Position of `TVs` in closest packed structure is//are

Options

A. Edge centre of unit cell
B. Two `TVs` on each body diagonal
C. Position of each `TV` from corner is `sqrt(3)a//4`.
D. Face centre of unit cell

Explanation

To solve the question regarding the position of tetrahedral voids in a closest packed structure, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
- The question refers to a closest packed structure, which is commonly a Face-Centered Cubic (FCC) structure.

2. **Determine the Number of Atoms**:
- In an FCC unit cell, there are 4 atoms per unit cell. This is calculated based on the contributions from the corners and face centers of the cube.

3. **Calculate the Number of Tetrahedral Voids**:
- The number of tetrahedral voids in a FCC unit cell is given by the formula \(2n\), where \(n\) is the number of atoms per unit cell. Therefore, for FCC, \(2 \times 4 = 8\) tetrahedral voids are present.

4. **Identify the Positions of Tetrahedral Voids**:
- Tetrahedral voids in an FCC structure are located along the body diagonals of the cube. Each body diagonal contains 2 tetrahedral voids.

5. **Calculate the Distance from the Corners**:
- The body diagonal of the cube can be expressed as \( \sqrt{3}a \), where \(a\) is the edge length of the cube.
- The tetrahedral voids are positioned at a distance of \( \frac{\sqrt{3}a}{4} \) from the corners along the body diagonal.

6. **Conclusion**:
- The tetrahedral voids are not located at the face centers or edge centers but specifically along the body diagonals of the FCC unit cell at the calculated distance.

### Final Answer:
The positions of tetrahedral voids in a closest packed structure (FCC) are located at a distance of \( \frac{\sqrt{3}a}{4} \) from the corners along the body diagonals.