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?

  • $\begingroup$ This might be a question rather for physics.stackexchange.com $\endgroup$
    – usernumber
    Dec 10, 2019 at 12:54

1 Answer 1


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. In astrophysics, this other particle will usually be a free electron, since these are much faster than atoms. 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. The difference 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 iozing 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 false form 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.

  • 1
    $\begingroup$ Thanks a lot for the reply! $\endgroup$
    – Rian
    Dec 10, 2019 at 17:21
  • $\begingroup$ So can we say that emission line is due to atoms and recombination line is due to ions? $\endgroup$
    – Rian
    Dec 11, 2019 at 3:05
  • $\begingroup$ @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. $\endgroup$
    – pela
    Dec 11, 2019 at 14:57

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