Raman Spectroscopy Introduction

You are already aware that photons interact with molecules to induce transitions between energy states. In the discussion of Raman spectroscopy, we use language from particle theory and we say that a photon is scattered by the molecular system. Most photons are elastically scattered, a process which is called Rayleigh scattering. In Rayleigh scattering, the emitted photon has the same wavelength as the absorbing photon. Raman Spectroscopy is based on the Raman effect, which is the inelastic scattering of photons by molecules. The effect was discovered by the Indian physicist, C. V. Raman in 1928. The Raman effect comprises a very small fraction, about 1 in 107, of the incident photons. In Raman scattering, the energies of the incident and scattered photons are different. A simplified energy diagram that illustrates these concepts is given below.

The energy of the scattered radiation is less than the incident radiation for the Stokes line and the energy of the scattered radiation is more than the incident radiation for the anti-Stokes line. The energy increase or decrease from the excitation is related to the vibrational energy spacing in the ground electronic state of the molecule and therefore the wavenumber of the Stokes and anti-Stokes lines are a direct measure of the vibrational energies of the molecule. A schematic Raman spectrum may appear as:

In the example spectrum, notice that the Stokes and anti-Stokes lines are equally displaced from the Rayleigh line. This occurs because in either case one vibrational quantum of energy is gained or lost. Also, note that the anti-Stokes line is much less intense than the Stokes line. This occurs because only molecules that are vibrationally excited prior to irradiation can give rise to the anti-Stokes line. Hence, in Raman spectroscopy, only the more intense Stokes line is normally measured.

Infrared (IR) and Raman spectroscopy both measure the vibrational energies of molecules but these method rely only different selection rules. Recall that for a vibrational motion to be IR active, the dipole moment of the molecule must change. Therefore, the symmetric stretch in carbon dioxide is not IR active because there is not change in the dipole moment. The asymmetric stretch is IR active due to a change in dipole moment.

For a transition to be Raman active, there must be a change in polarizability of the molecule.




Notice that the symmetric stretch in carbon dioxide is Raman active because the polarizability of the molecule changes. You can see when you compare the ellipsoid at the equilibrium bond length to the ellipsoid for the extended and compressed symmetric motions. For a vibration to be Raman active, the polarizability of the molecule must change with the vibrational motion. Thus, Raman spectroscopy complements IR spectroscopy.

Experimentally, we only observe the Stokes shift in a Raman spectrum. Recall that the Stokes lines will be at smaller wavenumbers (or higher wavelengths) than the exciting light. Since the Raman scattering is not very efficient, we need a high power excitation source such as a laser. Also, since we are interested in the energy (wavenumber) difference between the excitation and the Stokes lines, the excitation source should be monochromatic. This is another property of many laser systems.