# Tag Info

## Hot answers tagged raman

9

I think you are referring to a handheld prism based spectroscope with an eye-view/eyepiece. Using such spectroscope one does not observe Raman signal but only observe the dispersed light coming from an object giving information about the absorption characteristics of the object or emission characteristics (for light sources). The following is such device (...

9

Sir Martyn Poliakoff did the video about the spectroscope on his Periodic Videos channel on youtube. To my knowledge, he presented the original instrument but did not mention where to obtain a replica.

7

The main difference between Raman scattering and fluorescence is the excited state lifetime. Fluorescence excited states are longer-lived than the 'virtual' states associated with Raman scattering. In fluorescence, absorption of light excites an electron to a higher energy state. The lifetimes of these excited states are long enough where the geometry of ...

7

No, there is no automation for this, and there should not be. Using compound jobs is unfortunately very common among computational chemists, but they should always be viewed critically. If you are using any kind of IOp, then the probability that the second part of your job is doing what you want is low. Additionally, you do not really have any control about ...

5

When you perform a IR or Raman calculation, the primary step is the diagonalization of the mass-weighted Hessian: \begin{align} H_{ij,ab}^{\mathrm{mw}} &= \frac{1}{\sqrt{m_{i}m_{j}}} \frac{\partial^{2} E}{\partial a_{i} \partial b_{j}} \\ \mathbf{Hq} &= \lambda \mathbf{q} \end{align} where $H_{ij,ab}^{\mathrm{mw}}$ is a $3N\times3N$ Hermitian ...

4

As EdV said, when you have aqueous samples, Raman spectroscopy is the way to go. The problem is not the container (which should not be glass etc),the problem is water. Its absorption will hide the absorption of everything. If you ever open an FTIR instruments, you will see that there is nothing which is made of glass or quartz because they absorb too ...

4

I've been doing some digging and I think I have a possible answer so I'm gonna post it and if I'm misinterpreting information, I'm open to being told so! As I said in my comment, this vibration has to be coming from some dimer which is formed because if one of these modes were the $\ce{HF}$ fundamental it would be up around $4000\ cm^{-1}$. So, I found ...

3

Afaik it's not possible to do this directly since Gaussian does only allow geometry optimization during scans. The definitely easiest way to do this, but not the cheapest, would be running your scan, extracting the *.xyz coordinates of every (optimized) structure and then running the freq calculations separately. Extracting the *.xyz structures can be done ...

3

Since they're using ACN/ammonium acetate to elute from solid phase extraction cartridges and that's what they're injecting into their LC analysis, they probably want to use the same mixture to avoid problems with unmatched matrices with their standards. The ammonium acetate is the pH buffer. Malachite green has different protonation states at different pH, ...

3

okay, old question, but: recall that the virtual state can always be decomposed as a sum over all real states, and the contribution of every real state to the superposition is determined by the matrix element and the difference in energy from the real state to the photon energy; usual perturbation-theory sort of equation where you divide by energy; by the ...

2

There is no law that a transition must go to the ground state. Now, each molecule has a multitude of electronic states, and within each of them a multitude of vibrational states. Thus each transition between two pure electronic states is accompanied by a variety of related transitions between higher-to-lower and lower-to-higher vibrational states. Which of ...

2

You have asked question about vibrational modes and functional groups with respect to Raman spectra. General information: Specific vibrational modes are deduced by normal-coordinate analysis. This gives most probable observable frequency for certain vibration (or vice-versa). Normal coordinate analysis is core idea in spectral interpretation and tables ( ...

2

Chemists have tables and tables and tables that correlate functional groups to numbers. For example: http://www2.ups.edu/faculty/hanson/Spectroscopy/IR/IRfrequencies.html If we see a number on the graph, a certain type of functional group may be present. The sharp one at 3056 tells me aromatic hydrogen. The 2919 is the benzylic hyrodgens. Since I'm a ...

2

The reaction is given here. $$\ce{S + 2H2SO4 → 3SO2 + 2H2O}$$ Sulfur react with sulfuric acid to produce sulfur dioxide and water. Sulfuric acid should be concentrated solution. The reaction takes place in a boiling solution. $\ce{[S4]^2+}$, $\ce{[S8]^2+}$ or $\ce{[S19]^2+}$ is formed when sulfur is dissolved in oleum. The following is a paraphrase of an ...

2

In a review (PDF, CalTech) of the properties of elemental sulfur, the author notes that the most stable form of sulfur at STP is in fact cycloocta-sulfur, $\ce{S8}$, with other polyatomic species also found. Elemental sulfur is commonly formed as a molecular solid, with arrangements of $\ce{S_n}$ rings or chains, and the S-S bonds can be Raman active. The ...

2

Not in any way that you can use in practice for getting a working predictive model (= data analysis procedure). Depending on the molecular symmetry of your analyte(s), infrared and Raman spectra may share some bands, or not: a vibration can be infrared-active, Raman-active, or both. Near-infrared spectra show combinations and overtones of the infrared-...

2

Thank you for attaching your input file. The first place to start for improving these calculations is your choice of basis set. 6-31G is a rather small basis set and is likely a source of significant error here. If you're fond of using the Pople type basis then you would be better off using something like 6-31++G* so that you include polarization and diffuse ...

2

First of all, you are extremely lucky in asking this question for cyclohexane: cyclohexane serves as reference material for checking your intensity calibration, so it is one of the few substances for which integrated band intensity ratios are available in the literature. Integral intensity 2567 - 3068 cm$^{-1}$ is higher than integral intensity 700 - 900 (i....

1

from theory this should work what you have scatched here. When it comes to SERS a main point is, that you got a surface which is rough enough that you can create mini hot-spots for the electric field. This will give you the enhancement. Maybe you must put a little bit of work in you set up. The scattering is needed to arrive at the detector. maybe you try a ...

1

The starting point is to look at the potential point groups of the two molecules and work out if there are common vibrational mode symmetries between IR and Raman spectra, assuming that you are able to obtain both. In the 3rd and 4th columns of the point group are the $x, \,y, \,z$ and $xy, \,xz$ etc operators. The $x, \,y, \,z$ operators transform as does a ...

1

I guess for homogeneous read homo-nuclear. He is trying to explain how the electric field of the radiation induces a dipole by shoving the electrons around, hence reference to polarisability: this relates to the standard description. This induced oscillating field then radiates, hence scattering. The electron and proton centre is when the field is zero, i.e. ...

1

I think there might be a mistake in your question... Around $\pu{2680 cm-1}$, there is the $S_1$ band (attributed to overtone or combination scattering). I think the $2D$ band you are referring to is the band shouldering the $G$ band (also called $D'$) which is around $\pu{1680 cm-1}$. About the ratio, there are two ratios that are often used to characterise ...

1

You need to understand some details of using point group tables. The pyrazine belongs to $\ce{D_{2h}}$ point group and pyridazine belongs to $\ce{C_{2v}}$ In the third column of the point group are found x, y, z which are used to identify dipole transitions, e.g. vibrational symmetries that can exhibit IR transitions and in the fourth $x^2, z^2, xz$ etc. ...

1

In Raman spectroscopy the difference in frequency between the exciting line and the measured signal frequencies contains the information you want. These values, made positive if necessary, give the ro-vibrational transition frequencies. You are assuming that the molecule is vibrating with an anharmonic potential, such as a Morse potential. The energy levels ...

1

You have two options in that case. You can try to understand the frequency dependent (or dynamic) property in the formulation for Raman activity. This property is a bit complex and is known as polarizability. For Raman spectra, certain invariants of this tensor are involved in the intensity formulation. Be careful with this. You can then find the ...

1

It is really a matter of definition. Rayleigh scattering does not involve any change to the internal energy of the molecule whereas Raman transitions do. The Raman transitions always involve a change in the vibrational /rotational energy levels. If the molecule has sufficient vibrational energy in its ground state then anti-stokes scattering can be ...

Only top voted, non community-wiki answers of a minimum length are eligible