The volume of atom present in a face-centred cubic unit cell of a metal (`r` is atomic radius ) is

Options

A. `12/3 pir^3`
B. `16/3 pir^3`
C. `20/3 pir^3`
D. `24/3 pir^3`

Explanation

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

### Step 1: Determine the number of atoms in an FCC unit cell
In a face-centered cubic unit cell, atoms are located at:
- 8 corners of the cube
- 6 faces of the cube

Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell because each corner is shared by 8 unit cells.

Each face-centered atom contributes \( \frac{1}{2} \) of an atom to the unit cell because each face is shared by 2 unit cells.

Calculating the total contribution:
- Contribution from corners:
\[ 8 \text{ corners} \times \frac{1}{8} = 1 \text{ atom} \]
- Contribution from faces:
\[ 6 \text{ faces} \times \frac{1}{2} = 3 \text{ atoms} \]

Adding these contributions together:
\[ 1 + 3 = 4 \text{ atoms} \]

### Step 2: Calculate the volume of one atom
The volume \( V \) of a single atom (considered as a sphere) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the atomic radius.

### Step 3: Calculate the total volume of atoms in the FCC unit cell
Since there are 4 atoms in the FCC unit cell, the total volume \( V_{total} \) occupied by these 4 atoms is:
\[ V_{total} = 4 \times \left(\frac{4}{3} \pi r^3\right) \]
Calculating this gives:
\[ V_{total} = \frac{16}{3} \pi r^3 \]

### Final Answer
Thus, the volume of atoms present in a face-centered cubic unit cell of a metal is:
\[ \frac{16}{3} \pi r^3 \]