 Solution

For a photon gas in thermal and chemical equilibrium, photon number is not a conserved thermodynamic charge. Number-changing processes can create and absorb photons without a corresponding conserved-number constraint, so the equilibrium photon chemical potential is zero.

At $T=5\times 10^9\mathrm{K}$, $k_{\mathrm{B}}T\simeq 0.431\mathrm{MeV}$. The other light species are electrons, Positrons, three neutrino species and their antiparticles. Neutrinos are ultrarelativistic; electrons and Positrons are only moderately relativistic at this particular temperature because $m_e{c}^2=0.511\mathrm{MeV}$. The later simple counting formulas use the intended ultrarelativistic approximation for them. Baryons, muons and heavier particles are not an abundant relativistic bath at this temperature.

Pair equilibrium implies ${\mu }_{e^{-}}+{\mu }_{e^{+}}=2{\mu }_{\gamma }=0$. Approximate charge symmetry, with the net electron excess tied to the much smaller baryon asymmetry, then makes both electron and Positron chemical potentials small compared with $k_{\mathrm{B}}T$. For neutrinos one additionally assumes negligible lepton asymmetry. Their near-zero chemical potentials are not implied merely by their statistics. After neutrino decoupling, a distribution initially having negligible chemical potential retains a redshifted Fermi-Dirac distribution, with its own temperature.