What is the volume of the atoms having radius 'r' present in FCC unit cell?

\(\frac{16πr^3}{3}\)

In FCC unit cell, atoms are present at corners and at face-centres.

Total number of atoms per unit cell = \(\frac{1}{8}\) x 8 + 1 x 0 + \(\frac{1}{2}\) x 6 = 1 + 3 = 4

As atoms are considered to be spherical, 

Therefore, Volume of one atom = \(\frac{4}{3}\)πr³

Volume of two atoms in each BCC unit cell is given by = 4 x \(\frac{4}{3}\)πr³ = \(\frac{16πr^3}{3}\)