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According to thoughtco:

If fluorescent dyes weren't put in glow sticks, you wouldn't see any light at all. This is because the energy produced is usually invisible ultraviolet light. Fluorescent dyes may be added to light sticks to release colored light.

The dyes all have there own unique chemical structure, but how do these chemicals affect the amount of energy released in the chemical reaction? I am planning to do an investigation on the activation of energy of a lightstick, and I am wondering if the colors have anything to do with the reaction rate.

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  • $\begingroup$ The chemiluminescence of glow sticks has been investigated in great detail, so you can easily find lots of information on it. Definitely search on wikipedia and do a search. $\endgroup$ – Ed V Aug 22 at 12:46
  • $\begingroup$ Check this one out: chemistry.stackexchange.com/a/115635/79678 $\endgroup$ – Ed V Aug 22 at 13:20
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    $\begingroup$ One last thing: the activating reaction does not produce UV light. It produces an excited species that then transfers energy to the recipient dye. $\endgroup$ – Ed V Aug 22 at 14:18
  • $\begingroup$ @EdV Some sources say that there is a two-step process, and that first, the hydrogen peroxide and the phenyl oxalate ester combine to produce UV light, which is then absorbed by the dye and is excited, releasing visible light when dropping down to the ground state. Is this wrong? Is there an existence of UV light as an intermediate energy transferer between the chemical reaction and the excitation of the dye in a glowstick reaction? $\endgroup$ – Frédéric François Chopin Aug 23 at 1:20
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    $\begingroup$ I think it is wrong and I would be very surprised if UV was produced. Certainly the dyes can be excited by UV and I have used 405 nm laser light to excite the dyes in 3 different unbroken color dye capsules from glow sticks. Of course, the UV emission idea can be tested experimentally, and maybe I am simply wrong: that would be neat! What I think happens is just collisional transfer from whatever the excited species is to the dye molecule. As I say, I would be happy to be wrong! If UV is actually generated, it must be in the reported literature and can be detected easily. $\endgroup$ – Ed V Aug 23 at 1:33

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