Question

What is the percentage of the occupied space in a simple cubic unit cell?

Show that the percentage of the occupied space in a simple cubic unit cell is 52.4%.

Calculate the efficiency of packing in case of a metal crystal for a simple cubic unit cell.

Answer 1

A single atom is the sole component of a simple cubic unit cell. Assume that the radius of the atom is r and the edge length of a simple cubic unit cell is a.

The two spheres located at the extremities of a simple cubic unit cell are in contact with one another, as illustrated in Fig.

a = 2r    ...(i)

Number of atoms per unit cell = `8 xx 1/8 = 1`

Volume of an atom = `4/3 pi r^3`    ...(ii)

Volume of unit cell (cube) = a³

= (2r)³

= 8r³

∴ Packing fraction = `"Volume occupied by atoms"/"Volume of unit cell"`

= `((4/3)pi r^3)/(8 r^3)`

= `(1.34 pi r^3)/(8 r^3)`

= `(1.34 xx 3.14)/8   ...[∵ pi = 3.14]`

= 0.524

Hence, the percentage of the occupied space = 52.4%