Calculation of number of particles in a unit cell (Z)
In a crystal atoms located at the corner and face centre of a unit cell are shared by other cells and opnly a portion of such an atom actually lies within a given unit cell.
(i) A point that lies at the corner of a unit cell is share among eight unit cells and therefore, only one eighth of each such point lies within the given unit cell.
(ii) A point along an edge is shared by four unit cells and only one-fourth of it lies within any one cell.
(iii) A face-centred point is shared by two unit cells and only one half of it is present in a given unit cell.
(iv) A body-centred point lies entirely within the unit cell and contributes one complete point to the cell
|Type of Lattice point||Contribution to one unit cell|
Calculate the number of particles in a unit cell
|Type of unit cell||Lattice points at corner||Lattice points at face-centered||Lattice points at body centered||Z = no. of lattice points per unit cell|
Relation between edge length of unit cell and radius of atoms
(i) Simple cubic or Primitive
(a) edge length of unit cell a = 2r (b) atomic radius r = a/2
(ii) Body centre cubic
(a) edge length of unit cell =
(b) atomic radius, r =
(iii) Face centre cubic/cubic close packed
(a) edge length of unit cell a =
(b) atomic radius r =
Illustration: A solid has a cubic structure in which X atoms are located at the corners of the cube, Y atoms are at the cube centres and O atoms are at the edge centres. What is the formula of the compound:
Solution: Atoms of X are present at all the eight corners of the cube. Therefore, each atom of X at the corner makes 1/8 contribution towards the unit cell.
Number of atoms of X per unit cell =
Y atom is present at the body centre, thus contribution of Y towards unit cell = 1 × 1 = 1
O atom is present at each of the edge centre (number of edges of cube = 12)
And each O atom present at edge centre will make ¼ contribution towards the unit cell.
The number of O atoms per unit cell =
The formula of the compound is, therefore XYO3
Fraction of volume occupied by atoms in a cube
Packing Fraction = Volume of atoms in the cube/Total volume of cube
Density of unit cell
Density of unit cell (and hence density of a crystal)
Where a is edge of unit cell in pm
NA = Avogadro number (6.02 × 1023)
M = Atomic mass of element or formula mass of the compound
Z = No. Of toms present per unit cell or formula units e.g. for fcc, Z = 4, for bcc, Z = 2, for simple cubic, Z = 1
Illustration: Gold has a close-packed structure which can be viewed as-spheres occupying 0.74 of the total volume. If the density of gold is 19.3 g/cc, calculate the apparent radius of a gold ion in the solid
Solution: Gold has s close-packed structure with a packing fraction value of 0.74. This shows that it has a face-centred cubic cell. The number of ions in a face-centred unit cell is 4.
Now density =
Or 19.3 =
In a face-centred cubic cell.