Uranium ores contain one radium `-226 `atom for every `2.8 xx 10^(6)` uranium `-238` atoms. Calculate the half-life of `._(88)Ra^(226)` is `1600` years `(._(88)Ra^(226)` is a decay product of `._(92)U^(238))`.

Options

A. `1.75 xx 10^(3) years`
B. `1600 xx (238)/(92)years`
C. `4.5 xx 10^(9) years`
D. `1600 xx 238 years`

Explanation

To solve the problem, we need to calculate the half-life of uranium-238 based on the given information about the ratio of radium-226 to uranium-238 atoms and the half-life of radium-226.

### Step-by-Step Solution:

1. **Understand the Given Ratio**:
We are given that there is 1 radium-226 atom for every \(2.8 \times 10^6\) uranium-238 atoms. This can be expressed as:
\[ \frac{n_R}{n_U} = \frac{1}{2.8 \times 10^6} \]
where \(n_R\) is the number of radium-226 atoms and \(n_U\) is the number of uranium-238 atoms.

2. **Relate Activities to Number of Atoms**:
The activity (A) of a radioactive substance is proportional to the number of atoms (n) and inversely proportional to the half-life (t_half). Thus, we can write:
\[ \frac{A_R}{A_U} = \frac{n_R / t_{1/2,R}}{n_U / t_{1/2,U}} \]
Since the activities are proportional to the number of atoms, we can simplify this to:
\[ \frac{A_R}{A_U} = \frac{n_R}{n_U} \cdot \frac{t_{1/2,U}}{t_{1/2,R}} \]

3. **Set Activities Equal**:
Since the activities are equal (as there is a constant decay relationship), we can set:
\[ \frac{n_R}{n_U} = \frac{t_{1/2,U}}{t_{1/2,R}} \]
Rearranging gives us:
\[ t_{1/2,U} = \frac{n_U}{n_R} \cdot t_{1/2,R} \]

4. **Substitute Known Values**:
We know that:
- \(t_{1/2,R} = 1600\) years (half-life of radium-226)
- \(n_U = 2.8 \times 10^6\)
- \(n_R = 1\)

Substituting these values into the equation:
\[ t_{1/2,U} = \frac{2.8 \times 10^6}{1} \cdot 1600 \]

5. **Calculate the Half-Life of Uranium-238**:
\[ t_{1/2,U} = 2.8 \times 10^6 \times 1600 \]
\[ t_{1/2,U} = 4.48 \times 10^9 \text{ years} \]

6. **Final Answer**:
The half-life of uranium-238 is approximately:
\[ t_{1/2,U} \approx 4.5 \times 10^9 \text{ years} \]