If the cell edge of fcc unit cell of a metal is $8\sqrt{2}$ pm the radius of metal atom in 'pm' is :

In fcc lattice atoms at corners make contact with the face centred atom. So face diagonal length is equal to 4 × radius of atom.

$\therefore$if 'a' is the side length and the radius of atom is 'r'

$4\mathrm{r}=\mathrm{a}\sqrt{2}=8\sqrt{2}\times \sqrt{2}=16$

r = 4.