In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids ?

Options

A. `0.543`
B. `0.732`
C. `0.414`
D. `0.637`

Explanation

To find the relative ratio of the sizes of tetrahedral and octahedral voids in a face-centered cubic (FCC) arrangement of metallic atoms, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**:
- In a face-centered cubic (FCC) arrangement, atoms are located at each of the corners and the centers of each face of the cube.

2. **Identify the Radii of Voids**:
- The radius of a tetrahedral void (R_t) in an FCC structure is given as \( R_t = 0.225R \), where R is the radius of the metallic atom.
- The radius of an octahedral void (R_o) in an FCC structure is given as \( R_o = 0.414R \).

3. **Calculate the Ratio of the Radii**:
- To find the ratio of the sizes of tetrahedral voids to octahedral voids, we can set up the following ratio:
\[ \text{Ratio} = \frac{R_t}{R_o} = \frac{0.225R}{0.414R} \]

4. **Simplify the Ratio**:
- The R cancels out in the ratio:
\[ \text{Ratio} = \frac{0.225}{0.414} \]

5. **Perform the Calculation**:
- Now, we perform the division:
\[ \text{Ratio} \approx 0.543 \]

6. **Conclusion**:
- The relative ratio of the sizes of tetrahedral voids to octahedral voids is approximately \( 0.543 \).

### Final Answer:
The relative ratio of the sizes of tetrahedral voids to octahedral voids in a face-centered cubic arrangement is approximately \( 0.543 \).