Show that in a cubic close-packed structure, eight tetrahedral voids are present per unit cell.

In ccp, each unit cell consists atoms at face centres and at corners.

Thus, total atoms at face centers and at corners in a unit cell =
$(8\times \frac{1}{8})+(6\times \frac{1}{2})$

= $1+3$

= $4$

So, the total number of atoms in ccp unit cell is $4$.

Since tetrahedral voids are double the number of atoms in a unit cell, a ccp unit cell will contain $8$ tetrahedral voids.


Let's consider a single cubic lattice consisting of $8$ small cubic units. $2$ tetrahedral voids are present at each body diagonal, with each void at an equal distance apart from the corner.

There are 4 body diagonals in a unit cell.
So, the total number of tetrahedral voids
= $4\times 2=8$.