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Consider pure hydrogen around an isotropic central source with hydrogen-ionizing photon production rate $Q$. An empty cavity of physical radius $r_{\mathrm{in}}$ delivers an unattenuated direct photon flux $Q/(4\pi r_{\mathrm{in}}^2)$ to its inner boundary. Let $⟨\sigma ⟩$ be the photoionization cross section averaged over the incident ionizing photon-number spectrum. Then the direct photoionization rate per neutral atom is $q=Q⟨\sigma ⟩/(4\pi r_{\mathrm{in}}^2)$. With total hydrogen number density $n_H$ and ionized fraction $\zeta$, the electron and proton number densities are both $n_H\zeta$. In the local Case B recombination approximation, photoionization equilibrium givesThis uses a locally balanced photoionization and radiative recombination event rate; the on-the-spot treatment incorporates immediate reabsorption of ground-state recombination photons into ${\alpha }_B$.^[ref]

Solving the quadratic equation and rationalizing its physical root yieldsFor $\Theta \ll 1$, $\zeta \simeq 1-\Theta$; for $\Theta \gg 1$, $\zeta \simeq {\Theta }^{-1/2}$. Thus stronger incident photoionization drives the boundary closer to complete ionization, while a larger recombination rate leaves more neutral hydrogen.

Inferring $Q$ from a recombination line requires an additional, global photon-budget assumption. Suppose $F_{\ell }$ is the total extinction-corrected observed radiative flux in a recombination line of wavelength ${\lambda }_{\ell }$, at distance $D$. At fixed gas conditions let $y_{\ell }={\alpha }_{\ell }^{\mathrm{eff}}/{\alpha }_B$ be the line-photon yield per Case B recombination. The measured line then gives the absorbed hydrogen-ionizing rateFor a dust-free ionization-bounded nebula in steady photoionization equilibrium, with no competing absorbers, $Q_{\mathrm{abs},\mathrm{H}}=Q$. Under these assumptions,The angular ratio $r_{\mathrm{in}}/D$ makes the expression dimensionless and lets a resolved cavity and its integrated recombination line constrain the local ionized fraction.

If only a fraction $f_H$ of the source's ionizing photons produces hydrogen ionizations, then $Q_{\mathrm{abs},\mathrm{H}}=f_{H}Q$ and the last expression for $\Theta$ must be multiplied by $f_H$. A nonzero ionizing photon escape fraction or absorption by dust can therefore make the line underestimate the central $Q$. Correcting the escaping line for extinction alone does not recover ionizing photons that never produced a recombination line. A Dusty Strömgren sphere exemplifies the distinct absorption correction. The local boundary formula also requires an empty, transparent cavity; any attenuation before the gas changes its incident $q$.

Suppose an emitted spectral luminosity is $L_{\nu }=A{\nu }^{-s}$ for every $\nu \ge {\nu }_0$, where $A>0$ and $h{\nu }_0$ is an ionization threshold. The total emitted ionizing luminosity and photon rate use different weights:Thus a finite emitted mean photon energy for this unbounded power law requires the stronger condition $s>1$ and equalsFor $s=2$, the mean is twice the threshold energy. As $s\downarrow 1$, the high-frequency tail makes the mean diverge even though the photon count stays finite. For $0<s\le 1$, substituting into the finite-mean formula would therefore be invalid; for $s\le 0$ even the unbounded photon-rate integral diverges.

A finite high-frequency cutoff changes both integrals and removes these upper-tail divergences. The mean energy deposited in a nebula also need not equal the emitted mean: absorption weights the spectrum with its photoionization cross section, and energy above threshold can be distributed among heat, excitation and secondary ionizations. The excess $h\nu -h{\nu }_0$, rather than the whole photon energy, is the initial kinetic energy of the photoelectron in the simplest single-ionization event.

A photoionization cross section $\sigma (\nu )$ quantifies the probability that an incident photon of frequency $\nu$ ionizes a specified atomic or molecular target. For a weak monochromatic beam with photon number flux $F_{\gamma }$, the ionization rate per target is $\Gamma =F_{\gamma }\sigma$; the cross section has dimensions of area.

The cross section depends on photon energy, target state and accessible final channels. In an isotropic radiation field with mean specific intensity $J_{\nu }$, $\Gamma ={\int }_{{\nu }_0}^{\infty }[4\pi J_{\nu }/(h\nu )]\sigma (\nu )d\nu$. Energy-resolved photoelectron measurements probe the resulting channels; interpreting a count rate requires photon-flux normalization and detector calibration.^[ref]