Flurorescent Light bulb spectrum

I have read from sources that the curves are generated by the phosphors in the bulb, and the spikes are caused by the mercury vapor.

However, if the mercury vapor's release of uv particles combine with the phosphors to produce visible light, then how can they give out light in different wavelengths separately?

  • $\begingroup$ I think you should read something about electronic transitions in general. $\endgroup$ – Alchimista Aug 12 '17 at 19:01

A fluorescent lamp is filled with mercury (Hg) gas. When you switch on the lamp the gas starts emitting light. However in contrast to the sodium lamps which are widely used in street lighting (the orange lamps), mercury emits UV light (mainly 254 nm) which is not only invisible for us, but could also harm us in the same way UV light from the sun can harm us.

To overcome this problem scientists use phosphors. These phosphors are insulating materials to which impurities are added. These impurities absorb UV or blue light (depending on the material and the application) and reemit visible light (e.g. green or red). If you take a closer look at the lamp you will be able to see these phosphors since they are applied as a coating on the inside of the lamp and are responsible for the white color of the glass.

The most widely used phosphors contain lanthanides as impurities. In the case of your spectrum the red lines come from trivalent Europium whereas the green emission comes from trivalent Terbium. However on the above wikipedia link you will find a huge list of used phosphors containing also transition metal ions like Manganese and various others.

A final note which is not really related to your question but it is closely related to my research and nice to know: the glow in the dark materials work in a similar way, they only have the possibility to store the absorbed energy before reemitting it again.

  • $\begingroup$ So "the curves are generated by the phosphors in the bulb, and the spikes are caused by the mercury vapor" is an incorrect description? $\endgroup$ – paradox124 Aug 13 '17 at 3:33
  • $\begingroup$ Yes, only the peak around 430 nm is coming from Hg. The others are related to Eu and Tb. $\endgroup$ – David VdH Aug 13 '17 at 7:34
  • 1
    $\begingroup$ Does this answer your question or do you need some further explanation? $\endgroup$ – David VdH Aug 16 '17 at 20:46

Using google for help, the peak around 625 nm is from Europium which is added to the phosphor to produce red light. Several of the smaller peaks greater than 625nm are also attributable to Europium as well. The large peak at about 550 nm is due to mercury as are the smaller peaks at around 415 nm and 440 nm. the phosphor also contains terbium which accounts for several of the smaller peaks around 488nm to 600nm. The collection of small peaks around 575-600nm are also from Europium. This link https://commons.wikimedia.org/wiki/File:Fluorescent_lighting_spectrum_peaks_labeled_with_colored_peaks_added.png will help explain other peaks associated with your spectrum.

By having red, green and blue light, if they are of comparable intensity, you have produced white light.

  • $\begingroup$ Would "the curves are generated by the phosphors in the bulb, and the spikes are caused by the mercury vapor" be an incorrect description then, since all the spikes were generated by the phosphorous? $\endgroup$ – paradox124 Aug 13 '17 at 3:34
  • $\begingroup$ Yes because you have the large peak near 550 nm from mercury. $\endgroup$ – Natsfan Aug 13 '17 at 9:10
  • $\begingroup$ I'm sorry but the peak at 550 is a Terbium peak! A Tb spectrum generally consists of 4 peaks mostly green of which the last peak overlaps with a peak of Eu. $\endgroup$ – David VdH Aug 13 '17 at 20:28
  • $\begingroup$ The link I included listed that peak as a doublet with contributions from both Terbium and Mercury. The peaks are almost resolved in the link and appears wide, indicating a doublet, in the spectrum above. $\endgroup$ – Natsfan Aug 13 '17 at 20:35
  • $\begingroup$ According to nist the 254 nm line is about a 150 times more intense than the 545 nm implying the Tb peaks to be much brighter than Hg. $\endgroup$ – David VdH Aug 14 '17 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.