I am newly joined JRF and my research project is based on high pressure raman studies. I have just started reading about raman spectroscopy, I read that pure metals cannot show raman bands.
Dear Shradhanjali,
if you speak from Raman bands, you probably referring to the vibrational Raman effect (This is what usually is ment by using the word 'Raman'.). If you have an atom like in a pure metal you can not have vibrations. I hope that helps.
Best,
Thomas
For a material to show Raman bands, a change in polarization is required which is not quite possible in metals.
An interesting question - my first guess would be that the light is absorbed within the skin depth of the surface, and the plasmon/polaritons excited have little or no coupling to the lattice phonons and therefore there is no Raman scattering.
Because you need a change in the polarization of the molecules. :)
A change in polarizability during molecular vibration is an essential requirement to obtain Raman spectra of samples. The metals do not show the polarizability chnage and hence no Raman spectra are obtained.
I am trying to obtain Raman Spectra of a TiO2 layer which seems to be possible from literature. But I am not sure why that is possible but other metallic samples aren't. We are struggling to get useful data from our Raman spectroscope attached to a MIRA RISE SEM.
Try something simple first, like a silicon wafer (single peak at 520.17cm-1) or polystyrene.
Maybe your layer is too thin? Then you could try interference-enhanced Raman spectroscopy: Article Interference-Enhanced Raman Spectroscopy as a Promising Tool...
Article Interference-enhanced Raman scattering of F 16 CuPc thin films
Davide Donadio You are talking about a perfect conductor. Real metals have finite dielectric permittivites... here are the ones for gold which comes closest to the perfect conductor...
Aurelio Sanz Arranz how thick should be this oxide layer to be visible in Raman spectrum? I tried to get spectrum from oxides on pure Ti and CoCrMo alloy, but no success. Their oxides are about 4-10 nm.
Hello Dr. Tamás Váczi I am recently studying Raman effect. To follow up with your answer above: that the presence of optical branch is important to observe a Raman peak(s).
Why transitions between different acoustic branches cannot result in a Raman peak? For an example, I found the following dispersion relation for a BCC Fe (source: X. Sha and R. E. Cohen. (2006). Phys. Rev. B 73, 104303). At x~0.5 cm-1 (red line), we can see that there are LA and TA with gap of 50 cm-1. Can we say that this crystal Raman active, at least in principle?
Also, when people say that a crystal is Raman INACTIVE, do they refer to the case when the energy (frequency shift) is too small to be probed by the real instrument?
Thank you very much. I really appreciate any suggestions from you.
Galih R. Suwito I can only answer your second question. "Raman inactive" is something that results from symmetry analysis of the crystal and, for a perfect crystal, there is then no signal for whatever instrument you imagine (remember the Raman selection rule!). NaCl is a typical example... (see, e.g., https://arxiv.org/ftp/arxiv/papers/1903/1903.11824.pdf)
Thomas Mayerhöfer Thank you very much for your clarification on the terminology of "Raman inactive".
Also, I think I have got an answer to the prior question, I asked to Tamás Váczi (the first question). In the dispersion relation, only transitions near Brillouin zone (BZ) center can be probed. And at this BZ center, the gap between acoustic branches goes to zero. Hence, no Raman signal is generated from this "acoustic gap".
Let's put some values. We can consider a case of Raman spectrometer with blue-green laser light (~20,000 cm-1). Then, the generated crystal momentum would be ~2.5 E5 cm-1. This momentum is orders of magnitude different from the wave number at the BZ edge (~1 E8 cm-1).
Pure metals are highly reflective due to the presence of free electrons on their surface. As a result, the penetration of visible light (often the laser wavelength used in Raman) is very limited and thus the scattered signal is very weak. As Thomas mentioned, Raman inactive bands are a different story and refer to crystal symmetry (high symmetry such as NaCl). Even in such crystals though, there is second order Raman scattering taking place. Raman signal in metals can be very weak to observe but not necessarily inactive due to symmetry reasons, as in the case of NaCl for example.
Tamás Váczi Thanks so much for checking this. A very interesting phenomenon by itself.
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