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As a first estimate, neglect the small mass change and changes in geometry and vibrational populations. Removing an electron creates an unpaired spin, giving at least the spin-only electronic entropy increaseFor a more specific estimate, the actual cation ground term is $X{}^2{\Pi }_g$, rather than an orbitally nondegenerate doublet. Ignoring spin-orbit coupling, its two orbital components and two spin components give electronic degeneracy four andThe ground-state assignment is confirmed by state-resolved photoionization measurements.^[ref] Spin-orbit splitting and changed vibrational/rotational constants refine this estimate; the question's supplied partition functions alone do not determine those corrections. The $R\log 2$ answer is a spin-only estimate, while $R\log 4$ additionally retains the orbital degeneracy.

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]