Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 7M

Year: May 2014

• The diamond structure can be described as to be formed from two identical interpenetrating FCC sub-lattice.

• The lattice describes the repeat pattern separated by 1/4th of the width of the unit cell in each dimension.

• The lattice constant, a = 3.5$A^0$.

Atomic Radius:-

The length of the body diagonal = $\sqrt{3} a$

Also, 1/4th of body diagonal = 2r

$$∴ \frac{\sqrt{3} a}{4} = 2r$$

$$∴ r = \frac{\sqrt{3} a}{8}$$

Number of Atoms per Unit Cell:-

• Eight carbon atoms are at eight corners.

• Thus, contribution of corner atoms = 8 x 1/8 = 1

• There are six carbon atoms at the centers of each face.

• Thus, contribution of face centered atoms = 1/2 x 6 = 3.

• Four atoms are placed completely inside the unit cell along the diagonal in such a way that each of them has tetrahedral link with one corner atom.

∴ Numberof atom per unit cell, n = 1 + 3 + 4 = 8

Atomic Packing Factor (APF):-

Volume of a unit cell,

$$V_{Unit} = a^3$$

Now,

$$APF = \frac{n.\frac{4}{3} πr^3}{V_{Unit}}$$

$$∴ APF = \frac{8.\frac{4}{3}π(\frac{\sqrt{3}a}{8})^3}{a^3}$$

$$∴ APF = 0.3401$$