<p>There is nothing special about light emitted by an electron jumping orbits in a hydrogen atom and any other light at the same frequency. There is no way to distinguish the source of light by the perceived colour or by adding any kind of filter.</p>
<p>Every object (at some not too low temperature) emits light at every wavelength in the visible spectrum. In addition to that, some (usually hot and hydrogen-rich) objects emit light from excited hydrogen atoms which appears at very specific wavelengths only, including one that we perceive as red. The filter you have only lets a small range of wavelengths (3 nm as you write in the title) through, cutting away most of the light produced due to temperature, but lets all of the H-alpha light through, therefore enhancing it's share and making it visible more easily.</p>
<p>If what you see is &quot;normal&quot; light or emitted by hydrogen atoms is impossible to say, but you can make a scientific guess or make an analysis of the whole spectrum of light to get a more precise identification. One possibility for measurement would be to use a second filter, e.g. one with the same bandwidth but a filter range that is shifted to 3 nm longer wavelengths. Thermal radiation would produce roughly the same amount of light in both filters, but H<span class="math-container">$\alpha$</span> would show up in only one of them. The ratio of both intensities gives you a measure for the contribution of the H<span class="math-container">$\alpha$</span> emission to the light you perceive. Unless, of course, there is another emission line from some other atom unknown to you that only shows up in the second filter...</p>

CC BY-SA 4.0
2022-08-07T23:33:42.447Z
Great answer! Part of it is fundamental to science: Acknowledging the things we don’t know.
Pierre Paquette
CC BY-SA 4.0
2022-07-19T19:08:07.907Z
2022-08-08T07:54:39.737Z
2022-08-08T07:54:39.737Z
asdfex
<p><a href="http://www.astronomyknowhow.com/hydrogen-alpha.htm" rel="nofollow noreferrer">http://www.astronomyknowhow.com/hydrogen-alpha.htm</a></p>
<blockquote>
<p>The line that appears in the red part of the spectrum is created when an electron moves between the second and third orbit (N=2 and N=3) and the wavelength at which this occurs is 656nm. It is this line that is called the Hydrogen alpha line and hydrogen alpha filters are designed to block out as much of the spectrum as possible leaving only a very small bandwidth through which light can pass at the H-alpha frequency.</p>
</blockquote>
<p>I am seeing deep red when looking through the filter. Does this mean it is hydrogen alpha or just red light filtered from white light emitted by the sun passing at H-alpha frequency? If it is true, deep red color can be mistaken for H-alpha?</p>

CC BY-SA 4.0
2022-07-19T18:00:49.080Z
Your eyes could not see 3 nm anyway.
Uwe
CC BY-SA 4.0
2022-07-19T19:03:46.900Z
You should make it clear, is 3 nm a wavelength or a bandwidth?
Uwe
CC BY-SA 4.0
2022-07-20T21:06:40.533Z
@Uwe "Ha" or "Hα" is widely recognized as a wavelength specification. [H-alpha (Hα) is a specific deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm in air and 656.46 nm in vacuum...](https://en.wikipedia.org/wiki/H-alpha) so 3 nm will be widely recognized as a bandwidth.
uhoh
<observational-astronomy>
What am I looking at when seeing through a 3nm Hα filter?
would_like_to_be_anon
# What am I looking at when seeing through a 3nm Hα filter?

(Asked by: username_0 on 2022-07-19)

<http://www.astronomyknowhow.com/hydrogen-alpha.htm>

> The line that appears in the red part of the spectrum is created when an electron moves between the second and third orbit (N=2 and N=3) and the wavelength at which this occurs is 656nm. It is this line that is called the Hydrogen alpha line and hydrogen alpha filters are designed to block out as much of the spectrum as possible leaving only a very small bandwidth through which light can pass at the H-alpha frequency.

I am seeing deep red when looking through the filter. Does this mean it is hydrogen alpha or just red light filtered from white light emitted by the sun passing at H-alpha frequency? If it is true, deep red color can be mistaken for H-alpha?

**Tags:** observational-astronomy

> **Comments:**
> **username_1**: Your eyes could not see 3 nm anyway.  
> **username_1**: You should make it clear, is 3 nm a wavelength or a bandwidth?  
> **username_2**: @username_1 "Ha" or "Hα" is widely recognized as a wavelength specification. [H-alpha (Hα) is a specific deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm in air and 656.46 nm in vacuum...](https://en.wikipedia.org/wiki/H-alpha) so 3 nm will be widely recognized as a bandwidth.  

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## 1 answers

### Answer by username_3 on 2022-07-19. Score: 7 | Accepted answer

There is nothing special about light emitted by an electron jumping orbits in a hydrogen atom and any other light at the same frequency. There is no way to distinguish the source of light by the perceived colour or by adding any kind of filter.

Every object (at some not too low temperature) emits light at every wavelength in the visible spectrum. In addition to that, some (usually hot and hydrogen-rich) objects emit light from excited hydrogen atoms which appears at very specific wavelengths only, including one that we perceive as red. The filter you have only lets a small range of wavelengths (3 nm as you write in the title) through, cutting away most of the light produced due to temperature, but lets all of the H-alpha light through, therefore enhancing it's share and making it visible more easily.

If what you see is "normal" light or emitted by hydrogen atoms is impossible to say, but you can make a scientific guess or make an analysis of the whole spectrum of light to get a more precise identification. One possibility for measurement would be to use a second filter, e.g. one with the same bandwidth but a filter range that is shifted to 3 nm longer wavelengths. Thermal radiation would produce roughly the same amount of light in both filters, but H$\alpha$ would show up in only one of them. The ratio of both intensities gives you a measure for the contribution of the H$\alpha$ emission to the light you perceive. Unless, of course, there is another emission line from some other atom unknown to you that only shows up in the second filter...

> **Comments:**
> **username_4**: Great answer! Part of it is fundamental to science: Acknowledging the things we don’t know.  

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