Please help (not too sure where to put this)

[btbdance]

In the process called fluorescence, some substances absorb photons of one kind of radiation (e.g. ultraviolet radiation), and give out photons of a different kind (e.g. visible light). When infrared radiation falls on these materials, visible light is not emitted. Why not? (no further information about fluorescence to be included)

[YT2095]

because absorbtion and re-emission comsumes a little bit of energy, so it`s re transmit at a lower energy level, you can`t get more out than you put in as you would be Trying to do with IR to Visible

[woelen]

YT, in general this is true, but there are examples, where lower frequency input gives higher frequency output.

 

An example of the formation of singlet oxygen and the fallback to normal oxygen. Each molecule has energy for light at wavelengths of appr. 1200 nm. What happens, however, is that two molecules combine and that a single photon of 600 nm is ejected and then both molecules fall back to their normal state.

 

I have read even about a compound, which absorbs red light and emits green light. The compound then absorbs two photons and then it emits a single photon of higher frequency (more energy).

[YT2095]

but there`s no Law of thermodynamics broken there, you still operate at an energy loss Total.

2 photons gets you One.

 

I`ve PM`d Swansont about this also, this should be easy for him and more his Area than mine.

[swansont]

Energy must be conserved. You can get a higher energy photon out only if there is another source of energy — I have some IR cards in my lab that you can charge up with the UV from a fluorescent light, and will then fluoresce in the visible when IR is shined on them. But that's because you start in a metastable state, so there's already energy stored there.

 

Two photons forming one can certainly happen. In addition to chemical examples such as the one woelen gave, there is a whole slew of examples that are subsets of four-wave mixing that occur in nonlinear crystals. We do frequency doubling of telecom lasers at 1560 nm to get 780 nm light (that's the main Rb transition, to the first excited state) using this method.

 

But as far as the OP goes, it's simply conservation of energy. IR has less energy than visible. As YT noted, you can get energy loss as the excited molecule relaxes through other channels before emitting the photon, so you always expect the emitted photon to have a lower energy. The above examples are specific, carefully constructed exceptions, that don't happen with your run-of-the-mill atom or molecule that hasn't been specially prepared.

[btbdance]

It appears as if the answer to my question has been lost from the conversation.... could you please give me pointers.

[YT2095]

your question has be answered More than adequately, if IR hits, it gets sent back at a lower freq, is a freq lower than IR visible?

in fact is IR even visible?

 

NO.

[John Cuthber]

I can't help thinking that, if his question had been adequately answered, he wouldn't have sought clarification.

 

On the other hand, with the way the question is posed, I think it's from homework or an exam so I guess he will get an answer.

The simple answer (rather than invoking 2 photon processes and metastable states) is that an IR photon doesn't have much energy and so cannot excite the molecule up to a highly excited state from which visible light could be emitted.

 

BYW, if

" absorbtion and re-emission comsumes a little bit of energy,"

where does it go?

[swansont]

I can't help thinking that, if his question had been adequately answered, he wouldn't have sought clarification.

 

On the other hand, with the way the question is posed, I think it's from homework or an exam so I guess he will get an answer.

The simple answer (rather than invoking 2 photon processes and metastable states) is that an IR photon doesn't have much energy and so cannot excite the molecule up to a highly excited state from which visible light could be emitted.

 

Which is what I said in post #5. (IR has less energy than visible ... you always expect the emitted photon to have a lower energy)

 

BYW, if

" absorbtion and re-emission comsumes a little bit of energy,"

where does it go?

 

Short answer: The material heats up.

[John Cuthber]

That's not absorbtion or emision. That's internal transfer and happens later.