An element crystallises in a face-centred cubic (fcc) lattice with...

In an fcc unit cell, an octahedral void is located at the body centre, which has coordinates $\left(\dfrac{a}{2}, \dfrac{a}{2}, \dfrac{a}{2}\right)$.

Tetrahedral voids are located on the body diagonals at a distance of one-fourth of the body diagonal from each corner. The coordinates of a nearest tetrahedral void from the origin are $\left(\dfrac{a}{4}, \dfrac{a}{4}, \dfrac{a}{4}\right)$.

The shortest distance $d$ between the centre of an octahedral void and a tetrahedral void is the distance between these two points:

$d = \sqrt{\left(\dfrac{a}{2} - \dfrac{a}{4}\right)^2 + \left(\dfrac{a}{2} - \dfrac{a}{4}\right)^2 + \left(\dfrac{a}{2} - \dfrac{a}{4}\right)^2}$

$d = \sqrt{\left(\dfrac{a}{4}\right)^2 + \left(\dfrac{a}{4}\right)^2 + \left(\dfrac{a}{4}\right)^2} = \sqrt{3 \times \dfrac{a^2}{16}} = \dfrac{\sqrt{3}a}{4}$

Answer: $\dfrac{\sqrt{3}a}{4}$