The number of tetrahedral voids in the unit cell of a face centred cubic lattice of similar atoms is

Options

A. 4
B. 6
C. 8
D. 10

Explanation

To determine the number of tetrahedral voids in the unit cell of a face-centered cubic (FCC) lattice, we can follow these steps:

### Step-by-Step Solution:

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

2. **Calculate the Number of Atoms in FCC**:
- There are 8 corners in the cube, and each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 faces, and each face-centered atom contributes \( \frac{1}{2} \) of an atom to the unit cell.
- Therefore, the total number of atoms per unit cell in FCC can be calculated as follows:
\[ \text{Total atoms} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

3. **Determine the Number of Tetrahedral Voids**:
- The number of tetrahedral voids in a crystal lattice is given by the formula \( 2n \), where \( n \) is the number of atoms in the unit cell.
- Since we found that \( n = 4 \) for FCC, we can calculate the number of tetrahedral voids:
\[ \text{Number of tetrahedral voids} = 2n = 2 \times 4 = 8 \]

4. **Conclusion**:
- Therefore, the number of tetrahedral voids in the unit cell of a face-centered cubic lattice is **8**.

### Final Answer:
The number of tetrahedral voids in the unit cell of a face-centered cubic lattice is **8**.