# How does infrared light 'erase' phosphorescence on zinc sulfide?

I found some sheets of zinc sulfide in my basement that phosphoresce green for up to 24 hours or so after exposure to bright light in the violet range or shorter. One of the first things I tried was drawing on it with a violet laser pointer (405nm, 5mW) and as expected it draws bright lines. What I found more surprising was that I can 'erase' the phosphorescent lines if I focus my green laser onto them. I've determined with some color filters that it's infrared at 1064nm leaking from the green laser causing the erasure. This erasure does not appear to damage the panel in any way, once erased the erased spot can be lit back up to full brightness normally.

How is this infrared light erasing the panel, and how effective would other infrared wavelengths be at doing this?

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Try using a TV remote for more testing fun :) [note:you can detect the IR rays of a remote by pointing it at a camera] – Manishearth Mar 22 '12 at 10:52

Thanks to @Manishearth for the edit

In normal phosphorescence, the atoms are in a "metastable" state--where electrons are in a higher energy level, but do not immediately come to ground state due to partial stability. The electrons come down slowly, giving rise to the (relatively) long lasting glow.

The IR light frees away the electrons from the shallow metastable trap. It's like the usual chemical reaction, where the IR light provides enough activation energy to overcome the barrier. Basically, the IR light promotes the electrons to a higher, non-metastable "energy level" (virtual state). From here, the electron jumps down nearly immediately--dumping all its energy. So, instead of the electrons trickling down like in normal phosphescence, they all come down in a torrent.

The same experiment was performed by this guy: http://ajp.aapt.org/resource/1/ajpias/v29/i3/pxxv_s2

The effect has been explained by the assumption that the long-wave radiation frees electrons from shallow trapping centers into which they have fallen after being excited to the conduction band by the ultraviolet light. The freed electrons recombine quickly with ionized luminous centers with the emmision of light. More recent investigations indicate that probably other processes also are involved.

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so, IR $\approx$ thermal wavelength of the system is probably a coincidence in this case – pcr Mar 24 '12 at 14:19
investigation on the depth of the electron trap: rspa.royalsocietypublishing.org/content/184/999/365 – pcr Mar 24 '12 at 14:22
I hope all those journals will be freely accessible in <5 years... – pcr Mar 24 '12 at 14:54
So this is basically shaking a shallow dish that holds a marble, causing it to roll over the edge onto a smooth slope that drops it quickly to ground? I knew that luminous centers can trap the electrons for extraordinary lengths of time, but that gives me a quite different image of the individual steps in the process. Hmm.... I take it there is an intensity issue here also, since the infrared beam that @0x5f3759df (memorable!) notes is likely the pre-frequency-doubled deep infrared beam pumped optically by the barely-visible diode laser beam? (How do they make green lasers cheaply?) – Terry Bollinger Mar 25 '12 at 3:44
Yeah, that's for phosphorescence. You're probably right for @0x5f3759df's (more memorable than 0x5f375a86) IR beam is the pre-doubled freq of the laser. Just discovered this note on green laser (2nd harmonic generation, nonlinear != 2photon excitation): optics.arizona.edu/kost/Lecture%20notes/… – pcr Mar 25 '12 at 15:46

I'm not sure of this one, comments appreciated.

Update: See @pcr's answer--it links to the correct explanation. In this, the IR rays, instead of doing stimulated emission as I said, promote the trapped electron to a higher non-metastable state. From there, it comes back down. The rest is similar to my answer--substitute "lower intermediate state" with "higher state", and remove the stimulated emission part.

To me, it looks like stimulated emission is occuring.

Phosphorescence works by exciting electrons to an energy level where they sort of get stuck. They come down to the ground state slowly(as in only a few at a time), and each electron coming down to the ground state emits a photon of charateristic color. Due to the slowness of the whole thing, we get light for quite a while.

Here, you are shining light at it. What's happening is that photons of a certain wavelength are stimulating the "stuck" electron to come to some other state--not the ground state as you stated that the light you shone is IR and ZnS is usually in the visible range. From this other state, it jumps to the ground state. This entire process is a fast one, so basically all the trapped electrons are dumped in one go. I guess this should be accompanied by a flash of light, but it may be too weak compared to the laser to be seen.

If this is the case, then most of the other IR wavelengths will be ineffective. Only a wavelength corresponding to a transition between the "stuck" energy level and some lower level will work.

Note: As @TerryBollinger noted below, the mechanism may also involve a variant of two photon emission. It's not required to explain this phenomenon; the above mechanism seems to cover all bases, but it still is a possibility.

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Manishearth, might this be stimulated emission via a variant of two-photon excitation? The second paragraph of this Wikipedia section describes the idea in the context of fluorescence effects. – Terry Bollinger Mar 22 '12 at 23:31
@TerryBollinger: It could be, but I don't see how that would work... Two-photon excitation is, well "excitation". Excitation needs energy input. On the other hand, an excited atom doesn't need any excuses to emit. If its in a metastable state, then all it can get is an excuse to emit faster. We have already provided the excuse, no need to complicate it. Anyways, the reason I've provided is a multiphoton example: one photon to excite(sd usual), one to bump it off the metastable, one more that gets emitted in this bump, and one that gets emitted when it comes to ground. – Manishearth Mar 23 '12 at 0:57
@TerryBollinger: [Even I'm no expert here--the metastable/stimulated emission was the only thing that I could think of] Hmm, sounds interesting and impossible :P. Wouldn't it require some nifty timing though? AFAICT it would be completely impossible if the photons had even a small time difference. But yes, I guess two-photon excitation would be possible here as well, though I fail to see why it's required to explain this (Occam's razor). It would be better if someone more experienced in the field had answered this :/ – Manishearth Mar 24 '12 at 2:33
@Terry Yeah, I agree--after all, even though we are speculating, someone must already know the actual reason, and Occam's doesn't really hold for known stuff. I'll add that to the answer as well.. – Manishearth Mar 24 '12 at 8:00
@TerryBollinger we were both wrong, @pcr got it right.. Instead of being stimulated emission, the electron is sent up by the photon, to a non,etastable state--from which it comes down.. – Manishearth Mar 24 '12 at 14:56