<blockquote>
<p>but why is this? My thinking is that it is because there are minimum energy transitions of infrared, causing less light to be absorbed, but when I look at the hydrogen Bohr model there are clearly transitions that produce or absorb infrared light.</p>
</blockquote>
<p>In Astronomy, the  effects of &quot;dust&quot; is considered separately from that of gas or plasma (electrons).</p>
<p>So for example, if there is a lot of free atomic, or molecular hydrogen between you and the object you want to observe, you will indeed see absorption lines corresponding to those species. Because the atoms are far (on an atomic scale) from each other and mostly undisturbed, the absorption will happen at only those narrow bands. There will be some broadening due to collisions collisional broadening, finite temperature (thermal broadening) and if there are macroscopic structures moving in various directions, Doppler broadening. (for more on that start with <a href="https://en.wikipedia.org/wiki/Spectral_line#Broadening_due_to_non-local_effects" rel="nofollow noreferrer">Spectral line; Broadening due to non-local effects</a>).</p>
<p>But for all the spectral space in-between these absorption lines the light will still pass through and be available for imaging.</p>
<p><a href="https://en.wikipedia.org/wiki/Cosmic_dust" rel="nofollow noreferrer">Cosmic dust</a> is treated separately. These very small, solid particles might absorb some light, but they might also scatter even more. And this is where the wavelength becomes important. Even transparent particles will scatter light in many directions, thereby making direct imaging possible.</p>
<p>For dust particles <strong>smaller</strong> than a wavelength, we can look at <a href="https://en.wikipedia.org/wiki/Rayleigh_scattering" rel="nofollow noreferrer">Rayleigh scattering</a> for guidance. The scattering cross-section varies as <span class="math-container">$1/ \lambda^4$</span>, so the longer the wavelength of a photon, the much higher probability that it will pass a given particle without scattering.</p>
<p>And it seems that the dust particles in question must be smaller than a visible wavelength, because going to IR does allow for some transparency.</p>

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2024-05-20T05:24:07.567Z
2024-05-21T01:08:29.583Z
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uhoh
<p>Why do we use Infrared to view through clouds of gas and dust in space? I understand that it is because it can pass through without being absorbed, but why is this? My thinking is that it is because there are minimum energy transitions of infrared, causing less light to be absorbed, but when I look at the hydrogen Bohr model there are clearly transitions that produce or absorb infrared light.</p>
<p><a href="https://i.sstatic.net/3H8C6vlD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3H8C6vlD.png" alt="enter image description here" /></a></p>
<p>Are these transitions less probable? Is there another reason why Infrared passes through nebulae with minimal extinction?</p>

<light><spectroscopy><nebula><infrared><extinction>
Why does infrared pass through clouds of gas and dust with minimal extinction?
Astrovis
# Why does infrared pass through clouds of gas and dust with minimal extinction?

(Asked by: username_0 on 2024-05-20)

Why do we use Infrared to view through clouds of gas and dust in space? I understand that it is because it can pass through without being absorbed, but why is this? My thinking is that it is because there are minimum energy transitions of infrared, causing less light to be absorbed, but when I look at the hydrogen Bohr model there are clearly transitions that produce or absorb infrared light.

[![enter image description here](https://i.sstatic.net/3H8C6vlD.png)](https://i.sstatic.net/3H8C6vlD.png)

Are these transitions less probable? Is there another reason why Infrared passes through nebulae with minimal extinction?

**Tags:** light, spectroscopy, nebula, infrared, extinction

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## 1 answers

### Answer by username_1 on 2024-05-20. Score: 2 | Accepted answer

> but why is this? My thinking is that it is because there are minimum energy transitions of infrared, causing less light to be absorbed, but when I look at the hydrogen Bohr model there are clearly transitions that produce or absorb infrared light.

In Astronomy, the effects of "dust" is considered separately from that of gas or plasma (electrons).

So for example, if there is a lot of free atomic, or molecular hydrogen between you and the object you want to observe, you will indeed see absorption lines corresponding to those species. Because the atoms are far (on an atomic scale) from each other and mostly undisturbed, the absorption will happen at only those narrow bands. There will be some broadening due to collisions collisional broadening, finite temperature (thermal broadening) and if there are macroscopic structures moving in various directions, Doppler broadening. (for more on that start with [Spectral line; Broadening due to non-local effects](https://en.wikipedia.org/wiki/Spectral_line#Broadening_due_to_non-local_effects)).

But for all the spectral space in-between these absorption lines the light will still pass through and be available for imaging.

[Cosmic dust](https://en.wikipedia.org/wiki/Cosmic_dust) is treated separately. These very small, solid particles might absorb some light, but they might also scatter even more. And this is where the wavelength becomes important. Even transparent particles will scatter light in many directions, thereby making direct imaging possible.

For dust particles **smaller** than a wavelength, we can look at [Rayleigh scattering](https://en.wikipedia.org/wiki/Rayleigh_scattering) for guidance. The scattering cross-section varies as $1/ \lambda^4$, so the longer the wavelength of a photon, the much higher probability that it will pass a given particle without scattering.

And it seems that the dust particles in question must be smaller than a visible wavelength, because going to IR does allow for some transparency.


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