written 8.1 years ago by | • modified 8.1 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 7M
Year: May 2014
written 8.1 years ago by | • modified 8.1 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 7M
Year: May 2014
written 8.1 years ago by |
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,
$$APF = \frac{n.\frac{4}{3} πr^3}{V_{Unit}}$$
$$ ∴ APF = \frac{8.\frac{4}{3}π(\frac{\sqrt{3}a}{8})^3}{a^3}$$
$$ ∴ APF = 0.3401$$