The distance between an ocatahral and tetrahedral void in fcc lattice would be:

Options

A. `sqrt3 a`
B. `sqrt(3a)/(2)`
C. `sqrt(3a)/(3)`
D. `sqrt(3a)/(4)`

Answer

To find the distance between an octahedral void and a tetrahedral void in a face-centered cubic (FCC) lattice, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Positions of Voids:**
- In an FCC lattice, octahedral voids are located at the body center and at the centers of the edges.
- The tetrahedral voids are located at one-fourth the distance along the body diagonal.

2. **Determine the Length of the Body Diagonal:**
- The length of the body diagonal (d) in an FCC unit cell is given by the formula:
\[ d = \sqrt{3}a \]
where \(a\) is the edge length of the cube.

3. **Locate the Octahedral Void:**
- The octahedral void at the body center is located at:
\[ \text{Distance from corner} = \frac{\sqrt{3}}{2}a \]

4. **Locate the Tetrahedral Void:**
- The tetrahedral void is located at one-fourth the distance along the body diagonal:
\[ \text{Distance from corner} = \frac{1}{4} \times \sqrt{3}a = \frac{\sqrt{3}}{4}a \]

5. **Calculate the Distance Between the Voids:**
- To find the distance between the octahedral void (at \(\frac{\sqrt{3}}{2}a\)) and the tetrahedral void (at \(\frac{\sqrt{3}}{4}a\)), we subtract the distances:
\[ \text{Distance} = \left(\frac{\sqrt{3}}{2}a\right) - \left(\frac{\sqrt{3}}{4}a\right) \]
- To perform the subtraction, we need a common denominator:
\[ = \frac{2\sqrt{3}}{4}a - \frac{\sqrt{3}}{4}a = \frac{(2\sqrt{3} - \sqrt{3})}{4}a = \frac{\sqrt{3}}{4}a \]

6. **Final Result:**
- Thus, the distance between the octahedral void and the tetrahedral void in an FCC lattice is:
\[ \frac{\sqrt{3}}{4}a \]

### Conclusion:
The correct answer is \(\frac{\sqrt{3}}{4}a\).

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