Higher voltage [zener diodes](https://en.wikipedia.org/wiki/Zener_diode
) (above ~5.6 volts) are actually [avalanche breakdown diodes](https://en.wikipedia.org/wiki/Avalanche_breakdown
). That said, in avalanche breakdown the charge carrier [recombination can emit light](https://journals.aps.org/pr/abstract/10.1103/PhysRev.102.369
). And they do emit faint light. One problem is getting the diode to survive the power dissipation during reverse breakdown. This is done all the time though, for some types of diode. The real main problem is that of efficiency.
>The spectral distribution of the light is continuous with a long tail extending to photon energies greater than 3.3 ev. It is concluded that recombination between free electrons and free holes within the junction region is responsible for the light at the shorter wavelengths, the carrier energies in excess of the energy gap being supplied by the field. At longer wavelengths there appears to be a considerable contribution to the emission from intraband transitions.
>**A tentative figure for the emission efficiency over the visible spectrum is one photon for every 10^8 electrons crossing the junction. The recombination cross section required is reasonable, being about 10^−22 cm^2.**
To do what you want:
>would it be able to create visible light and then during an avalanche breakdown emit another non-visible wavelength
Is typically done by putting two different LED chips in the same case. [Wired back to back.](http://datasheet.octopart.com/LTL-14CHJ-Lite-On-datasheet-46704.pdf
) So in one polarity one lights up. Reverse it and the other lights up.
Any diode will have a reverse breakdown voltage, it's just not as well defined as a real zener diode. So if you characterize the reverse breakdown of an LED, you can call it a zener if you want to. But it will only emit light with current in the forward direction.