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LED Spectrum

CFL Spectrum

What is the theory behind this? Incandescents produce smooth curves as they are similar to black bodies, but for these non-blackbody bulbs, what creates their unique spectral curves?

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    $\begingroup$ In general and not specific to the bulbs from which the spectra you show came from. Different to metal filament (almost black bodies radiation), cold "bulbs" as led and cfl are based on luminescence. The light they emit has wavelengths that correspond to the various electronic transitions of the specific luminophore(s). Do you need a more in depth explanation? $\endgroup$ – Alchimista Aug 12 '17 at 12:25
  • $\begingroup$ If you could provide a more in depth explanation that'd be great, as I'm trying to understand this but I am struggling to find answers to this question online. Currently, all I know is that LED lights actually have blue lights in them that combine with a coating that's yellow and give off green-ish light? Is that right? Is there some sort of dye in LED lights? And are the spikes in CFL graphs caused by mercury vapor? How do they cause spikes? $\endgroup$ – paradox124 Aug 12 '17 at 13:49
  • $\begingroup$ I do not know every kind of bulbs in details but I can help by a few links. In case you want to start with incandescent filament : en.m.wikipedia.org/wiki/Thermal_radiation AND en.m.wikipedia.org/wiki/Thermal_radiation $\endgroup$ – Alchimista Aug 12 '17 at 13:56
  • $\begingroup$ Then for light sources based on luminescence phenomena look at Wikipedia again en.m.wikipedia.org/wiki/Luminescence $\endgroup$ – Alchimista Aug 12 '17 at 13:58
  • $\begingroup$ A LED lamp could be in principle of any colour, which will depend on on the semiconductor or molecule that actually electroluminesce. The lighting application is probably limited yet to some material in combination with phosphors (absorbing light and re-emitting at longer wavelength) or eventually coloured filters (coating, etc) to modulate the final colour. The same is with old neon.... Mercury lamps are based on luminescence due to arc discharge. You get the lines of Hg, while the hue can be again modulate by phosphors. $\endgroup$ – Alchimista Aug 12 '17 at 14:10
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You question involves several physical processes/technologies. I will sum them up below.

In the fluorescent lamp you have a gas, usually mercury (Hg). And when a voltage is applied to this gas, i.e. when you switch on the lamp, the Hg starts to emit light. The wavelength of this light depends on the exact quantum mechanical solution, but that is not important here. What is important, is that the spectrum of this gas is an atomic spectrum and will thus consist of narrow lines. The most important line in this Hg-atomic spectrum is located at 254 nm (in the UV), which is not useful for lighting.

To overcome this problem, scientists use phosphors. These phosphors are insulators (i.e., materials with a wide band gap, e.g. aluminum oxide) containing impurities. These impurities will absorb the high energy UV-light and convert it into low energy visible light. The impurities that are most used today are lanthanides like Europium (Eu) and Terbium (Tb). In the case of the spectrum you show, the Eu and Tb will absorb the 254 nm light and convert it into green light (for trivalent Tb, lines at around 500, 550 and 590 and 630 nm) and red light (for trivalent Eu, lines at 590, 620 and 710 nm).

In case of the LED you have another mechanism. The emission from the LED-chip itself will exist of a relatively broad blue band. This is the band you can see in your spectrum. Of course, we do not wish to have blue lights in our homes, so once again we have to use phosphors to convert part of the blue light into other colors. In this case the phosphor that is used is YAG:Ce (Cerium is just another lanthanide).

To find an explanation why the spectrum of trivalent (i.e. 3+) Eu and Tb consist of sharp peaks while that of trivalent Ce or divalent (i.e. 2+) Eu comprise of broad bands, we have to look at the nature of the electronic transitions giving rise to the luminescence. In the case of the former we have transitions from a 4f to a 4f electronic configuration while in the latter we have emission from the 5d to the 4f electronic configuration. Due to selection rules that follow from quantum mechanics, the 4f-4f transition is forbidden and the 5d-4f is allowed. The 4f electrons are located closer too the atomic nucleus than the 5d or 6s electrons and are thus shielded from the environment by the other electrons. This shielding leads to less interaction with the ions surrounding the impurity in the phosphor and together with the forbidden nature of the transition (4f to 4f) this results in a narrow emission spectrum. The involvement of the 5d electron in case of Ce means that there will be more interaction with the environment in this interaction together with the allowed nature of the transition, will lead to the narrow spectrum being smeared out into a broad band.

It has to be said that the broad spectrum of the LED will result in a better color quality of the items under the lamp, since an object can only have a color if light can reflect off it and due to the many gaps in the fluorescent lamp's spectrum not every color of light is present, so this isn't always possible. This is why objects can have very different colors inside the store and outside in the sun.

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