<p>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).</p>
<p>When you work through the atomic physics, work out the emission coefficients for radiative transmission from n=3 to n=2 (H<span class="math-container">$\alpha$</span>) and for n=4 to n=2 (H<span class="math-container">$\beta$</span>) etc., one finds a fairly standard ratio for H<span class="math-container">$\alpha$</span>/H<span class="math-container">$\beta$</span> of about 2.8 and for H<span class="math-container">$\beta$</span>/H<span class="math-container">$\gamma$</span> 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 &quot;<a href="https://www.wiley.com/en-us/Interpreting+Astronomical+Spectra-p-9780471976790" rel="nofollow noreferrer">Interpreting Astronomical Spectra</a>&quot; by D. Emerson; or &quot;<a href="https://uscibooks.aip.org/books/astrophysics-of-gaseous-nebulae-and-active-galactic-nuclei-second-edition/" rel="nofollow noreferrer">The Astrophysics of Gaseous Nebulae and Active Galactic Nuclei</a>&quot; by Osterbrock &amp; Ferland).</p>
<p>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<span class="math-container">$\alpha$</span>.</p>

CC BY-SA 4.0
2024-07-30T21:19:17.233Z
According to Wikipedia during recombination there's about a half chance of the n=3 to n=2 transition occuring
Astrovis
CC BY-SA 4.0
2024-07-30T21:55:23.773Z
@Astrovis unclear what your comment means. Half compared to what?
ProfRob
CC BY-SA 4.0
2024-07-30T22:46:05.527Z
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)
Astrovis
CC BY-SA 4.0
2024-07-31T07:17:00.310Z
@Astrovis as opposed to doing what? The answer to this determines the line ratios.
ProfRob
CC BY-SA 4.0
2024-08-06T00:38:17.923Z
to my understanding it's opposed to not directly transitioning from n=3 to n=2 and emitting H alpha radiation,.
Astrovis
CC BY-SA 4.0
2024-07-30T14:37:44.210Z
2024-08-05T21:52:40.057Z
2024-08-05T21:52:40.057Z
ProfRob
<p>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 &quot;stronger&quot;.
But what does that exactly mean? Is it just more frequent?</p>

<observational-astronomy><photography><nebula><balmer-series>
Why is said that H-alpha emission lines are stronger than the other of the Balmer series?
jack_O'Dim
# Why is said that H-alpha emission lines are stronger than the other of the Balmer series?

(Asked by: username_0 on 2024-07-30)

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?

**Tags:** observational-astronomy, photography, nebula, balmer-series

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## 1 answers

### Answer by username_1 on 2024-07-30. Score: 10

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](https://www.wiley.com/en-us/Interpreting+Astronomical+Spectra-p-9780471976790)" by D. Emerson; or "[The Astrophysics of Gaseous Nebulae and Active Galactic Nuclei](https://uscibooks.aip.org/books/astrophysics-of-gaseous-nebulae-and-active-galactic-nuclei-second-edition/)" 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$\alpha$.

> **Comments:**
> **username_2**: According to Wikipedia during recombination there's about a half chance of the n=3 to n=2 transition occuring  
> **username_1**: @Astrovis unclear what your comment means. Half compared to what?  
> **username_2**: 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)  
> **username_1**: @Astrovis as opposed to doing what? The answer to this determines the line ratios.  
> **username_2**: to my understanding it's opposed to not directly transitioning from n=3 to n=2 and emitting H alpha radiation,.  

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