1. Objectives Packing fraction in: Simple cubic unit cell
Face Centred Cubic (BCC)
Body Centred Cubic (BCC)
Hexagonal Closed Packing(hcp)

2. Packing fraction for simple cubic unit cell
The fraction of total volume of a cube occupied by constituent particles.
Effective number of atoms =

3. Packing fraction for fcc unit cell
The number of effective atoms/anions/cations = 4
For fcc unit cell,
By the definition of packing fraction,
Taking the value of ‘a’ , we get

4. Packing fraction for bcc unit cell
The number of effective atoms/anions/cations = 2
For bcc unit cell,
By the definition of packing fraction,

5. Packing fraction of hcp
Volume of the unit cell=Base area x height
Base area of regular hexagon =Area of six equilateral triangles each with side 2r and altitude 2rsin600
First we will calculate the distance between base atom surrounded by 6 other atoms and the centre of equilateral triangle formed by three atoms just above base atoms.

6. Packing fraction of hcp

7. Packing fraction of hcp

8. Calculations Involving Unit Cell Dimensions
From the unit cell dimensions, it is possible to calculate the volume of the unit cell. Knowing the density of the metal. We can calculate the mass of the atoms in the unit cell. The determination of the mass of a single atom gives an accurate determination of Avogadro constant.
Suppose edge of unit cell of a cubic crystal determined by X – Ray diffraction is a, d is density of the solid substance and M is the molar mass, then in case of cubic crystal

9. Volume of a unit cell = a3
Mass of the unit cell = no. of atoms in the unit cell × mass of each atom = Z × m
 Here Z = no. of atoms present in one unit cell
 m = mass of a single atom
 Mass of an atom present in the unit cell = m/NA
∴ Density d = mass of unit cell / volume of unit cell = Z.m / a3
d = Z.M. / a3 × NA

10. llustration 1. An element having atomic mass 60 has face centred cubic unit cell. The edge length of the unit cell is 400 pm. Find out the density of the element?
Solution:          Unit cell edge length = 400 pm
    = 400 × 10–10 cm
    Volume of unit cell = (400 × 10-10)3 = 64 × 10-24 cm3
Mass of the unit cell = No. of atoms in the unit cell × mass of each atom
        No. of atoms in fcc unit cell = 8 × 1/8 + 6 × 1/2 = 4
                      ∴ Mass of unit cell = 4 × 60 / × 1023
       Density of unit cell = mass of unit cell / volume of unit cell = 4 × 60 / × 1023× 64 × 10–24 = 6.2 g/cm3

11. llustration 2. An element has a body centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element?
Solution:         
Volume of unit cell = (288×10–10)3 cm3 = 2.39 × 10-23cm3
Volume of 208 g of the element = mass / volume = 208 / 7.2 = 28.88cm3
No of unit cells in this volume = / 2.39 × 10–23 = × 1023
Since each bcc unit cell contains 2 atoms
  ∴ no of atom in 208 g = 2 × 12.08 × 1023 = 24.16 × 1023 atom

12. llustration 3.
 A substance has density of 2 kg dm-3 & it crystallizes to fcc lattice with  edge-length equal to 700pm, then the molar mass of the substance is
  (A) 74.50gm mol-1        (B) 103.30gm mol-1
  (C) 56.02gm mol (D) 65.36gm mol-1
Solution:  
(B)p = n × Mm / NA × a3  2 = 4 × Mm / × 1023 × (7 × 10–8)3
  (since, effective number of atoms in unit      cell = 4)
  On solving we get Mm =  gm / mol