Aluminium metal forms a cubic face centred closed packed crystal structure. Its atomic radius is `125xx10^(-12)`m.
(a) Calculate the length of the side of the unit cell.
(b) How many unit cells are there in `1.0m^(3)` of aluminium?

Explanation

(a) For a face centred cubic lattice (fcc).
Radius (r) `=(a)/(2sqrt2)`
`a=rxx2sqrt2=125xx10^(-12)xx2sqrt2m`
`=125xx2xx1.414xx10^(-12)=354xx10^(-12)m`
(b) Volume of unit cell `(a)^(3)=(354xx10^(-12))^(3)m^(3)=4.436xx10^(-29)m^(3)`
No. of unit cells in `1.0m^(3)` of Al `=((1.0m^(3)))/((4.436xx10^(-29)m^(3)))=2.25xx10^(28)`