# Why are O III lines so prominent in the spectra of emission nebulae when the amount of oxygen relative to hydrogen is a million times smaller?

Looking at spectra of emission nebulae like the Lagoon Nebula, the $[\text{O III}]$ lines are prominent in intensity. However, the abundance of oxygen is minuscule compared to hydrogen. How then are the $[\text{O III}]$ spectral lines of such high intensity?

This is an excellent question. Think about the way in which emission happens. $\text{H}\alpha$ emission happens when an electron makes a transition from the third energy level to the second, emitting a photon in the process with the energy equivalent to the difference of the energies of the two states, roughly $1.9\text{ eV}$. There's a lot of hydrogen in emission nebulae - and in nebulae in general - and so it plays an important role in the emission spectrum.
Emission from $[\text{O III}]$ comes from a different process: collisions. Free electrons with energies of $\sim1\text{ eV}$ - about the amount of the oxygen's low excitation potential - collide with oxygen atoms. This produces forbidden lines that would otherwise have been unlikely to be created, as the ionizing photons typically would not have the right energies. Therefore, doubly-ionized oxygen (in particular) becomes extremely important in spectra, generally second only in intensity to $\text{H}\alpha$.
Given that $[\text{O III}]$ emission is collision-based, it makes sense that there is a temperature dependence - which there is, as is the case with $[\text{N II}]$ lines and is responsible for their high intensities. Conditions in many hot nebulae (in this case, $>10,000\text{ K}$) are right for such collisions and thus forbidden line emission. Hot host stars can sometimes help achieve high electron temperatures.
• @ArvindH See the edit: It's that there's a strong temperature dependence, and therefore at higher temperatures $[\text{O III}]$ becomes more important, while $\text{H}\alpha$ stays roughly the same. – HDE 226868 Jan 8 '17 at 19:36