1. Given Data:

• Ratio of mass of $ \mathrm{Pb}^{206} $ to $ \mathrm{U}^{238} $ = $ \frac{7}{1} $


• Mass of $ \mathrm{Pb}^{206} $ = 7 g


• Mass of $ \mathrm{U}^{238} $ = 1 g


• Half-life of $ \mathrm{U}^{238} $ = $ 4.5 \times 10^9 $ years


• $ \log_e 2 = 0.693 $

1. Calculate Moles:

• Moles of $ \mathrm{Pb}^{206} $:

$ \text{Moles of } \mathrm{Pb}^{206} = \frac{7}{206} = 34 \times 10^{-3} \text{ moles} $

• Moles of $ \mathrm{U}^{238} $:

$ \text{Moles of } \mathrm{U}^{238} = \frac{1}{238} = 4.2 \times 10^{-3} \text{ moles} $

1. Initial Moles of $ \mathrm{U}^{238} $ ( $ N_0 $ ):

$ N_0 = \text{Moles of } \mathrm{U}^{238} \text{ initially} = \text{Moles of } \mathrm{Pb}^{206} + \text{Moles of remaining } \mathrm{U}^{238} $

$ N_0 = 34 \times 10^{-3} + 4.2 \times 10^{-3} = 38.2 \times 10^{-3} \text{ moles} $

1. Calculate the Time:

• Using the decay formula:

$ \lambda t = \ln \frac{N_0}{N_t} $

• Where $ \lambda $ (decay constant):

$ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9 \text{ years}} $

• Calculate $ \ln \frac{N_0}{N_t} $:

$ \ln \frac{N_0}{N_t} = \ln \frac{38.2}{4.2} = \ln 9.09 $

• Simplify $ \ln 9.09 $:

$ \ln 9.09 \approx 2.21 $

1. Plug in the Values:

$ t = \frac{4.5 \times 10^9 \text{ years}}{0.693} \times \ln 9.09 $

$ t = 4.5 \times 10^9 \text{ years} \times \frac{2.21}{0.693} $

$ t \approx 4.5 \times 10^9 \times 3.184 \approx 14.33 \times 10^9 \text{ years} $

1. Convert to $ P \times 10^8 $:

$ t = P \times 10^8 \text{ years} $

$ P \times 10^8 = 14.33 \times 10^9 $

$ P = 143.28 $

Thus, the value of $ P $ is $ \boxed{143.28} $.