The atomic radius for a face-centered cubic cell of lattice parameter ' $a$' is

Consider the following diagram to calculate the atomic radius of the fcc cell:

$\text{ In }∆\mathrm{ABC}$,

${\mathrm{AC}}^2={\mathrm{b}}^2={\mathrm{BC}}^2+{\mathrm{AB}}^2=a^2+a^2=2a^2$

or $b=\sqrt{2a}$

If r is the radius of the sphere, then we have:

$b=4\mathrm{r}=\sqrt{2}a$

$\mathrm{r}=\frac{\sqrt{2}}{4}a$

Therefore, the correct option is A).