<p>A recombination line is a special case of an emission line.</p>
<h3>Emission lines</h3>
<p>An emission line is any spectral feature that rises above the <em>continuum</em> — i.e. the average amplitude of the spectrum (in some wavelength region) — and is due to atomic transitions (where &quot;atomic&quot; include atoms in molecules and dust grains, and &quot;transitions&quot; may be electronic, vibrational, or rotational).</p>
<p>These transitions may arise from absorption of light, or collisions of an atom with another particle. If the gas emitting is hot and dilute, this other particle will usually be a free electron, since these are much faster than atoms. If it's cold and dense, it will likely be an atom. The absorption of energy puts the atom in an excited state, and when the atom de-excites, a photon is emitted.</p>
<h3>Recombination lines</h3>
<p>Recombination is the process of a free electron being &quot;caught&quot; by an atom missing an electron, i.e. an <em>ion</em>. The ion, in turn, has been produced previously by being ionized by some high-energy process, e.g. an ionizing photon or a violent collision (by &quot;ionizing&quot; we mean &quot;a photon with sufficiently high energy that it is able to ionize an atom&quot;; this typically means <span class="math-container">$\gtrsim\!10\,\mathrm{eV}$</span>). The difference (roughly) between the kinetic energy of the electron before the encounter, and the energy of the state it goes to, is emitted as a photon.</p>
<p>The electron may go directly to the ground state, in which case another ionizing photon is just produced. But it may also go to an intermediate state, emitting a lower-energy photon. The electron is then in an excited state, from which it will soon de-excite, possibly through several levels until it reaches the ground state. This process is called <em>cascading</em>.</p>
<p>The photons that are emitted during this cascade are called recombination lines.</p>
<h3>Recombination lines from gas around stars</h3>
<p>In the vicinity of very hot stars (O and B stars) which produce many ionizing photons (the Lyman continuum; LyC), you will have a region of neutral hydrogen gas (HI) being ionized. Timescales for recombination in these regions are quite small, so the LyC is &quot;almost immediately&quot; reprocessed into recombination lines.</p>
<p>Quantum mechanics tells you the probabilities of ending up in the various states, and it turns out that, for instance, for each LyC photon you get roughly 0.68 Lyman <span class="math-container">$\alpha$</span> photons (with a small dependence on temperature), i.e. photons emitted when the electron falls from the first excited to the ground state. In terms of energy, this translates into ~1/3 of the total power.</p>
<p>This is actually quite amazing: It means that 1/3 of the total continuum of photons more energetic than the 13.6 eV needed to ionize hydrogen is converted into a <em>single</em> emission line! For this reason, galaxies — in particular young galaxies that still host many hot stars — are often quite luminous in Lyman <span class="math-container">$\alpha$</span>; sometimes even <em>only</em> in Lyman <span class="math-container">$\alpha$</span>.</p>
<h3>Absorption lines</h3>
<p>Emission lines are related to <em>absorption</em> lines: In the first case, you have a continuum of light (e.g. from thermal processes) with some physical process &quot;adding&quot; some extra light to the total spectrum. In the latter case, you have <em>the same</em> physical process <em>removing</em> light from a continuum.</p>
<p>For instance, if you observe a (non-ionizing) source through an HI cloud, you'll see <em>absorption</em> lines at the wavelengths corresponding to the various transitions of neutral hydrogen. If you were able to observe the same cloud from another angle, you'd see <em>emission</em> lines at the same wavelengths.</p>
<p><a href="https://i.sstatic.net/4aDavOvL.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4aDavOvL.png" alt="spectra" /></a>
<sup>A luminous source, e.g. a star (<em>left</em>) emits light at all wavelengths. If observed directly (<em>top eye</em>), the spectrum therefore shows all colors. An intervening gas cloud (<em>middle</em>) absorbs light at certain wavelengths, given by its chemical composition, and thus observing the source through this cloud (<em>middle eye</em>) lacks light at these wavelengths. This is called absorption lines. When the gas cloud de-excites, light is emitted in all directions, so when observing only the cloud (<em>bottom eye</em>), <em>emission</em> lines are seen at the same distinct wavelengths.</sup></p>

CC BY-SA 4.0
2019-12-10T17:21:13.257Z
Thanks a lot for the reply!
Rian
CC BY-SA 4.0
2019-12-11T03:05:18.957Z
So can we say that emission line is due to atoms and recombination line is due to ions?
Rian
CC BY-SA 4.0
2019-12-11T14:57:22.217Z
@Rian I wouldn't put it that way, since ions can also produce e.g. collisional emission lines, and since in principle you can have a recombination of a non-ionized atom (e.g. HI + e- → H-). In astronomy, we often use the term "atom" for both neutral and ionized atoms.
pela
CC BY-SA 4.0
2019-12-10T13:57:28.837Z
2024-12-27T21:12:12.483Z
2024-12-27T21:12:12.483Z
pela
<p>When an electron jumps from higher level to lower level a photon is emitted. This is seen as an emission line. But what exactly is a recombination line? I found them similar. Can you please tell me what's the difference between these two?</p>

CC BY-SA 4.0
2019-12-10T12:54:37.657Z
This might be a question rather for physics.stackexchange.com
usernumber
<observational-astronomy><spectroscopy><star-formation><stellar-astrophysics>
What is the difference between emission line and recombination line?
Rian
# What is the difference between emission line and recombination line?

(Asked by: username_0 on 2019-12-10)

When an electron jumps from higher level to lower level a photon is emitted. This is seen as an emission line. But what exactly is a recombination line? I found them similar. Can you please tell me what's the difference between these two?

**Tags:** observational-astronomy, spectroscopy, star-formation, stellar-astrophysics

> **Comments:**
> **username_1**: This might be a question rather for physics.stackexchange.com  

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## 1 answers

### Answer by username_2 on 2019-12-10. Score: 9

A recombination line is a special case of an emission line.

### Emission lines

An emission line is any spectral feature that rises above the *continuum* — i.e. the average amplitude of the spectrum (in some wavelength region) — and is due to atomic transitions (where "atomic" include atoms in molecules and dust grains, and "transitions" may be electronic, vibrational, or rotational).

These transitions may arise from absorption of light, or collisions of an atom with another particle. If the gas emitting is hot and dilute, this other particle will usually be a free electron, since these are much faster than atoms. If it's cold and dense, it will likely be an atom. The absorption of energy puts the atom in an excited state, and when the atom de-excites, a photon is emitted.

### Recombination lines

Recombination is the process of a free electron being "caught" by an atom missing an electron, i.e. an *ion*. The ion, in turn, has been produced previously by being ionized by some high-energy process, e.g. an ionizing photon or a violent collision (by "ionizing" we mean "a photon with sufficiently high energy that it is able to ionize an atom"; this typically means $\gtrsim\!10\,\mathrm{eV}$). The difference (roughly) between the kinetic energy of the electron before the encounter, and the energy of the state it goes to, is emitted as a photon.

The electron may go directly to the ground state, in which case another ionizing photon is just produced. But it may also go to an intermediate state, emitting a lower-energy photon. The electron is then in an excited state, from which it will soon de-excite, possibly through several levels until it reaches the ground state. This process is called *cascading*.

The photons that are emitted during this cascade are called recombination lines.

### Recombination lines from gas around stars

In the vicinity of very hot stars (O and B stars) which produce many ionizing photons (the Lyman continuum; LyC), you will have a region of neutral hydrogen gas (HI) being ionized. Timescales for recombination in these regions are quite small, so the LyC is "almost immediately" reprocessed into recombination lines.

Quantum mechanics tells you the probabilities of ending up in the various states, and it turns out that, for instance, for each LyC photon you get roughly 0.68 Lyman $\alpha$ photons (with a small dependence on temperature), i.e. photons emitted when the electron falls from the first excited to the ground state. In terms of energy, this translates into ~1/3 of the total power.

This is actually quite amazing: It means that 1/3 of the total continuum of photons more energetic than the 13.6 eV needed to ionize hydrogen is converted into a *single* emission line! For this reason, galaxies — in particular young galaxies that still host many hot stars — are often quite luminous in Lyman $\alpha$; sometimes even *only* in Lyman $\alpha$.

### Absorption lines

Emission lines are related to *absorption* lines: In the first case, you have a continuum of light (e.g. from thermal processes) with some physical process "adding" some extra light to the total spectrum. In the latter case, you have *the same* physical process *removing* light from a continuum.

For instance, if you observe a (non-ionizing) source through an HI cloud, you'll see *absorption* lines at the wavelengths corresponding to the various transitions of neutral hydrogen. If you were able to observe the same cloud from another angle, you'd see *emission* lines at the same wavelengths.

[![spectra](https://i.sstatic.net/4aDavOvL.png)](https://i.sstatic.net/4aDavOvL.png)
A luminous source, e.g. a star (*left*) emits light at all wavelengths. If observed directly (*top eye*), the spectrum therefore shows all colors. An intervening gas cloud (*middle*) absorbs light at certain wavelengths, given by its chemical composition, and thus observing the source through this cloud (*middle eye*) lacks light at these wavelengths. This is called absorption lines. When the gas cloud de-excites, light is emitted in all directions, so when observing only the cloud (*bottom eye*), *emission* lines are seen at the same distinct wavelengths.

> **Comments:**
> **username_0**: Thanks a lot for the reply!  
> **username_0**: So can we say that emission line is due to atoms and recombination line is due to ions?  
> **username_2**: @username_0 I wouldn't put it that way, since ions can also produce e.g. collisional emission lines, and since in principle you can have a recombination of a non-ionized atom (e.g. HI + e- → H-). In astronomy, we often use the term "atom" for both neutral and ionized atoms.  

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