Hello! Why the width of bands in the FTIR absorption spectra of minerals is always greater than that in the Raman spectra?
If you talk about FTIR spectra of minerals, I suppose you mean reflectance spectra? In contrast to transmittance spectra, the band width of the former is proportional to the oscillator strength. The onset of the band on the low wavenumber side is usually close to the oscillator position, and the inflection point at the high wavenumber side is the maximum of the negative imaginary part of the inverse dielectric function (the so-called longitudinal optical mode position, cf.Preprint Understanding longitudinal optical oscillator strengths and mode order
). The latter is the more blueshifted the stronger the oscillator is. For stronger oscillators with reflectance close to unity, the bands are called restrahlenbands. Transmittance spectra (powder in KBr) usually feature bands with much smaller halfwidths situated in between the inflection points of the reststrahlenband.
Dear Tomas, thank you for your answer. I mean, why in absorption (in KBr) some bands (v3 in carbonates, or asymmetric stretch in SiO2) is so broad? In IR absorption of tiny plates these bands are sharp.
In thin films transmission is mostly depending on the damping constant, which is in a first approximation the halfwidth at full maximum. The thicker the films are, the more reflectance sets in and dictates the halfwidth. In KBr, strong bands lead to the problem that light often does often not even penetrate into the particle. Therefore you often have the bulk modes plus Fröhlich modes. In particular if you have a distribution of particle sizes and shapes, those strong bands lead to the same broad features like in reflectance spectra...
Dear Tomas! Does it mean IR absorption spectra in KBr do not coincide with the averaging of spectra in all orientations of the thin plate?
In general, the spectra of non-oriented films in transmission are definitely different from those in KBr. Actually, in addition to the reasons I already discussed above, two more things need to be considered. For inorganic materials you need extremely thin films, otherwise a band generates additional minima in the transmittance, partly due to interference effects:
Article Messungen am NaCl und KCl im Spektralbereich ihrer ultrarote...
Secondly, the dielectric function of the embedding material (KBr, CsI etc.) shifts absorption bands and also change somewhat their shapes: Article Infrared study of surface phonon modes in α-Fe2O3 microcrystals
Dear Tomas! Here are spectra of calcite CaCO3 in KBr (blue) in reflection (red) and of thin (approx 0.01mm) plate . What is the main reazon for broadening of v3 (1433 cm-1) line in KBr? It seems to be the new world, if use thin plates instead of popular KBr?
It seems that the particle size distribution is responsible for the broadening of the v3 together with the very high oscillator strength. See e.g. Article EXPRESS: On the Widths of Bands in the Infrared Spectra of Oxyanions
(but the theoretical treatment in this paper neglects some things like broadening through orientational averaging). The only valuable method for such materials is to evaluate the reflectance spectra of single crystals by dispersion analysis like this was done in the papers under e.g. https://www.researchgate.net/project/Infrared-spectroscopy-and-optics-of-anisotropic-and-randomly-oriented-materials
I believe that the difference in the width of IR absorption and Raman features a general phenomenon, which we highlighted at the end of this paper, comparing the spectra of the same skin samples Figure 8. We try to discuss this in terms of the differences between electric dipole transitions (IR) versus a scattering process (Raman) although not in a robust theoretic approach.
At the time, I could not find any literature discussing this, although noted that the width of resonantly enhanced Raman features are generally broader (D.A. Long, Raman Spectroscopy).