A face-centered cubic unit cell contains 8 ‘X’ atoms at the corners of the cell and 6 ‘Y’ atoms at the faces. If the empirical formula of the solid is `X_(a)Y_(b)` then, what is `(b)/(a)`?

Explanation

To solve the problem, we need to determine the contributions of the 'X' and 'Y' atoms in a face-centered cubic (FCC) unit cell and derive the empirical formula \(X_aY_b\).

### Step-by-Step Solution:

1. **Identify the number of atoms at corners and faces:**
- In a face-centered cubic unit cell, there are 8 'X' atoms located at the corners and 6 'Y' atoms located at the faces.

2. **Calculate the contribution of 'X' atoms:**
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell because each corner atom is shared by 8 adjacent unit cells.
- Since there are 8 corners, the total contribution of 'X' atoms is:
\[ \text{Total contribution of } X = 8 \times \frac{1}{8} = 1 \text{ atom} \]

3. **Calculate the contribution of 'Y' atoms:**
- Each face atom contributes \( \frac{1}{2} \) of an atom to the unit cell because each face atom is shared by 2 adjacent unit cells.
- Since there are 6 faces, the total contribution of 'Y' atoms is:
\[ \text{Total contribution of } Y = 6 \times \frac{1}{2} = 3 \text{ atoms} \]

4. **Determine the empirical formula:**
- From the contributions calculated, we have:
- \( a = 1 \) (for 'X')
- \( b = 3 \) (for 'Y')
- Therefore, the empirical formula can be expressed as \( X_1Y_3 \).

5. **Calculate the ratio \( \frac{b}{a} \):**
- We need to find the ratio \( \frac{b}{a} \):
\[ \frac{b}{a} = \frac{3}{1} = 3 \]

### Final Answer:
The ratio \( \frac{b}{a} \) is \( 3 \).

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