Question

Atoms of an element crystalize in an FCC lattice that has an atomic diameter of $\sqrt{6}$ Å. What is the minimum distance between a tetrahedral void and an octahedral void (in Å)?

Answer 1

Octahedral voids are located at the centre and the mid-points of the edges of the cube. Tetrahedral voids are located at a distance of 1/4th from the corner of the body diagonal. Consider the octahedral void at the centre of the cube for the required calculation. The distance is simply between the centre and the body diagonal that has the tetrahedral void.
Therefore, the minimum distance between a tetrahedral and an octahedral void would be $\frac{\sqrt{3}\times a}{4}$
In an FCC lattice= a = $2\sqrt{2}\times r$
Substituting,
Minimum distance = $\frac{\sqrt{3}}{4}\times 2\sqrt{2}\times \frac{\sqrt{6}}{2}$
=1.5 Å