Statistical Thermodynamics and Rate Theories/Raman spectroscopy

The Raman Effect

Examples of energy transitions for each type of Raman line

The Raman effect is a form of light scattering by a material. It can be observed when a sample is exposed to an intense, high energy, monochromatic light source, such as a laser. Most photons of this light that are scattered by the material will be scattered elastically, meaning that there will be no change in the frequency of light. This results in an intense line at the same frequency of the source, known as the Rayleigh line.

Photons can also undergo Raman scattering, where they are inelastically scattered. This means that they will emerge at a higher or lower frequency. This type of scattering is less common than elastic scattering, with only 1 in 10⁷ photons experiencing Raman scattering. Stokes lines occur when the scattered light emerges at a lower frequency. Anti-stokes lines occur when the scattered light emerges at a higher frequency.

The Raman effect is applicable even when the frequency of radiation does not correspond to a transition between molecular energy levels. This cannot be explained by standard absorption or emission and therefore must be the Raman effect.

Raman Spectroscopy

Raman spectroscopy is a spectroscopic technique which measures inelastic scattering. The gain or loss of energy by incident light corresponds to transitions between the energy levels of the molecules. These transitions are both rotational and vibrational. A high energy photon excites a molecule to a "virtual state" (v₁), following this the molecule returns to a different state (v₂) on emission. A different frequency of light is emitted due to the difference in initial and final energy levels.

The Gross Selection Rule for Raman Spectroscopy

The gross selection rule for Raman spectroscopy states that a molecule must be anisotropically polarizable to have a Raman spectrum. Anisotropic indicates that something is different in some directions, in this case the electron density of the molecule must be polarized heterogeneously allowing for it to be observed via Raman spectroscopy.

Examples of some molecules which are spherical tops

This selection rule explains why Raman spectroscopy can successfully analyze homonuclear diatomics, such as H₂ and N₂, when rotational and vibrational spectroscopic techniques cannot.

The Raman effect is present for any molecule that can be polarized anisotropically. Molecules that are spherical tops cannot be anisotropically polarized, this is because they have three equal moments of inertia. Therefore these molecules cannot be observed using Raman spectroscopy.

The Specific Selection Rule of Rotational Raman Spectroscopy

The specific selection rule for Raman spectroscopy of linear molecules is $\Delta J=0,\pm 2$. The Raman spectrum has regular spacing of lines, as seen previously in absorption spectra, but separation between the lines is doubled. The evenly spaced lines have a separation of: ${\widetilde {v}}=4{\widetilde {B}}(J+1)$ where: ${\widetilde {B}}={\frac {h}{8\pi ^{2}cI}}$

The rotational energy of the molecule can be calculated using the following equation: $E_{rot}={\frac {h^{2}}{4\pi ^{2}\mu r_{e}^{2}}}J(J+1)$

The Specific Selection Rule of Vibrational Raman Spectroscopy

The specific selection rule for vibrational Raman spectroscopy states that only Δv = ±1 transitions are allowed. This is the same as for vibrational absorption spectroscopy. Vibrational Raman transitions occur simultaneously with rotational Raman transitions, this results in branching caused by rotational transitions in the Δv = ±1 peaks. The vibrational transitions result in 3 branches with fine structure from rotational transitions. In the spectrum each line corresponds to a change in quantum numbers v, J, or both.

Branch	Energy Level Transition
O	ΔJ=-2
Q	ΔJ=0
S	ΔJ=2

Raman Spectroscopy Example 1

Calculation of rotational energy levels of N₂ from pure rotational Raman spectrum.

For a molecule to have a Raman spectrum it must be anisotropically polarizable and thus have a different electron cloud distortion along different axis. Given the Pure Rotational Raman Spectrum of a molecule it is possible to calculate the rotational energy levels based on the spectrum itself. Transitions of $\Delta J=\pm 1$ can be observed in the spectrum and used to find the energy levels.

A sample calculation to find the rotational energy levels of Nitrogen can be computed using the spacings and J values of the Raman spectrum below. This calculation will be done in three steps:

1. Finding the rotational constant, ${\widetilde {B}}$.

2. Finding the moment of Inertia, $I$.

3. Finding the rotational energy, $E_{rot}$.

Step 1: Computing the rotational constant B:

As seen from the figure above, the spacing, or v, is approximately 8.00 cm^-1 which corresponds to the quantum number J = 1. Using these values it is possible to compute B by re-arranging the formula below:

${\widetilde {\nu }}=4{\widetilde {B}}(J+1)$

${\widetilde {B}}={\frac {\widetilde {\nu }}{4(J+1)}}$

${\widetilde {B}}={\frac {8.00cm^{-1}}{4(1+1)}}$

${\widetilde {B}}=1.00cm^{-1}$

Step 2: Compute the moment of Inertia I using the formula below and the constant B determined above:

${\widetilde {B}}={\frac {h}{8\pi ^{2}cI}}$

$I={\frac {h}{8\pi ^{2}c{\widetilde {B}}}}$

Note: h is Planck's constant $6.62607004\times 10^{-34}J/s$ and c is the speed of light in cm/s: $2.99792458\times 10^{10}cm/s$. When using frequencies in units of wavenumber ($cm^{-1}$), the speed of light in units of cm/s should be used in order for them to cancel out during the calculation.

$I={\frac {6.626\times 10^{-34}J/s}{8\pi ^{2}(2.998\times 10^{10}cm/s)(1.00cm^{-1})}}$

$I=2.799\times 10^{-46}kg/m^{2}$

Step 3: Compute the rotational energy level using the value of I from above and the following two equations:

$I=\mu r_{e}^{2}$

$E_{rot}={\frac {h}{4\pi \mu r_{e}^{2}}}J(J+1)$

Note: Substituting the equation for I into the energy equation will give:

$E_{rot}={\frac {h^{2}}{4\pi ^{2}I}}J(J+1)$

$E_{rot}={\frac {(6.626\times 10^{-34}J/s)^{2}}{4\pi ^{2}(2.799\times 10^{-46}kg/m^{2})}}1(1+1)$

$E_{rot}=3.94\times 10^{-23}J$

Note: Since the calculated value of $E_{rot}$ was done using $J=1$, this corresponds to the energy of the $J=1$ energy level. This calculation can be repeated using different values of $J$ to determine the different energies of each rotational energy level.