Metallic gold crystallizes in fcc lattice. The length of the cubicunit cell is 4.07 Å(a) What is the closest distance between gold atoms ?(b) How many 'nearest neighbours' does each gold atom have at the distance calculated in (a) ?(c ) What is the density of gold ? (d) Prove that the packing fraction of gold is 0.74.

Metallic gold crystallizes in fcc lattice. The length of the cubicunit

1.	Metallic gold crystallizes in fcc lattice. The length of the cubicunit cell is 4.07 Å(a) What is the closest distance between gold atoms ?(b) How many 'nearest neighbours' does each gold atom have at the distance calculated in (a) ?(c ) What is the density of gold ? (d) Prove that the packing fraction of gold is 0.74.
Answer» Solution :(a) For fcc lattice nearst distance between TWO neighbours = 2r As we know `4r = sqrt(2)a` or`2r=(sqrt(2))/(2)a` `= (sqrt(2))/(2)xx4.07 Å` `= 2.88 Å` (b)If we consider a face centred gold ATOM, it has four comers and eight other adjacent face centre atoms present at `a//sqrt(2)` distance. Therefore, there are 12 nearest neighbours. (c ) Density `(rho) = (nM_(m))/(N_(o)xx a^(3))=(4xx197)/(6.023xx10^(23)xx(4.07xx10^(-8))^(3))=19.4` g/cc. (d) Packing fraction `= (4xx4//3pi R^(3))/((4r//sqrt(2))^(3))~~ 0.74`