The atomic ratio between the uranium isotopes `.^(238)U and.^(234)U` in a mineral sample is found to be `1.8xx10^(4)`. The half life of `.^(234)U` is `T_((1)/(2))(234)=2.5xx10^(5)` years. The half - life of `.^(238)U` is -

Options

A. `4.5xx10^(9)" years"`
B. `5.4xx10^(9)" years"`
C. `4.5" years"`
D. `5.4" years"`

Answer

To find the half-life of uranium-238 (U-238) given the atomic ratio of uranium isotopes U-238 and U-234 and the half-life of U-234, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: In a mineral sample, the isotopes U-238 and U-234 are in radioactive equilibrium. This means that the decay rates of the two isotopes are related by their respective decay constants and the number of atoms present.

2. **Define the Decay Constants**: The decay constant (λ) is related to the half-life (T₁/₂) by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
For U-234:
\[ \lambda_2 = \frac{\ln(2)}{T_{1/2}(234)} \]
For U-238:
\[ \lambda_1 = \frac{\ln(2)}{T_{1/2}(238)} \]

3. **Set Up the Equilibrium Condition**: The equilibrium condition can be expressed as:
\[ \lambda_1 N_1 = \lambda_2 N_2 \]
Where \(N_1\) and \(N_2\) are the number of atoms of U-238 and U-234, respectively.

4. **Substitute the Decay Constants**: Substitute the expressions for the decay constants:
\[ \frac{\ln(2)}{T_{1/2}(238)} N_1 = \frac{\ln(2)}{T_{1/2}(234)} N_2 \]
The \(\ln(2)\) cancels out:
\[ \frac{N_1}{N_2} = \frac{T_{1/2}(238)}{T_{1/2}(234)} \]

5. **Use the Given Atomic Ratio**: We know from the problem that:
\[ \frac{N_1}{N_2} = 1.8 \times 10^4 \]
Therefore:
\[ 1.8 \times 10^4 = \frac{T_{1/2}(238)}{2.5 \times 10^5} \]

6. **Solve for T₁/₂(238)**: Rearranging the equation gives:
\[ T_{1/2}(238) = 1.8 \times 10^4 \times 2.5 \times 10^5 \]

7. **Calculate T₁/₂(238)**:
\[ T_{1/2}(238) = 1.8 \times 2.5 \times 10^{4 + 5} = 4.5 \times 10^9 \text{ years} \]

### Final Answer:
The half-life of U-238 is \(4.5 \times 10^9\) years.