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Saha-Boltzmann Hydrogen Ionization

What you are seeing: the ionization fraction x=n+/ntotx = n_+/n_\text{tot} of a pure-hydrogen plasma as a function of temperature, at the user-set total number density. The Saha equation x2/(1−x)=Saha(T)/ntotx^2/(1-x) = \mathrm{Saha}(T) / n_\text{tot} with Saha(T)=(2πmekBT/h2)3/2exp⁡(−χ/kBT)\mathrm{Saha}(T) = (2\pi m_e k_B T / h^2)^{3/2} \exp(-\chi / k_B T) is solved as a closed-form quadratic in xx.

For solar-photosphere densities (n∼1023n \sim 10^{23} m−3^{-3}, T=5800T = 5800 K), hydrogen is only about 10−410^{-4} ionized; for the chromosphere (T∼104T \sim 10^4 K) it crosses 50 percent. The classic ratio χ/kB≈158000\chi / k_B \approx 158000 K is deceptive because the prefactor ∝T3/2/n\propto T^{3/2}/n pushes the half-ionization temperature down by an order of magnitude.

Figure 1. Saha ionization fraction x(T,n)x(T, n) for hydrogen. Method: closed-form quadratic in xx from the Saha-Boltzmann balance.
log10 n (1/m^3)20.00
T (K)8000

WHAT TO TRY

  • Raise the temperature T: the hydrogen ionization fraction x climbs through a sharp transition, half-ionized around 10000 to 15000 K. The Saha equation balances ionization against recombination.
  • Lower the total density (log10 n): the ionization curve shifts to lower temperature, because a thinner gas recombines less often, so it ionizes more easily. Density and temperature trade off.
  • Read T_ion in the rail: it marks where x = 1/2. This is why the Sun photosphere at 5800 K is mostly neutral while hotter stars show ionized hydrogen.