In an ore containing Uranium, the ratio of `U^(238)` to `Pb^(206` nuceli is `3`. Calculate the age of the ore, assuming that alll the lead present in the ore is the final stable, product of `U^(238)`. Take the half-like of `U^(238)` to be `4.5 xx 10^(9)` years. In `(4//3) = 0.288`.

Answer

To calculate the age of the ore containing Uranium-238 (U-238) and Lead-206 (Pb-206), we can follow these steps:

### Step 1: Understand the Given Ratio
We are given the ratio of U-238 to Pb-206 nuclei as 3:1. This means that for every 3 nuclei of U-238, there is 1 nucleus of Pb-206.

### Step 2: Set Up the Initial and Final Amounts
Let the initial number of U-238 nuclei be \( N_0 \). After some time \( t \), the number of U-238 nuclei will be \( N \), and the number of Pb-206 nuclei will be \( N_{Pb} \). According to the problem:
- \( N_{Pb} = N_0 - N \)
- Given the ratio \( \frac{N}{N_{Pb}} = 3 \), we can express this as:
\[ N = 3N_{Pb} \]

### Step 3: Relate U-238 and Pb-206
From the relationship \( N_{Pb} = N_0 - N \):
\[ N_{Pb} = N_0 - 3N_{Pb} \]
This simplifies to:
\[ 4N_{Pb} = N_0 \implies N_{Pb} = \frac{N_0}{4} \]
Thus, we can express \( N \) in terms of \( N_0 \):
\[ N = 3N_{Pb} = 3 \left(\frac{N_0}{4}\right) = \frac{3N_0}{4} \]

### Step 4: Use the Law of Radioactive Decay
According to the law of radioactive decay:
\[ \frac{N_0}{N} = 2^{\frac{t}{t_{1/2}}} \]
Substituting our values:
\[ \frac{N_0}{\frac{3N_0}{4}} = 2^{\frac{t}{t_{1/2}}} \]
This simplifies to:
\[ \frac{4}{3} = 2^{\frac{t}{t_{1/2}}} \]

### Step 5: Take Logarithms
Taking logarithms on both sides:
\[ \log\left(\frac{4}{3}\right) = \frac{t}{t_{1/2}} \log(2) \]
Rearranging gives:
\[ t = t_{1/2} \cdot \frac{\log\left(\frac{4}{3}\right)}{\log(2)} \]

### Step 6: Substitute the Half-Life
Given the half-life \( t_{1/2} = 4.5 \times 10^9 \) years, we can substitute this value:
\[ t = 4.5 \times 10^9 \cdot \frac{\log\left(\frac{4}{3}\right)}{\log(2)} \]

### Step 7: Calculate the Logarithms
Using the values:
- \( \log\left(\frac{4}{3}\right) \approx 0.1249 \)
- \( \log(2) \approx 0.3010 \)

Now substituting these values:
\[ t = 4.5 \times 10^9 \cdot \frac{0.1249}{0.3010} \]

### Step 8: Final Calculation
Calculating the right side:
\[ t \approx 4.5 \times 10^9 \cdot 0.415 \]
\[ t \approx 1.867 \times 10^9 \text{ years} \]

### Conclusion
The age of the ore is approximately \( 1.867 \times 10^9 \) years.

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