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

Options

A. `20//3 pir^(3)`
B. `24//3 pir^(3)`
C. `12//3 pir^(3)`
D. `16//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-by-Step Solution:

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

2. **Determine the Number of Atoms in FCC:**
Each corner atom is shared among eight unit cells, and each face-centered atom is shared between two unit cells. Therefore, the total number of atoms in one FCC unit cell is:
- Corner atoms: 8 corners × 1/8 atom per corner = 1 atom
- Face-centered atoms: 6 faces × 1/2 atom per face = 3 atoms
- Total = 1 + 3 = 4 atoms

3. **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.

4. **Calculate the Total Volume of Atoms in the FCC Unit Cell:**
Since there are 4 atoms in the FCC unit cell, the total volume \( V_{total} \) of the atoms in the unit cell is:
\[ V_{total} = 4 \times \left(\frac{4}{3} \pi r^3\right) = \frac{16}{3} \pi r^3 \]

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