Chemistry - Solid State Question with Solution | TestHub

The minimum distance between two tetrahedral voids in an FCC lattice is $a/2$.

Given $a/2=407$ pm, so $a=814$ pm.

Tetrahedral voids lie at $(1/4,1/4,1/4)$ and $(3/4,3/4,3/4)$ along a body diagonal.

Distance between them is $\sqrt{(3/4-1/4{)}^2+(3/4-1/4{)}^2+(3/4-1/4{)}^2}\times a=\sqrt{(1/2{)}^2\times 3}\times a=\frac{\sqrt{3}}{2}a$.

Distance $=\frac{\sqrt{3}}{2}\times 814\approx 704.8$ pm.

The final answer is $\boxed{704pm}$.