Dhristi JEE 2022-24

Introduction to Stokes Lines

The formation of stokes lines is an important part of the Raman spectrograph. We know that Raman spectra incorporate many important aspects regarding the scattering of light. According to Raman scattering, when a light ray is allowed to p[ass through a transparent medium, the particles of the molecules will scatter a part of light in all possible directions. Raman scattering is based on inelastic scattering of the monochromatic light, usually a LASER light in the UV region, near the Infrared region, and near the UV range.


A few parts of the scattered light have lower frequencies (higher wavelengths) than incident light, giving rise to the origin of the Stokes lines in a Raman spectrum. This occurs when a photon transfers part of its total energy to the molecule and, since it encounters a loss of energy, its frequency will be shifted towards lower frequencies. The molecule absorbs energy to undergo the transition to higher vibrational states. Thus, the stroke lines are those lines whose wavelengths are longer than that of the incident light. In this article, we will look into the concept of stokes line with comparison to anti-stokes lines, detailed explanation for stokes & anti stokes lines, calculating stokes and antistokes, and Raman frequency. So, without any further ado, let us understand the stokes & anti-stokes lines.


Stokes and Antistokes Lines

A typical Raman spectrum is centered at the frequency of incident light (normally a laser light in the visible part of the electromagnetic spectrum), with very high intensity owing to Rayleigh scattering. Stokes lines can be observed towards the lower frequencies and anti-Stokes can be seen towards the higher frequencies, later they are mirrored at the center of the Raman spectrum. Nevertheless, Stokes lines are more intense in comparison with the anti-Stokes counterparts, because the vibrational ground state is more populated than excited states. Let us understand the anti-stokes line and stroke lines in Raman spectra in detail as follows.


When a beam of monochromatic light is passed through the transparent liquid or gasses known as the scatterer, a small fraction of it is scattered due to the collision between the molecules of the scatterer and the photons of the light. It is explained by assuming that an energy level is formed like a complex between the molecule and the photon during a very short (About ~10⁻15 s interval of collision. 


The photons have energy hv, which is insufficient for molecules to make a transition from the ground electronic state with low vibrational and rotational states to the higher electronic state.


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Two cases are noticed depending upon whether the collision between the molecules and the photons are elastic or inelastic.


1. If the Collision is Elastic:

Suppose that the collision between the molecules and the photons is an elastic collision, then the total energy of the molecules and the photons will remain unchanged. This explains that the molecule will not get scattered completely without absorbing photon energy. In other words, the collision will merely induce forced oscillations in the molecule. This will result in unmodified lines and this kind of scattering of monochromatic light is known as the Rayleigh scattering. 


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2. If the Collision is Inelastic:

If the collision between the molecule and the photon is inelastic in nature then, there will be the exchange or transfer of energy between the molecule and the photons. This case is further studied by subdividing into two special cases that will lead to the explanation of the formation of stokes and anti-stokes lines.


1. The molecule is in the lower vibrational and rotational level before the collision and is present in any one of higher rotational- vibrational states of the same level (as allowed by selection rules) after Excitation and emission of a photon. Thus, the molecule absorbs or gains energy equal to the difference in the two energy levels involved. Because the energy is supplied by a photon of the incident light, the frequency of the emitted photon is less. The above explanation can be understood by the following diagram given below.


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2. Before the collision, the molecule is present in one of the excited vibrational and rotational energy states but after the emission of a photon, it occupies one of the lower vibrational energy states of the same level according to the selected role. Since the molecule imparts some of its intrinsic energy to the incident, thus the incident photon has higher energy than the incident photon. The above explanation can be understood by the following diagram given below.


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These two special cases constitute the Raman scattering and hence gives an idea about Raman spectra. Such scattering was observed by Sir C V Raman in 1928. 


Raman Frequency

The frequency of the scattered light or the emitted photon which either lower or higher than the frequency of the incident photon is known as the Raman Frequency, denoted by vR


and it is given by the formula:

⇒ v\[_{R}\] = v\[_{0}\] + \[\frac{\epsilon' - \epsilon'' }{h}\] ……….(1)




\[ v_{0}\] - The frequency of the incident photon


\[ v_{R} \] - The frequency of scattered photon or Raman frequency


\[ \epsilon’\] - The energy of the molecule before the collision or before impact

\[ \epsilon’’ \] - The energy of the molecule after a collision or after impact

Now, if:

  • \[ \epsilon’\] = \[ \epsilon’’ \] then we have \[v_R = v_0\], This explains the Rayleigh scattering and formation of unmodified lines. 

  • \[ \epsilon’\] > \[ \epsilon’’ \], then we have \[v_R < v_0\], i.eThe frequency of scattered light is less than the frequency of the incident light, this results in the formation of Stokes lines in Raman spectra.

  • \[ \epsilon’\] > \[ \epsilon’’ \], then we have  \[v_R > v_0\], i.eThe frequency of scattered light is more than the frequency of the incident light, this results in the formation of Anti Stokes lines in Raman spectra.


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Anti Stokes Lines

Anti Stokes lines are those modified lines obtained with wavelengths shorter than the incident line. It is possible for the molecule to transfer energy to the incident photon and shift the frequency of scattered light to higher values (lower wavelengths). When this occurs, the molecule that is in an excited vibrational state undergoes the transition to the lower state or even to the ground state. A set of lines occurs in a Raman spectrum called anti-Stokes lines, owing to this process.


When the molecule makes a transition with ΔJ = -2, the scattered photon emerges with increased energy. These transitions account for the anti-Stokes lines of the Raman spectrum. The anti Stokes lines appear at displacements of 6B, 10B, 14B...for J = 2,3,4…. And the high-frequency side of the incident radiation. Note that J = 2 is the lowest state that can contribute to the spectrum under the selection rule ΔJ = -2.


Did You Know?

Sir C V Raman won the Nobel prize for his extraordinary contribution to the theory of scattering and he won the Nobel prize for Raman scattering. Sir C V Raman was the first Indian scientist to win the Nobel prize in his honor every year on February 28th National science day is celebrated. The Raman scattering experiment was first demonstrated on February 28, 1928.

FAQs on Stokes Lines

1. Why are the Strokes Lines More Intense than the Anti-stokes Line?

Even though Raman scattering is observed at very low intensity, the Stokes line's scattered radiation is more intense than the anti-Stokes scattered radiation. The reason for this is that very few molecules would exist at the exciting level as compared to the ground state before the absorption of radiation.

2. Which Lines are More Intense Among the Stokes Line and the Anti-stokes Line?

Stokes lines are considerably more intense than the anti-stokes lines.

3. How does increasing the temperature affect the intensity of Stokes and anti-Stokes lines?

The population of the ground state would decrease as the temperature rose, while the number of higher energy vibrational states increased. As a result, as the temperature rises, the intensity of the Stokes lines decreases while the intensity of the anti-Stokes lines rises. The Stokes lines, on the other hand, would have a higher intensity than the anti-Stokes lines.


The scattered photons are measured at 90o to the incident radiation since scatter happens in all directions. Furthermore, Raman scattering is a generally unfavorable process that results in a weak signal.

4. Among the Stokes and anti-Stokes, which set of lines is weaker?

The anti-Stokes lines are much weaker than the Stokes lines. The reason is, there are many more molecules in the ground state as compared to excited vibrational states.

5. Why is the intensity of Stoke lines more than that of anti-Stokes lines?

The intensity of spectral lines is always proportional to the number of atoms or molecules present in any given condition.


Stoke lines in Raman spectra are mainly caused by the excitation of molecules that are in the ground state. Anti-Stroke lines, on the other hand, are formed when molecules in the excited state are de-excited and returned to the ground state. In general, the concentration of molecules in the ground state is higher than the concentration in the excited state. As a result, the Raman scattering mechanism produces more intense Stoke lines than anti-Stroke lines.

6. Why does the anti-stokes to stokes intensities ratio increase with the sample?

As the temperature rises, the number of molecules at the first vibrational level of the ground state increases. So, as the temperature rises, the ratio of anti-stokes to stokes intensity rises as well.

7. Why are anti-stroke lines less intense?

The anti-stokes lines are less intense than the Stokes line. The anti-Stokes line can only be produced by molecules that have been vibrationally stimulated prior to irradiation.

8. What kind of source would be best for measuring Raman spectra?

The more incident photons that enter the sample, the more likely the molecules will be in the proper virtual state for Raman scattering to occur. Because the signal is measured with no background, a high-power source is required. As a result, a laser would be a better choice for measuring the Raman spectra. Because a laser's emission is highly monochromatic, we can more precisely determine the frequency of the Stokes lines in the resulting spectrum. An array detector is also preferable since it enables simultaneous monitoring of all scattered radiation.

9. What are the basic principles of Raman Spectroscopy?

The chemical and structural information is provided by the shift in wavelength of inelastically scattered radiation. Depending on the vibrational state of the molecule under study, Raman shifted photons might be either of higher or lower energy. 


Stokes radiation has a lower energy (longer wavelength) than Rayleigh radiation, while anti-Stokes radiation has higher energy (shorter wavelength). The observed Raman shift of the Stokes and anti-Stokes features is a direct measure of the molecule's vibrational energies since The rise or decrease in energy is proportional to the vibrational energy levels in the molecule's ground electronic state.


For the Stokes line, the dispersed radiation has less energy than the incident radiation, and for the anti-Stokes line, the scattered radiation has more energy than the incident radiation. The vibrational energy gap in the ground electronic state of the molecule is related to the energy increase or decrease from excitation, hence the wavenumber of the Stokes and anti-Stokes lines are a direct measure of the molecule's vibrational energies.


We hope this article on Stoke lines was helpful to the students in gaining knowledge. The article is made by subject experts with thorough knowledge. The students got to understand stroke lines, anti-stroke lines, & Raman frequency. These concepts are important in physics. If students understand them properly, then they can score well in their exams. For more such important concepts, visit the website of Vedantu. If you have any questions or suggestions, please leave them in the box below. We will surely reply to your questions.