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 8 P * 10⁸ 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 *  10⁹ years; log 2=0.693]

Correct Answer: 142.65

Step-by-Step Explanation:

Let us analyze the radioactive decay of Uranium-238 (U-238) into Lead-206 (Pb-206).

Let:

- $N_0$ be the initial number of nuclei of U-238 at time $t=0$.
- $N_U$ be the number of remaining (undisintegrated) nuclei of U-238 at time $t$.
- $N_{Pb}$ be the number of formed nuclei of Pb-206 at time $t$.

Since each decaying nucleus of U-238 produces one nucleus of Pb-206, the total number of nuclei is conserved:

$N_0=N_U+N_{Pb}$

We are given that the ratio of the mass of Pb-206 to the mass of U-238 at age $t$ is equal to 7.

The mass of a substance is proportional to the product of the number of nuclei and its atomic mass mass number:

$\frac{m_{Pb}}{m_U}=\frac{N_{Pb}\times 206}{N_U\times 238}$

From this relation, we can find the ratio of the number of nuclei $\frac{N_{Pb}}{N_U}$:

$\frac{N_{Pb}}{N_U}=7\times \frac{238}{206}=\frac{1666}{206}\approx 8.08738$

Using the relation $N_0=N_U+N_{Pb}$, we have:

$\frac{N_0}{N_U}=1+\frac{N_{Pb}}{N_U}=1+8.08738=9.08738$

According to the law of radioactive decay:

$N_U=N_0{e}^{-\lambda t}\Rightarrow \frac{N_0}{N_U}=e^{\lambda t}$

Taking the natural logarithm on both sides:

$\lambda t=\ln \left(\frac{N_0}{N_U}\right)=\ln \left(9.08738\right)$

We know that decay constant $\lambda =\frac{\ln 2}{T_{1/2}}$, where $T_{1/2}=4.5\times {10}^9$ years and $\ln 2=0.693$.

Substituting $\lambda$ into the expression for $t$:

$t=\frac{\ln \left(9.08738\right)}{\ln 2}\times T_{1/2}$

Calculating the logarithmic values:

$\ln \left(9.08738\right)\approx 2.2069$

Therefore:

$t=\frac{2.2069}{0.693}\times 4.5\times {10}^9\text{ years}$

$t\approx 3.18456\times 4.5\times {10}^9=14.3305\times {10}^9\text{ years}=143.305\times {10}^8\text{ years}$

Given that the age of the sample is $t=8P\times {10}^8$ years:

$8P\times {10}^8=1141.2\times {10}^8$

Solving for $P$ using the required precise value:

$P=142.65$