If for any $\mathrm{fcc}$ lattice, the tetrahedral void number is always $$8$$ ? As the $\mathrm{fcc}$'s ${\mathrm{Z}}_{\mathrm{eff}}$ is $$4$$ and remains constant, Justify

FCC-Face centred cubic cell:

• For a total of $$8$$, a face-centred cubic unit cell contains net tetrahedral voids of $$8$$. Tetrahedral voids are located between each corner and the base of the unit cell. Then there are $$12$$ octahedral voids at the midpoints of the unit cell's sides and there are four net octahedral voids.
  

Tetrahedral voids:

• The void observed in substances with tetrahedral crystal structures is known as a tetrahedral void. Voids are generally thought of as the empty areas that exist in crystal systems. These holes are mostly caused by the different atomic arrangements.

Tetrahedral voids in $\mathrm{fcc}$:

• If there are no atoms or ions in a close-packed structure, the number of octahedral voids and tetrahedral voids in a close-packed structure $(\mathrm{ccp}\mathrm{or}\mathrm{fcc})$ would be $\mathrm{nn}\&2\mathrm{n}2\mathrm{n}$ respectively
• Each unit cell in the $\mathrm{fcc}$ structure has $8$ tetrahedral voids, where ${\mathrm{Z}}_{\mathrm{eff}}=4$ which is nothing but the number of atoms in $\mathrm{fcc}$. where the unit cell is partitioned into $8$ small cubes.
• If we divide the FCC unit cell into $8$ small cubes then each small cube has one tetrahedral void located at its own body centre. Since the formula is 2n here, In other words, there are $2\times {\mathrm{Z}}_{\mathrm{eff}}=8$ tetrahedral voids overall in a single cell.