Age of Sample from Uranium-238 to Lead-206 Radioactive Decay

1. Relating Mass Ratio to Mole / Atom Ratio:
   Let $N_0$ be the initial number of $^{238}\text{U}$ nuclei at $t = 0$.
   At time $t$, let $N$ be the remaining number of $^{238}\text{U}$ nuclei.
   The number of $^{206}\text{Pb}$ nuclei formed by decay is:
   $N_{\text{Pb}} = N_0 - N$

The mass of $^{206}\text{Pb}$ and $^{238}\text{U}$ in the sample are:
$m_{\text{Pb}} = N_{\text{Pb}} \times 206 = (N_0 - N) \times 206$
$m_{\text{U}} = N \times 238$

Given that the ratio of mass of $^{206}\text{Pb}$ to $^{238}\text{U}$ is $7$:
$\frac{m_{\text{Pb}}}{m_{\text{U}}} = \frac{(N_0 - N) \times 206}{N \times 238} = 7$
$\frac{N_0 - N}{N} = 7 \times \frac{238}{206} = \frac{1666}{206} = \frac{833}{103}$
$\frac{N_0}{N} = 1 + \frac{833}{103} = \frac{936}{103} \approx 9.0874$

1. Radioactive Decay Kinetics:
   Radioactive decay follows first-order kinetics:
   $N = N_0 e^{-\lambda t} \implies \lambda t = \ln\left(\frac{N_0}{N}\right)$
   The decay constant $\lambda$ is given by:
   $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9\text{ years}}$

2. Calculating Age of the Sample ($t$):
   $t = \frac{\ln\left(\frac{936}{103}\right)}{\ln 2} \times t_{1/2}$
   $\ln\left(\frac{936}{103}\right) \approx \ln(9.0874) \approx 2.2069$
   $t = \frac{2.2069}{0.693} \times (4.5 \times 10^9\text{ years}) \approx 3.1845 \times 4.5 \times 10^9\text{ years} \approx 14.33 \times 10^9\text{ years}$
   $t = 143.3 \times 10^8\text{ years} \approx 143 \times 10^8\text{ years}$

Comparing with $P \times 10^8\text{ years}$:
$P = 143$

Therefore, the value of $P$ is 143.