written 8.2 years ago by | • modified 8.2 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 7M
Year: Dec 2012
written 8.2 years ago by | • modified 8.2 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 7M
Year: Dec 2012
written 8.2 years ago by |
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,
$$V_{Unit} = (Area \ \ of \ \ Hexagon) × (Height \ \ i.e \ \ c)$$
$$V_{Unit} = 6 × (Area \ \ of \ \ Equilateral \ \ Triangles) × \sqrt{\frac{8}{3}}a $$
$$V_{Unit} = 6 × (\frac{1}{2} a × a.sin(60)) × \sqrt{\frac{8}{3}}a $$
$$V_{Unit} = 3\sqrt{2a}^3$$
$$APF = \frac{n.\frac{4}{3}\pi r^3}{V_{Unit}}$$
$$APF = \frac{6.\frac{4}{3}\pi(a/2)^3 }{3\sqrt{2a}^3}$$
$$APF = 0.7405$$