Chemistry Labs

Problem 2

Dating historical events using 210Pb^{210}\ce{Pb}. Nathan Thompson, an early settler of Lord Howe Island, planted European deciduous trees; the year is unknown. Pollen of European oak and elm (which pollinate in their first year) accumulated in lake sediment together with radioactive 210Pb^{210}\ce{Pb} (half-life = 22.0 years). In 1995 a sediment core was taken: oak/elm pollen first occurs at 50 cm depth, where the 210Pb^{210}\ce{Pb} activity is 1.40 Bq kg⁻¹, versus 356 Bq kg⁻¹ at the top of the core. (2.1) In what year were the seeds planted? (2.2) 210Pb^{210}\ce{Pb} is a daughter of X238X22238U\ce{^{238}U}, which stays in the earth's crust: X238X22238U→X234X22234U→X230X22230Th→X226X22226Ra→X222X22222Rn→(X218X22218Po,X214X22214Bi)→X210X22210Pb→X206X22206Pb\ce{^{238}U} \rightarrow \ce{^{234}U} \rightarrow \ce{^{230}Th} \rightarrow \ce{^{226}Ra} \rightarrow \ce{^{222}Rn} \rightarrow (\ce{^{218}Po}, \ce{^{214}Bi}) \rightarrow \ce{^{210}Pb} \rightarrow \ce{^{206}Pb} (stable). Which step explains how 210Pb^{210}\ce{Pb} reaches rainwater while its parent X238X22238U\ce{^{238}U} remains in the crust?
Step 1 of 2: Counting half-lives
n=log⁡23561.40=ln⁡254ln⁡2≈8 half-lives;t=8×22.0=176 years ⇒ 1995−176=1819n = \log_2\frac{356}{1.40} = \frac{\ln 254}{\ln 2} \approx 8\ \text{half-lives};\qquad t = 8 \times 22.0 = 176\ \mathrm{years}\ \Rightarrow\ 1995 - 176 = 1819
Analysis

Each half-life halves the activity: 356 → 178 → 89 → 44.5 → 22.2 → 11.1 → 5.6 → 2.8 → 1.4 Bq kg⁻¹ — exactly eight halvings reach the pollen horizon. The 50 cm layer was therefore deposited 176±2176 \pm 2 years before 1995, dating the planting to about 1819.