ObjectivesObjectives

Packing fraction in:

Simple cubic unit cell

Face Centred Cubic (BCC)

Body Centred Cubic (BCC)

Hexagonal Closed Packing(hcp)

Packing fraction for simple cubic unit cellPacking fraction for simple cubic unit cell

Volume of total number of atoms in a unit cellSo packing fraction = Total volume of unit cell

3 3

3 3

4 41× r r3 3= = = 0.52 or 52%a 2r

1 ×8 =18Effective number of atoms =

The fraction of total volume of a cube occupied by constituent particles.

Packing fraction for fcc unit Packing fraction for fcc unit cellcellThe number of effective atoms/anions/cations = 4

a

r

By the definition of packing fraction,

Taking the value of ‘a’ , we get

3

3

44 r 2 2 1 3.143PF 0.7432(4r)

For fcc unit cell, 4ra2

3

3

44 r3PFa

Packing fraction for bcc unit cellPacking fraction for bcc unit cell

The number of effective atoms/anions/cations = 2

4ra3

For bcc unit cell,

33

38 r × 3 3PF = = ×3.14=0.67983× 4r

3

3

44× r3PF =a

By the definition of packing fraction,

Packing fraction of hcpPacking 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 atomsurrounded by 6 other atoms and the centre of equilateral triangle formed by three atoms just abovebase atoms.

346 r3PF Base area Height

rh

2r c

or 3cos30h 22rh3

2

2 2 2 2

22 2 2

2r2r h c c3

4r 2c 4r 4r3 32c 2r 3

2Height of unit cell 2c 4r 3

Packing fraction of hcpPacking fraction of hcp

Packing fraction of hcpPacking fraction of hcp

2 3

Volume of unit cell26 3r 4r 24 2 r3

3

33

4Volume of six spheres 6 r38 rpacking fraction

24 2 r

% volume occupied 74%% volume empty space 26%

Calculations Involving Unit Cell DimensionsFrom 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

∴ Density d = mass of unit cell / volume of unit cell = Z.m / a3

d = Z.M. / a3 × NA

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

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 / 6.023 × 1023

        Density of unit cell = mass of unit cell / volume of unit cell = 4 × 60 / 6.023 × 1023× 64 × 10–24 = 6.2 g/cm3

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 = 28.88 / 2.39 × 10–23 = 12.08 × 1023

Since each bcc unit cell contains 2 atoms  ∴ no of atom in 208 g = 2 × 12.08 × 1023 = 24.16 × 1023 atom

 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

llustration 3.

  (A) 74.50gm mol-1        (B) 103.30gm mol-1

  (C) 56.02gm mol-1 (D) 65.36gm mol-1

Solution:  (B)p = n × Mm / NA × a3 2 = 4 × Mm / 6.023 × 1023 × (7 × 10–8)3

  (since, effective number of atoms in unit      cell = 4)  On solving we get Mm = 103.03 gm / mol