`.^(238)U` decays with a half-life of `4.5 xx10^(9)` years, the decay series eventually ending at `.^(206)Pb`, which is stable. `A` rock sample analysis shows that the ratio of the number of atoms of `.^(206)Pb` to `.^(238)U` is 0.0058. Assuming that all the `.^(206)Pb` is produced by the decay of `.^(238)U` and that all other half-lives on the chain are negligible, the age of the rock sample is `(ln 1.0058 =5.78 xx10^(-3))`.

Options

A. `38 xx 10^(8) years`
B. `38 xx 10^(6) years`
C. `19 xx 10^(8) years`
D. `19 xx 10^(6) years`

Explanation

To determine the age of the rock sample based on the decay of Uranium-238 to Lead-206, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
- Uranium-238 (\(^{238}U\)) decays into Lead-206 (\(^{206}Pb\)). The half-life of \(^{238}U\) is given as \(4.5 \times 10^9\) years.

2. **Establishing the Relationship**:
- Let \(N_0\) be the initial number of \(^{238}U\) atoms.
- After time \(t\), the number of \(^{238}U\) atoms remaining, \(N_U\), can be expressed as:
\[ N_U = N_0 e^{-\lambda t} \]
- The number of \(^{206}Pb\) atoms produced, \(N_{Pb}\), can be expressed as:
\[ N_{Pb} = N_0 (1 - e^{-\lambda t}) \]

3. **Setting Up the Ratio**:
- We are given the ratio of \(N_{Pb}\) to \(N_U\):
\[ \frac{N_{Pb}}{N_U} = 0.0058 \]
- Substituting the expressions for \(N_{Pb}\) and \(N_U\):
\[ \frac{N_0 (1 - e^{-\lambda t})}{N_0 e^{-\lambda t}} = 0.0058 \]
- Simplifying this gives:
\[ \frac{1 - e^{-\lambda t}}{e^{-\lambda t}} = 0.0058 \]
- Rearranging leads to:
\[ 1 - e^{-\lambda t} = 0.0058 e^{-\lambda t} \]
- Therefore:
\[ 1 = 0.0058 e^{-\lambda t} + e^{-\lambda t} = (0.0058 + 1)e^{-\lambda t} \]
- This simplifies to:
\[ e^{-\lambda t} = \frac{1}{1.0058} \]

4. **Finding \(\lambda\)**:
- The decay constant \(\lambda\) is related to the half-life (\(T_{1/2}\)) by:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
- Substituting the half-life:
\[ \lambda = \frac{\ln(2)}{4.5 \times 10^9 \text{ years}} \]

5. **Substituting into the Equation**:
- From the expression \(e^{-\lambda t} = \frac{1}{1.0058}\), we take the natural logarithm:
\[ -\lambda t = \ln\left(\frac{1}{1.0058}\right) = -\ln(1.0058) \]
- Thus:
\[ t = \frac{\ln(1.0058)}{-\lambda} \]

6. **Calculating \(t\)**:
- Substitute \(\lambda\):
\[ t = \frac{\ln(1.0058)}{-\frac{\ln(2)}{4.5 \times 10^9}} \]
- This simplifies to:
\[ t = \frac{4.5 \times 10^9 \cdot \ln(1.0058)}{-\ln(2)} \]

7. **Final Calculation**:
- Using the provided value \(\ln(1.0058) = 5.78 \times 10^{-3}\):
\[ t = \frac{4.5 \times 10^9 \cdot 5.78 \times 10^{-3}}{\ln(2)} \]
- Approximating \(\ln(2) \approx 0.693\):
\[ t \approx \frac{4.5 \times 10^9 \cdot 5.78 \times 10^{-3}}{0.693} \approx 37.6 \times 10^6 \text{ years} \]

8. **Conclusion**:
- Therefore, the age of the rock sample is approximately:
\[ t \approx 38 \times 10^6 \text{ years} \]