The minimum distance between an octahedral and a tetrahedral void in fcc lattice is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understand the FCC Structure**:
- In an FCC lattice, atoms are located at the corners and the centers of each face of the cube.
- The octahedral voids are located at the body center and at the edge centers of the cube.
- The tetrahedral voids are located at specific positions along the body diagonals.

2. **Identify the Positions of the Voids**:
- The octahedral void (OHV) is located at the body center of the cube, which is at coordinates (a/2, a/2, a/2).
- The tetrahedral voids (THV) are located at positions along the body diagonal. There are two tetrahedral voids along each body diagonal.

3. **Determine the Body Diagonal Length**:
- The length of the body diagonal of a cube with edge length 'a' is given by the formula:
\[ \text{Body Diagonal} = \sqrt{3}a \]

4. **Calculate the Distance Between OHV and THV**:
- The tetrahedral voids are located at positions along the body diagonal. For the calculation, we can consider one of the tetrahedral voids located at (0, 0, 0) and the octahedral void at (a/2, a/2, a/2).
- The distance between these two points can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
- Plugging in the coordinates:
\[ d = \sqrt{(a/2 - 0)^2 + (a/2 - 0)^2 + (a/2 - 0)^2} = \sqrt{(a/2)^2 + (a/2)^2 + (a/2)^2} = \sqrt{3 \cdot (a/2)^2} = \sqrt{3} \cdot \frac{a}{2} \]

5. **Find the Minimum Distance**:
- The minimum distance between the octahedral void and the tetrahedral void is along the body diagonal, which we have calculated as:
\[ d = \frac{\sqrt{3}a}{2} \]
- However, since there are multiple tetrahedral voids, we need to consider the closest one, which occurs at a quarter of the body diagonal. Therefore, we divide this distance by 2:
\[ \text{Minimum Distance} = \frac{\sqrt{3}a}{4} \]

### Final Answer:
The minimum distance between an octahedral void and a tetrahedral void in an FCC lattice is:
\[ \frac{\sqrt{3}a}{4} \]