Chemistry - Distance between successive tetrahedral voids in FCC

Solution 1:

As shown in figure, if we divide a FCC unit cell into 8 small cubes, then each small cube has 1 Tetrahedral void at its own body centre. Thus, there are total 8 Tetrahedral voids in one unit cell. It can also be seen from the figure that the nearest distance between two Tetrahedral voids is a/2.

Solution 2:

The distance between any two successive (adjacent) tetrahedral voids. To calculate it, draw a line joining a tetrahedral void to the corresponding vertex. We know that this distance is $\frac{\sqrt{3}a}{4}$, now take the projection of this on any of the edges. The angle between body diagonal and an edge is cos inverse of $\frac{1}{\sqrt{3}}$, so the projection of the line on an edge is $a/4$, Similarly the other tetrahedral void will also be $a/4$ away from its vertex on this edge. So the distance between these two is $a - a/4 - a/4 = a/2$