Packing Efficiency of Close Packed Structure - 1
Both ccp and hcp are highly efficient lattice; in terms of packing. The packing efficiency of both types of close packed structure is 74%, i.e. 74% of the space in hcp and ccp is filled. The hcp and ccp structure are equally efficient; in terms of packing.
The packing efficiency of simple cubic lattice is 52.4%. And the packing efficiency of body centered cubic lattice (bcc) is 68%.
Calculation of pacing efficiency in hcp and ccp structure:
The packing efficiency can be calculated by the percent of space occupied by spheres present in a unit cell.
Thus, packing efficiency (in %) `= text(Volume of spheres in unit cell)/text(Total volume of unit cell)xx100`
Since there are 4 atoms in the unit cell of hcp or ccp structure
Therefore, packing efficiency of hcp or ccp structure
`= text(Volume of 4 spheres in unit cell)/text(Total volume of unit cell)xx100`
Let the side of an unit cell = a
And diagonal AC = b
Now, in ∆ ABC,
AB is perpendicular, DC is base and AC is diagonal
Let r is the radius of sphere, so `b=4r`
Or, `a=2r\sqrt2` ………(1)
Now, volume of cube =Side3=a3
Substituting the value of a from equation (i) we get
Volume of cube `=(2r\sqrt2)^3` …………….(ii)
Now, volume of sphere `=4/3πr^3`
Since one unit cell of ccp or hcp contains 4 atoms, i.e. 4 spheres
Therefore, volume of 4 atoms, i.e. 4 spheres `=4xx4/3πr^3` ……………….(iii)
Now, packing efficiency (in %) `=text(volume of 4 spheres in unit cell)/text(total volume of unit cell)xx100`
Thus,packing efficiency of hcp or ccp structure=74%
Packing efficiency of body centered cubic (bcc) structure:
In body centered cubic unit cell, one atom is present in body center apart from 4 atoms at its corners. Therefore, total number of atoms present in bcc unit cell is equal to 2.
Let a unit cell of bcc structure with side a.
Let FD (diagonal) = b and diagonal AF = c
Let the radius of atom present in unit cell = r
Now, in ∆EFD
`FD^2=ED^2+EF^2`Or, `b^2=a^2+a^2=2a^2` ………………..(iv)
Or, `b=a\sqrt2` ……………(v)
Now, in ΔAFD,
Or, `c^2=2a^2+a^2` (from equation (iv))
Since c is equal to 4r
Or, `a=(4r)/(sqrt3)` …………………..(vi)
Volume of cube =Side3 = a3`
After subtituting the value of a from equation (vi) we get
Volume of cube `=((4r)/sqrt3)^3`
Volume of 2 atoms present in bcc structure `=2xx4/3πr^3`
Now, packing efficiency in percentage `=text(Volume of 2 sphere in unit cell)/text(Total volume of unit cell)xx100`
Thus,packing efficiency of bcc structure=68%