A unit cell of simple cubic lattice contains one atom.
Let the side of a simple cubic lattice is ‘a’ and radius of atom present in it is ‘r’.
Since, edges of atoms touch each other, therefore, a = 2r
Volume of cube =Side3 `=a^3=(2r)^3`
Volume of one atom `=4/3πr^3`
Packing Efficiency `=text(Volume of 1 sphere in unit cell)/text(Total volume of unit cell)xx100`
Thus,packing efficiency of bcc structure=52.4%
The edge of a unit cell is ‘a’.
The density of unit cell is ‘d’
Molar mass of unit cell is ‘M’.
Number of atoms present in unit cell is ‘z’.
Mass of each atoms present in unit cell is ‘m’.
Mass of one atom in unit cell `m=M/(N_A)`
Therefore, mass of unit cell = Number of atoms present in unit cell `xx` Mass of one atom
Or, Mass of unit cell `=zm`
Or, Mass of unit cell `=z\xx(M)/(N_A)`
We know that density `d=text(Mass)/text(Volume)`
Therefore, density d of unit cell `=text(Mass of unit cell)/text(Volume of unit cell)`
Where, d is density, z is number of atoms present in unit cell, a is length of edge, and NA is Avogadro constant.
Above expression has five parameters, d, z, a, m and NA . By knowing any four of them fifth can be calculated.
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