Diamond Cubic (DC) Structure

DIAMOND CUBIC (DC) STRUCTURE

The diamond cubic structure is a very important crystal structure. Besides diamond, the elemental semiconductors silicon and germanium also have this structure. The unit cell of diamond cubic structure is shown in fig. 1.24.

This structure is a combination of two interpenetrating face-centred - cubic (FCC) sub lattices.

One sub- lattice has its origin at (0, 0, 0) (atom X). The other sub- lattice has its origin (atom Y) quarter of the way along the body diagonal ie., at the point (a/4, a/4, a/4)

It is loosely packed structure since each atom has only four nearest neighbours (ie., coordination number is 4).

1. Number of atoms per unit cell

In the unit cell of diamond, the carbon atoms are present at three different positions of the unit cell as shown in fig. 1.24.

(i) Corner atoms represented by 'C'.

(ii) Face centred atoms represented by 'F'.

(iii) Four atoms present fully fully represented as 1, 2, 3 and 4.

The positions of atoms projected on a cube face is as shown in fig. 1.25

(i) Number of corner atoms per unit cell

Each corner atom is shared by 8 unit cells. There are 8 corner atoms in the unit cell.

Therefore, the number of atoms due to corner atoms per unit cell = (1/8) x 8 = 1 atom

(ii) Number of face centred atoms per unit cell

Each face centred atom is shared by 2 unit cells. We have 6 face centred atoms.

Number of face centred atom per unit cell = (1/2) x 6 = 3 atoms

(iii) Number of atoms inside unit cell

Inside the unit cell, we have 4 atoms represented by 1, 2, 3, 4 in fig. 1.26 which are not shared by any other sourrounding unit cells.

Total number of atoms per unit cell = 1 + 3 + 4 = 8

2. Atomic radius

The corner atoms do not touch each other. Similarly the face centred atoms also do not touch each other.

But both face centred atoms and corner atoms touch with the atoms (1, 2, 3, 4) situated inside the unit cell as shown in fig. 1.26.

For example, the nearest two neighbours which have direct contact (shown by double line) are atoms 'X' and 'Y as shown in fig 1.26.

A perpendicular is drawn to Y atom which meets the unit cell at a point 'Z' as shown in fig. 1.26 which is at a distance own in fig. 1.26 which is at a distance of a/4.

XY² = XZ² + ZY²

XY² = XT² + TZ² + ZY² [XZ² = XT² + TZ²]

Substituting for XT, TZ and ZY, we have

XY² = (a/4)² + (a/4)² + (a/4)²

XY² = a²/4² + a²/4² + a²/4²

XY² = a²/16 + a²/16 + a²/16

XY² = 3a²/16

Taking square root on both sides, we have

Since XY = 2r, we have

2r = √3 a/4

r = (√3 a)/(4 ×2)

r = √3 a/8

Atomic radius r = √3 a/8

3. Coordination number

From fig. 1.26 the number of nearest atoms (shown by double line) for Y atom is 4. Therefore, the coordination number for diamond is 4.

Note: The coordination number is found to be same even if it is calculated with respect to atoms say (2), (3), (4), corner (or) face centred atoms.

4. Packing factor

We know that Packing factor (PF)

= Volume occupied by the atoms per unit cell (v) / Volume of the unit cell (V) .................(1)

Volume occupied by 1 atom = (4/3) πr³

In diamond, we have 8 atoms per unit cell

Volume occupied by all the '8' atoms per unit cell (v) = 8 x (4/3) πr³

We know that atomic radius for diamond structure

r = √3 a / 8

Volume occupied by the atoms per unit cell

Volume occupied by the atoms per unit cell

Since diamond has cubic structure, the volume of the unit cell

V = a³

Substituting the equations (2) and (3) in (1) we get

Packing factor = 0.34 = 0.34 x 100 %

Packing factor = 34%

Thus only 34% volume of the unit cell in diamond cubic structure is occupied by atoms and the remaining 66% volume is vacant.

Since the packing density is very low, it is termed as very loosely packed structure.

Note: Packing factor of diamond cubic = 1/2 of packing factor of BCC