The total of tetrahedral voids in the face centred unit cell is …………. .

Options

A. 6
B. 8
C. 10
D. 12

Explanation

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

### Step-by-Step Solution:

1. **Identify the number of atoms in the FCC unit cell**:
- In a face-centered cubic (FCC) lattice, there are 4 atoms per unit cell. This is because there are 8 corner atoms (each contributing 1/8 to the unit cell) and 6 face-centered atoms (each contributing 1/2 to the unit cell).
- Calculation:
\[ \text{Total atoms} = \left( \frac{1}{8} \times 8 \right) + \left( \frac{1}{2} \times 6 \right) = 1 + 3 = 4 \]

2. **Determine the number of octahedral voids**:
- The number of octahedral voids in a unit cell is equal to the number of atoms in that unit cell. Therefore, in an FCC unit cell, there are 4 octahedral voids.

3. **Calculate the number of tetrahedral voids**:
- The number of tetrahedral voids is double the number of atoms in the unit cell. Hence, if there are 4 atoms, the number of tetrahedral voids will be:
\[ \text{Tetrahedral voids} = 2 \times \text{Number of atoms} = 2 \times 4 = 8 \]

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

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

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