A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb-206 while the rest of it remains undisintegrated. When the age of the sample is P x10⁸ years, the ratio of mass of Pb-206 to that of U-238 in the sample is found to be 7. The value of P is _____.

[Given: Half-life of U-238 is 4.5 x10⁹ years; log_e2 = 0.693]

The correct answer is 143.

Step-by-step Explanation:

Let the initial number of nuclei of Uranium-238 (${}_{92}^{238}\text{U}$) in the sample be $N_0$ at time $t=0$.

At time $t$ (the age of the sample), let:
- $N_{\text{U}}$ be the number of active Uranium-238 nuclei remaining.
- $N_{\text{Pb}}$ be the number of Lead-206 nuclei formed by radioactive decay.

Since each decayed ${}^{238}\text{U}$ nucleus eventually becomes a ${}^{206}\text{Pb}$ nucleus, the total number of nuclei is conserved:
$N_0=N_{\text{U}}+N_{\text{Pb}}$

The mass of Uranium-238 and Lead-206 in the sample can be written as:
$m_{\text{U}}=N_{\text{U}}\times 238\text{amu}$
$m_{\text{Pb}}=N_{\text{Pb}}\times 206\text{amu}$

We are given that the ratio of the mass of Pb-206 to that of U-238 is 7:
$\frac{m_{\text{Pb}}}{m_{\text{U}}}=7$

Substituting the expressions for the masses:
$\frac{N_{\text{Pb}}\times 206}{N_{\text{U}}\times 238}=7$

Solving for the ratio of the number of nuclei:
$\frac{N_{\text{Pb}}}{N_{\text{U}}}=7\times \frac{238}{206}=\frac{1666}{206}\approx 8.0874$

Now, find the ratio of the initial number of nuclei ($N_0$) to the remaining number of nuclei ($N_{\text{U}}$):
$\frac{N_0}{N_{\text{U}}}=\frac{N_{\text{U}}+N_{\text{Pb}}}{N_{\text{U}}}=1+\frac{N_{\text{Pb}}}{N_{\text{U}}}\approx 1+8.0874=9.0874$

According to the law of radioactive decay:
$N_{\text{U}}=N_0{e}^{-\lambda t}\Rightarrow \frac{N_0}{N_{\text{U}}}=e^{\lambda t}$

Taking the natural logarithm on both sides:
$\lambda t={\log }_e\left(9.0874\right)$

Using the relation for decay constant $\lambda =\frac{{\log }_{e}2}{T_{1/2}}$:
$t=\frac{{\log }_e\left(9.0874\right)}{{\log }_{e}2}\times T_{1/2}$

Given:
- Half-life $T_{1/2}=4.5\times {10}^9\text{years}$
- ${\log }_{e}2=0.693$
- ${\log }_e\left(9.0874\right)\approx 2.2069$

Substitute these values to calculate the age of the sample $t$:
$t=\frac{2.2069}{0.693}\times 4.5\times {10}^9\text{years}$
$t\approx 3.1846\times 4.5\times {10}^9\text{years}$
$t\approx 14.33\times {10}^9\text{years}=143.3\times {10}^8\text{years}$

Comparing this with the given format $P\times {10}^8\text{years}$, we find:
$P\approx 143$