Question 11

Let $N_0$ be the initial number of moles of U-238 and $N_t$ be the number of moles at time $t$. Let $N_{Pb}$ be the moles of Pb-206 formed. $N_0 = N_t + N_{Pb}$. Given mass ratio: $\frac{\text{mass of Pb-206}}{\text{mass of U-238}} = 7$ $\frac{N_{Pb} \times 206}{N_t \times 238} = 7 \implies \frac{N_{Pb}}{N_t} = 7 \times \frac{238}{206} \approx 8.087$ $\frac{N_0}{N_t} = \frac{N_t + N_{Pb}}{N_t} = 1 + \frac{N_{Pb}}{N_t} = 1 + 8.087 = 9.087$ Using first-order decay formula: $t = \frac{T_{1/2}}{\ln 2} \ln\left(\frac{N_0}{N_t}\right)$ $t = \frac{4.5 \times 10^9}{0.693} \times \ln(9.087)$ $t = \frac{4.5 \times 10^9}{0.693} \times 2.2068 \approx 1.433 \times 10^{10}$ years $t = 143.3 \times 10^8$ years. Therefore, $P = 143.3$.