I am researching BaTiO3 ceramics and i find peaks in the wavenumber [1000-2000cm-1].How can we separate between the fluorescent effect and Second-Order vibration in the wavenumber [1000-2000cm-1] in Raman Spectroscopy?
you must compare spectra recorded with different lasers.
Features shifted are not Raman scattering
Technically you could thing to time-resolved (time-gated) detection but both signals (fluorescence and Raman emission) are fast so that you need relatively sophisticated instrumentation.
To flag the Raman 2nd order emission bands, simply change the Raman excitation wavelength. Fluorescence signals will not shift, whereas Raman emission bands will, proportionally to the change in excitation energy (i.e. excitation wavelength expressed in cm-1), which is what Philippe Colomban was saying.
1. Raman peaks will not change if you change the excitation wavelength, but features associated with fluorescence will shift and change as a function of the excitation source wavelength.
2. Raman peaks follow Raman selection rules, so using an analyzer one may test how some of the lines may disappear as orientation of analyzer is changed.
3. Raman peaks are sharper and "usually" symmetric in nature, while peaks associated with fluorescence are much broader (i.e., FWHM is much broader than Raman peaks)
Sorry but I do not concur with 1. Raman is an inelastic scattering where the emitted photons have the energy of the excitation frequency minus the energy of a vibration. Therefore changing the excitation frequency will shift the bands accordingly. On the other hand, fluorescence is emitted from an electronic excited state. Provided the same state remains excited, shifting the excitation frequency will not affect the position of the fluorescent band, but only its intensity (reflecting the absorption cross-section of the excited state).
On the other hand, point 2 is indeed a clever way of distinguishing Raman bands, provided that the symmetry of the material is not too low.
Dear Jean-Claude G Bünzlito reiterate my comment for item 1 above, refers to the fact that Raman shifts represent the change in energy of the photon due to creation or annihilation of Phonons or hf = h f (incident) +/- h f (scattered). Therefore, Raman peaks show a constant shift with respect to the laser line, regardless of the laser frequency used as the excitation source, because Raman spectra represent conservation of energy and momentum that represent the energy of the phonon created or annihilated.
For item 2 in my previous comments, I can provide a reference on polarized Raman Scattering such as:
Uma Ramabadran, and Bahram Roughani, “Intensity Analysis of Polarized Raman Spectra for off Axis Single Crystal Silicon”, Mat. Sci. and Eng. B, 230, pp 31–42 (April 2018). https://doi.org/10.1016/j.mseb.2017.12.040. Also, a featured review about this paper entitle; “Shining light on the off-axis single crystalline Si properties” was published at Advances in Engineering (https://advanceseng.com/shining-light-off-axis-single-crystalline-si/).
Dear Bahram Roughani, thank you, I perfectly agree. You are referring to the Raman shift, while I am referring to the observed frequency if the Raman band.!
With a 266 nm laser, you can separate the Raman from fluorescence given that fluorescence start to appear after 300 nm (4200 cm-1 Raman Shift).
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