Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is ___.

Diamond has a face-centred cubic (FCC) unit cell with additional carbon atoms occupying 50% of the tetrahedral voids.

The number of carbon atoms from the FCC lattice points per unit cell: In an FCC arrangement, there are 8 corner atoms (each shared by 8 unit cells) and 6 face-centred atoms (each shared by 2 unit cells).

$$\text{FCC atoms} = \frac{8}{8} + \frac{6}{2} = 1 + 3 = 4$$

The number of tetrahedral voids in an FCC unit cell is twice the number of lattice points:

$$\text{Tetrahedral voids} = 2 \times 4 = 8$$

50% of these voids are occupied by additional carbon atoms:

$$\text{Extra C atoms} = 0.5 \times 8 = 4$$

Therefore, the total number of carbon atoms per unit cell of diamond is:

$$4 + 4 = \mathbf{8}$$