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What is the distance between successive tetrahedral voids in a FCC crystal structure?
The distance between a face centered atom and a tetrahedral void is $\sqrt{\frac{3a}{4}}$ if the cube's edge length is $a$. Now If I know the angle made, between the line joining a tetrahedral void and the face centered atom, and the successive tetrahedral void and face centered atom, I can find the distance. But how do I calculate that distance since I don't know the angle.
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 $a - a/4 - a/4 = a/2$
See.. The centre of the tetrahedral void is same as the centre of the smaller cube whose vertices the tetrahedron shares. Use this simple fact to calculate the distance between centres of 2 consecutive tetrahedral void.
Now , clearly distance between centres is a/4 + a/4 = a/2.
