Will a cloud of atoms become ionized in space at 1 au? How hot will the cloud be?

Question

I'm exploring this question out of personal curiosity. If I take a cloud of atoms of a given element and release it in space at a distance of Earth's orbit from the sun (but not so close to Earth as to be affected by Earth's atmosphere/gravity/mag-field/radiation belts, etc.), will the neutral cloud turn into an ionized plasma? Of course a few of the atoms will always be neutral regardless of the radiation field intensity, so for specificity let's say an ionization fraction of at least 95% or so. I think that I could use the Saha equation for this:

$$ \frac{N_{i+1}}{N_i}=\frac{2 Z_{i+1}}{n_e Z_i}\left(\frac{2\pi m_e k T}{h^2}\right)^{3/2} \exp\left(\frac{-\chi_i}{kT}\right) $$

Naturally it depends on the electron density, $n_e$, and the ionization energy, $\chi_i$. But to use it I also need to know the temperature of the cloud of atoms. How would I determine this? I know the diffuse solar wind has an electron temperature ~140000 K, would the cloud equilibrate with the electrons? Or would it be driven toward another temperature?

And is the Saha equation even the correct approach given that for some elements the solar radiation will be able to directly photo-eject electrons?

Answer 1

I did some more thinking and digging:

I think the Saha equation is not necessarily the right way to answer this question, as it requires the gas be in thermal equilibrium, which may or may not be the case in a space environment. A better back-of-the-envelope method might be to try to follow the recombination rate = ionizing photon rate method that is often used in calculating HII regions. (See for instance: lecture notes on HII regions.) This too, is an approximation, as it doesn't consider all sources and sinks of the ionized species. Energetic electrons in the solar wind, for example, can also be significant sources of ionization (they have enough kinetic energy to liberate other electrons from neutral atoms.) Turning to the literature, I found this paper, which discusses neutral hydrogen in the solar system's interplanetary medium. With their more detailed analysis, they find a neutral to ionized hydrogen ratio of less than $10^{-6}$ at 1 au (shown in their Fig. 2.) That's plenty ionized enough for me, and since hydrogen has one of the highest ionization energies, I can assume that a cloud composed of most elements (those with lower ionization energies) will be ionized.

Answer 2

First of all, without something to contain it, any cloud would quickly disperse in seconds due to the thermal speed of the atoms. And if you put it in an actual container, you would block either any ionizing photons or particles or both. So realistically, you really need gravity to contain the cloud if you want to have it ionized.

Now as for the dominant source of ionization in your case:

the solar wind at 1 AU has a plasma density of the order of $10/cm^3$ and a velocity of about $400 km/sec$. This amounts to an electron flux of $4*10^8/cm^2/sec$. Now this is negligible compared to the solar EUV photon flux at 1 AU (which is about $10^{11}/cm^2/sec$), considering that collisional ionization by electrons with a temperature of $140,000K$ has about the same cross section as photoionization, about $10^{-17} cm^2$).

So any cloud would be ionized by the ultraviolet radiation of the sun only. The degree of ionization in a state of ionization/recombination equilibrium can be calculated from the equation $$(1)\:\:n_p^2 \alpha = \sigma F n_n$$

where $n_p$ is the plasma density, $\alpha$ the recombination coefficient, $\sigma$ the ionization cross section, F the ionizing flux (in this case the EUV photon flux) and $n_n$ the density of neutrals (or only partially ionized) atoms. Now if we define $$N =n_p+n_n$$ we have $$n_p =N q$$ $$n_n = N (1-q)$$ where q is the degree of ionization. Inserting this into (1) gives a quadratic equation for q which has the solution

$$(2)\:\:q = \frac{\sigma F}{2 \alpha N} ( \sqrt{ 1+\frac{4 \alpha N}{\sigma F}} -1)$$

With $$F=10^{11}/cm^2/sec$$ $$\sigma=10^{-17}cm^2$$ $$\alpha=10^{-9}cm^3/sec$$

we have in this case

$$q = \frac{500}{N} ( \sqrt{ 1+4*10^{-3}*N} -1)$$

This means for instance that for standard atmospheric pressure ($N=2.7*10^{19}/cm^3$) the degree of ionization would be less than $10^{-8}$; for the earth's ionosphere ($N=10^8/cm^3$) it is $3*10^{-3}$. To get a degree of ionization higher than 90%, the density N has to be $100/cm^3$ or less. If you do a Taylor expansion of the square root in Eq.(2) for small values of the fraction term, you get

$$q\approx 1-\frac{\alpha N}{\sigma F}$$

so you see that for an ionization rate close to 1 you need either a very high ionizing flux or a very low density. With the ionizing flux as given by the solar radiation at 1AU, the required density is so low that that the 'cloud' would practically blend into the interplanetary medium. You also have to take into account that the time $1/\sigma F$ required on average to ionize an atom is $10^6$ sec = 11 days here, so it takes a very long time before such a high degree of ionization is reached under these circumstances.