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Why pure metals cannot show raman bands?
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Lou Heaton
Why pure metals cannot show raman bands?
Dear Shradhanjali, It is not fully true that metals in general don't have first-order Raman bands. This is a structure/symmetry question: if the primitive unit cell (smallest translational unit cell) contains only a single atom, the material will only have so-called acoustic phonons. This is the case for fcc and bcc pure metals (e.g. gold, silvel, aluminum, alpha iron, chromium etc. etc.). If the primitive unit cell contains at least two atoms, optical phonons will also exist. This is the case for hcp metals (zirconium, magnesium, alpha titanium etc.). In Raman spectroscopy, only optical phonons are observed, acoustic phonons have practically zero frequencies in the long-wavelength limit. Therefore, fcc/bcc metals do not show Raman spectra. In contrast, hcp metals do actually have a Raman spectrum, though it is experimentally demanding to observe. See e.g. http://journals.aps.org/prb/pdf/10.1103/PhysRevB.56.10751.
Dear Shradhanjali, It is not fully true that metals in general don't have first-order Raman bands. This is a structure/symmetry question: if the primitive unit cell (smallest translational unit cell) contains only a single atom, the material will only have so-called acoustic phonons. This is the case for fcc and bcc pure metals (e.g. gold, silvel, aluminum, alpha iron, chromium etc. etc.). If the primitive unit cell contains at least two atoms, optical phonons will also exist. This is the case for hcp metals (zirconium, magnesium, alpha titanium etc.). In Raman spectroscopy, only optical phonons are observed, acoustic phonons have practically zero frequencies in the long-wavelength limit. Therefore, fcc/bcc metals do not show Raman spectra. In contrast, hcp metals do actually have a Raman spectrum, though it is experimentally demanding to observe. See e.g. http://journals.aps.org/prb/pdf/10.1103/PhysRevB.56.10751.
Raman intensity is given by the change of the dielectric constant (for IR) with respect to phonons, but metals do not have finite dielectric permittivity for frequencies below the plasma frequency.
Raman intensity is given by the change of the dielectric constant (for IR) with respect to phonons, but metals do not have finite dielectric permittivity for frequencies below the plasma frequency.
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)
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)
A practical advice: be careful of metal passivation. It's possible to obtain a Raman spectra of a "pure" aluminium, cooper... sample. But that sample is covered by a thin, natural layer of oxide. And Raman is a "superficial" technique, so it analyses the composition of the sample surface. It's possible to mistake (I did it, ups) the metal oxide Raman spectra with the (no-existing) Raman spectra of the metal...
A practical advice: be careful of metal passivation. It's possible to obtain a Raman spectra of a "pure" aluminium, cooper... sample. But that sample is covered by a thin, natural layer of oxide. And Raman is a "superficial" technique, so it analyses the composition of the sample surface. It's possible to mistake (I did it, ups) the metal oxide Raman spectra with the (no-existing) Raman spectra of the metal...
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...
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...
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.
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.
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.
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.
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.
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 Mrs. Oksana, I think your oxides are to thin for a regular microRaman device. Focusing with a 100x objective on the sample, a few micrometers of sample thickness would be enough to get its Raman spectrum.
Hello Mrs. Oksana, I think your oxides are to thin for a regular microRaman device. Focusing with a 100x objective on the sample, a few micrometers of sample thickness would be enough to get its Raman spectrum.
Dear Shradhanjali,
It is not fully true that metals in general don't have first-order Raman bands. This is a structure/symmetry question: if the primitive unit cell (smallest translational unit cell) contains only a single atom, the material will only have so-called acoustic phonons. This is the case for fcc and bcc pure metals (e.g. gold, silvel, aluminum, alpha iron, chromium etc. etc.). If the primitive unit cell contains at least two atoms, optical phonons will also exist. This is the case for hcp metals (zirconium, magnesium, alpha titanium etc.). In Raman spectroscopy, only optical phonons are observed, acoustic phonons have practically zero frequencies in the long-wavelength limit.
Therefore, fcc/bcc metals do not show Raman spectra. In contrast, hcp metals do actually have a Raman spectrum, though it is experimentally demanding to observe. See e.g. http://journals.aps.org/prb/pdf/10.1103/PhysRevB.56.10751.
Dear Shradhanjali,
It is not fully true that metals in general don't have first-order Raman bands. This is a structure/symmetry question: if the primitive unit cell (smallest translational unit cell) contains only a single atom, the material will only have so-called acoustic phonons. This is the case for fcc and bcc pure metals (e.g. gold, silvel, aluminum, alpha iron, chromium etc. etc.). If the primitive unit cell contains at least two atoms, optical phonons will also exist. This is the case for hcp metals (zirconium, magnesium, alpha titanium etc.). In Raman spectroscopy, only optical phonons are observed, acoustic phonons have practically zero frequencies in the long-wavelength limit.
Therefore, fcc/bcc metals do not show Raman spectra. In contrast, hcp metals do actually have a Raman spectrum, though it is experimentally demanding to observe. See e.g. http://journals.aps.org/prb/pdf/10.1103/PhysRevB.56.10751.
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Raman intensity is given by the change of the dielectric constant (for IR) with respect to phonons, but metals do not have finite dielectric permittivity for frequencies below the plasma frequency.
Raman intensity is given by the change of the dielectric constant (for IR) with respect to phonons, but metals do not have finite dielectric permittivity for frequencies below the plasma frequency.
More
VOTE
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)
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)
More
VOTE
A practical advice: be careful of metal passivation. It's possible to obtain a Raman spectra of a "pure" aluminium, cooper... sample. But that sample is covered by a thin, natural layer of oxide. And Raman is a "superficial" technique, so it analyses the composition of the sample surface. It's possible to mistake (I did it, ups) the metal oxide Raman spectra with the (no-existing) Raman spectra of the metal...
A practical advice: be careful of metal passivation. It's possible to obtain a Raman spectra of a "pure" aluminium, cooper... sample. But that sample is covered by a thin, natural layer of oxide. And Raman is a "superficial" technique, so it analyses the composition of the sample surface. It's possible to mistake (I did it, ups) the metal oxide Raman spectra with the (no-existing) Raman spectra of the metal...
More
VOTE
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...
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...
More
VOTE
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.
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.
More
VOTE
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
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
More
VOTE
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.
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.
More
VOTE
Thomas Mayerhöfer I interpreted the way the question was formulated as referring to ideal conductors
Thomas Mayerhöfer I interpreted the way the question was formulated as referring to ideal conductors
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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.
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.
More
VOTE
Try something simple first, like a silicon wafer (single peak at 520.17cm-1) or polystyrene.
Try something simple first, like a silicon wafer (single peak at 520.17cm-1) or polystyrene.
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Hello Mrs. Oksana, I think your oxides are to thin for a regular microRaman device. Focusing with a 100x objective on the sample, a few micrometers of sample thickness would be enough to get its Raman spectrum.
Hello Mrs. Oksana, I think your oxides are to thin for a regular microRaman device. Focusing with a 100x objective on the sample, a few micrometers of sample thickness would be enough to get its Raman spectrum.
More
VOTE