**Packing fraction**

The ratio of the volume of the atoms occupying the unit cell to the volume of the unit cell relating to that structure is called packing fraction.

i.e. ** Packing factor (f) = volume of the atoms per unit cell (v) / Volume of the unit cell (V)**

**Atomic radius :**

The atomic radius is the half of the distance among neighbours in a crystal of a natural element.

### 1. Crystal structure of Simple cubic lattice (sc):

i) It contains 8 corner atoms touching each other

ii) Number of atoms unit cell: 8 x 1/8 = 1

as each corner atom has contribution to a unit cell is 1/8

iii) Atomic radius:

Let the distance between two neighbours i.e lattice constant = a and atomic radius = r

Then 2r = a or r = a/2

iv) Volume of unit cell = a³

v) Volume of the atoms in the unit cell =

vi) Packing fraction = volume of one atom (v) / Volume of the unit cell (V)

Co-ordination number = 6 – as each atom is surrounded by 6 nearest neighbours.

Example – Polonium (Po)

### 2. Crystal structure of Face centred cubic lattice (fcc) :

i) It contains 8 corner atoms and 6 face centred atoms.

ii) Number of atoms per unit cell is = 1/8 x 8 + 1/2 x 6 = 4

as each corner atom has contribution to a unit cell is 1/8 and each face centred atom has contribution to a unit cell is 1/2

iii) Atomic radius:

Let the distance between two neighbours i.e lattice constant = a and atomic radius = r

Thus,

iv) Volume of unit cell = a³

v) Volume of each atom in the unit cell =

vi) Packing fraction = volume of atoms (v) in a unit cell / Volume of unit cell (V)

Examples: Cu, Ag, Au, Ca, Al, Pb, Pt.

### 3. Crystal structure of Body centred cubic lattice (bcc):

i) It contains 8 corner atoms and a body centred atom.

ii) Number of atoms per unit cell : 8 x 1/8 + 1 = 2 – as each corner atom has contribution to a unit cell is 1/8 and central atom has whole contribution.

iii) Atomic radius:

Let the distance between two neighbours i.e lattice constant = a and atomic radius = r

Then

Again

iv) Volume of unit cell = a³

v) Volume of the atoms in the unit cell =

vi) Packing fraction = volume of atoms in the unit cell / Volume of unit cell (V)

Examples : Li, Na, K, Cr, Mo, etc.

**CRYSTAL DIFFRACTION**

*Bragg’s law :*

*Bragg’s law :*

The atoms in a crystal are arranged in a regular manner and the spacing between them is comparable to the wavelength of x-rays.

W.L. Bragg presented a simple explanation of the observed angles of the diffracted beams from a crystal. Consider a series of parallel rows in which the atoms are arranged in a given plane of the crystal. Suppose parallel beams of x-rays is incident in a direction making a glancing angle θ with the surface of the plane.

Each atom acts as the centre of disturbance and sends spherical wave fronts by Huygen’s construction. (As x-rays are much more penetrating than ordinary light, it is necessary to consider the rays reflected not from a single layer but from several layers together.

Superposition of x-rays, reflected from two successive parallel planes, will give diffraction pattern. We get construction if the two beam met in the same phase and destruction if in opposite phase.

Derivation –

Let, two parallel rays LMN and PQR which are reflected by two atoms M and Q in adjacent layers, separated by d. Path distinction among superposing waves –

= AQ + BQ = dsinθ + dsinθ = 2dsinθ

Let be the wavelength of the wave, then condition of maxima, path difference = nλ

2dsinθ = nλ

This is known as Bragg’s equation.

**Von Laue equation of scattering vector : (Using Laue equation derivation of Bragg’s law)**

**Von Laue equation of scattering vector : (Using Laue equation derivation of Bragg’s law)**

**Laue equation :**

Let the x-rays scattered by two identical scattering centres A and B separated by a distance **r**. Here the incident radiation is assumed to be a parallel beam and the scattered beam is assumed to be detected at a long distance away i.e they are also parallel.

Let **So** be the unit vector along the direction of incident beam.

**S** be the unit vector along the direction of scattered beam.

Thus, path difference =

The vector (**So** –** S**) represents the direction normal to the plane that reflects the incident direction to the scattered direction i.e scattering normal

If 2θ be the angle between **So** and** S**

Then

So phase difference between the radiations scattered from A and B,

The condition of diffraction maxima is satisfied by neighbour atoms if the scattered radiations by them superpose in same phase.

We know nearest neighbour atoms are separated by primitive translation distances a, b and c. Hence replacing r by a,b,c and applying condition of diffraction maxima we get,

Where h’, K’, I’ are any integers. These integers may form a smallest set of integers in such a way that they may

be identical with Miller indices of a common plane. Considering the common factor as n,

where h, k, l are the Miller indices.

Thus ,

Let α, β and γ be the angles of a,b and c with scattering normal (**So** –** S**) respectively,

Comparing equation 3 and 4, we get

These are called; Laue’s equations

**Derivation of Bragg’s law**

The direction cosines of the scattering normal (**So** –** S**) are proportional to h/a, k/b and I/c. We also know that direction cosines of the normal to hkl plane are proportional to h/a, k/b and I/c .

If d be the spacing between two adjacent planes of a family (hkl) then,

Put it in equation 6 , we get –

2dsinθ = nλ

That is the Bragg’s law.

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