I'm new to all things Spectroscopy so be gentle :-). I'm puzzled to find the difference between fluorescence and most variations of Raman scattering. They both involve exciting atoms from their ground state to some excited state, then fluoresce back down to a ground vibrational state, emitting radiation of lesser energy than the former (at least for Stokes Raman).
1. What am I missing?
2. Should I assume the difference in energy between initial and final radiation is lost in the vibrational state, so as to conserve energy?
3. Yeah... and the virtual energy states involved in Raman scattering? That's just too weird for now.
So, in simple English, what are they really? Wiki explanations are so over the top to be readily understood by a newbie like me.
Thanks,
David
Raman spectroscopy is all about inelastic scattering. If the collision between a photon and a molecule is perfectly elastic, there will be no exchange of energy, giving the Rayleigh line. But, inelastic scattering is associated with a collision which causes some exchange of energy, equal to the difference b/w 2 allowed states of the molecule, and thus it makes a frequency-shift of the scattered photon.
In the fluorescence emission process, a molecule is excited from the ground state (E0) to one of the vibrational states in the electronic excited state (E1) ac to the Franck-Condon Principle. Through vibrational relaxation, the molecule relaxes back to the lowest excited state by losing excess energy as heat. Then, the molecule transits from the lowest excited state to one of the vibrational states in E0 with the emission of a photon. The energy of the emitted photon is lower than that of the incident photon because of the energy loss during vibrational relaxation. The fluorescence emission spectrum is broad band covering the same wavelength band as the Stokes Raman signal. Since the cross section for fluorescence is larger than for Raman scattering, detection of Raman scattering is very difficult when strong fluorescence emission is present.
Raman spectroscopy is all about inelastic scattering. If the collision between a photon and a molecule is perfectly elastic, there will be no exchange of energy, giving the Rayleigh line. But, inelastic scattering is associated with a collision which causes some exchange of energy, equal to the difference b/w 2 allowed states of the molecule, and thus it makes a frequency-shift of the scattered photon.
In the fluorescence emission process, a molecule is excited from the ground state (E0) to one of the vibrational states in the electronic excited state (E1) ac to the Franck-Condon Principle. Through vibrational relaxation, the molecule relaxes back to the lowest excited state by losing excess energy as heat. Then, the molecule transits from the lowest excited state to one of the vibrational states in E0 with the emission of a photon. The energy of the emitted photon is lower than that of the incident photon because of the energy loss during vibrational relaxation. The fluorescence emission spectrum is broad band covering the same wavelength band as the Stokes Raman signal. Since the cross section for fluorescence is larger than for Raman scattering, detection of Raman scattering is very difficult when strong fluorescence emission is present.
Fluorescence follows an electric dipole transition from a ground to an excited electronic state. The absorption probability/cross-section is described by the imaginary component of the polarisability (absorption). Excitation only occurs at wavelengths which are resonant with the transition. In principle, the electronic states can be of an atom, a molecule, or a solid. A fluoresence transition has a an associated lifetime .
Raman is an inelastic scattering process, which depends on the polarisability of a a bond (therefore does not happen in an atom) which is related to the real component of the polarisibility (refractive index). It is an instantaneous process, which results in the coupling of energy to/from a photon to/from molecular vibrational states. It is sometimes represented as a transition to an intermediate "virtual state", which by definition does not exist, and has a 0 lifetime.
Fluorescence involves the relaxation of the vibration, while Raman is sensitive to different vibrational modes. Therefore Raman is more sensitive than fluorescence and is called the fingerprint of the molecules. ... IR sensitive requires µ changes as the molecule vibrates, while Raman requires the change
Welcome to spectroscopy!
I also had to learn these things as part of my graduate studies.
The answers from Hugh J Byrne and Souvik Bhattacharjee are very good so I will only answer briefly, so you also have a "short-answer":
1. In Raman the molecule is only excited to a virtual state, it doesn't make the "full leap" to excited. It is easier to imagine it as "stretched" rather than excited.
There is no "absorbance". And yes, it's very weird.
2. Because of (1), the Raman effect is much rarer, meaning the signals are usually orders-of-magnitude smaller.
Which is why, if you do Raman spectroscopy, fluorescence signals are usually hard to get rid of.
I highly recommend these two books:
Reading the first chapter of each will cover you very well about the theory of how and why Raman and Fluorescence spectroscopies work.
- Larkin 2011: Infrared and Raman Spectroscopy
https://www.sciencedirect.com/science/article/pii/B9780123869845100011
- Lakowicz 2006: Principles of Fluorescence Spectroscopy https://link.springer.com/chapter/10.1007/978-0-387-46312-4_1
In simple words, in Fluorescence exciting photons as well as emitted photons have exactly same wavelengths. However in Raman scattering the emitted photons have slightly different wavelengths which is usually given by wavenumber difference.
Vishal Kaushik Fluorescence emission is shifted from the absorption by the Stokes shift - this is due to nuclear relaxation in the excited state.
Hi! I will just add with the practicalities of the differences that are very well explained in previous answers (Souvik Bhattacharjee and Hugh Byrne) :
- As Hugh Byrne said, fluorescence involves an absorbance and emission processes, whereas Raman is instantaneous (scattering process), meaning for instance that if you only acquire for a very short period after excitation, you can obtain only Raman and not fluorescence, for example. Fluorescence signal is also way stronger than Raman as it is more likely to happen.
- As you said, the virtual state is super weird, and that is just because it is a way of explaining scattering processes in Jablonsky diagrams, but it actually DOES NOT EXIST. This means that the energy between the base and the virtual state is not fixed, and you can choose to do Raman with different excitation lasers! They will all give you (more or less) the same signal (subject to other considerations). If you choose a laser light so that the virtual state falls close to an excited state, then you can increase the probability of the process and increase your signal intensity (Resonant Raman).
Wish you luck with your spectroscopy experiments!
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