# Explain the Classical Theory of Raman Effect

## Classical Theory of Raman Effect: Molecular Polarizability

The basic concept of classical theory of the Raman effect depends on the polarizability of a molecule and the applied Electric field.

When a molecule is placed in a static electric field, a distortion takes place in it because of the attraction of positively charged nuclei towards negative pole of the field and of electron towards positive pole. This separation of charge centers causes an induced electric dipole moment in the molecule and the molecule becomes polarized. The magnitude of induced dipole (𝜇) depends both on magnitude of the applied field (E) and on the case with which the molecule can be distorted. Thus,

𝜇 = 𝛼E

where α is the polarizability of the molecule.

In case of hydrogen molecule, the polarizibility is anisotropic i.e. the electron that form the bond are more easily
displaced by the field along the bond axis than along the one across this direction.

The polarizability of a molecule in various directions can be conventionally represented by polarizability ellipsoid as shown below-

The ellipsoid is a three-dimensional surface whose distance from the electrical centre of the molecule (in H_{2} this is also the centre of gravity) is proportional to 1/√𝛼_{i}, where 𝛼_{i} is the polarizability along the line joining a point 'i' on the ellipsoid with the electrical centre.

In case of diatomic molecule, the polarizibility in the same for all directions at right abgels to the bond axis, the ellipsoid has a circular cross section in this direction, so it is shaped like a tangerine. The shape of ellipsoid of all diatomic molecules like oxygen, HCl and of polyatomic linear molecule if similar to this except that the relative sizes of their major and minor axis are different.

When a diatomic molecules or a linear polyatomic molecule is subjected to a beam of radiation of frequency '𝜈' the electric field experienced by each molecule varies according to the equation-

E = E_{o} sin 2𝜋νt

and thus the induced dipole also undergoes oscillations of frequency '𝜈'.

𝜇 = 𝛼E

or, 𝜇 = 𝛼E_{o} sin 2𝜋𝜈t

Such an oscillating dipole gives out radiations of its own oscillation frequency and so it explains the Rayleigh scattering. We have not taken into consideration the vibration and rotation of molecule while deriving this equation. These also affect the polarizability of the molecule.

**Effect of Vibration**

If the molecule undergoes vibrational motion, the oscillating dipole will have superimposed upon it vibrational oscillation. The change in polarizibility due to a frequency of 𝜈_{vib} can be expressed as-

𝛼 = 𝛼_{0} + β 2𝜋𝜈_{vib}t

where, 𝛼_{0} is the equilibrium polarizability and β represents the rate of change of polarizability with the vibration.

Since, 𝜇 = 𝛼E

∴ 𝜇 = (𝛼_{0} + β 2𝜋𝜈_{vib}t)E_{0} sin 2𝜋𝜈t

We know the trigonometric expression-

sin A sin B = ½[cos (𝐴 − 𝐵) − cos(𝐴 + 𝐵)]

𝜇 = (𝛼_{0} + 𝛽 sin 2𝜋𝜈_{vib}t)E_{0} sin 2𝜋𝜈𝑡

𝜇 = 𝛼_{0}E_{0} sin 2𝜋𝜈𝑡 + 𝛽 sin2𝜋𝜈_{vib}t · E_{0} sin 2𝜋𝜈𝑡

𝜇 = 𝛼_{0}E_{0} sin 2𝜋𝜈𝑡 + ½ 𝛽E_{0}[cos 2𝜋(𝜈 − 𝜈_{vib})𝑡 − cos 2𝜋(𝜈 + 𝜈_{vib})𝑡]

So the oscillating dipole has frequency components 𝜈 ± 𝜈_{vib} as well as exciting frequency 𝜈. So, the Raman shift will be equal to 𝜈_{vib}.

Raman shift = ( 𝜈 + 𝜈_{vib} − 𝜈) = 𝜈_{vib}

**Effect of Rotation**

Again we consider a diatomic molecule, when such a molecule rotates its orientation varies with respect to the electric field of rotation. When such a molecule is not optically isotropic, the polariation will vary with time. In such a case the variation of '𝛼' can be expressed by the equation-

𝛼 = 𝛼_{0} + 𝛽 sin 2𝜋𝜈_{rot}t

where, 𝜈_{rot} is the frequency of rotation.

This equation is identical with the equation obtained while considering the effect of vibration.

Substituting the above equation for dipole , i.e.

𝜇 = 𝛼E

𝜇 = 𝛼E_{0} sin 2𝜋𝜈𝑡 + &frac;12 𝛽'E_{0}[cos 2𝜋(𝜈 − 2𝜈_{rot})𝑡 − cos 2𝜋(𝜈 + 𝜈_{rot})𝑡]

Thus the frequency of Raman lines will be (𝜈 + 2𝜈_{rot}) and (𝜈 − 𝜈_{rot}) and Raman shift will be 2𝜈. Thus Raman shift in this case will be twice the frequency of rotation of molecule.

Is should be clearly understood that if the vibration does not change polarizability of a molecule the 𝛽 = 0 and in such case dipole oscillates only at the frequency of the incident radiation. Similarly, if the rotation does not change polarizability of a molecule the 𝛽' = 0 and again the dipole oscillates only at the frequency of the incident radiation. Thus, we get the following general rules-

__For a molecule to be raman active, the molecular rotation or vibration must produce some change in a component of molecular polarizability__.

A change in molecular polarizability is reflected by a change in either the magnitude or the direction of the polarizability ellipsoid.