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%