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

NaCl, 216 `(au)^(3)`

Explanation

ccp = fcc. For fcc, `r = a/(2sqrt(2))or a = 2 sqrt2r = 2 xx 1.414 xx 125 xx 10^(-2) m = 354 xx10^(-12) m =354 "pm"`
No. of unit cell in ` 1 m^(3) = 1/(4.436 xx 10^(-29))= 2.25 xx 10^(-29) m^(3)`
No. of unit cells in ` 1m ^(3) = 1/( 4.436 xx 10^(-29)) = 2.25 xx 10^(28)`