# Solid State

## 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

AC^2=AD^2+DC^2

Or, b^2=a^2+b^2

Or, b^2=2a^2

Or, b=a\sqrt2

Let r is the radius of sphere, so b=4r

Thus, b=4r=a\sqrt2

Or, a=(4r)/sqrt2

Or, a=(4r)/sqrt2xx\sqrt2/sqrt2

Or, a=(4r\xx\sqrt2)/2

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

=(4xx4/3πr^3)/(2r\sqrt2)^3xx100

=(4xx4/3πr^3)/(8xx\r^3xx2sqrt2)xx100

=(4xx4xx3.14xx\r^3)/(3xx8xx\r^3xx2sqrt2)xx100

=(3.14xx100)/(3xx1.414)=74%

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,

AF^2=FD^2+AD^2

Or, c^2=b^2+a^2

Or, c^2=2a^2+a^2 (from equation (iv))

Or, c^2=3a^2

Or, c=a\sqrt3

Since c is equal to 4r

Therefore, 4r=a\sqrt3

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

=(2xx4/3πr^3)/(((4r)/(sqrt3))^3)xx100

=((8πr^3)/2)/((64r^3)/(3sqrt3)xx100

=(8πr^3xx3xx1.732)/(2xx64r^3)xx100

=(8xx3.14xx1.732)/(2xx64)xx100=68%`

Thus,packing efficiency of bcc structure=68%