Question

7a. Imagine that a Raman spectrum of a hypothetical molecule is going to be used to make a temperature measurement by comparing Stokes and Anti-Stokes scattering intensities. If the vibrational states of the molecule are separated by 0.10 eV and the laser is providing photons with energy E=2.00eV, what photon energies can one expect to find in the Raman scattered laser light? Use a simple energy level diagram to illustrate the scattering process, and then explain which of the scattered photon energies should be scattered at a higher intensity and which at a lower intensity in this situation. How can that information be used to measure a vibrational temperature?

Answer 1

Stokes scattering will correspond to an emission of a photon of energy E = 2.0 - 0.1 = 1.9 eV where the molecule first gets excited to a higher energy level and later emits a photon of energy less than the incident energy of the photon.

Anti-Stokes scattering will correspond to an emission of a photon of energy E = 2.0 + 0.1 = 2.1 eV where the molecule which was already in an excited state, gets excited to a higher energy level and returns to the ground state by emission of a photon of energy less than the incident photon energy.

So, 1.9eV, 2.0eV and 2.1 eV are the photon energies which are expected to find in the Raman scattered light.

Stokes scattered photon will be much more intense in the signal than the Anti-Stokes signal because the molecule is very less likely to be in the excited state (Maxwell-Boltzmann's condition) when the incident photon impinges on the molecule.

These energies can be equated to the thermal energies at temperature T as E = kT for measuring vibrational temperature.