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Offline Schrödinger

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Absorption spectra
« on: May 04, 2013, 11:55:29 PM »
Hey guys!

I don't want to sound stupid, but I guess with a question like this, it is inevitable. I was going through the topic of absorption spectra in complexes and I was confounded with a doubt that left me feeling stupid.

When compounds absorb light of a particular frequency, we see the colour of the compound as resulting from the reflection or transmittance of whatever wavelengths were NOT absorbed.

What I don't get is, once an electron is excited to a higher energy level, it is bound to come back to the original level, is it not? When it does that, aren't we supposed to see the same wavelength that was absorbed, since the energy of the upward and downward transitions are the same? (Unless of course once absorbed, the only way for it to come back is through relaxation through a number of pathways. Although I found a similar post from ages ago, my question is : Is this the case ALWAYS?)
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Offline Corribus

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Re: Absorption spectra
« Reply #1 on: May 05, 2013, 01:47:25 AM »
I'm not sure I quite understand what the question is.  But you certainly don't sound stupid for asking it.

A photoexcited electron does return to its ground state, but unless you're talking about isolated atoms/ions, it usually does so via a different pathway.

When an electron in a molecular orbital absorbs a photon of light (of some energy in the UV or visible) and is excited to another electronic state, there is a natural driving force to return to a state of low potential energy.  One of the ways this can happen is for the electron to re-emit a photon or nearly the same energy that took to excite it in the first place.  This phenomenon we call fluorescence.

However most molecules do not manifest observable fluorescence.  This is because there is usually coupling between the excited electronic state and the ground electronic state mediated by molecular vibrational wavefunctions, and the rate of fluorescence is typically slower than the rate of conversion of electronic excitation to vibrational excitation, a process called internal conversion.  Essentially what happens is that instead of emitting a photon of light, the molecule disperses all this excess energy through heat loss to the surrounding environment.  Internal conversion is almost always fast compared to fluorescence - except under the condition of high rigidity and under the condition of very large energy gaps between the ground and excited electronic states.  This is why there are far more blue-fluorophores than red fluorophores, and it is why most fluorophores tend to be large flat molecules without a lot of "floppy" functional groups.  (Amino groups, for example, are notorious for quenching fluorescence.)  The specific reasons that rigidity and high energy gap play a role in the fluorescence (quantum) yield are fairly complicated but I would be happy to elaborate if you are interested.  Do note that no matter whether the molecule is highly fluorescent or not, there is always an interplay between fluorescence and internal conversion - the relative rates of these two processes determine the overall fluorescence yield (number of photons emitted vs. number of photons absorbed).  Like most things it's a matter of probability.

(Also note that fluorescence and internal conversion aren't the only processes that can deactivate the excited electronic state.  Photochemistry can, as can intersystem crossing to an excited triplet surface; triplet states can also relax to the ground state via internal conversion or by emission of a light photon.  The latter process is called phosphorescence and due to the forbidden nature of this transition it is typically prolonged and can linger seconds or minutes after the light sources is removed.  This is how glow-in-the-dark stickers work.  Ironically phosphorous, after which the process is named, does not emit light via phosphorescence.) 

Having said all that, even in a molecule with a quantum yield of 100% (every photon absorbed is emitted as a photon) - and there are some, like rhodamine, that can come pretty close to this - the light emitted never is the same color as the light absorbed.  For one thing, absorption can result in population of many electronic states, but typically fluorescence only occurs from one of them.  This is called Kasha's Rule.

http://en.wikipedia.org/wiki/Kasha%27s_rule 

Although in grad school we cheekily liked to call it Kasha's Good Idea, because it's not always obeyed.  This is why an absorption spectrum can have many peaks but a fluorescence spectrum typically only has one.  So for instance if you irradiate a molecule with blue light of 400 nm, you may actually be populating a high-lying singlet state rather than the lowest-lying excited electronic singlet.  That state will relax to the lowest lying excited singlet within a few femtoseconds and from there you get fluorescence of red light.

Even if you specifically excite into the lowest energy absorption band, though, the emission wavelength is still red-shifted due to a Stokes shift.  This may only be a few nanometers or can be very large, depending on the molecule and the solvent environment.  This happens because when you excite an electron in a molecule, all of the electron density in the molecule changes.  Usually this results in polarization of the electron cloud, which can result in very different polarity of molecules when they are in their excited states versus their ground states.  Light absorption is nearly instantaneous compared to nuclear motion (embodied in the Franck-Condon principle, which also explains why vibronic transitions vary in intensity), but if the excited state lives long enough, nearby solvent molecules will reorganize themselves around the polarized, excited molecule, which will in turn stabilize the excited state.  This results in a shrinking of the energy gap, which manifests itself is a lower-energy emission wavelength than the absorption wavelength.  In addition, structural relaxation of the excited molecule itself can occur, which can lower the excited-state energy even more.

So, to sum.  Excited state relaxation dynamics are complicated but almost always result in a lowering of the energy of the excited-state prior to emission, so the fluorescence color is almost always different from the absorption color.  That's if there is fluorescence at all, which most times there isn't.  The exception to this rule about red-shifting is atomic fluorescence, which have practically zero Stokes shift because there is no structural relaxation in an atom.  Some laser dyes and other highly rigid molecules have very small Stokes shifts (and high quantum yields), but even in these cases you can see a color difference with even the most basic spectrophotometric equipment.  And even in the idealized case where the emission color was EXACTLY the same color at the absorption color, fluorescence is in a random direction.  So if you were looking at a sample that did emit 100% of its absorbed photons at EXACTLY the same energy, the substance would STILL appear colored to your eye because you'd only detect a small fraction of the emitted photon with your eye.

Well that was a shotgun approach to your question.  Maybe I hit it with something in there.  If not, please clarify the question and I'll try to narrow the answer. :)
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Offline curiouscat

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Re: Absorption spectra
« Reply #2 on: May 05, 2013, 02:11:49 AM »
I thought it was an excellent reply. Good post. I learnt something new too! Never knew Kasha's Rule. :)

Offline curiouscat

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Re: Absorption spectra
« Reply #3 on: May 05, 2013, 02:19:44 AM »
Having said all that, even in a molecule with a quantum yield of 100% (every photon absorbed is emitted as a photon) - and there are some, like rhodamine, that can come pretty close to this - the light emitted never is the same color as the light absorbed. 

Is the fluorescence always at a lower freq. (longer wavelength?) than the absorption. Just wondering....

Or are there any exceptions.

Offline Borek

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Re: Absorption spectra
« Reply #4 on: May 05, 2013, 05:39:34 AM »
Note that my answer given in the previous thread is still valid - even if the light emitted has exactly the same color, its intensity, if compared with the intensity of other wavelengths, will be still lower, as it gets emitted in all directions, not just following the original path.
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Offline curiouscat

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Re: Absorption spectra
« Reply #5 on: May 05, 2013, 05:44:34 AM »
Note that my answer given in the previous thread is still valid - even if the light emitted has exactly the same color, its intensity, if compared with the intensity of other wavelengths, will be still lower, as it gets emitted in all directions, not just following the original path.

Does the notion of direction have any meaning during atomic scale emission / absorption?

From the point of view of atoms successive incoming photons must also come in from several directions, no matter how well collimated your source? OTOH, any single outgoing photon must go out in  a single direction, right?

Offline Borek

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Re: Absorption spectra
« Reply #6 on: May 05, 2013, 06:01:28 AM »
Does the notion of direction have any meaning during atomic scale emission / absorption?

Photon direction is a vector, isn't it?

Quote
From the point of view of atoms successive incoming photons must also come in from several directions, no matter how well collimated your source? OTOH, any single outgoing photon must go out in  a single direction, right?

There is a difference between photons being almost parallel (in a collimated light) and being shot in all possible directions.

Shine a collimated light on a glass cube (perpendicularly to the side, to make things simpler) - basically all energy (photons) going into a cube side exits it through the other side.

Now put something dispersing the light in all directions in the center of the cube. Now energy leaves the cube through all sides, and each side emits 1/6 of the original amount of energy (probably not, but it is about a general idea) - including the side through which we expected 100% to pass without the dispersion.

If only a single wavelength is dispersed, only this wavelength intensity will be lower.
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Offline Arkcon

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Re: Absorption spectra
« Reply #7 on: May 05, 2013, 06:44:24 AM »
Having said all that, even in a molecule with a quantum yield of 100% (every photon absorbed is emitted as a photon) - and there are some, like rhodamine, that can come pretty close to this - the light emitted never is the same color as the light absorbed. 

Is the fluorescence always at a lower freq. (longer wavelength?) than the absorption. Just wondering....

Or are there any exceptions.

Yes.  Fluorescence is always less energy emission than excitation energy.  So the shift is always to the red.  Now, there are phosphor screens -- you expose them to z-rays or gamma rays, and you give them a sort of latent image".  You can scan that with a red laser, and get blue emission where the radiation hit.  These screens essentially replace film for casual x-rays that are going to be digitized and thrown away anyway.  And of course, there are CCD chips sensitive to IR that output a visible signal for your standard IR "night vision" goggles, but many photons are used to make one output photon, in that case.

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Offline curiouscat

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Re: Absorption spectra
« Reply #8 on: May 05, 2013, 08:46:40 AM »
Does the notion of direction have any meaning during atomic scale emission / absorption?

Photon direction is a vector, isn't it?

Quote
From the point of view of atoms successive incoming photons must also come in from several directions, no matter how well collimated your source? OTOH, any single outgoing photon must go out in  a single direction, right?

There is a difference between photons being almost parallel (in a collimated light) and being shot in all possible directions.

Shine a collimated light on a glass cube (perpendicularly to the side, to make things simpler) - basically all energy (photons) going into a cube side exits it through the other side.

Now put something dispersing the light in all directions in the center of the cube. Now energy leaves the cube through all sides, and each side emits 1/6 of the original amount of energy (probably not, but it is about a general idea) - including the side through which we expected 100% to pass without the dispersion.

If only a single wavelength is dispersed, only this wavelength intensity will be lower.

Ok, maybe you are right. But:

(1) Does it not matter that the size of the collimated beam (say dimension of cross section) is much much larger (several orders of magnitude) than a emitting atom? It's like pointing a firehose at a dust speck.

(2) For a typical flurescence won't there be multiple reflections once the photon enters your sample and before it interacts with a suitable atom? So also for emitted light?

Offline Corribus

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Re: Absorption spectra
« Reply #9 on: May 05, 2013, 09:37:18 AM »
Just to be clear - fluorescence is typically in a random direction but photon emission doesn't necessarily have to be.  If this wasn't true, lasers wouldn't exist.

Also, things get more complicated when the incoming light is polarized.

EDIT: I also want to add that direction of fluorescence radiation isn't quite random, even though it's common to say as much.  It follows a sin-squared relationship to the dipole polarization direction.  So while there is some degree of random character, there isn't equal probability of emission in all directions (and some directions have zero probability of emitting light).  That said, in most practical cases light isn't polarized, so there's no photoselection happening, and molecules in any case are rotating with respect to the frame of reference (of the absorbed light) on timescales competitive to the fluorescence lifetime, so this can randomize the emission direction even more.  In the limiting case where molecules are rotating very fast (because they're very small, or the solution viscosity is low) or the radiative lifetime is long, then the fluorescence direction is, for all intents and purposes, random.
« Last Edit: May 05, 2013, 09:50:31 AM by Corribus »
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Offline Borek

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Re: Absorption spectra
« Reply #10 on: May 05, 2013, 10:14:14 AM »
(1) Does it not matter that the size of the collimated beam (say dimension of cross section) is much much larger (several orders of magnitude) than a emitting atom? It's like pointing a firehose at a dust speck.

Solution doesn't contain single atom (molecule) and those existing in the solution are dispersed evenly. Many photons do have a chance to hit a molecule on their way through the solution.

Quote
(2) For a typical flurescence won't there be multiple reflections once the photon enters your sample and before it interacts with a suitable atom? So also for emitted light?

Have you ever tried to shine a torchlight through the fish tank? Light goes through almost as if there were no tank at all.

In the limiting case where molecules are rotating very fast (because they're very small, or the solution viscosity is low) or the radiative lifetime is long, then the fluorescence direction is, for all intents and purposes, random.

I am far from stating it is always perfectly random. Thing is, even if the new direction is not perfectly random, it is still changed, so the intensity of the beam leaving the solution in the original direction is lower.
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Offline Corribus

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Re: Absorption spectra
« Reply #11 on: May 05, 2013, 10:19:23 AM »
Quote
am far from stating it is always perfectly random. Thing is, even if the new direction is not perfectly random, it is still changed, so the intensity of the beam leaving the solution in the original direction is lower.
Yes, absolutely true.
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Offline Corribus

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Re: Absorption spectra
« Reply #12 on: May 05, 2013, 11:03:52 AM »
Is the fluorescence always at a lower freq. (longer wavelength?) than the absorption. Just wondering....

Or are there any exceptions.
Sorry, missed this.

Of course there are exceptions. :) The only rule to which there are seemingly no exceptions is that energy has to be conserved.

If the molecule has a sizable 2-photon absorption cross-section, it can absorb two photons of low energy to promote an electron to the excited state, followed by fluorescence at the usual wavelength.  For instance, suppose the lowest energy electronic absorption band is at 500 nm and the emission is at 520 nm.  Instead of absorbing a single 500 nm photon, it could absorb two 1000 nm photons, and then emit at 520 nm.  Note that conservation of energy is not violated in this case - one 520 nm photon still has less energy than two 1000 nm photons.  The difference is heat dispersal.
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Offline curiouscat

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Re: Absorption spectra
« Reply #13 on: May 05, 2013, 03:35:44 PM »
Is the fluorescence always at a lower freq. (longer wavelength?) than the absorption. Just wondering....

Or are there any exceptions.
Sorry, missed this.

Of course there are exceptions. :) The only rule to which there are seemingly no exceptions is that energy has to be conserved.

If the molecule has a sizable 2-photon absorption cross-section, it can absorb two photons of low energy to promote an electron to the excited state, followed by fluorescence at the usual wavelength.  For instance, suppose the lowest energy electronic absorption band is at 500 nm and the emission is at 520 nm.  Instead of absorbing a single 500 nm photon, it could absorb two 1000 nm photons, and then emit at 520 nm.  Note that conservation of energy is not violated in this case - one 520 nm photon still has less energy than two 1000 nm photons.  The difference is heat dispersal.

Ah! Very interesting. So there are indeed 2-photon co-operative fluroscences.

Know any molecules that do this?

Getting ambitious now, are there 3 and 4 photon fluroscences too?

Offline Corribus

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Re: Absorption spectra
« Reply #14 on: May 05, 2013, 03:55:23 PM »
The phenomenon is being exploited a lot these days in two-photon excitation microscopy, especially of living cells.  IR light has higher tissue penetration, and 2-photon absorption microscopy has in principle better spatial resolution than the traditional one-photon analog.

http://en.wikipedia.org/wiki/Two-photon_excitation_microscopy

You can have higher order photon absorption as well, but I believe the cross sections decrease substantially as the order increases.

Any molecule which has a high fluorescence yield as well as nonlinear optical properties will work.  Porphyrin derivatives typically satisfy this requirement.  The two photon cross section typically is higher in molecules with high polarizabilities.  The classic examples are push-pull chromophores like p-nitroaniline.  (Electron withdrawing group on one side, electron donating group on the other.)  The larger you can make the polarization, typically the larger is the two photon cross section.  So large conjugated oligomers with an electron withdrawing group on one end and an electron donating group on the other end are fairly ideal.
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