Questions: A moon rock collected by a U.S. Apollo mission is estimated to be 3.50 billion years old by uranium/lead dating. Assuming that the rock did not contain any lead when it was formed, what is the current mass of ^206 Pb in the rock, if it currently contains 1.115 g of ^238 U? The half-life of ^238 U is 4.47 x 10^9 years. ^206 Pb mass: g

Solution Steps

The decay constant (\(\lambda\)) is related to the half-life (\(t_{1/2}\)) by the formula:

\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]

Given that the half-life of \({ }^{238} \mathrm{U}\) is \(4.47 \times 10^9\) years, we can calculate \(\lambda\):

\[ \lambda = \frac{\ln(2)}{4.47 \times 10^9} \approx 1.5513 \times 10^{-10} \, \text{year}^{-1} \]

The decay of \({ }^{238} \mathrm{U}\) to \({ }^{206} \mathrm{Pb}\) can be described by the equation:

\[ N(t) = N_0 e^{-\lambda t} \]

where \(N(t)\) is the current amount of \({ }^{238} \mathrm{U}\), \(N_0\) is the initial amount, and \(t\) is the time elapsed (3.50 billion years).

Rearranging for \(N_0\):

\[ N_0 = N(t) e^{\lambda t} \]

Substituting the known values:

\[ N_0 = 1.115 \, \text{g} \times e^{(1.5513 \times 10^{-10} \times 3.50 \times 10^9)} \]

\[ N_0 \approx 1.115 \, \text{g} \times e^{0.542955} \approx 1.115 \, \text{g} \times 1.7213 \approx 1.9183 \, \text{g} \]

The mass of \({ }^{206} \mathrm{Pb}\) formed is the difference between the initial and current mass of \({ }^{238} \mathrm{U}\):

\[ \text{Mass of } { }^{206} \mathrm{Pb} = N_0 - N(t) = 1.9183 \, \text{g} - 1.115 \, \text{g} = 0.8033 \, \text{g} \]

Final Answer