A rock is `1.5 xx 10^(9)` years old. The rock contains `.^(238)U` which disintegretes to form `.^(236)U`. Assume that there was no `.^(206)Pb` in the rock initially and it is the only stable product fromed by the decay. Calculate the ratio of number of nuclei of `.^(238)U` to that of `.^(206)Pb` in the rock. Half-life of `.^(238)U` is `4.5 xx 10^(9). years. `(2^(1/3) = 1.259)` .

Explanation

To solve the problem, we need to calculate the ratio of the number of nuclei of \(^{238}U\) to that of \(^{206}Pb\) in the rock after \(1.5 \times 10^9\) years, given that the half-life of \(^{238}U\) is \(4.5 \times 10^9\) years.

### Step-by-Step Solution

1. **Identify the Variables**:
- Let \(N_0\) be the initial number of nuclei of \(^{238}U\).
- Let \(N_U\) be the number of nuclei of \(^{238}U\) after time \(t\).
- Let \(N_{Pb}\) be the number of nuclei of \(^{206}Pb\) after time \(t\).
- The time \(t = 1.5 \times 10^9\) years.
- The half-life \(T_{1/2} = 4.5 \times 10^9\) years.

2. **Calculate the Number of Half-Lives**:
\[ n = \frac{t}{T_{1/2}} = \frac{1.5 \times 10^9 \text{ years}}{4.5 \times 10^9 \text{ years}} = \frac{1}{3} \]

3. **Calculate the Remaining Nuclei of \(^{238}U\)**:
The formula for the remaining nuclei after \(n\) half-lives is:
\[ N_U = N_0 \left(\frac{1}{2}\right)^n = N_0 \left(\frac{1}{2}\right)^{1/3} \]

4. **Calculate the Number of Nuclei of \(^{206}Pb\)**:
Since \(^{206}Pb\) is the decay product of \(^{238}U\) and there was no \(^{206}Pb\) initially, the number of \(^{206}Pb\) nuclei formed is:
\[ N_{Pb} = N_0 - N_U = N_0 - N_0 \left(\frac{1}{2}\right)^{1/3} = N_0 \left(1 - \left(\frac{1}{2}\right)^{1/3}\right) \]

5. **Calculate the Ratio of \(^{238}U\) to \(^{206}Pb\)**:
The ratio \(R\) of the number of nuclei of \(^{238}U\) to that of \(^{206}Pb\) is:
\[ R = \frac{N_U}{N_{Pb}} = \frac{N_0 \left(\frac{1}{2}\right)^{1/3}}{N_0 \left(1 - \left(\frac{1}{2}\right)^{1/3}\right)} \]
Simplifying this gives:
\[ R = \frac{\left(\frac{1}{2}\right)^{1/3}}{1 - \left(\frac{1}{2}\right)^{1/3}} \]

6. **Substituting the Value of \(\left(\frac{1}{2}\right)^{1/3}\)**:
Given that \(2^{1/3} = 1.259\), we have:
\[ \left(\frac{1}{2}\right)^{1/3} = \frac{1}{1.259} \approx 0.7937 \]
Thus,
\[ R = \frac{0.7937}{1 - 0.7937} = \frac{0.7937}{0.2063} \approx 3.84 \]

### Final Answer
The ratio of the number of nuclei of \(^{238}U\) to that of \(^{206}Pb\) in the rock is approximately \(3.84\).