How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred
(a) Simple cubic: In a simple cubic lattice, particles are present only at the corners and they touch each other along the edge.
Let ,the edge length = a
Radius of each particle = r.
Thus, a = 2r
Volume of spheres = πr3(4/3)
Volume of a cubic unit cell = a3 = (2r)3 = 8r3
We know that the number of particles per unit cell is 1.
Therefore, Packing efficiency = Volume of one particle/Volume of cubic unit cell = [πr3(4/3)] /8r3 = 0.524 or 52.4 %
(b) Body-centred cubic:
From ∆FED, we have:
b2 = 2a2
b = (2a)1/2
Again, from ∆AFD, we have :
c2 = a2 + b2
=> c2 = a2 + 2a2
c2 = 3a2
=> c = (3a)1/2
Let the radius of the atom = r.
Length of the body diagonal, c = 4r
=> (3a)1/2 = 4r
=> a = 4r/ (3)1/2
or , r = [ a (3)1/2 ]/ 4
Volume of the cube, a3 = [4r/ (3)1/2 ]3
A BCC lattice has 2 atoms.
So, volume of the occupied cubic lattice = 2πr3(4/3) = πr3(8/3)
Therefore, packing efficiency = [ πr3( 8/3) ]/ [ { 4r/(3)1/2 }3 ] = 0.68 or 68%
(iii) Face-centred cubic:
Let the edge length of the unit cell = a
Let the radius of each sphere = r
Thus, AC = 4r From the right angled triangle ABC, we have :
AC = (a2 + a2)1/2 = a(2)1/2
Therefore, 4r = a(2)1/2
=> a = 4r/( 2)1/2
Thus, Volume of unit cell =a3 = { 4r/(2)1/2}3
a3 = 64r3/2(2)1/2 = 32r3/ (2)1/2
No. of unit cell in FCC = 4
Volume of four spheres = 4 × πr3(4/3)
Thus, packing efficiency = [πr3(16/3)] / [32r3/(2)1/2] = 0.74 or 74 %