Raman scatterring and fluorescence

[vigneshwaran 0]

Can any on explain what is the difference between Raman scattering and fluorescence???

[swansont 7432]

Can any on explain what is the difference between Raman scattering and fluorescence???

 

Fluorescence is a description where there has been an excitation to an excited state, which then decays. The excited state has a lifetime associated with it and is real, so the emission always has a specific frequency. Raman scattering involves a virtual excited state; only the initial and final states are real. It can happen for any frequency light (it's non-resonant); it's the frequency shift that is fixed by the target's energy level structure.

[studiot 2108]

A simpler answer than swansont's is that flourescence is a light emission and Ramam scattering is the result of light subtraction (absorbtion).

 

With flourescence some of the incident light is first absorbed by the substance.

But what is left of the incident beam is simply reduced in intensity, its frequency/wavelength is unchanged.

That is the individual photon's energy remains unchanged, there are just fewer of them left in the incident beam.

Each absorbed photon is fully absorbed.

Additionally other light is now observed, at particular frequencies, different from those of the incident light.

This additional light is due to the absorbed light being re-emitted, almost instantaneously and is called the flourescence.

These are new photons due to the emission.

 

With Raman scattering some of the incident light is again absorbed but in a different manner so that the photons actually loose some energy, but none are fully absorbed.

So this time the intensity remains the same (ie number of photons) but they have lost some energy so have lower frequency/longer wavelength.

Further they are partly deflected in direction hence so the beam is broadened or scattered.

None of the observed photons are new photons as with flourescence.

[John Cuthber 3813]

As a philosophical point, how can you say that a photon with a different wavelength and (in some cases) polarisation is not "new"?

 

(Also, learn to spell fluorescence).

[studiot 2108]

 

As a philosophical point, how can you say that a photon with a different wavelength and (in some cases) polarisation is not "new"?

 

(Also, learn to spell fluorescence).

 

 

Because I am not drawing Feynman diagrams

 

How is that caustic comment helpful to the thread?

[swansont 7432]

 

With Raman scattering some of the incident light is again absorbed but in a different manner so that the photons actually loose some energy, but none are fully absorbed.

 

 

The photon energy can be shifted higher or lower. Lower is Stokes and higher is anti-Stokes; it depends on the relative positioning of the initial and final states of the system. Generally speaking these are always portrayed as new photons, since you get into trouble if you start thinking that a photon can be partially absorbed, especially (as JC points out), when the angular momentum state and wavelength change. That's not consistent with the models that work so well.

[studiot 2108]

 

It can happen for any frequency light (it's non-resonant);

 

I note that the OP is a geoinformatics engineer and is perhaps looking for more of an operational view.

I am guessing that some satellite is gathering data using one of these techniques.

Anyway it is good that several interpretations are available.

 

 

There are also techniques known as resonance fluorescence and resonance Raman spectroscopy.

 

Indeed here is a quote from Whiffen : spectroscopy

 

 

If the molecule is in the gas phase at low pressure, the re emission may occur from an excited vibrational level as the molecule is unable to share its vibrational energy with the surroundings by collision before re-emission. This case is called resonance fluorescence.

Also if the molecule is in a solid and unable to rotate the emission may be partially polarized and emerge unequally at different angles with respect to the direction and polarization of the incident light. There is thus a close connection with Raman spectra, there being a main line of unchanged frequency (compare Rayleigh line) with flanking bands on the low frequency side (compare Stokes lines) separated from it by the vibrational intervals of the ground state.

Indeed there is no sharp distinctive break in the series (a) Raman spectrum (b) resonance Raman spectrum ( c) resonance fluorescence.

Qualitatively these cases correspond to the incoming light being of a frequency at which the molecule is (a) transparent, (b) weakly absorbing in the wings of a band, ( c) strongly absorbing.

The general intensity of the emission varies ( c) > (b) > (a)

 

Edit Why are computers so stupid?

 

On the posting entry editor I can write a c in parenthesis, and see it that way there on the editing editor afterwards.

 

Yet it appears in my full screen as the copyright symbol.

I make more than enough genuine spelling mistakes, without the system inventing them.

Edited June 17, 2014 by imatfaal
fixed copyright symbol auto"correct"

[Sensei 1033]

 

Edit Why are computers so stupid?

 

On the posting entry editor I can write a c in parenthesis, and see it that way there on the editing editor afterwards.

 

Yet it appears in my full screen as the copyright symbol.

I make more than enough genuine spelling mistakes, without the system inventing them.

 

Computers are not stupid. They just do what programmer told them to do.

In this case programmer that made SFN forum script made predefined array of symbols that are automatically converted to html entities.

 

I am able to enter (c) (select "c" then pick up font size 14). Procedure searching for shortcuts won't be able to find it and convert.

Edited June 15, 2014 by Sensei

[studiot 2108]

Thank you Sensei, yes I know it was programmed to not do what it was instructed.

 

Would you expect your children to think they know better and not do what they were told?

[mississippichem 456]

An annoying technical note that may or may not be helpful to the OP:

 

Strictly speaking, in order to be considered fluorescence the decay must be from the first singlet excited state to the ground state. If intersystem crossing occurs then we are talking about phosphorescence instead of fluorescence.

 

Jablonski diagrams may be a helpful tool in understanding the difference between all the different types of molecular electronic decays.

Edited June 16, 2014 by mississippichem