Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: Dec 2012

Ionic Polarization:-

• Ionic Polarization occurs due to relative displacements between positive and negative ions in an ionic crystal.

• If a crystal or molecule consists of atoms of more than one kind, the distribution of charges around an atom in the crystal or molecule leans to positive or negative.

• As a result, when lattice vibrations or molecular vibrations induce relative displacements of the atoms, the centers of positive and negative charges are also displaced

• The locations of these centers are affected by the symmetry of the displacements.

• When the centers don't correspond, polarizations arise in molecules or crystals.

• The displacement is independent of temperature.

Polarizability:-

Consider the above figure showing ionic crystal NaCl. Let $x_1$ and $x_2$ be the displacements of $Na^+$ and $Cl^-$ ions respectively when an electric field (E) is applied.

The resultant dipole moment per unit cell is given by,

$μ_i = e(x_1 + x_2).......(i)$

Where, e = Magnitude of Charge

When field is applied, due to displacement of ions, a restoring force is produced which tends to bring the ions back into their original position.

For positive ion,

$$F_1 ∝ x_1 \ \ \ OR \ \ \ F_1 = k_1 x_1$$

Similarly for negative ion,

$$F_2 = k_2 x_2$$

Where, $k_1$ and $k_2$ are constants which depend upon the mass of the ions and the angular frequency of vibration of molecule.

Let ω0 be the angular frequency and $m_1$ and $m_2$ be the masses of Cation and Anion respectively.

Then,

$$k_1 = m_1 ω_0^2 \ \ and \ \ k_2 = m_2 ω_0^2$$

Therefore,

$$F_1 = m_1 ω_0^2 x_1 \ \ \ and F_2 = m_2 ω_0^2 x_2$$

We Know that, $ F = eE$

Hence,

$$eE = m_1 ω_0^2 x_1 \ \ and \ \ eE = m_2 ω_0^2 x_2$$

$$x_1 =\frac{eE}{m_1 ω_0^2} \ \ and \ \ x_2 = \frac{eE}{m_2 ω_0^2}$$

Substituting above values in Equation No. 1, we get;

$$μ_i = e(\frac{eE}{m_1 ω_0^2} + \frac{eE}{m_2 ω_0^2})$$

$$ ∴ μ_i = \frac{e^2}{ω_0^2}(\frac{1}{m_1} + \frac{1}{m_2})E$$

$$ ∴ μ_i = α_i E$$

Where, $α_i$ is ionic polarizability.

$$α_i = \frac{e^2}{ω_0^2}(\frac{1}{m_1} + \frac{1}{m_2})$$