The Saha Ionization Equation

Derivation

The Saha Ionization Equation

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Statement

For an ideal gas in thermal and chemical equilibrium at temperature \(T\), the number densities of two adjacent ionization stages \(i\) and \(i+1\) of a chemical species, together with the free-electron density \(n_e\), obey \[ \frac{n_{i+1}\,n_e}{n_i} = \frac{2\,Z_{i+1}}{Z_i}\left(\frac{2\pi m_e k_B T}{h^2}\right)^{3/2} e^{-\chi_i/(k_B T)}, \] where \(Z_i,\,Z_{i+1}\) are the internal partition functions of the two stages, \(m_e\) the electron mass, \(\chi_i\) the ionization energy of stage \(i\), and the factor \(2\) is the free-electron spin degeneracy.

Why it matters

The Saha equation is the bridge between the microscopic energy \(\chi_i\) and the macroscopic state of a plasma. It tells us how far a gas is ionized given only its temperature and electron density, and it underlies the entire enterprise of stellar spectroscopy: it is why the same element produces radically different spectra at different depths and in different stars, and it is what Cecilia Payne used to show that stars are made mostly of hydrogen.

Beyond astrophysics it governs ionization equilibrium in laboratory plasmas, flames, semiconductor dopant statistics (in modified form), and cosmological recombination — the epoch at which the Universe became transparent.

Assumptions

Local thermodynamic equilibrium (LTE) at a single temperature \(T\).If dropped, the same \(T\) no longer governs translation, excitation and ionization simultaneously; one must solve coupled statistical-equilibrium rate equations (non-LTE), as in stellar coronae and diffuse nebulae.

The electrons and ions form a classical, nondegenerate ideal gas obeying Maxwell–Boltzmann statistics.If dropped (electrons degenerate, \(n_e\lambda_e^3\gtrsim 1\)), the chemical potential must come from Fermi–Dirac statistics; the simple \(\lambda_e^{-3}\) factor is wrong, as in white-dwarf and neutron-star matter.

Particles are non-interacting: no Coulomb coupling between charges, isolated energy levels.If dropped, one needs continuum lowering, Debye–Hückel corrections and pressure ionization; the internal partition function must be truncated because an isolated Coulomb series diverges.

Only two adjacent stages are coupled and the electron is nonrelativistic, \(k_B T \ll m_e c^2\).If dropped, multiple ionization stages and, at \(k_B T\sim m_e c^2\), electron–positron pair creation enter, changing the electron reservoir entirely.

Derivation

\[ \mathrm{A}_{i} \;\rightleftharpoons\; \mathrm{A}_{i+1} + e^{-},\qquad \mu_i = \mu_{i+1} + \mu_e \]

Treat ionization as a reversible reaction. At fixed \(T,V\) the Helmholtz free energy is stationary against transferring one electron, which is exactly the equality of chemical potentials for reactants and products. B

\[ n_s = \frac{Z_s}{\lambda_s^{3}}\,e^{(\mu_s - E_s)/(k_B T)},\qquad \lambda_s \equiv \frac{h}{\sqrt{2\pi m_s k_B T}} \]

For a nondegenerate ideal gas the number density of species \(s\) follows from the grand-canonical (or classical) partition function: \(Z_s\) is the internal partition function measured from that species' ground state, \(E_s\) is the ground-state energy on a common absolute scale, and \(\lambda_s^{-3}=(2\pi m_s k_B T/h^2)^{3/2}\) is the translational quantum concentration. C

\[ e^{\mu_s/(k_B T)} = \frac{n_s\,\lambda_s^{3}}{Z_s}\,e^{E_s/(k_B T)} \]

Solve Step 2 for the fugacity of each species; this is pure algebra, isolating \(\mu_s\) so the equilibrium condition can be imposed multiplicatively. A

\[ \frac{n_i\lambda_i^{3}}{Z_i}e^{E_i/(k_B T)} = \frac{n_{i+1}\lambda_{i+1}^{3}}{Z_{i+1}}e^{E_{i+1}/(k_B T)}\cdot\frac{n_e\lambda_e^{3}}{Z_e}e^{E_e/(k_B T)} \]

Insert Step 3 into \(\mu_i=\mu_{i+1}+\mu_e\), i.e. \(e^{\mu_i/k_BT}=e^{\mu_{i+1}/k_BT}\,e^{\mu_e/k_BT}\). The exponential of a sum becomes a product. A

\[ \frac{n_{i+1}\,n_e}{n_i} = \frac{Z_{i+1}Z_e}{Z_i}\,\frac{\lambda_i^{3}}{\lambda_{i+1}^{3}\lambda_e^{3}}\; e^{-(E_{i+1}+E_e-E_i)/(k_B T)} \]

Rearrange Step 4 to collect the densities of the products over the reactant. The energy exponent is the net energy cost of the reaction. A

\[ m_{i+1}\simeq m_i \Rightarrow \frac{\lambda_i^{3}}{\lambda_{i+1}^{3}}\simeq 1,\quad Z_e=2,\quad \lambda_e^{-3}=\left(\frac{2\pi m_e k_B T}{h^2}\right)^{3/2},\quad \chi_i \equiv E_{i+1}+E_e-E_i \]

The ion mass differs from the atom mass only by \(m_e/m_i\sim 10^{-4}\), so their thermal wavelengths cancel. A free electron has no internal states except its two spin orientations, so \(Z_e=2\). The reaction energy is by definition the ionization energy \(\chi_i\). Substituting gives the Saha equation. B

Result

\[ \boxed{\;\frac{n_{i+1}\,n_e}{n_i} = \frac{2\,Z_{i+1}(T)}{Z_i(T)}\left(\frac{2\pi m_e k_B T}{h^2}\right)^{3/2} e^{-\chi_i/(k_B T)}\;} \]

Reading. The degree of ionization rises with temperature both through the Boltzmann factor \(e^{-\chi_i/k_B T}\) (more particles clear the ionization barrier) and through the phase-space factor \(T^{3/2}\) (a hotter electron gas has more accessible translational states, so ionized states are entropically favoured). Crucially, \(n_e\) sits on the left: raising the electron density at fixed \(T\) drives the equilibrium back toward the neutral stage, because recombination becomes more likely. Ionization is therefore a competition between temperature and density, not temperature alone.

Units check. The left side has units of number density, \(\mathrm{m^{-3}}\). On the right, \((2\pi m_e k_B T/h^2)^{3/2}\) has units \(\big[\mathrm{kg\cdot(J\,K^{-1})\cdot K / (J\,s)^2}\big]^{3/2}=[\mathrm{kg/(J\,s^2)}]^{3/2}\); with \(\mathrm{J=kg\,m^2 s^{-2}}\) this is \([\mathrm{m^{-2}}]^{3/2}=\mathrm{m^{-3}}\). The partition-function ratio, the factor \(2\), and the exponential are dimensionless, so the right side is \(\mathrm{m^{-3}}\) as well. Consistent.

Limiting cases

Low temperature, \(k_B T \ll \chi_i\): \(e^{-\chi_i/k_B T}\to 0\), the right side vanishes and the gas is essentially fully neutral.
High temperature, \(k_B T \gg \chi_i\): the exponential \(\to 1\) and the right side grows as \(T^{3/2}\); the stage \(i\) is driven to exhaustion and the gas becomes fully ionized (until the next stage takes over).
High electron density: at fixed \(T\), increasing \(n_e\) forces \(n_{i+1}/n_i\) down — dense gas recombines. This is why a tenuous nebula can be more highly ionized than a far hotter but dense stellar photosphere.
Validity edge: the classical treatment holds only while \(n_e\lambda_e^{3}\ll 1\); as this product approaches unity the electrons degenerate and the equation must be replaced.

Breaks when

The electron gas degenerates (\(n_e\lambda_e^{3}\gtrsim 1\)): Maxwell–Boltzmann statistics fail and the electron chemical potential must be taken from Fermi–Dirac statistics. This regime governs white-dwarf interiors and the cores of massive stars, where pressure ionization proceeds even at low temperature.
The plasma is strongly coupled / dense (Coulomb interaction energy comparable to \(k_B T\)): isolated-atom energy levels no longer exist. The bound-state series is truncated by the ambient microfields (continuum lowering), the partition function must be regularized, and pressure ionization (the Mott transition) sets in — as in the solar interior and inertial-confinement plasmas.
Local thermodynamic equilibrium fails (radiation field decoupled from matter, collision rates too low): in stellar coronae, planetary nebulae and the interstellar medium, photoionization and radiative/dielectronic recombination — not collisions at temperature \(T\) — set the level populations, and one must solve explicit rate (statistical-equilibrium) equations instead.

Failure modes

Dropping the electron spin factor of \(2\) (\(Z_e=2\)), giving ionization fractions low by a factor of two.
Using ground-state degeneracies \(g\) when excited states are populated — for metals such as Fe, Ca and Ti at stellar temperatures the full temperature-dependent \(Z(T)\) matters and the \(g\)-approximation is wrong by large factors.
Confusing the ionization energy \(\chi_i\) with an excitation energy of a bound level; the Saha exponent is the energy to reach the continuum, not to reach an excited state.
Forgetting that \(n_e\) is on the left and concluding that ionization depends on temperature alone; density is co-equal.
Mixing units: using the SI prefactor \(2.415\times10^{21}\,T^{3/2}\,\mathrm{m^{-3}}\) while quoting densities in \(\mathrm{cm^{-3}}\), a \(10^{6}\) error.
Treating \(n_e\) as an external constant when in a pure element it is fixed self-consistently by charge neutrality (\(n_e=n_{i+1}+\dots\)), so the Saha relation is actually a quadratic (or higher) equation.

Discussion

The Saha equation is, at bottom, a Boltzmann factor dressed with the entropy of the freed electron. The exponential \(e^{-\chi_i/k_B T}\) is the familiar energetic penalty for creating a bound-continuum transition, exactly as in the Boltzmann distribution; the novelty is the phase-space prefactor \((2\pi m_e k_B T/h^2)^{3/2}\), which counts how many translational cells the liberated electron can occupy per unit volume. Ionization is favoured not because it is energetically cheap — it is not — but because a free electron has enormous translational entropy. This is why the density appears: crowding the electrons removes that entropic advantage.

Historically the equation transformed astronomy. Because \(\chi_i\) and \(Z(T)\) differ between elements, two stars of different temperature show different lines even with identical composition. Cecilia Payne-Gaposchkin's 1925 application of Saha's relation showed that the near-absence of hydrogen lines in cool stars reflects excitation and ionization statistics, not chemistry, and hence that hydrogen is overwhelmingly the dominant element — one of the pivotal results of twentieth-century astrophysics.

The same relation, run in reverse, describes cosmological recombination: as the expanding Universe cooled through \(T\sim 3000\,\mathrm{K}\), the free-electron fraction dropped and photons decoupled, producing the cosmic microwave background. The Saha estimate gives the right temperature scale, though the actual freeze-out is delayed because expansion outpaces the recombination rate — an instructive case where LTE is only approximately valid.

Formally the cleanest route is the grand canonical ensemble: each species is a reservoir at chemical potential \(\mu_s\), and \(n_s=(Z_s/\lambda_s^3)e^{\mu_s/k_BT}\) is the \(\log Z\)-derivative in the nondegenerate limit of \(n=g\,\lambda^{-3}\!\int\! [\,e^{(\epsilon-\mu)/k_BT}\pm1\,]^{-1}\). Chemical equilibrium \(\sum_r\nu_r\mu_r=0\) then generates a law of mass action; Saha is that mass-action law for the reaction \(\mathrm{A}_i\rightleftharpoons\mathrm{A}_{i+1}+e^-\). Detailed balance guarantees the same populations arise from equating collisional ionization and three-body recombination rates, which is why LTE and kinetic equilibrium coincide when collisions dominate — and why they diverge (non-LTE) when radiation does not share the matter temperature.

Common misconceptions. "Hotter always means more ionized" is false at fixed pressure or where density rises with temperature; the density term can dominate. And the equation gives equilibrium populations, not rates: it says nothing about how fast equilibrium is reached, which is exactly what breaks cosmological and nebular applications.

Worked examples

Hydrogen in the solar photosphere: \(T=5800\,\mathrm{K}\), \(n_e=2\times10^{19}\,\mathrm{m^{-3}}\), \(\chi=13.6\,\mathrm{eV}\), \(Z_{\mathrm{I}}\simeq2,\ Z_{\mathrm{II}}=1\).

Set up the ratio \(n_{\mathrm{II}}/n_{\mathrm{I}}\) from Saha; the electron density is given. A

\[ \frac{2Z_{\mathrm{II}}}{Z_{\mathrm{I}}}=\frac{2\times1}{2}=1,\qquad k_B T = (1.381\times10^{-23})(5800)=8.01\times10^{-20}\,\mathrm{J}=0.500\,\mathrm{eV} \]

Evaluate the degeneracy prefactor and the thermal energy in electron-volts to match \(\chi\). A

\[ \left(\frac{2\pi m_e k_B T}{h^2}\right)^{3/2}=2.415\times10^{21}\,T^{3/2}\,\mathrm{m^{-3}} =2.415\times10^{21}(5800)^{3/2}=1.07\times10^{27}\,\mathrm{m^{-3}} \]

Use the standard grouped constant \(2.415\times10^{21}\,\mathrm{m^{-3}\,K^{-3/2}}\); \((5800)^{3/2}=4.42\times10^{5}\). A

\[ \frac{\chi}{k_BT}=\frac{13.6}{0.500}=27.2,\quad e^{-27.2}=1.5\times10^{-12};\qquad \frac{n_{\mathrm{II}}n_e}{n_{\mathrm{I}}}=1\cdot(1.07\times10^{27})(1.5\times10^{-12})=1.6\times10^{15}\,\mathrm{m^{-3}} \]

Multiply prefactor, degeneracy and Boltzmann factor. A

\[ \frac{n_{\mathrm{II}}}{n_{\mathrm{I}}}=\frac{1.6\times10^{15}}{2\times10^{19}}\approx 8\times10^{-5} \]

Divide by the given \(n_e\). A

\[ \frac{n_{\mathrm{II}}}{n_{\mathrm{I}}}\approx 8\times10^{-5}\ \Rightarrow\ \text{hydrogen is }\sim99.99\%\text{ neutral} \]

Reading. Despite a temperature of nearly 6000 K, the photosphere is almost entirely neutral hydrogen — the 13.6 eV barrier is \(27\,k_BT\) high. The tiny free-electron population actually comes mostly from metals with lower \(\chi\), which is why they, not hydrogen, are the electron donors here.

The same hydrogen gas raised to \(T=10\,000\,\mathrm{K}\) with \(n_e=1\times10^{19}\,\mathrm{m^{-3}}\).

Recompute to see the ionization transition. Same formula, new \(T\) and \(n_e\). A

\[ k_BT=(1.381\times10^{-23})(10^4)=1.381\times10^{-19}\,\mathrm{J}=0.862\,\mathrm{eV},\qquad \frac{\chi}{k_BT}=\frac{13.6}{0.862}=15.8 \]

Thermal energy in eV and the exponent. A

\[ \left(\frac{2\pi m_e k_B T}{h^2}\right)^{3/2}=2.415\times10^{21}(10^4)^{3/2}=2.415\times10^{27}\,\mathrm{m^{-3}},\quad e^{-15.8}=1.4\times10^{-7} \]

\((10^4)^{3/2}=10^{6}\); evaluate the Boltzmann factor. A

\[ \frac{n_{\mathrm{II}}n_e}{n_{\mathrm{I}}}=1\cdot(2.415\times10^{27})(1.4\times10^{-7})=3.4\times10^{20}\,\mathrm{m^{-3}},\qquad \frac{n_{\mathrm{II}}}{n_{\mathrm{I}}}=\frac{3.4\times10^{20}}{1\times10^{19}}=34 \]

Assemble, then divide by \(n_e\). A

\[ x=\frac{n_{\mathrm{II}}}{n_{\mathrm{I}}+n_{\mathrm{II}}}=\frac{34}{35}=0.97 \]

Convert the ratio into an ionization fraction. A

\[ x\approx 0.97\ \Rightarrow\ \text{hydrogen is }\sim97\%\text{ ionized} \]

Reading. A factor \(1.7\) rise in temperature swings hydrogen from \(0.02\%\) to \(97\%\) ionized. The transition is extraordinarily sharp because \(\chi/k_BT\) sits in the exponent: the ionization "switch" is narrow, which is precisely why spectral-line strengths are such sensitive thermometers of stellar atmospheres.

Problems

Evaluate the quantum-concentration prefactor \((2\pi m_e k_B T/h^2)^{3/2}\) at \(T=6000\,\mathrm{K}\) from first principles (do not simply quote the grouped constant).
Solution
\(2\pi m_e k_B/h^2 = 2\pi(9.109\times10^{-31})(1.381\times10^{-23})/(6.626\times10^{-34})^2\). Numerator \(=7.90\times10^{-53}\); denominator \(=4.39\times10^{-67}\); ratio \(=1.80\times10^{14}\,\mathrm{m^{-2}\,K^{-1}}\). Multiply by \(T=6000\): \(1.08\times10^{18}\,\mathrm{m^{-2}}\). Raise to \(3/2\): \((1.08\times10^{18})^{3/2}=1.12\times10^{27}\,\mathrm{m^{-3}}\). (Equivalently \(2.415\times10^{21}\times6000^{3/2}=1.12\times10^{27}\,\mathrm{m^{-3}}\).)
A pure hydrogen gas at \(T=8000\,\mathrm{K}\) has total hydrogen density \(n_{\mathrm{tot}}=n_{\mathrm{I}}+n_{\mathrm{II}}=10^{20}\,\mathrm{m^{-3}}\). Using \(n_e=n_{\mathrm{II}}\) (charge neutrality), find the ionization fraction \(x=n_{\mathrm{II}}/n_{\mathrm{tot}}\).
Solution
Saha: \(n_{\mathrm{II}}n_e/n_{\mathrm{I}}=S\), with \(S=1\cdot(2.415\times10^{21})(8000)^{3/2}e^{-\chi/k_BT}\). \(k_BT=(1.381\times10^{-23})(8000)=1.105\times10^{-19}\,\mathrm{J}=0.690\,\mathrm{eV}\); \(\chi/k_BT=13.6/0.690=19.7\); \(e^{-19.7}=2.8\times10^{-9}\). \((8000)^{3/2}=7.16\times10^{5}\), so prefactor \(=1.73\times10^{27}\), and \(S=1.73\times10^{27}\times2.8\times10^{-9}=4.8\times10^{18}\,\mathrm{m^{-3}}\). With \(n_e=n_{\mathrm{II}}=x\,n_{\mathrm{tot}}\) and \(n_{\mathrm{I}}=(1-x)n_{\mathrm{tot}}\): \(\dfrac{x^2 n_{\mathrm{tot}}}{1-x}=S\Rightarrow \dfrac{x^2}{1-x}=\dfrac{S}{n_{\mathrm{tot}}}=\dfrac{4.8\times10^{18}}{10^{20}}=0.048.\) Solve \(x^2+0.048x-0.048=0\): \(x=(-0.048+\sqrt{0.048^2+0.192})/2=(-0.048+0.441)/2=0.20\). So \(x\approx0.20\); the gas is about \(20\%\) ionized — the \(19.7\,k_BT\) barrier keeps most hydrogen neutral even though the density is low.
At the conditions of Worked Example 1 (\(T=5800\,\mathrm{K}\)), compare the ionization ratios of hydrogen (\(\chi=13.6\,\mathrm{eV}\)) and neutral sodium (\(\chi=5.14\,\mathrm{eV}\), take \(2Z_{\mathrm{II}}/Z_{\mathrm{I}}\approx1\)). By what factor is sodium more ionized?
Solution
The prefactors and \(n_e\) are identical, so the ratio of ionization ratios is just the ratio of Boltzmann factors: \(\exp[-(\chi_{\mathrm{Na}}-\chi_{\mathrm H})/k_BT]\). With \(k_BT=0.500\,\mathrm{eV}\), the exponent difference is \((13.6-5.14)/0.500=16.9\). Thus \(\frac{(n_{\mathrm{II}}/n_{\mathrm I})_{\mathrm{Na}}}{(n_{\mathrm{II}}/n_{\mathrm I})_{\mathrm H}}=e^{+16.9}=2.2\times10^{7}\). Sodium is about \(2\times10^{7}\) times more ionized — which is why low-ionization-potential metals, not hydrogen, supply the photospheric electrons.
Holding \(T\) fixed, a cloud's density is increased so that \(n_e\) rises by a factor of \(100\). By what factor does the ionization ratio \(n_{i+1}/n_i\) change, and why?
Solution
The right side of Saha depends only on \(T\), so the product \(n_{i+1}n_e/n_i\) is unchanged. Hence \(n_{i+1}/n_i \propto 1/n_e\): a \(100\times\) rise in \(n_e\) lowers \(n_{i+1}/n_i\) by a factor of \(100\). Physically, more free electrons make recombination proportionally more frequent, shifting equilibrium toward the neutral stage. This is the density (Le Chatelier) effect and explains why diffuse gas is more ionized than dense gas at equal temperature.
Estimate the temperature at which hydrogen is 50% ionized in a gas with electron density fixed at \(n_e=10^{19}\,\mathrm{m^{-3}}\). (Take \(2Z_{\mathrm{II}}/Z_{\mathrm I}=1\); solve by iteration.)
Solution
At \(x=0.5\), \(n_{\mathrm{II}}/n_{\mathrm{I}}=1\), so Saha requires \(n_e = (2.415\times10^{21})T^{3/2}e^{-\chi/k_BT}\), i.e. \(10^{19}=2.415\times10^{21}\,T^{3/2}e^{-1.578\times10^{5}/T}\) (using \(\chi/k_B=13.6\,\mathrm{eV}/k_B=1.578\times10^{5}\,\mathrm{K}\)). Rearranged: \(T^{3/2}e^{-1.578\times10^{5}/T}=4.14\times10^{-3}\). Iterate: try \(T=9000\): \(T^{3/2}=8.5\times10^{5}\), \(e^{-17.5}=2.5\times10^{-8}\), product \(=2.1\times10^{-2}\) (too high). Try \(T=8000\): \(7.16\times10^{5}\times e^{-19.7}=7.16\times10^{5}\times2.8\times10^{-9}=2.0\times10^{-3}\) (too low). Try \(T=8500\): \(7.84\times10^{5}\times e^{-18.6}=7.84\times10^{5}\times8.4\times10^{-9}=6.6\times10^{-3}\) (slightly high). Try \(T=8300\): \(7.56\times10^{5}\times e^{-19.0}=7.56\times10^{5}\times5.6\times10^{-9}=4.2\times10^{-3}\approx4.14\times10^{-3}\). So \(T\approx8300\,\mathrm{K}\). Note this half-ionization temperature is far below \(\chi/k_B=1.6\times10^{5}\,\mathrm{K}\): the huge phase-space prefactor lets ionization proceed at \(k_BT\approx\chi/19\).