# Plasma recombination rate

Let's say I ionize a gas and create a plasma. I know this doesn't last very long because they recombine. But I don't understand why. Why do electrons give up their energy in the form of light and recombine with the atom. If they have sufficient energy, then why don't they just stay that way?

So, the question I have in my mind is: If I have a moving charged plasma, and I add some electrons, what would the recombination rate depend on?

I was thinking that if the electrons have low velocity components relative to the moving plasma, it would be faster, and also if the temperature of the plasma is low. Density? Etc.

I was wondering if there was a method or equation governing this phenomenon to calculate the plasma recombination rate considering all these factors. (Any research papers along with the answer would be appreciated.).

Thanks!

• Can you link some of the answers (assuming they're from this site) you've seen before? – Kyle Kanos Feb 10 '17 at 12:15
• – honeste_vivere Feb 10 '17 at 14:45
• I've read them. The first one was the one I couldn't understand. And the second is not my question – Chandrahas Feb 11 '17 at 2:44
• @Chandrahas: Both electrons and ions loose energy through collisions with the neutrals in a high-density, low-ionisation plasma, that you seem to be talking about. Because they're ionized you have an electric field, and through collisions the kinetic energy of ionized particles soon is lower than the energy stored in the electric field, which will force them to follow field lines and recombine. Sorry,I can't give recombination rates though. I think that was addressed in the other answers. – AtmosphericPrisonEscape Feb 12 '17 at 19:37
• Have you looked at https://en.wikipedia.org/wiki/Saha_ionization_equation? – honeste_vivere Feb 14 '17 at 13:29

## Why does recombination occur at all?

The recombination of a free electron with a positive charge results in the emission of a photon. This is the time-reversed process of photoionization. Generally speaking, if nature allows for some process to occur, then it will also allow for its time-reversed process. One reason to see why is to consider a system in thermal equilibrium. Such a system will be in detailed balance. A consequence is that the cross-section for a process like photoionization ($$\sigma_{bf}$$) will be related to the cross-section for its inverse process of recombination ($$\sigma_{fb}$$), and we can use thermal equilibrium to relate their values. In this specific case of photoionization and recombination, this is expressed in the Milne relations. Since these cross sections are intrinsic properties of photons and electrons, we can then apply these values to conditions outside of thermal equilibrium.

Incidentally, finding the correct values of $$\sigma_{bf}$$ (and hence $$\sigma_{fb}$$ via the Milne relation) is perhaps the biggest obstacle to getting quantitatively accurate predictions for ionization and recombination. Naturally the situation is best understood for hydrogen, but the situation gets murkier for many atomic levels for many heavier elements.

## Dependence on density and temperature

Consider a number density of electrons $$n_e$$, traveling at velocity $$v$$, a number density of positive charges $$n_i$$, and the recombination cross section $$\sigma_{fb,nl}(v)$$ to a bound atomic level with quantum numbers $$n$$ and $$l$$ . Then the rate of recombinations per unit volume is

$${\cal R_{v,nl}} = n_e n_i \sigma_{fb,nl}(v) v$$

However, in a typical plasma the electrons will have a thermal distribution of velocities, specified by the Maxwell-Boltzmann distribution, $$f(v;\,T)$$. Here my notation is meant to suggest that at a given temperature $$T$$, $$f$$ will provide you with a distribution for the velocities. To account for this range of speeds, we must perform an integral over all possible speeds, weighted by $$f(v; \, T)$$ to find the actual recombination rate in the plasma to this atomic level.

$${\cal R_{nl}} = n_e n_i \int \sigma_{fb,nl}(v) \, v \, f(v; \,T) \,dv$$

This integral (not including the factors of $$n_e$$ and $$n_i$$ out front) has units of volume per time and is known as the recombination coefficient. You can see that it is temperature dependent, and it is often denoted as $$\alpha_{nl}(T)$$. Explicitly,

$${\cal R_{nl}} = n_e n_i \alpha_{nl}(T)$$

Higher temperatures cause $$f(v, T)$$ to give more weight to electrons moving at faster velocity. This leads to two competing effects. First, higher velocities mean the electrons will encounter more positive charges per unit time -- this is accounted for by the factor of $$v$$ in the first equations. On the other hand, the recombination cross section $$\sigma(v)$$ decreases at higher velocities, as you correctly hypothesized in your question -- faster moving electrons have an easier time flying right by a positive charge without being captured.

The second effect tends to dominate, and for a wide range of temperatures $$\alpha$$ tends to scale roughly as $$T^{-1/2}$$.

The number of positive charges in the plasma will usually be proportional to the number of negative charges, since most material is net neutral. So we can summarize by saying that the recombination rate scales roughly as density squared and temperature to -1/2 power.

## Plugging in some numbers

For a hydrogen plasma at $$5000$$ Kelvin, $$\alpha$$ for recombination to the ground state is about $$2.3 \times 10^{-13}$$ cm$$^{-3}$$ $${\rm s}^{-1}$$. If the temperature is raised by a factor of 4 to $$20,000$$ K, $$\alpha$$ drops to about $$1.1 \times 10^{-13} {\rm cm}^{-3}$$ s$$^{-1}$$

The typical time for an electron to recombine to level of interest is then $$1/[ n_e \alpha_{nl}(T)]$$. So for an electron density of $$10^{23}$$ particles per cubic centimeter, at $$5000$$ K electrons recombine to the ground state in a time of about $$10^{-10}$$ seconds. But that doesn't mean that the gas will necessarily become totally neutral in that time, because thermal processes (photons or electron collisions) will also be providing a competing ionization rate. There might also be an external source of ionizing photons that maintain the plasma's ionization state in photoionization equilibrium, as in a nebula.

There are more complications to consider. We've been considering recombination to one specific atomic level. For a complete calculation, you'd need to find $$\alpha(T)$$ for each level, and sum their contributions. Moreover, for recombinations to the ground state, the photon that is released can go on to ionize other atoms. For a sufficiently dense plasma, a useful short cut in this case is to sum the contributions of recombination to all levels except for the ground state and use this as the effective recombination coefficient. This is known as "case B" recombination, or the "on-the-spot" approximation, since it assumes that all ionizing photons released during recombination are re-absorbed right away, effectively cancelling out their contribution.

• So, what we have to do is find the recombination coefficient for all levels except for the ground state and sum them up. But can't an electron have an infinite number of exited states? And can we approximate the cross section to just depend on the principal quantum number (Just use the atomic radius)? – Chandrahas Feb 16 '17 at 5:54
• Can I just consider the velocity distribution to be constant if the electrons didn't have sufficient time to attain thermal equilibrium. Approximately how will it take to attain thermal equilibrium? Thanks – Chandrahas Feb 16 '17 at 5:57
• Good questions. 1) There are indeed an infinite number of excited states, but the recombination coefficients decrease rapidly enough at higher energy states that the sum converges to a finite value. For ionized hydrogen at 5000 K, this sum is $6.8 \times 10^{-13}$, and subtracting out the ground state leaves you with $4.5 \times 10^{-13}$. 2) The electron-electron collision cross section is generally much larger than the electron-ion recombination cross section, so the electrons thermalize rapidly and it is usually appropriate to assume a thermal distribution for their velocities. – kleingordon Feb 16 '17 at 17:18
• Regarding quantum numbers: it depends on what level of detail you are looking for. If you care about tracking everything exactly, it turns out there is a difference in the rate of recombination to different $l$ states at a given $n$ value. – kleingordon Feb 16 '17 at 17:21
• I also forgot to mention that, at sufficiently high densities, collisional ionization (electron collides with an atom/ion, knocking off a bound electron) will also be playing an important role in keeping the plasma ionized. Depending on your application, either photoionization or collisional ionization can be the dominant effect that keeps the plasma ionized. To find the equilibrium situation, you need to balance the appropriate ionization rate with the recombination rate, which you calculate as I've discussed in my answer – kleingordon Feb 16 '17 at 23:06