A sample of `U^(238) ("half life" = 4.5 xx 10^(9)yr)` ore is found to contain `23.8 g" of " U^(238) ` and `20.6g ` of `Pb^(206)`. Calculate the age of the ore

Answer

To calculate the age of the ore sample containing Uranium-238 and Lead-206, we can use the concept of radioactive decay and the relationship between the parent isotope (Uranium-238) and its daughter isotope (Lead-206). Here’s a step-by-step solution:

### Step 1: Determine the number of moles of U-238 and Pb-206

1. **Calculate the number of moles of U-238:**
\[ \text{Number of moles of } U^{238} = \frac{\text{mass}}{\text{molar mass}} = \frac{23.8 \, \text{g}}{238 \, \text{g/mol}} \approx 0.1 \, \text{mol} \]

2. **Calculate the number of moles of Pb-206:**
\[ \text{Number of moles of } Pb^{206} = \frac{20.6 \, \text{g}}{206 \, \text{g/mol}} \approx 0.1 \, \text{mol} \]

### Step 2: Determine the initial amount of U-238

Since each decay of U-238 produces one atom of Pb-206, the total number of moles of U-238 originally present (before decay) can be calculated as follows:

- Let \( N_0 \) be the initial number of moles of U-238.
- After some time, the remaining moles of U-238 is \( N \) and the moles of Pb-206 produced is equal to the moles of U-238 that have decayed.

From the data:
\[ N_0 = N + \text{moles of } Pb^{206} \]
\[ N_0 = 0.1 + 0.1 = 0.2 \, \text{mol} \]

### Step 3: Use the half-life to calculate the age

The relationship between the remaining amount of a radioactive substance and its half-life can be expressed using the formula:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Where:
- \( N \) = remaining amount of U-238 = 0.1 mol
- \( N_0 \) = initial amount of U-238 = 0.2 mol
- \( t_{1/2} \) = half-life of U-238 = \( 4.5 \times 10^9 \) years
- \( t \) = age of the ore

Rearranging the formula gives:
\[ \frac{N}{N_0} = \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Substituting the values:
\[ \frac{0.1}{0.2} = \left( \frac{1}{2} \right)^{\frac{t}{4.5 \times 10^9}} \]
\[ 0.5 = \left( \frac{1}{2} \right)^{\frac{t}{4.5 \times 10^9}} \]

### Step 4: Solve for \( t \)

Taking logarithm on both sides:
\[ \log(0.5) = \frac{t}{4.5 \times 10^9} \log(0.5) \]
Since \( \log(0.5) \) cancels out:
\[ 1 = \frac{t}{4.5 \times 10^9} \]
Thus,
\[ t = 4.5 \times 10^9 \, \text{years} \]

### Conclusion

The age of the ore is approximately \( 4.5 \times 10^9 \) years.

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