Chemistry help

[Geoff23]

In the process called fluorescence, some substances absorb photons of one kind of radiation (eg, ultraviolet radiation), and give out photons of a different kind (eg, visible light). When infrared radiation falls on these material, visible light is not emitted, why not? Chemistry is not my strong point so any help gratefully recieved, thanks.

[InfiniteNow]

Let me preface by saying that chemistry is not my strong suit either, so take this with a grain of salt. (PLEASE SEE MB's POINT BELOW FOR SOME CLARIFICATION OF POINTS).

 

Generally speaking, ultraviolet (above purple) light is higher energy. It's pointier/bouncier than the other (visibile) lights. Like it's too caffienated.

 

Infrared (below red) light is generally lower energy. It's wavier/smoother/like it's been drinking too much alcohol or is tired. Check out the visual below.

 

My thought is that, in flourescence, there is a certain amount of energy required to make the material "warm up" enough to emit those photons, and that the infrared light doesn't zap it with enough energy to do this. However, the ultraviolet light does. Ultraviolet hits it "harder" and knocks more off.

 

Put a wine glass in front of a concert speaker. Play a quiet sound (which we will say in this example is the infrared light) and glass barely moves. Keep the wine glass in front of the concert spearker, but this time play a loud sound (which we will say in this example is the ultraviolet light) and the glass will wobble and shake and even shatter

 

In fluorescence, you need to direct enough energy to the material to get it to "shake" off some photons. It's those photons "shaking off" that make it flourece. :esmoking:

 

 

[Mercedes Benzene]

Whenever some sort of EM radiation strikes a substance, in short, the atoms eject a photon which is what we percieve as fluorescence. InifiniteNow is correct in saying that ultraviolet radiation is higher in energy than infrared, which means that the atoms are receiving higher energy. The resulting increase in energy is followed by a transition back to the ground state. This releases a photon in some form of radiation.

The only point of InfiniteNow's that I woudl argue is the following:

In fluorescence, you need to direct enough energy to the material to get it to "shake" off some photons. It's those photons "shaking off" that make it flourece.

This is not entirely correct. Regardless of the energy provided, some emission of radiation will occur. The problem: Is that photon in the visible range? So there is no "shaking off", but rather a transition from an energetic state to a ground state.

 

It is highly unlikely that any infrared radiation will cause the emission of a photon in the visible range, but I would not say that it is impossible. It all depends on the wavelength of the initial radiation, and the substance that is being exposed.

I'll try to do some research on infrared exposure-visible emission and get back to you. :esmoking: