An atom crystallizes in a fcc unit cell having radius of $100\sqrt{2}pm$.
Calculate the distance between the two tetrahedral voids which are formed on a same body diagonal

The correct option is A $200\sqrt{3}pm$
In a fcc unit cell, 2 tetrahedral voids are present at each body diagonal.
One such body diagonal is shown in the figure:


In the body diagonal AD, tetrahedral voids are present at B and C.
$\text{Distance of a tetrahedral void from corner}\left(x\right)=\frac{\sqrt{3}a}{4}$
$\text{Length of body diagonal in fcc}\left(AD\right)=4x=\sqrt{3}a$
$\text{Distance between two tetrahedral voids}\left(2x\right)=\frac{\sqrt{3}a}{2}$
In a fcc lattice:
$\sqrt{2}a=4R\Rightarrow a=\frac{4\times 100\sqrt{2}}{\sqrt{2}}a=400pm$
Since,
$\text{Distance between two tetrahedral voids (BC)}=\left(2x\right)=\frac{\sqrt{3}a}{2}\Rightarrow \text{BC}=\frac{\sqrt{3}\times 400}{2}=200\sqrt{3}pm$