0
27kviews
Explain ionic polarization and obtain polarizability

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: Dec 2012

1 Answer
1
972views

enter image description here

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 x1 and x2 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,

μi=e(x1+x2).......(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,

F1x1   OR   F1=k1x1

Similarly for negative ion,

F2=k2x2

Where, k1 and k2 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 m1 and m2 be the masses of Cation and Anion respectively.

Then,

k1=m1ω20  and  k2=m2ω20

Therefore,

F1=m1ω20x1   andF2=m2ω20x2

We Know that, F=eE

Hence,

eE=m1ω20x1  and  eE=m2ω20x2

x1=eEm1ω20  and  x2=eEm2ω20

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

μi=e(eEm1ω20+eEm2ω20)

∴ 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})

Please log in to add an answer.