Metallic gold crystallizes in the fcc lattice.
The length of the cubic unit cell is `a = 4.242 A`.
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? (Aw of Au `= 197.0 g mol^(-1)`)
d. Prove that the packing factor for gold is `0.74`.

Explanation

(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`