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The fraction of a specified population of chemical atoms that has become ionized. For singly ionized atoms with no external charge source, charge neutrality gives equal free-electron and ion number densities. Its equilibrium value depends on both temperature and total number density through the Saha ionization equation.

For classical nondegenerate electrons, the free-particle Boltzmann occupancy has the form $z_e\exp [-p^2/(2mkT)]$, with chemical potential measured relative to its rest energy. Counting the two spin states in phase space gives $n_e=2z_e(2\pi mkT/h^2{)}^{3/2}$. A bound level of energy $-I_n$ has relative occupancy $(g_n/g_{+})z_e{e}^{I_n/kT}$ against its parent ion. Eliminating $z_e$ gives the level-specific Saha ionization equation.

For monatomic species, there are no molecular rotational or vibrational factors. Assume excited electronic states are negligible, so the internal partition sums are the ground-state degeneracies. A free electron has two spin states, givingwith the analogous expression for the ion. Since losing one electron changes the atomic mass negligibly, $(m_{{\mathrm{M}}^{+}}/m_{\mathrm{M}}{)}^{3/2}\simeq 1$. HenceThis is the ground-state form of the Saha ionization equation. Its assumptions include a dilute, nondegenerate ideal mixture, negligible interparticle interactions, thermal and chemical equilibrium, no additional ionization stages, and consistent treatment of internal degeneracies. Nuclear-spin factors may be omitted when they cancel between the atom and ion; they must not be counted on only one side.

The Rosseland mean opacity is the transport opacity appropriate to an optically thick region close to local thermodynamic equilibrium. If ${\kappa }_{\nu }$ is monochromatic transport opacity and $B_{\nu }(T)$ the Planck function, thenIt is a weighted harmonic mean: low-opacity frequency windows carry much of the diffusive flux. One must combine the monochromatic processes before averaging, rather than adding their separately averaged opacities.

In hot fully ionized material, electron-scattering opacity supplies an almost density- and temperature-independent contribution at fixed composition. In the nonrelativistic Thomson scattering limit, ${\kappa }_{\mathrm{es}}={\sigma }_T{n}_e/\rho \simeq 0.2(1+X)\mathrm{c}{\mathrm{m}}^2{\mathrm{g}}^{-1}$. Free-free opacity, the absorption counterpart of thermal bremsstrahlung, comes from photon absorption during encounters between free electrons and ions. Its approximate mean obeys Kramers' opacity law, ${\kappa }_{\mathrm{ff}}\propto \rho T^{-7/2}$, with composition and quantum correction factors.

When ions retain bound electrons, bound-free absorption is photoionization; edges, occupations and ionization fractions strongly affect its frequency dependence. Bound-bound absorption produces absorption lines. Numerous metal lines can close low-opacity windows and produce important opacity features in partially ionized layers. Line populations depend on excitation as well as ionization, while Doppler broadening, pressure broadening and overlapping lines change the transport windows. Stimulated inverse processes must be included consistently with detailed balance and the local thermal radiation field.

In cooler hydrogen-rich atmospheres, negative hydrogen ion opacity from both bound-free and free-free processes is often important. Its strong temperature dependence helps determine convective surface relations. In still cooler layers, molecular bands, and at sufficiently low temperatures dust absorption/scattering, may be important. These processes cannot be represented by one Kramers power law over the whole density-temperature plane.

An opacity calculation therefore needs the chemical abundances, the equation of state, electron density and bound-state populations. Saha ionization equation and thermal excitation populations give useful dilute LTE estimates; high-density pressure ionization, nonideal effects and altered level populations require corrections. Composition matters even when one writes only $\kappa (\rho ,T)$: hydrogen, helium and the metal abundance pattern are additional fixed inputs.

In dense degenerate cores, thermal conduction by electrons can dominate energy transport. An equivalent conductive opacity may be introduced so that $1/{\kappa }_{\mathrm{eff}}=1/{\kappa }_R+1/{\kappa }_{\mathrm{cond}}$, because the two diffusive fluxes add; conduction is not itself a photon absorption process. Very hot regimes also require departures from the simple nonrelativistic scattering approximation. Near the optically thin photosphere, a Rosseland diffusion description alone is inadequate, and frequency-dependent radiative transfer with appropriate boundary conditions is required. These regime changes are why realistic stellar opacity tables, rather than one global power law, enter stellar structure.

The Boltzmann distribution for two atomic levels at thermodynamic equilibrium iswhere $\omega$ is the statistical weight of an atomic level. For a resolved level of total angular momentum $J$, $\omega =2J+1$ when magnetic substates are unresolved.

To extend this counting to the continuum, assume an ideal, nondegenerate plasma with classical free electrons and neglect the small difference between the translational masses of the two heavy ionic species. Let $z_e=e^{{\mu }_e/kT}$ be the electron fugacity in the Saha equation, with the electron chemical potential measured relative to its rest energy. The two electron spin states and the phase-space cell volume $h^3$ giveChemical equilibrium for capture equates the bound species' chemical potential to the sum for its parent ion and free electron. Thus applying Boltzmann factors to a bound level at energy $-I_n$ gives $N_n/N_{+}=({\omega }_n/{\omega }_{+})z_e{e}^{I_n/kT}$. Eliminating the electron fugacity proves the level-specific Saha ionization equation:The factor two is the electron spin multiplicity; the factor with power $3/2$ is the inverse free-electron translational state density. A total-ionization Saha equation sums the bound states into an internal partition function, whereas this expression refers to one specified level.

For dielectronic recombination, write the entrance ion as $X^{(z+1)+}$ in ground state $i$, and the intermediate state of $X^{z+}$ as $d=(j,n\ell )$. Capture excites the core from $i$ to $j$ while binding the incident electron. The resonance energy relative to the entrance continuum is $\bar{E}=E_{ij}-I_{n\ell }>0$. The doubly excited state can eject the electron by autoionization, or undergo radiative stabilization to a bound state.

Let $A_{di}^a$ be the partial autoionization rate back to the entrance ion, $A_a$ the total autoionization rate, $A_r$ the total radiative decay rate, and $A_s$ the rate of radiative decays that stabilize into bound levels. For an isolated narrow resonance, the Saha-Boltzmann equilibrium ratio has the bound-state energy replaced by the positive resonance energy:At equilibrium, inverse dielectronic capture and autoionization fluxes balance: $N_e{N}_i{C}_d=N_d^{\mathrm{eq}}{A}_{di}^a$. This establishes the capture coefficient even when the actual plasma is not in local thermodynamic equilibrium. In the low-density, weak-radiation recombining plasma, the intermediate level instead has the stationary balanceMultiplying its population by the successful stabilization rate gives the actual recombination coefficientHere the equilibrium population was used only to determine the capture rate by detailed balance; it was not assumed to equal the intermediate population when photons escape. This is precisely why the competing decay probabilities appear.

Use the supplied core-transition relation between the Einstein coefficients and atomic oscillator strength,Here $\alpha$ is the fine-structure constant and $a_0$ the Bohr radius, not a recombination coefficient. Factor out $A_{ji}$ and defineSubstitution yieldsThis definition accommodates several entrance and decay channels. In the common spectator-electron approximation with one autoionization channel and stabilization by the core transition, $A_s=A_r=A_{ji}$ and $A_{di}^a=A_a$. If all coupled substates of the captured $n\ell$ electron are summed, $g_d=2(2\ell +1)g_j$, givingFor an intermediate state resolved by fine structure, retain its actual $g_d$ instead of that summed weight. The stabilization probability itself is $A_s/(A_a+A_r)$; the requested ${\beta }_{j,n\ell }$ is not just this branching probability, because the prefactor has already factored out the core radiative rate and statistical weights of atomic levels. Radiative transitions to another still-autoionizing state require that state's eventual survival probability, rather than automatically counting every emitted photon as stabilization.

The total dielectronic recombination coefficient sums all accessible positive-energy resonances. In this isolated-resonance approximation it has the formwith atomic factors included in $c_d$. The resonance-temperature dependence of dielectronic recombination follows by differentiating a single term's logarithm:Thus low temperatures suppress a positive-energy resonance, although resonances very near threshold can still matter; at sufficiently high temperature a fixed set of resonances falls as $T^{-3/2}$. Several core-excitation series can produce several peaks. When autoionization is much faster than radiation, the capture-and-survival product is limited by the stabilizing radiative rate; when capture is weak, autoionization is the limiting inverse-capture rate. For a fixed Rydberg angular channel, autoionization typically decreases at high principal quantum number while the core radiative rate changes little. Capture into very high levels is vulnerable to subsequent collisional or field ionization, so finite-density survival can suppress the total rate. dielectronic recombination is resonant and strongly temperature-selective; its magnitude can be important in coronal ionization balance.

For $p+e\rightleftharpoons H+\gamma$, chemical equilibrium and the vanishing photon chemical potential imply ${\mu }_H={\mu }_p+{\mu }_e$. Taking the ratio of the three nonrelativistic Maxwell--Boltzmann number densities cancels these chemical potentials:The ground-state hydrogen atom has four combined electron and proton spin states if the small hyperfine splitting is ignored, whereas $g_p=g_e=2$. Hence $g_H/(g_p{g}_e)=1$. With Hydrogen binding energy $B=M_p+M_e-M_H$ and $M_H\simeq M_p$, the Saha ionization equation givesThe approximation in the prefactor neglects relative corrections of order $(M_e+B)/M_p$; the Hydrogen binding energy must be retained in the exponential.

Write $n_b=n_p+n_H$ for the cosmological baryon density in this hydrogen-only model, and $x_e=n_p/n_b$ for the hydrogen ionization fraction. By charge neutrality, $n_e=n_p$, so the Saha ionization equation becomesThe baryon-to-photon ratio is $\eta =n_b/n_{\gamma }$, and the equilibrium photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. At an order-one hydrogen ionization fraction, neglecting factors of order one therefore givesReplacing $T$ inside the slowly varying logarithm by $B$ givesWith $B=13.6\mathrm{eV}$, $M_e=5\times 10^5\mathrm{eV}$ and $\eta =10^{-10}$, this is $T\simeq 0.351\mathrm{eV}\simeq 4.1\times 10^3\mathrm{K}$. Solving the preceding implicit approximation instead gives $T\simeq 0.306\mathrm{eV}\simeq 3.5\times 10^3\mathrm{K}$. Both are reasonable estimates of the recombination temperature at the accuracy requested. The small baryon abundance delays hydrogen recombination: even when $T\ll B$, the high-energy tail of the abundant photons can still ionize the much rarer hydrogen atoms.

Assume dilute nondegenerate nonrelativistic particles, a common temperature, chemical equilibrium, charge neutrality $n_p=n_e$, and photons with zero chemical potential. The reaction balance gives ${\mu }_p+{\mu }_e={\mu }_H$. Dividing the two charged-species equilibrium densities by the neutral density cancels their chemical-potential factors and leavesFor ground-state hydrogen, $g_p=g_e=2$ and $g_H=4$, and $m_H\simeq m_p$ in the thermal prefactor. The rest-mass difference is $I/c^2$. These approximations give the Saha ionization equationExcited levels or different degeneracy conventions would change the prefactor; the stated assumptions give the printed normalization.

Write the right side as $S(T)$. Since $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the hydrogen ionization fraction obeys $X_e^2/(1-X_e)=S(T)/n_B$. With $n_B=\eta n_{\gamma }$, the given blackbody photon density then yieldsThere are vastly more photons than baryons. Even when the typical photon energy is below the ionization energy, the high-energy tail can contain enough ionizing photons to keep nearly every baryon ionized. Neutral hydrogen forms only when that tail becomes sufficiently depleted relative to the very small baryon-to-photon ratio. This entropy/number-density effect explains why cosmological recombination occurs at a temperature far below $I$, rather than simply when $kT$ first falls below the binding energy. The Saha ionization equation describes the equilibrium estimate; late kinetic freeze-out is a separate limitation of that approximation.

Chemical equilibrium imposes ${\mu }_e+{\mu }_p={\mu }_H+{\mu }_{\gamma }$. For blackbody photons, ${\mu }_{\gamma }=0$. Define the positive Hydrogen binding energy by $I=(m_e+m_p-m_H)c^2$, approximately $13.6\mathrm{eV}$. Multiplying the two Maxwell-Boltzmann distribution number densities and dividing by the neutral-hydrogen density givesUse $g_e{g}_p/g_H=1$ and $m_p/m_H\simeq 1$ to obtain the Saha ionization equationThe assumptions are nonrelativistic, dilute, nondegenerate species at a common temperature, ground-state hydrogen with the stated degeneracy factors, negligible mass correction in the prefactor, and sufficiently rapid reactions to maintain chemical equilibrium.

In chemical equilibrium, the reaction $p+e\leftrightarrow H+\gamma$ obeys chemical-potential balance for ionizationbecause a photon gas in equilibrium has photon chemical potential ${\mu }_{\gamma }=0$. Applying the given nonrelativistic Maxwell-Boltzmann distribution to each massive species givesThe Hydrogen binding energy is $I=m_e+m_p-m_H$ in units with $c=1$. Thus Saha's equation isFor ground-state hydrogen, $g_e=g_p=2$ and $g_H=4$, so the degeneracy factor is one. Since $m_p\simeq m_H$, the translational factor is commonly written $(2\pi m_e{k}_{B}T/h^2{)}^{3/2}$.

Charge neutrality gives $n_e=n_p$. Write $n_B=n_e+n_H\simeq \eta n_{\gamma }$. Thenand henceCombining this with the stated photon density produces

In our universe the baryon-to-photon ratio is tiny, so there are roughly $10^9$ photons per baryon. Even when the mean photon energy is far below $I$, the high-energy blackbody radiation tail contains enough ionizing photons to suppress neutral hydrogen. The recombination temperature is reached only after the Boltzmann factor overwhelms this large photon-to-baryon ratio, at $k_{B}T\simeq 0.3\mathrm{eV}$.

For a universe with $\eta =1$, take the degeneracy factor as one, $m_H\simeq m_p$, and define $x=I/(k_{B}T)$. Neutral formation occurs when the boxed ratio is of order one, giving approximatelyWith $I=13.6\mathrm{eV}$ and $m_e=5.11\times 10^5\mathrm{eV}$, this gives $x\simeq 19$ and thereforeto order unity. This is much hotter than recombination in our universe because there is no enormous excess of photons, but it remains below $I$ because the electron's large translational phase space still favours the ionized state.

Chemical equilibrium for $p+e^{-}\rightleftharpoons H+\gamma$ gives ${\mu }_p+{\mu }_e={\mu }_H$, because the photon chemical potential vanishes. Applying the nonrelativistic number-density formula to all three massive species and taking the ratio givesUse charge neutrality $n_p=n_e$, the Hydrogen binding energy $E_{\mathrm{bind}}=m_p{c}^2+m_e{c}^2-m_H{c}^2$, $m_H\simeq m_p$, and the ground-state statistical weights $g_H=g_p{g}_e$. ThenThe approximations are a dilute, nondegenerate ideal gas in thermal and chemical equilibrium, nonrelativistic electrons, protons and hydrogen, negligible excited hydrogen states, and negligible corrections from the proton-to-hydrogen mass ratio. This relation is the Saha ionization equation.

Write $n_b=n_e+n_H\simeq \eta n_{\gamma }$. Charge neutrality gives $n_p=n_e$, and thereforeSubstitution into the Saha ionization equation yieldsUsing the given photon density,

The baryon-to-photon ratio is extremely small. Even when $k_{B}T\ll I$, the high-energy tail of the enormous photon population contains enough ionizing photons per baryon to delay cosmological recombination. The exponential factor must become large enough to overcome both $\eta \sim 10^{-9}$ and the phase-space factor, which occurs near $k_{B}T\simeq 0.3\mathrm{eV}$ rather than near $I$.

With $X_e=n_e/n_b$, charge neutrality and $n_b=n_p+n_H$ giveMultiplying part a by $n_b=\eta n_{\gamma }$ and using $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ yields the Saha ionization equationAlthough $T\ll B_H$ suppresses ionization of an individual atom, the baryon-to-photon ratio is only about $10^{-9}$. The enormous number of photons per baryon leaves enough photons in the high-energy thermal tail to ionize hydrogen until $T\simeq 0.3\mathrm{eV}$, far below $B_H$.

Chemical equilibrium for $p+e\leftrightarrow H$ givesand charge neutrality gives $n_p=n_e$. Insert the nonrelativistic Maxwell-Boltzmann number densities, use $m_H{c}^2=m_p{c}^2+m_e{c}^2-E_{\mathrm{bind}}$, and neglect the proton correction in the reduced mass. The spin degeneracies cancel in the ground-state approximation. One obtains Saha equationEquivalently, its reciprocal is the ionization ratio. The exponential favors neutral hydrogen as the temperature falls, while the translational phase-space factor favors ionization at high temperature.

In the nonrelativistic regime,The assumption $E-\mu \gg k_{B}T$ puts the Fermi-Dirac distribution in its Maxwell-Boltzmann limit, soTherefore the nonrelativistic Maxwell--Boltzmann number density is

For $p+e^{-}\leftrightarrow H+\gamma$, chemical equilibrium and the vanishing photon chemical potential implyApplying the number-density formula to each massive species and eliminating the chemical potentials givesUse charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_e{g}_p/g_H$. With the Hydrogen binding energywe obtain the Saha ionization equationThe assumptions are thermal equilibrium and chemical equilibrium before cosmological recombination, a nonrelativistic nondegenerate gas, zero photon chemical potential, charge neutrality, negligible proton-electron mass correction in the translational prefactor, and the stated degeneracy-factor approximation.

Chemical equilibrium for $e^{-}+p\leftrightarrow H+\gamma$ gives ${\mu }_e+{\mu }_p={\mu }_H$, since ${\mu }_{\gamma }=0$. Insert the supplied nonrelativistic equilibrium densities. Neglecting $m_e/m_p$ in the translational reduced mass and using the stated degeneracy convention givesCharge neutrality gives $n_p=n_e$, soThis is the hydrogen Saha ionization equation.

Assume a dilute, nondegenerate, nonrelativistic ideal gas in local thermodynamic equilibrium, with ground-state atomic hydrogen, protons and electrons, and neglect excited atomic levels and interactions. Let the conserved number of hydrogen nuclei per volume be $n_0$, so $n_e=n_p=\zeta n_0$ and $n_H=(1-\zeta )n_0$. Equality of chemical potentials for $H\rightleftharpoons p+e$ gives the Saha ionization equationThe factor two is electron spin; $g_p$ and $g_H$ must be defined consistently with nuclear spin. The small difference between the atomic and proton translational partition functions has been neglected. ThereforeFor the ground-state convention $g_p=1$, $g_H=2$, the statistical prefactor $2g_p/g_H$ is one. At fixed volume $n_0$ is constant, whereas an isobaric heating process would not have a constant $A$.

One mole of original atoms contains $(1-\zeta )$ moles of neutral atoms, $\zeta$ moles of protons and $\zeta$ moles of electrons. All are monatomic translational species, so their combined thermal energy is $\frac{3}{2}RT(1+\zeta )$. Taking the neutral ground state as zero internal energy, ionization stores $\zeta N_A{E}_I$. ThusRadiation, excited states and nonideal interactions have been excluded consistently with the elementary Saha ionization equation.

At fixed volume, logarithmic differentiation of the Saha ionization equation givesDifferentiation the internal energy then yieldsAt low temperature $\zeta \to 0$ and $C_V\to 3R/2$. At high nonrelativistic temperature the gas is fully ionized, $\zeta \to 1$, and $C_V\to 3R$. The large intermediate excess reflects energy spent increasing ionization, as well as heating and creating additional translational particles. “Very high” here remains below regimes where relativistic, radiative or pair-creation contributions invalidate the model.

Put $x=E_I/(k_{B}T)$. With $\zeta =1/2$, the heat capacity formula becomesTaking $E_I=13.6\mathrm{eV}$ gives $\boxed{T=1.02\times 10^4\mathrm{K}}$. At half ionization, $n_e{n}_p/n_H=n_e$, so the Saha ionization equation directly givesThis uses $2g_p/g_H=1$ as specified in part (a); a different internal-state convention must be carried consistently into the numerical electron density. The original nuclei density is $n_0=2n_e\simeq 10^{21}{\mathrm{m}}^{-3}$.