Th} = \dfrac{4.64 \times 10^{-7}}{232} = 2.00 \times 10^{-9}\ \text{mol}$ Initial Th-232: $N_0 = n_{Th} + n_{Th,\ \text{decayed}} = 2.00 \times 10^{-9} + 0.50 \times 10^{-9} = 2.50 \times 10^{-9}\ \text{mol}$ Step 4 - Calculate Age $$\frac{N_0}{N} = \frac{2.50}{2.00} = 1.25$$ $$t = \frac{t_{1/2}}{\ln 2} \ln\!\left(\frac{N_0}{N}\right) = \frac{1.39 \times 10^{10}}{0.693} \times \ln(1.25) = 2.005 \times 10^{10} \times 0.2231 \approx \boxed{4.5 \times 10^9\ \text{yr}}$$ This is closest to option **(C) $4.6 \times 10^9$ years**. $$\boxed{\text{Answer: C — }4.6 \times 10^9\ \text{years}}$$