Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 7M

Year: Dec 2012

1) Co-ordination Number:-

Each atom is positioned in the empty space formed by

• Three adjacent atoms of the top layer

• Three adjacent atoms in the bottom level and

• Surrounded by six neighboring atoms.

• Thus, 12 atoms are in contact with each atom.

  Hence, Co-ordination Number = 12

2) Atomic Radius:-

The atoms are in contact along the edge of the hexagon.

$∴r= \frac{a}{2}$

3) Number of Atoms Per Unit Cell:-

• Each Corner atom of hexagonal face is shared by 6 unit cells i.e. they contribute 1/6th of the mass. There are two such hexagonal faces i.e. 12 such atoms.

• Thus, contribution of corner atoms = 2 x 6 x 1/6 = 2

• The atom at the center of the hexagonal faces is shared by two cells each.

• Therefore, contribution of face centered atoms = 2 x 1/2 = 1

• Three atoms are contained within a single cell, hence their contribution is 3.

  ∴ Number of atom per unit cell, n = 2 + 1 + 3 = 6

4) Atomic Packing Factor (APF):-

Theoretical ratio of c/a = $\sqrt{8/3}$

Volume of a unit cell,

$$V_{Unit} = (Area \ \ of \ \ Hexagon) × (Height \ \ i.e \ \ c)$$

$$V_{Unit} = 6 × (Area \ \ of \ \ Equilateral \ \ Triangles) × \sqrt{\frac{8}{3}}a $$

$$V_{Unit} = 6 × (\frac{1}{2} a × a.sin(60)) × \sqrt{\frac{8}{3}}a $$

$$V_{Unit} = 3\sqrt{2a}^3$$

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

$$APF = \frac{6.\frac{4}{3}\pi(a/2)^3 }{3\sqrt{2a}^3}$$

$$APF = 0.7405$$