In a face centred cubic unit cell, what is the volume occupied ?

Options

A. `4/3 pi r^3`
B. `8/3 pi r^3`
C. `16/3 pi r^3`
D. `(64 r^3)/(3sqrt(3))`

Explanation

To find the volume occupied in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of FCC**:
- In a face-centered cubic unit cell, atoms are located at each of the eight corners and at the centers of each of the six faces.

2. **Calculate the Contribution of Corner Atoms**:
- There are 8 corners in the unit cell.
- Each corner atom contributes \( \frac{1}{8} \) of its volume to the unit cell.
- Therefore, the total contribution from the corner atoms is:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \text{ atom} \]

3. **Calculate the Contribution of Face-Centered Atoms**:
- There are 6 faces in the unit cell.
- Each face-centered atom contributes \( \frac{1}{2} \) of its volume to the unit cell.
- Therefore, the total contribution from the face-centered atoms is:
\[ \text{Contribution from faces} = 6 \times \frac{1}{2} = 3 \text{ atoms} \]

4. **Total Number of Atoms in FCC Unit Cell**:
- The total number of atoms in the FCC unit cell is:
\[ \text{Total atoms} = 1 \text{ (from corners)} + 3 \text{ (from faces)} = 4 \text{ atoms} \]

5. **Calculate the Volume Occupied by One Atom**:
- The volume occupied by a single atom (assuming it is a sphere) is given by the formula:
\[ V_{\text{atom}} = \frac{4}{3} \pi r^3 \]
- Here, \( r \) is the radius of the atom.

6. **Calculate the Total Volume Occupied by All Atoms in the Unit Cell**:
- Since there are 4 atoms in the unit cell, the total volume occupied by the atoms in the FCC unit cell is:
\[ V_{\text{total}} = 4 \times V_{\text{atom}} = 4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3 \]

### Final Answer:
The volume occupied in a face-centered cubic unit cell is:
\[ \frac{16}{3} \pi r^3 \]