I've read that everything from fire to LEDs can produce UV radiation. Generally, unless intended otherwise, lightbulbs will have a phosphor coating to prevent UV radiation from escaping the bulb. Although this technically creates a light source that emits a low amount of UV rays, I'm wondering if there exists any possibly way to create raw, unfiltered light that does not emit ultraviolet radiation?
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$\begingroup$ Perhaps it's possible with lasers? i.e. stimulated emission $\endgroup$– DanDan面Commented Oct 30, 2020 at 20:06
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2$\begingroup$ I'm pretty sure diodes don't produce any UV unless intended to.. can you reference where you read what you are claiming? $\endgroup$– user9413641Commented Oct 30, 2020 at 20:12
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3$\begingroup$ Indeed, the classic red GaAs LED isn’t putting out any UV... $\endgroup$– Jon CusterCommented Oct 30, 2020 at 20:27
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2$\begingroup$ Thermal sources emit the full spectrum of EM radiation, including UV. Non-thermal sources like LEDs and lasers can produce light of one wavelength and effectively zero light in the UV range. $\endgroup$– S. McGrewCommented Oct 30, 2020 at 20:55
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$\begingroup$ premierltg.com/do-led-lights-produce-uv-led-tanning-beds/…. This is where I read that even diodes emit UV $\endgroup$– xoristCommented Oct 30, 2020 at 22:10
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2 Answers
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Common LEDs do not produce UV light unless designed to. Note that tungsten filament light bulbs do not have a phosphor coating inside them to convert UV into visible light- the only light bulbs that do are compact fluorescent light bulbs.
answered Oct 30, 2020 at 21:37
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Well, anything with non zero temperature produces light over the whole spectrum, but cool anything low enough and you get a pretty much zero emission of UV (or visible light for that matter, things don't get visible for us until hundreds of degrees)
Otherwise, you can get something to emit light of only specific frequencies if it works by making an electron go from one shell to an other in an atom. In that case the energy of the emitted photon, is exactly the energy difference between the two shells, and from E = hf, this also means exact frequency. Thus find the right atom, and excite it with the right method and you get light with no UV's.
Hope i answered your question, don't hesitate to ask if it needs clarifications
