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I know a fluorescent lamp works my emitting UV first and then the specific substance inside absorbs UV and finally emits visible light. And an object may emit infrared under sunlight due to heating. But those examples are absorbing short wavelength of EM and then reemitting a longer one. Is there a substance that can do the opposite (eg: absorb visible light and then emit UV)?

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    $\begingroup$ Maybe not possible in theory (like claimed in the answers) but in practice many fluorophores (and perhaps some phosphorophores) will emit a small percentage of photons with a shorter wavelength than the absorbed photon. Normally some energy is lost in vibrational energy level crossings (see Pritt's answer), but a photon can also be absorbed by a molecule that's already vibrationally excited, giving it the energy to emit a photon with more energy (from the incoming photon + some vibrational energy). Of course it's not easy / common because energy flows downhill (to longer wavelenghts). $\endgroup$ – VonBeche Jul 7 '17 at 11:13
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    $\begingroup$ Take any green laser pointer and you've got an example. The core laser is in the IR, and it's frequency-doubled via second-harmonic generation. $\endgroup$ – Emilio Pisanty Jul 9 '17 at 18:15
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It is possible, but impractical. You can "upconvert" light. This effect, as Wikipedia relays, is due to the sequential absorption of two photons per emitted photon.

This is certainly possible to make in a fluorescent / phosphorescent light scenario, but the engineering challenges (or, more specifically - the cost) involved in such a solution would be steep.

There is a chemical model of such a compound in the two-photon-absorbance wiki site, but I have not the nomenclature skills to name it for you. I even tried using online search for the structure (and it took me a while. I hate nested benzene groups - probably why I chose inorganic chemistry) Hell in a handbasket

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    $\begingroup$ Does this happen in other fractions as well? 3 in 2 out, or even 5 in 3 out? $\endgroup$ – tuskiomi Jul 7 '17 at 19:04
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    $\begingroup$ @tuskiomi There is no reason why not, it is just that the 2-into-1 version is rare enough $\endgroup$ – Stian Yttervik Jul 7 '17 at 21:19
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    $\begingroup$ Is this the sane thing that's happening In a Nd:YAG doubling crystal? rp-photonics.com/frequency_doubling.html $\endgroup$ – tuskiomi Jul 7 '17 at 22:29
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    $\begingroup$ @tuskiomi No. Well. You could call frequency doubling a version of 2-absorption-1-emission, but there are different mechanisms at play. And you should think of them as separate phenomenon. I was always in a permanent state of confusion when learning about qm and lasers, but I see there are answers below that covers it. $\endgroup$ – Stian Yttervik Jul 7 '17 at 23:49
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Yes it is possible for molecules to absorb a single photon at a longer wavelengths than some of those that they emit at. If the molecules are thermally isolated from their surroundings then the result is that these molecules will be cooled. The effect is, however, small.

The effect is caused when the fluorescence and absorption spectra overlap to a significant extent. This can be seen at the longest absorption wavelengths and shortest fluorescence ones. This occurs in almost all molecules but particularly whose excited state geometry is rather similar to that of the ground state, e.g many type of dye molecules, porphyrins, chlorins etc. which means molecule with large planar aromatic rings.

The first figure gives two examples of spectral overlap, but the class of molecule in which this is most important are the chlorophylls and the overlap of absorption and emission is essential for efficient light harvesting in photosynthesis. In fact without it photosynthesis would be incredibly inefficient.

overlap Fig 1. The figure shows absorption and fluorescence spectra for trans-stilbene and anthracene. The peaks in the anthracene spectrum are due to vibrational transitions in addition to the electronic transition forming the excited state.

The second figure shows, in schematic form only, the potential energy of a molecule in its ground and first excited state and a few vibrational energy levels. Because molecules have vibrational (and rotational) energy levels at room temperature it is possible by thermal means alone for a few vibrational levels to be excited.

If excitation is at long wavelength, shown in the figure by the red arrow, absorption occurs from the $v = 1$ ground state into the $v = 0$ level of the excited state. Subsequent emission to a lower level ($v=0$) in the ground state produces a more energetic photon than occurs in the absorption process. (Fluorescence emission can also occur to $v=1,2, 3$, in the ground state as determined by the Franck-Condon factors).

The black arrow indicates, additionally, that if the excited state is relatively long lived, a few nanoseconds, for example, in solution, that it is possible for thermal excitation in the excited state to populate $v=1$ and so also produce higher energy emission than absorption (green line).

overlap-potential

Fig 2. Schematic of vibrational energy levels in the ground and excited state of a molecule. Only transitions to lower levels are illustrated.

(Note. If the molecules are in the gas phase then collisions with inert gases can promote energy to higher vibrational levels in the excited state; this effect was described decades ago in naphthalene, for example; see Chem. Phys. Letts v22, 235, 1973, & Proc. Roy Soc. (Lond). Ser A, v340, 519, 1974)

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As noted in Pritt Balagopal's answer, this is not trivially possible. A quantum of light has an energy determined by its wavelength (or, equivalently, frequency), thus going from longer wavelength to shorter is forbidden by conservation of energy.

However, if you also count frequency converters as a material, you can combine photons to obtain a shorter wavelength.

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    $\begingroup$ Why wouldn't one count frequency converters as a material? $\endgroup$ – Dmitry Grigoryev Jul 7 '17 at 10:13
  • $\begingroup$ The example I was thinking of is used by the NIF (en.wikipedia.org/wiki/National_Ignition_Facility) and consists of single crystals of $\ce{KH2PO4}$. I consider that more than just a mere material. $\endgroup$ – TAR86 Jul 7 '17 at 18:57
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There are indeed: we use them in "Laser Viewing Cards" to be able to see IR Laser. An example of such a card is presented below: enter image description here

When illuminated with IR light, it emits visible light. The principle is that the molecule are trapped in an excited state that requires an IR photon to de-excite. These cards need prior charging but it last several hours.

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    $\begingroup$ That's not really what he's talking about because of the prior charging requirement. If they could charge from the laser they are detecting it would be what he's asking for. $\endgroup$ – Loren Pechtel Jul 9 '17 at 4:28
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What about (pure) rotational Raman spectroscopy?

The specific selection rule is $\Delta J = \pm 2$. The anti-Stokes lines, with $\Delta J = -2$, arise from the emission of a photon with shorter wavelength (higher energy) than the incident photon.

Rotational Raman spectrum of H2

Here's an idealised Raman spectrum of $\ce{H2}$, to illustrate the point (from Oxford lecture notes). The wavenumber scale at the bottom refers to the shift in wavenumber of the emitted photon relative to the incident photon. If this number is positive, the emitted photon has a higher wavenumber, i.e. higher energy/shorter wavelength.

Edit: As porphyrin has kindly pointed out - Raman spectroscopy is a scattering phenomenon. As such it is not technically an absorption + emission; although I guess (I hope) it is in the spirit of the question.

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    $\begingroup$ This is a nice example of energy being taken from or given to a molecule via the radiation field but it does not involve absorption or emission as it is a scattering process and need not be resonant with an absorption. $\endgroup$ – porphyrin Jul 7 '17 at 17:45
  • $\begingroup$ @porphyrin True that, it did cross my mind, but unfortunately only after posting my answer; it's technically not an excitation + decay. $\endgroup$ – orthocresol Jul 7 '17 at 17:57
  • $\begingroup$ Doesn't Raman scattering still involve absorption and desorption (emission) of a photon? I cannot think of another way how the photon ought to impart energy to the analyte. Even though the excitation is to a virtual state, the photon "before" should be different from the photon "after" (regardless of the change in frequency). Is this wrong? $\endgroup$ – Linear Christmas Jul 11 '17 at 20:43
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It is possible and is applied in optical microscopy. This method is called two-photon microscopy. Two photons of longer wavelength are absorbed by a dye molecule which then emits one photon of shorter wavelength. As mentioned before, this process is highly unlikely, but this property can be used to reduce the size of the fluorescent spot in confocal mocroscopy. (Confocal microscopy is a techique where a fluorescent sample is scanned with a focused laser spot while fluorescence emitted by this spot is detected. The resolution of the acquired image depends on the spot size.) Especially the extension of the spot along the optical axis of the microscope is reduced. Furthermore, excitation with a longer wavelength is beneficial in thick biological samples (tissues) because red light is scattered less and is less harmful than, say, blue light.

I have used this technique with the fluorescent dyes Alexa Fluor 488 and Alexa Fluor 594, but I guess there are many more. We excited them with far red light at 800 nm and detected fluorescence in the green and orange range.

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    $\begingroup$ "Especially the extension of the spot along the optical axis of the microscope is reduced." Can you elaborate it? It would be helpful if you add a link to Confocal Microscopy. $\endgroup$ – Mockingbird Jul 8 '17 at 9:43
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For the sake of completeness: One way is the already explained chemical two-photon absorption, another option is the second harmonic generation of non-linear optics.

If you have a stereo and you increase the loudness, after a specific level the speakers begin to sound displeasing. The problem is that the speakers cannot amplify anymore without disturbances. These disturbances can be approximated by sums of frequency part, the base frequency part, the second harmonic, the third harmonic etc. The interesting thing is now if very strong light goes through material, the electrons do not react linearly anymore. So if you send infrared light with a wavelength of 1064 nm from a Neodym-YAG laser with enough intensity, you see the doubled frequency light, a green with 532 nm.

enter image description here Picture by kkmurray, CC BY 3.0, 1997-04-11

To be fair, the conversion ratio is very, very low. CORRECTION: The double frequency crystals are quite efficient, the KTP crystal for Nd-YAG reaches 80%. The thing is the conversion rate is AFAIK low for lasers in the range of hard UV or X-rays which cannot be produced directly.

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  • $\begingroup$ It bears mentioning that this mechanism is in widespread use in just about all green laser pointers - they're IR lasers with a frequency-doubling SHG step at the end. $\endgroup$ – Emilio Pisanty Jul 9 '17 at 18:15
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I don't think so.

The type of absorption and remission is called phosphorescence. The way phosphorescence works is, the shorter wavelength of light is absorbed by a suitable material and electrons are excited to many states higher above the ground state.

However, unlike most excitations, these do not de-excite immediatly. The decay back to regular states involve spin-flips, a forbidden transition. As a result, the decay can take over hours to decay step by step, with each step having a longer radiation due to lower energies.

For more information about Phosphorescence, check out the Wikipedia page.

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    $\begingroup$ Nitpick: it doesn't have to be phosphorescence. Fluorescence also (nearly always) occurs at longer wavelengths than excitation. $\endgroup$ – orthocresol Jul 7 '17 at 4:49
  • $\begingroup$ @Pritt Great answer. Can you suggest me a link to understand spin-flips? $\endgroup$ – Mockingbird Jul 8 '17 at 9:34
  • $\begingroup$ @Mockingbird I'm no expert on that topic, but see if this helps you: astronomy.swin.edu.au/cosmos/S/Spin-flip+Transition $\endgroup$ – Pritt Balagopal Jul 8 '17 at 9:35
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Quantum dots can be used to convert infrared light to visible light

Their approach was to use two films placed on top of a clear plate of glass. The film layer on the bottom was made using a type of quantum dot—an inorganic semiconductor lead sulphid which had been coated with a single layer of fatty acids to make the surface passive. The top film layer was crystalline and made of rubrene, an organic molecule.

In the two film approach, the top film absorbs infrared light, and the energy from it is then transferred to the bottom film. That energy, which exists in the form of excitons is then diffused as it passes through the rubrene—a process called triplet–triplet annihilation. They tested the films by shining an infrared laser at the finished product and found it glowed with visible light—it works, the team notes, because the collision of two low-energy excitons create high-energy excitons, i.e. singlets, which can emit visible light. The team reports that the films are quite efficient at converting the infrared light to visible light.

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