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To my understanding, the sun contains large amounts of ionized hydrogen compared to the amount of hydrogen in the first excited state. I am curious about the explanation, at the quantum scale, for how this large amount of ionization happens.

I have seen the potential explanation that there are infinite degeneracies for ionized states, thus, statistically, it is most likely that these states are occupied. However, this seems dubious to me. I have not been able to find another explanation.

What is the mechanism for this process?

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    $\begingroup$ Have you considered the temperatures in the Sun? $\endgroup$
    – Ghoster
    Oct 12, 2023 at 5:47
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    $\begingroup$ The answer to your question does not furnish a mechanism of the frequent ionization of hydrogen on the surface of the sun. @Ghoster $\endgroup$ Oct 12, 2023 at 7:21
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    $\begingroup$ @SillyGoose Thermal equilibrium is the same regardless of the mechanism by which it is achieved. I’ll let the OP contemplate the multiple mechanisms that are exciting the hydrogen. They may be interesting but they’re irrelevant. $\endgroup$
    – Ghoster
    Oct 12, 2023 at 8:01
  • $\begingroup$ I think I am asking for the mechanism. @Ghoster $\endgroup$ Oct 12, 2023 at 8:45

2 Answers 2

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The hydrogen is radiatively ionised - i.e. the photoelectric effect - though there will be some collisional ionisation too. Once you are even a few 100 km below the photosphere, the local radiation field is very close to a blackbody, with an average photon energy of $\sim 3k_B T$, where $T$ is the local temperature.

The ionisation energy of hydrogen is just 13.6 eV. The average photon energy exceeds this for $T>50\,000$ K, which is achieved around 1% of the way into the Sun. However, even for temperatures above $\sim 10\,000$ K there are easily enough photons with energies above 13.6 eV to keep the hydrogen ionised.

The hydrogen at the photosphere isn't mainly ionised, it is mainly neutral. Is it possible you are asking why there is more ionised hydrogen than hydrogen in the $n=2$ level at the low temperatures found in the photosphere? Well, only photons with a very small range of energies are capable of radiatively exciting hydrogen from the ground state into the $n=2$ states. There are far more photons that are capable of ionising hydrogen from the ground state or from the first excited state or from the second excited state etc. Then, if the plasma electron density is low, the rate at which those ions and electrons can recombine to form atoms is also very low and this allows a significant fraction of ionised hydrogen even at (surprisingly) low temperatures. i.e. The ionisation rate is proportional to the number of atoms, but the recombination rate is proportional to the number of ions multiplied by the number of electrons. For a given ionisation fraction, the recombination rate depends quadratically on density, while the ionisation rate is linear. This means lower densities require lower temperatures to give the same ionisation fraction.

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  • $\begingroup$ Thanks for the answer! If it is just that there is enough energy in the sun to propel the photoelectric effect, then this explains why there is so much ionized hydrogen, but not why there is so much more than the first state, which I understand to be extremely rarely populated. As far as I understand, also not all of the hydrogen is ionized, so I don't think it can be that the reason is that it's all ionized. $\endgroup$ Oct 12, 2023 at 23:04
  • $\begingroup$ wow, very cool answer. Or perhaps hot, at least ~10, 000K. -NN $\endgroup$ Oct 13, 2023 at 21:59
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In local thermodynamic equilibrium, the number of particles in two states $n_0, n_1$ with energy $E_0, E_1$ satisfies the Boltzmann equation $$\frac{n_1}{n_0}=\frac{g_{1}}{g_{0}} \exp \left[-\left(E_{1}-E_{2}\right) / k_B T\right]$$

For Sun, the degree of ionization of a gas in local thermal equilibrium is described by the Saha equation (which is a special case of the Boltzmann equation). This is a function of the temperature $T$, density $n$, and ionization energies of the atoms ($\epsilon$). The equation for hydrogen gas takes the form

$$ \frac{n_{e}^{2}}{n-n_{e}}=\frac{2}{\lambda^{3}} \frac{g_{1}}{g_{0}} \exp \left[\frac{-\epsilon}{k_{B} T}\right]$$ where

\begin{array}{l}- (n - n_e) \text { is the density of unionised H atoms } \\ -g_{i} \text { is the degeneracy of states for the } i \text {-ions } \\ \text {- } n_{e} \text { is the density of ionized gas } \\ \text { - } \lambda \text { is the thermal de Broglie wavelength of an electron }\end{array}

The mechanism for ionization is just the old photo-electric effect. Now if you sub in the relevant temperature of the sun, then you will find that nearly all hydrogen has to be in the ionized state. This is because $\epsilon$ is very tiny in comparison.

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