Total volume of atoms present in a face centred cubic unit cell of a metal is (r = radius of atom).

Correct option: (B)

In a face-centered cubic arrangement, the atoms are present at corners and at the center of the face.

Contribution of atoms located at the corner of cube $=\frac{1}{2}$

Contribution of atoms located at the face of cube $=\frac{1}{2}$

$Z_{eff}$ for FCC lattice $=8\times \frac{1}{8}+6\times \frac{1}{2}=4$

So, the number of atoms present in FCC unit cell $=4$

Let $r$ is the radius of an atom

The volume of one atom =$V=\frac{4}{3}\pi r^3$

Total volume of atoms present in a face centred cubic unit cell
V $=4\times \frac{4}{3}\pi r^3=\frac{16}{3}\pi r^3$