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?

Answer

(a)354 pm (b)`2.25xx10^28`

Explanation

ccp=fcc. For fcc, `r=a/(2sqrt2)` or `a=2sqrt2r=2xx1.414xx125xx10^(-2) m =354xx10^(-12) m`=354 pm
Volume of one unit cell =`(354xx10^(-12)m)^3=4.436xx10^(-29) m^3`
`therefore` No. of unit cell in 1 `m^3=1/(4.436xx10^(-29))=2.25xx10^28`