H-alpha lines are usually the ones you see when observing most nebulae. If you search why is that, why are H-alpha lines more visible than H-beta, H-gamma or H-delta, the usual answer is that H-alpha is "stronger". But what does that exactly mean? Is it just more frequent?
1 Answer
It's just a consequence of the atomic physics. The usual setup in most nebulae is that the temperatures are high enough to produce ionised hydrogen; the electrons then recombine into bound energy levels and emit photons to get down to the ground state. Lyman alpha is almost always absorbed (Case B recombination) and thus, in terms of what is emitted by the nebula, the cascade ends in the emission of a Balmer photon (radiative transition to n=2).
When you work through the atomic physics, work out the emission coefficients for radiative transmission from n=3 to n=2 (H$\alpha$) and for n=4 to n=2 (H$\beta$) etc., one finds a fairly standard ratio for H$\alpha$/H$\beta$ of about 2.8 and for H$\beta$/H$\gamma$ of about 2. These ratios are only slightly temperature and density dependent and are known as the Balmer decrement (I can't find a good internet reference - you need to look in something like "Interpreting Astronomical Spectra" by D. Emerson; or "The Astrophysics of Gaseous Nebulae and Active Galactic Nuclei" by Osterbrock & Ferland).
Variations from the theoretical values above are often used as a tracer of dust extinction, since the bluer Balmer lines are much more affected by extinction than H$\alpha$.
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1$\begingroup$ According to Wikipedia during recombination there's about a half chance of the n=3 to n=2 transition occuring $\endgroup$– AstrovisCommented Jul 30 at 21:19
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1$\begingroup$ @Astrovis unclear what your comment means. Half compared to what? $\endgroup$– ProfRobCommented Jul 30 at 21:55
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1$\begingroup$ Sorry, I messed up my wording a bit. I meant that there's a half chance that the n=3 to n=2 transition occurring (the election has a 50% chance to move between n=3 to n=2 during recombination) $\endgroup$– AstrovisCommented Jul 30 at 22:46
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1$\begingroup$ @Astrovis as opposed to doing what? The answer to this determines the line ratios. $\endgroup$– ProfRobCommented Jul 31 at 7:17
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$\begingroup$ to my understanding it's opposed to not directly transitioning from n=3 to n=2 and emitting H alpha radiation,. $\endgroup$– AstrovisCommented Aug 6 at 0:38