Question

Show that the packing fraction in a face-centred cubic unit cell is 0.74.

Answer 1

In a face-centred cubic (FCC) unit cell, the atoms are present at 8 corners and the centres of 6 faces. The atoms touch each other along the face diagonal, so face diagonal = 4r

The face diagonal can also be expressed in terms of the edge length a as `sqrt2a` = 4r

a = `(4r)/sqrt 2`

a = `2 sqrt 2 r`

Volume of the unit cell (V_cell) = a³

= `(2 sqrt2 r)^3`

= `16 sqrt 2 r^3`

An FCC unit cell contains 4 atoms (from 8 corners and 6 face centres).

Volume of 4 atoms (V_atoms) = `4(4/3 pi r^3)`

= `16/3 pi r^3`

Packing Fraction = `(V_"atoms")/(V_"cell")`

= `(16/3 pi r^3)/(16 sqrt2 r^3)`

= `(16/3 pi r^3) xx 1/(16 sqrt2 r^3)`

= `pi/(3 sqrt2)`

= `3.14/(3 xx 1.414)`

= `3.14/4.242` 

= 0.74 or 0.74%