Question

Obtain a relation for the density of the unit cell and radius of atom or sphere for the following :

(1) Simple cubic (scc) crystal 

(2) Body centred cubic (bcc) crystal 

(3) Face centred cubic (fcc) crystal.

Answer 1

(1) Consider a unit cell of a simple cubic crystal. It has 8 atoms at 8 corners of the unit cell.

∴ Total number of atoms in unit cell = \(\frac{1}{8}\) x 8 = 1

If a is the length of edge of cubic unit cell and r is the radius of the atom, then r = a/2 or a = 2r.

Volume of the unit cell = a³= (2r)³ = 8r³

If M is atomic mass of the element, then mass of one atom is M/N_A where N_A is Avogadro number. If there are V atoms in one unit cell then,

Mass of unit cell = n × Mass of one atom = n x \(\frac{M}{N_A}\)

Since there is one atom present in one unit cell,

(2) Consider a unit cell of body centred cubic (bcc) crystal. It has 8 atoms at 8 comers and one additional atom at the centre of body of unit cell.

Number of atoms due to 8 corners = \(\frac{1}{8}\) x 8 = 1

Body centred atom, wholly belong to the unit cell. Hence total number of atoms in the unit cell is two. If M is atomic mass of an element then M/N_A is mass of one atom where N_A is Avogadro number.

Mass of unit cell = Mass of 2 atoms in unit cell = 2 M/N_A

If a is the edge length or lattice parameter then,

Volume of cubic unit cell = a³

(3) Consider a unit cell of face centred cubic (fcc) crystal. It has 8 atoms at 8 comers and 6 atoms at 6 face centres. 

∴ Total number of atoms in unit cell = \(\frac{1}{8}\) x 8 + \(\frac{1}{2}\) x 6 = 1 + 3 = 4

If M is the atomic mass of an element, then mass of one atom is M/N_A, where N_A is Avogadro number. Mass of unit cell = Mass of 4 atoms 

= 4 x \(\frac{M}{N_A}\)

If a is the edge length of this cubic unit cell then, volume of unit cell = a³