In contrast to other conventional branches of spectroscopy, Raman spectroscopy deals with the **scattering of light **& not with its absorption.

# Raman Effect

**Chandrasekhar Venkat Raman** discovered in 1928 that *if light of a definite frequency is passed through any substance in gaseous, liquid or solid state, the light scattered at right angles contains radiations not only of the original frequency (Rayleigh Scattering) but also of some other frequencies which are generally lower but occasionally higher than the frequency of the incident light.*

The phenomenon of scattering of light by a substance when the frequencies of radiations scattered at right angles are different (generally lower and only occasionally higher) from the frequency of the incident light, is known as **Raman Scattering or Raman effect**.

The lines of lower frequencies as known as **Stokes lines** while those of higher frequencies are called anti-stokes lines.

If $f$ is the frequency of the incident light & $f’$ that of a particular line in the scattered spectrum, then the difference $f-f’$ is known as the **Raman Frequency.** This frequency is independent of* the frequency of the incident light*. It is constant and is characteristic of the substance exposed to the incident light.

A striking feature of Raman Scattering is that * Raman Frequencies are identical*, within the limits of experimental error, with those obtained from rotation-vibration (infrared) spectra of the substance.

Here is a home-made video explaining the Raman Scattering of Yellow light:

And here is another video guide for Raman Scattering:

## Advantage of Raman Effects

- Raman Spectroscopy can be used not only for gases but also for liquids & solids for which the infrared spectra are so diffuse as to be of little quantitative value.
- Raman Effect is exhibited not only by polar molecules but also by non-polar molecules such as $O_2, N_2, Cl_2$ etc.
- The rotation-vibration changes in non-polar molecules can be observed only by Raman Spectroscopy.
- The most important advantage of Raman Spectra is that it involves measurement of frequencies of scattered radiations, which are only slightly different from the frequencies of incident radiations. Thus, by appropriate choice of the incident radiations, the scattered spectral lines are brought into a convenient region of the spectrum, generally in the visible region where they are easily observed. The measurement of the corresponding infrared spectra is much more difficult.
- It uses visible or ultraviolet radiation rather than infrared radiation.

### Uses

- Investigation of biological systems such as the polypeptides and the proteins in aqueous solution.
- Determination of structures of molecules.

C. V. Raman was awarded the 1930 Physics Nobel Prize for this.

# Classical Theory of Raman Effect

The classical theory of Raman effect, also called the polarizability theory, was developed by G. Placzek in 1934. I shall discuss it briefly here. It is known from electrostatics that the electric field $E$ associated with the electromagnetic radiation induces a dipole moment $\mu$ in the molecule, given by

$$ \mu = \alpha E \ldots (1)$$ where $\alpha$ is the polarizability of the molecule. The electric field vector $E$ itself is given by $$E = E_0 \sin \omega t = E_0 \sin 2\pi \nu t \ldots (2)$$

where $E_0$ is the amplitude of the vibrating electric field vector and $\nu$ is the frequency of the incident light radiation.

Thus, from equations (1) & (2),

$$ \mu= \alpha E_0 \sin 2\pi \nu ldots (3)t$$

Such an oscillating dipole emits radiation of its own oscillation with a frequency $\nu$, giving the **Rayleigh scattered beam**. But as, the polarizability varies slightly with molecular vibration, we can write $$\alpha =\alpha_0 + \frac {d \alpha} {dq} q \ldots (4)$$

where the coordinate $q$ describes the molecular vibration. We can define $q$ as:

$$ q=q_0 \sin 2\pi \nu_m t \ldots (5)$$

Where $q_0$ is the amplitude of the molecular vibration and $\nu_m$ is its (molecular) frequency. From equations 4 & 5, we have

$$ \alpha =\alpha_0 + \frac {d\alpha} {dq} q_0 \sin 2\pi \nu_m t \ldots (6)$$

Substituting for alpha in (3), we have

$$ \mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {d\alpha}{dq} q_0 E_0 \sin 2\pi \nu t \sin 2\pi \nu_m t \ldots (7)$$

Making use of the trigonometric relation $ \sin x \sin y = \frac{1}{2} [\cos (x-y) -\cos (x+y) ]$ this equation reduces to:

$$ \mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {1}{2} \frac {d\alpha}{dq} q_0 E_0 [\cos 2\pi (\nu – \nu_m) t – \cos 2\pi (\nu+\nu_m) t] \ldots (8)$$

Thus, we find that the oscillating dipole has three distinct frequency components:

The exciting frequency $\nu$ with amplitude $\alpha_0 E_0 \ldots (9)$ and frequencies $\nu – \nu_m$ & $\nu + \nu_m$ with very small amplitudes of $\frac {1}{2} \frac {d\alpha}{dq} q_0 E_0$.

Hence, the Raman spectrum of a vibrating molecule consists of a relatively intense band at the incident frequency and two very weak bands at frequencies slightly above and below that of the intense band.

If, however, the molecular vibration does not change the polarizability of the molecule then $$(\dfrac{d\alpha}{dq})=0$$ so that the dipole oscillates only at the frequency of the incident (exciting) radiation. The same is true for the molecular rotation. We conclude that for a molecular vibration or rotation to be active in the Raman Spectrum, it must cause a change in the molecular polarizability, i.e., $d\alpha/dq \ne 0$

Homo-nuclear diatomic molecules such as $\mathbf {H_2 , N_2 , O_2}$ which do not show IR Spectra since they don’t possess a permanent dipole moment, do show Raman spectra since their vibration is accompanied by a change in polarizability of the molecule. As a consequence of the change in polarizability, there occurs a change in the induced dipole moment at the vibrational frequency.

REFERENCES:-

- Principles in Physical Chemistry, Puri, Sharma & Pathania
- Physics Chemistry, Atkins
- Spectroscopy, Raj Kumar

Last updated on **February 28, 2017 **for National Science Day!

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