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. `8pir^(3)`
C. `4pir^(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 1: Understand the FCC Structure
In a face-centered cubic unit cell, atoms are located at each of the eight corners and at the center of each of the six faces.

### Step 2: Calculate the Effective Number of Atoms (Z effective)
- Each corner atom contributes \( \frac{1}{8} \) of its volume to the unit cell because it is shared among eight unit cells.
- There are 8 corner atoms, so the contribution from corners is:
\[ 8 \times \frac{1}{8} = 1 \]
- Each face-centered atom contributes \( \frac{1}{2} \) of its volume to the unit cell because it is shared between two unit cells.
- There are 6 face-centered atoms, so the contribution from faces is:
\[ 6 \times \frac{1}{2} = 3 \]
- Therefore, the total effective number of atoms per unit cell (Z effective) is:
\[ Z_{\text{effective}} = 1 + 3 = 4 \]

### Step 3: Calculate the Volume of a Single Atom
Assuming each atom is spherical, the volume \( V \) of a single atom can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the atomic radius.

### Step 4: Calculate the Total Volume of Atoms in the Unit Cell
Now, to find the total volume of atoms in the FCC unit cell, we multiply the effective number of atoms by the volume of a single atom:
\[ \text{Total Volume} = Z_{\text{effective}} \times V = 4 \times \left( \frac{4}{3} \pi r^3 \right) \]
\[ \text{Total Volume} = \frac{16}{3} \pi r^3 \]

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