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 $$P \times 10^8$$ 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 $$4.5 \times 10^9$$ years; $$\log_e 2$$ = 0.693]

Let the sample contain $$N_0$$ atoms of $$^{238}U$$ at the start (time $$t = 0$$).
After a time $$t$$, some of these nuclei have decayed to $$^{206}Pb$$ while the rest, $$N$$, are still $$^{238}U$$.

The radioactive decay law is
$$N = N_{0}\,e^{-\lambda t}$$
where $$\lambda$$ is the decay constant of $$^{238}U$$.

The number of $$^{206}Pb$$ atoms produced is the number of decayed $$^{238}U$$ atoms:
$$N_{Pb} = N_{0} - N = N_{0}\left(1 - e^{-\lambda t}\right)$$.

The ratio given in the statement is a ratio of MASSES, not of atom numbers.
If $$m_U$$ and $$m_{Pb}$$ are the masses of $$^{238}U$$ and $$^{206}Pb$$ present after time $$t$$, then

$$m_U = N \times 238 \ (\text{atomic mass units})$$
$$m_{Pb} = N_{Pb} \times 206$$.

We are told that
$$\frac{m_{Pb}}{m_U} = 7$$.

Inserting the expressions for the masses:
$$\frac{206\,(N_{0} - N)}{238\,N} = 7$$.

Simplify by cancelling $$N_0$$ (write $$N = N_0 e^{-\lambda t}$$):
$$\frac{206\left(1 - e^{-\lambda t}\right)}{238\,e^{-\lambda t}} = 7$$.

Bring the denominator to the right-hand side:
$$206\left(1 - e^{-\lambda t}\right) = 7 \times 238\,e^{-\lambda t}$$.

Divide both sides by 206:
$$1 - e^{-\lambda t} = 7\left(\frac{238}{206}\right)e^{-\lambda t} = 7\left(\frac{119}{103}\right)e^{-\lambda t}$$.

Calculate the numerical factor:
$$\frac{119}{103} \approx 1.15534 \quad\Rightarrow\quad 7 \times 1.15534 \approx 8.0874.$$

So
$$1 - e^{-\lambda t} = 8.0874\,e^{-\lambda t}.$$ Move the exponential term to one side:
$$1 = 9.0874\,e^{-\lambda t}.$$

Hence
$$e^{-\lambda t} = \frac{1}{9.0874} \approx 0.1100.$$

Take natural logarithm:
$$-\lambda t = \ln(0.1100) \approx -2.207.$$
Therefore
$$\lambda t = 2.207.$$

The decay constant is related to the half-life $$t_{1/2}$$ by $$\lambda = \dfrac{0.693}{t_{1/2}}$$.
Given $$t_{1/2} = 4.5 \times 10^{9}\,\text{years}$$, we have

$$\lambda = \frac{0.693}{4.5 \times 10^{9}} = 1.54 \times 10^{-10}\,\text{year}^{-1}.$$

Substitute into $$\lambda t = 2.207$$:
$$t = \frac{2.207}{\lambda} = \frac{2.207}{1.54 \times 10^{-10}} \text{ years}.$$

Compute:
$$t \approx 1.4329 \times 10^{10}\ \text{years}.$$

Express the age as $$P \times 10^{8}$$ years:
$$1.4329 \times 10^{10} = 143.29 \times 10^{8}.$$ Taking the nearest integer (since data are given to two-three significant figures), $$P = 143.$$

Final Answer: 143