Define fractional ionization as x=np/(nH+np) and use electrical neutrality. For x=0.01 and ne=109m−3, nH=(1−x)ne/x=99ne. Thus the density form of the Saha equation requiresSolving this monotone equation, conveniently after taking its logarithm, givesThe thermal energy is only about 2.2% of the 13.6 eV ionization energy. Ionization is not decided by setting kBT=R: the high-energy tail and the large translational state-count factor in chemical equilibrium allow a residual ionized fraction at a much lower temperature. This is the equilibrium estimate for the stipulated density and fraction, without inferring an expansion age from it.
(h22πmekBT)3/2e−13.6eV/(kBT)=99ne≃1.01×107m−3.
T≃3.48×103K,kBT≃0.300eV.
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