March 01, 2024, 09:19:58 AM
Forum Rules: Read This Before Posting


Topic: Absorption spectra  (Read 10604 times)

0 Members and 1 Guest are viewing this topic.

Offline Schrödinger

  • Chemist
  • Sr. Member
  • *
  • Posts: 1162
  • Mole Snacks: +138/-98
  • Gender: Male
Re: Absorption spectra
« Reply #15 on: May 07, 2013, 12:29:22 AM »
This is the best thread I've been involved in so far! Extremely informative :)

Another question though. When a photon does excite the electron to an excited state, it de-excites almost always via vibrational relaxations to the lowest vibrational level of an excited singlet electronic state. This works well when trying to explain why a red coloured object absorbs violet light, since it's of a higher energy.

But what about the reverse? When a molecule absorbs red light, it gets excited, but when it de-excites, how come it does so via the emission of violet light? Red light is of a lesser energy than violet, is it not?

Basically, why do we see complementary colours? i.e., why is a solid violet coloured when what it absorbs is red?
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
- William Jennings Bryan

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3466
  • Mole Snacks: +523/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Absorption spectra
« Reply #16 on: May 07, 2013, 09:55:16 AM »
Color we see from a substance rarely has anything to do with fluorescence.  If a molecule absorbs red light, it can only fluoresce red or near-IR light (with some rare exceptions).  Energy must be conserved.

When molecules absorb light - and assuming there is no fluorescence - what we see is everything that is NOT absorbed.  If the light source is white (dispersion of light of all wavelengths).  So if a molecule absorbs blue photons, we see red light because that's all that is left.  If a molecule absorbs red light, we see blue-violet color because that's all that is left. 

So - water appears blue because water vibrational overtones absorb just a tiny bit of red light (there's one really weak peak @ around 800 nm and a slightly stronger one at 1000 nm).  If you hold up a glass of water to light, it seems clear.  But look at the deep ocean and it is blue - Beer's Law!  Chlorophyll a, one of the primary pigments in plants, absorbs blue light very strongly (λmax ~ 430 nm) and red light fairly (λmax ~ 662 nm) strongly as well. Chlorophyll b, the other one, has similar absorption characteristics. What neither one absorbs well is green light, and so that's what we see when we look at leaves. Beta-carotene, on the other hand, absorbs just about everything between 400-500 nm, which is blue, green, and some yellow light.  What remains?  Orange and red.  Hence, carrots are orange.  (Probably they are more orange than red looking because most "white" light sources, including the sun, put out more orange light and red light).

And so on.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline blaisem

  • Regular Member
  • ***
  • Posts: 87
  • Mole Snacks: +5/-0
Re: Absorption spectra
« Reply #17 on: May 08, 2013, 07:49:20 PM »
Quote
Schrödinger
Basically, why do we see complementary colours? i.e., why is a solid violet coloured when what it absorbs is red?

What solid does this?  I'm just curious to know.

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

Or are there any exceptions.

Cool to read about the two photon absorption concept.  But no one mentioned Anti-Stokes shifts as a possibility for higher frequency fluorescence.  Is it just too uncommon to be even worth mentioning?

Quote
Corribus

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 happens because when you excite an electron in a molecule, all of the electron density in the molecule changes...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... In addition, structural relaxation of the excited molecule itself can occur, which can lower the excited-state energy even more.

Is all of this solvent/structural reorganization likely at all on a fluorescence timescale of nanoseconds, or does it require very special conditions to prolong the excited state?  In other words, in the normal time frame of fluorescence, is it usual to see the energy of an excited state lessen noticeably due to reorganizational stabilization?

Is our scale here a 1-2 nm difference between absorption and fluorescence, or is it something else?

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3466
  • Mole Snacks: +523/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Absorption spectra
« Reply #18 on: May 08, 2013, 09:48:27 PM »
Cool to read about the two photon absorption concept.  But no one mentioned Anti-Stokes shifts as a possibility for higher frequency fluorescence.  Is it just too uncommon to be even worth mentioning?
You're confusing two concepts named after the same person.  The Stokes shift is the difference in wavelength (energy) between the lowest energy electronic absorption band and the highest energy fluorescence band.  It is ALWAYS to the red.

Stokes and Anti-Stokes scattering are seen in Raman spectroscopy and occur due to the various energy differences between vibrational levels of a molecule. 

Quote
Quote
Corribus

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 happens because when you excite an electron in a molecule, all of the electron density in the molecule changes...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... In addition, structural relaxation of the excited molecule itself can occur, which can lower the excited-state energy even more.

Is all of this solvent/structural reorganization likely at all on a fluorescence timescale of nanoseconds, or does it require very special conditions to prolong the excited state?  In other words, in the normal time frame of fluorescence, is it usual to see the energy of an excited state lessen noticeably due to reorganizational stabilization?
Solvent polarization and reorientation typically happens at the femtosecond and single-digit picosecond timescales.  You pretty much need an ultrafast, pump-probe transient absorption setup to see it.  Structural reorganization of excited chromophores usually is a bit slower, because they have more mass than solvent molecules.  Excited biphenyl, if I recall, relaxes on a 6-8 ps timescale.  Bigger molecules are considerably slower.  In grad school I worked on porphyrin chromophores, and structural relaxation there was 30-50 ps, and sometimes even slower.  You can sometimes see these processes with time-resolved fluorescence experiments.  Depends on what kind of laser system you are using.

Quote
Is our scale here a 1-2 nm difference between absorption and fluorescence, or is it something else?
I don't understand the question.  A Stokes shift this small is fairly uncommon, but not unheard of in very rigid molecules that just don't have very much excited state structural relaxation.  The Stokes shift is also dependent on the solvent polarity and the polarity of the chromophore.  Water tends to give rise to large Stokes shift because it is highly polar and reorganization of water molecules around a polarized excited state chromophore can give rise to large energy stabilization of the excited state.  This also tends to greatly decrease the fluorescence yield, which is why finding bright, water-soluble fluorophores is a tough challenge.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline blaisem

  • Regular Member
  • ***
  • Posts: 87
  • Mole Snacks: +5/-0
Re: Absorption spectra
« Reply #19 on: May 09, 2013, 05:33:59 AM »
Thanks for the prompt reply!

Regarding Stokes Shifts, I was under the impression that a Stokes Shift was, as you said, a red-shifted fluorescence spectrum with respect to the absorption spectrum caused by vibrational relaxation in the excited state and suggested by Kasha's "Good Idea" :)

I thought an Anti-Stokes Shift was the result of electrons already in an excited state--primarily in a vibrationally excited state--absorbing photons to a higher singlet state.  The absorbed photon could then have a lower frequency than the subsequently emitted photon.

Here is a picture of what I mean.

I have a feeling we misunderstand each other.  This was only in response to Curiouscat's question whether an emitted photon is always longer wavelength than a corresponding absorbed photon.

Quote
Is our scale here a 1-2 nm difference between absorption and fluorescence, or is it something else?

I don't understand the question.

I was simply wondering how much solvent/structural reorganization is expected to redshift an emission spectrum on that timescale.  But, I didn't realize it happened so quickly.  That was definitely interesting to read, thanks for that info!

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3466
  • Mole Snacks: +523/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Absorption spectra
« Reply #20 on: May 09, 2013, 10:10:23 AM »
Regarding Stokes Shifts, I was under the impression that a Stokes Shift was, as you said, a red-shifted fluorescence spectrum with respect to the absorption spectrum caused by vibrational relaxation in the excited state and suggested by Kasha's "Good Idea" :)
Not vibrational relaxation - structural relaxation.  When an electron absorbs a photon and is promoted to an excited state, this happens almost instantaneously, long before the nuclei have a chance to adjust to the new electronic environment.  However as I've mentioned the excited state electron density distribution is usually quite a bit different than the ground electronic state, so once the electron is in its excited state, the nuclei will slowly reposition themselves.  The entire average structure of the molecule changes!  This results in a stabilization of the excited electronic state.  Since fluorescence is usually slower than this structural relaxation, photon emission primarily happens from this relaxed state.  The wavelength of absorption is related to the energy gap between the ground state and the excited state.  And wavelength of fluorescence is related to the energy gap between the excited state and the ground state.  However while the ground state is pretty much the same in both processes, the excited state is lower energy in the fluorescence process.  The result is lower energy wavelength for emission.  The difference in energy between absorption and emission is called the Stokes shift.

Stokes and anti-stokes scattering is a completely different phenomenon related to transitions between vibrational levels, not electronic levels.  (Fluorescence as we refer to above is mostly an electronic process, although vibronic structure can be resolved.)  Raman scattering is when a vibrating molecule is promoted to a high lying "virtual" vibrational state and then relaxes back to a different low-lying vibrational state in a concerted-type process. In a way this is like fluorescence but since it is scattering the process is nearly instantaneous so is not subject to relaxation dynamics as we see in the electronic absorption described above.  If the lowest lying vibrational state is v1 and the next lowest is v2, most scattering events result in absorption to a virtual excited state and then relaxation to the same origin state - that is, the scattered photon has the same energy as the incident photon.  This is called Raleigh scattering.  A small portion of scattered photons do not have the same energy as the incident photons, however.  This is called Raman scattering.  Raman scattering results in excitation to an excited state and then relaxation to a different vibrational state.  Considering only the two states v1 and v2, Raman scattering can result in a transition between these states in two ways.  Either the molecule starts in v1 and ends up in v2 or it starts in v2 and ends up in v1.  In the former process the scattered photon has less energy than the incident photon and in the latter process the scattered photon has more energy than the incident photon.  The former process is called Stokes scattering and the latter process is called anti-Stokes scattering.  anti-Stokes scattering requires pre-population of a higher lying vibrational state (it has to start in v2).  Raman scattering is a non-linear (two photon) effect and the selection rules are different.  A dipole moment change during the transition is not required, which is why IR-silent molecules like nitrogen gas give Raman signals.  However the same vibrational states are interrogated by IR and Raman experiments, so they are complimentary techniques.  Both Stokes and anti-Stokes scattering a very small effects - typically the signal is at least 6-7 orders of magnitude weaker than Raleigh scattering.

The Raman effect can also be observed between rotational states and I believe electronic states, but I stress again it is a different process than simple electronic absorption and fluorescence.  The use of Stokes in these two areas is coincidental.

Quote
I have a feeling we misunderstand each other.  This was only in response to Curiouscat's question whether an emitted photon is always longer wavelength than a corresponding absorbed photon.
In electronic spectroscopy and fluorescence, it is.  In light scattering, where vibrational levels (or other levels) are involved, scattered photons can be higher energy than incoming photons.

Quote
Is our scale here a 1-2 nm difference between absorption and fluorescence, or is it something else?
I was simply wondering how much solvent/structural reorganization is expected to redshift an emission spectrum on that timescale.  But, I didn't realize it happened so quickly.  That was definitely interesting to read, thanks for that info!
The degree of shift doesn't necessarily have anything to do with the timescale of reorganization.  It has more to do with the difference in polarity between the ground and excited state and the polarity of the solvent.  Where reorganization timescale comes into play is if it is long enough that it begins to compete with fluorescence.  If structural relaxation takes place on a 100 ps timescale, for example, and fluorescence is 500 ps timescale, you can start to see fluorescence from both the unrelaxed and relaxed states, even in an experiment that isn't time-resolved.  (You will see this is a broadening of the fluorescence signature.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Sponsored Links