The Raman effect is the inelastic scattering of a photon by molecules of a substance, resulting in a shift in frequency and wavelength of the scattered light. Discovered by C. V. Raman in 1928, it provides direct evidence for the quantum nature of light and molecular energy levels.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of Raman effect, Rayleigh scattering vs Raman scattering, Stokes and anti-Stokes lines, experimental arrangement, and qualitative wave-number shift.
PGT extension
Quantum mechanical treatment of polarizability, rotational-vibrational Raman spectra, selection rules, calculation of moment of inertia and bond length from rotational Raman lines.
Prerequisites
Quantum theory of radiation and photon concept
Molecular energy levels (rotational and vibrational states)
Polarizability of molecules
Phenomenology of Raman Scattering
When monochromatic light of frequency ν0 is incident on a substance, a major part of the scattered light has the same frequency ν0 (Rayleigh scattering).
A minute fraction of the scattered light possesses discrete frequencies different from ν0, known as Raman frequencies.
Scattered lines with frequencies lower than ν0 (ν0 - νm) are called Stokes lines.
Stokes and Anti-Stokes Lines
Scattered lines with frequencies higher than ν0 (ν0 + νm) are called anti-Stokes lines.
Stokes lines originate from molecules in the ground vibrational or rotational state absorbing energy, leaving the molecule in an excited state (ΔE > 0).
Anti-Stokes lines originate from molecules already in an thermally excited state dropping to a lower energy state by giving up energy to the scattered photon (ΔE < 0).
Intensity Distribution and Temperature Dependence
The intensity of Stokes lines is always greater than the intensity of anti-Stokes lines at room temperature because more molecules populate the lower energy ground state according to the Boltzmann distribution.
As temperature increases, the population of excited states increases, leading to a relative increase in the intensity of anti-Stokes lines.
The Raman shift (Δν) is independent of the incident frequency ν0 and is a characteristic property of the scattering molecule.
Classical Polarizability Theory
An incident electric field $E = E_{0}\cos(\omega_{0}t) induces an electric dipole moment p = \alphaE in a molecule$, where α is the polarizability.
If the molecule vibrates with frequency ωm, its polarizability varies as $\alpha = \alpha_{0} + (\partial \alpha/\partial q)_{0} q_{0}\cos(\omega_{m}t)$.
Substituting α into the dipole moment expression yields terms containing frequencies ω0, (ω0 - ωm), and (ω0 + ωm), corresponding to Rayleigh and Raman scattering.
Derivations
Classical Derivation of Raman Frequencies via Polarizability
Conditions: Harmonic approximation of molecular vibration under an incident electromagnetic wave.
Let the incident electric field be given by E(t) = E_0 \cos(\omega_0 t).
The induced dipole moment is p(t) = \alpha(t) E(t), where \alpha is the molecular polarizability.
Assume the polarizability varies with normal vibrational coordinate $q = q_0 \cos(\omega_m t) such that \alpha(t) = \alpha_0 + \left(\frac{\partial \alpha}{\partial q}\right)_0 q_0 \cos(\omega_m t)$.
Substitute \alpha(t) and E(t) into the dipole moment equation: p(t) = \left[ \alpha_0 + \left(\frac{\partial \alpha}{\partial q}\right)_0 q_0 \cos(\omega_m t) \right] E_0 \cos(\omega_0 t).
Result: The dipole moment oscillates at frequencies \omega_0 (Rayleigh), \omega_0 - \omega_m (Stokes), and \omega_0 + \omega_m (anti-Stokes), proving the existence of Raman shifts.
Rotational Raman Energy Levels and Selection Rules
The rotational energy of a rigid diatomic molecule is given by $E_J = \frac{h^2}{8 \pi^2 I} J(J+1) = B hc J(J+1)$, where B is the rotational constant and J is the rotational quantum number.
The selection rule for pure rotational Raman transitions is \Delta J = 0, \pm 2.
For Raman scattering, the transition J \rightarrow J + 2 gives the energy change \Delta $E = E_{J+2} - E_J = B hc [(J+2)(J+3) - J(J+1)] = B hc (4J + 6)$ for J = 0, 1, 2...
The corresponding wave number shift is \Delta \tilde{\nu} = \pm B (4J + 6).
N = Population, g = Degeneracy, $k_B = Boltzmann constant$, T = Temperature Dimensionless
Thermal equilibrium
Rotational Constant Expression
$$B = \frac{h}{8 \pi^2 I c}$$
h = Planck's constant, I = Moment of inertia, c = Speed of light cm^{-1}
Diatomic molecule
Moment of Inertia of Diatomic Molecule
$$I = \mu r_0^2$$
\mu = Reduced mass, $r_0 = Equilibrium bond length$ kg m^2
Diatomic molecule fixed at center of mass
Graphs and experiments
Raman Spectrum Intensity Distribution
Draw: A central intense Rayleigh line at ν0, flanked symmetrically by weaker Stokes lines on the low-frequency side and very weak anti-Stokes lines on the high-frequency side.
Interpret: Stokes lines are more intense than anti-Stokes lines because the ground state population exceeds the excited state population at room temperature.
Exam use: Frequently asked in identifying relative intensities and positions of Rayleigh, Stokes, and anti-Stokes lines.
Raman Spectrometer Experimental Setup
Draw: A monochromatic light source (laser or mercury arc), a sample tube (Raman tube) containing the liquid/gas, a collection lens system at right angles, a spectrograph/monochromator, and a detector (photomultiplier or CCD).
Interpret: Right-angle scattering configuration is used to minimize the entry of direct incident unscattered beam into the detector.
Exam use: MCQs regarding optical components and geometric arrangement in UP TGT/PGT exams.
Rotational Raman Spectrum Line Series
Draw: Equidistant spectral lines on either side of the Rayleigh line separated by 6B, 10B, 14B, etc., starting from a gap of 6B from the center.
Interpret: The spacing between adjacent rotational Raman lines is 4B, allowing direct determination of the rotational constant B and bond length.
Exam use: Numerical problems calculating moment of inertia or bond length from spectral line separation.
Solved numericals
TGT core
Light of wavelength 435.8 nm is incident on a scattering medium. Calculate the wavelength of the Stokes line for a vibrational frequency shift of 1200 cm-1.
The spacing between adjacent lines in the rotational Raman spectrum of a diatomic molecule is 1.6 cm-1. Find the rotational constant B and the moment of inertia of the molecule. (Given $h = 6.626 \times 10^{-34} J s$, $c = 3 \times 10^{8} m s^{-1})$
The separation between adjacent rotational Raman lines is given by \Delta \tilde{\nu} = 4B.
Answer: $B = 0.4 cm^-1$, $I = 6.99 x 10^-47 kg m^2$
PGT extension
In a Raman spectrometer using a laser of wavelength 500 nm, an anti-Stokes line is observed at 480 nm. Calculate the frequency of the molecular vibration responsible for this shift in Hz.
The classical polarizability theory assumes small oscillations of nuclei compared to intermolecular distances (harmonic approximation), which fails at high vibrational excitation.
Interaction between vibration and rotation (centrifugal distortion) is neglected in simple rigid rotor approximations.
Common misconceptions
Common error: Raman scattering is caused by the absorption and subsequent re-emission of a photon, similar to fluorescence.
Correct: Raman scattering is an instantaneous two-photon scattering process involving virtual energy states, not real electronic transitions.
Fluorescence involves real absorption to an excited electronic state with a finite lifetime (~10^-8 s), whereas Raman scattering is prompt.
Common error: The Raman shift depends upon the wavelength or frequency of the incident light source.
Correct: The Raman frequency shift (Δν or Δν˛) is strictly a constant for a given molecular transition and is independent of the incident photon frequency.
The absolute scattered frequencies change with incident frequency, but the difference (ν0 - νs) remains constant.
Quick revision
Raman effect is inelastic scattering where scattered frequency differs from incident frequency by molecular vibrational/rotational frequencies.
Stokes Frequency Condition: \n$u_s = \nu_0 - \nu_m$
Must remember
Raman scattering is an instantaneous scattering process involving virtual energy states, unlike fluorescence which involves real absorption and re-emission.
Stokes lines have frequencies lower than the incident frequency ($\nu_0 - \nu_m$), while anti-Stokes lines have higher frequencies ($\nu_0 + \nu_m$).
The Raman shift ($\Delta \tilde{\nu}$) is entirely independent of the incident frequency $\nu_0$ and represents characteristic molecular vibrational or rotational frequencies.
At room temperature, the intensity of Stokes lines is greater than anti-Stokes lines because the ground state population exceeds the thermally excited state population.
Classically, Raman scattering arises from the modulation of molecular polarizability ($\alpha$) by molecular vibrations at frequency $\omega_m$.
Glossary
Rayleigh ScatteringElastic scattering of light where the scattered photon has the same frequency $\nu_0$ as the incident photon.
Stokes LinesInelastic scattered spectral lines with lower frequency ($\nu_0 - \nu_m$) than the incident light.
Anti-Stokes LinesInelastic scattered spectral lines with higher frequency ($\nu_0 + \nu_m$) than the incident light, originating from thermally excited states.
Polarizability ($\alpha$)The measure of ease with which an electron cloud of a molecule is distorted by an external electric field.
Raman Shift ($\Delta \tilde{\nu}$)The difference between the wave number of the incident light and the scattered light, expressed in $\text{cm}^{-1}$.
Stokes vs Anti-Stokes Raman Lines
Parameter
Stokes Lines
Anti-Stokes Lines
Frequency
$\nu_s = \nu_0 - \nu_m$
$\nu_{as} = \nu_0 + \nu_m$
Energy Transition
Ground to Excited state ($\Delta E > 0$)
Excited to Ground state ($\Delta E < 0$)
Intensity at Room Temp
Higher
Lower
Temperature Dependence
Decreases slightly or remains stable
Increases significantly with rising temperature
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following scattering processes corresponds to elastic scattering of light?
Explanation: Rayleigh scattering is elastic scattering where the scattered light has the exact same frequency $\nu_0$ as the incident light.
2. Stokes lines in Raman spectra are characterized by which frequency relation?
Explanation: Stokes lines have lower frequencies than the incident light, given by $\nu_s = \nu_0 - \nu_m$.
3. What is the primary reason why Stokes lines are more intense than anti-Stokes lines at room temperature?
Explanation: According to the Boltzmann distribution, more molecules reside in the ground state at room temperature, making upward transitions (Stokes) much more probable.
4. How does the intensity of anti-Stokes lines change when the temperature of the scattering substance is increased?
Explanation: As temperature increases, thermal excitation populates higher energy states more densely, leading to an increase in the intensity of anti-Stokes lines.
5. The Raman shift ($\Delta \tilde{\nu}$) depends on which of the following?
Explanation: The Raman shift is independent of the incident frequency and is strictly a characteristic property of the scattering molecule's internal energy levels.
6. In the classical polarizability theory, the induced dipole moment is expressed as $p = \alpha E$. Here $\alpha$ represents:
Explanation: In dielectric response theory, $\alpha$ is the molecular polarizability relating the induced dipole moment $p$ to the electric field $E$.
7. Which of the following processes resembles Raman scattering in being prompt, but differs by involving real electronic absorption?
Explanation: Fluorescence involves real absorption to an excited electronic state with a finite lifetime, unlike instantaneous Raman scattering via virtual states.
8. What is the unit of Raman shift when expressed in wave numbers?
Explanation: Raman shift in terms of wave numbers is typically measured in $\text{cm}^{-1}$ or $\text{m}^{-1}$.
9. For a rigid diatomic rotor, what is the selection rule for rotational Raman transitions?
Explanation: Rotational Raman spectroscopy follows the selection rule $\Delta J = \pm 2$ for linear rigid rotor molecules.
10. The expression for the moment of inertia of a diatomic molecule with reduced mass $\mu$ and bond length $r_0$ is:
Explanation: The moment of inertia for a diatomic system about its center of mass is given by $I = \mu r_0^2$.
11. Anti-Stokes lines originate from molecules that are:
Explanation: Anti-Stokes lines result when molecules in thermally excited states drop to lower energy levels, giving up energy to the scattered photon.
12. Which physical parameter links the rotational constant $B$ with the moment of inertia $I$?
Explanation: The rotational constant in wave numbers is defined as $B = \frac{h}{8 \pi^2 I c}$.
13. What kind of energy states are predominantly involved during the instantaneous scattering event in the Raman effect?
Explanation: Raman scattering is a two-photon process that proceeds through intermediate virtual energy states rather than real absorption levels.
14. If the incident light wave number is $\tilde{\nu}_0$ and the Stokes scattered wave number is $\tilde{\nu}_s$, the Raman shift is given by:
Explanation: The Raman shift in wave numbers is defined as $\Delta \tilde{\nu} = \tilde{\nu}_0 - \tilde{\nu}_s$.
15. Which molecular property must change during a vibration for the vibration to be Raman active according to classical theory?
Explanation: For a vibration to be Raman active, the polarizability of the molecule must change with respect to the normal vibrational coordinate ($\partial \alpha / \partial q \neq 0$).
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. In a Raman spectrometer, if the incident laser line has a wave number of $20480\text{ cm}^{-1}$ and a molecular vibration produces a Raman line at $19520\text{ cm}^{-1}$, what is the Raman shift?
Explanation: The Raman shift is $\Delta \tilde{\nu} = \tilde{\nu}_0 - \tilde{\nu}_s = 20480 - 19520 = 960\text{ cm}^{-1}$.
2. Which of the following statements regarding Rayleigh and Raman scattering is correct?
Explanation: Rayleigh scattering involves no change in photon frequency (elastic), whereas Raman scattering involves energy exchange resulting in modified frequencies (inelastic).
3. Why are anti-Stokes lines generally much weaker than Stokes lines in a normal Raman spectrum?
Explanation: Thermal Boltzmann population ensures that fewer molecules occupy excited vibrational states at room temperature, yielding weaker anti-Stokes scattering.
4. According to classical polarizability theory, the modulation of molecular polarizability by a vibration of frequency $\omega_m$ gives rise to scattered frequencies equal to:
Explanation: Substituting the time-varying polarizability into the dipole moment equation generates terms at the Rayleigh frequency $\omega_0$ and Raman sidebands $\omega_0 - \omega_m$ and $\omega_0 + \omega_m$.
5. In rotational Raman scattering of a diatomic molecule, the separation between adjacent Stokes lines corresponding to $\Delta J = +2$ transitions is approximately proportional to:
Explanation: The rotational Raman shift for Stokes lines is $\Delta \tilde{\nu} = 4B(J + \frac{3}{2})$, leading to a constant spacing of $4B$ between adjacent rotational Raman lines.
6. A common misconception is that Raman scattering is identical to fluorescence. The fundamental difference is that:
Explanation: Fluorescence features real electronic transitions with a measurable lifetime, whereas Raman scattering is an instantaneous scattering event through virtual states.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following diatomic molecules will exhibit a pure rotational Raman spectrum?
Explanation: Pure rotational Raman scattering requires an anisotropic polarizability change. Homonuclear diatomic molecules like $\text{H}_2$ lack a permanent dipole moment but possess anisotropic polarizability, making them active in rotational Raman scattering.
2. If the temperature of a sample is raised significantly, what happens to the ratio of intensities of anti-Stokes to Stokes lines ($I_{as} / I_s$)?
Explanation: Higher temperatures populate excited states more heavily according to Boltzmann statistics, increasing the relative population and thus the intensity ratio $I_{as} / I_s$.
3. In the classical theory of Raman scattering, if polarizability $\alpha$ does not change with the normal coordinate $q$ (i.e., $\partial \alpha / \partial q = 0$), then:
Explanation: For a vibration to be Raman active, the polarizability must change during the vibration ($\partial \alpha / \partial q \neq 0$). If it is zero, no Raman sidebands are generated.
4. What is the physical significance of the virtual state in Raman scattering?
Explanation: Virtual states are not real stationary eigenstates of the molecule; they represent short-lived polarization states induced during the instantaneous photon-molecule interaction.
5. Calculate the rotational Raman shift ($\text{in cm}^{-1}$) for a transition from $J = 2$ to $J = 4$ given a rotational constant $B = 2.0\text{ cm}^{-1}$ (using $\Delta \tilde{\nu}_{Stokes} = 2B(2J + 3)$ for initial $J$):
6. Which scattering phenomenon is primarily responsible for the blue color of the sky?
Explanation: Rayleigh scattering (elastic scattering by particles smaller than the wavelength of light) heavily favors shorter (blue) wavelengths, causing the blue sky.
7. The Boltzmann population ratio between an excited vibrational state and the ground state depends exponentially on:
Explanation: The Boltzmann factor governing thermal population is $\exp(-\Delta E / k_B T)$.
8. If an incident photon of wave number $25000\text{ cm}^{-1}$ undergoes Stokes Raman scattering corresponding to a vibrational frequency of $1500\text{ cm}^{-1}$, what is the wave number of the scattered photon?
Explanation: Stokes wave number is $\tilde{\nu}_s = \tilde{\nu}_0 - \nu_m = 25000 - 1500 = 23500\text{ cm}^{-1}$.
9. Which of the following molecules has a zero dipole moment yet exhibits vibrational Raman spectra?
Explanation: Carbon dioxide ($\text{CO}_2$) is symmetric and non-polar (zero dipole moment), but its symmetric stretch alters its polarizability, making it Raman active.
10. In Raman spectroscopy, why is a monochromatic and intense light source (such as a laser) typically required?
Explanation: Raman scattering is an extremely weak phenomenon (about $1$ in $10^6$ photons is Raman scattered), requiring an intense, monochromatic laser source for detection.