Saha equation / hydrogen

Smythe

The Saha equation for the hydrogen atom can be written as

log(N+/N) = log(u+/u) + (5/2)logT - log(Pe) - χionӨ - 0.18

where Ө=5040/T
χion is measured in electron volts (eV).

Calculate the number of negative hydrogen ions (H-) in the solar photosphere relative to neutral hydrogen (H) for a temperature of T = 6,000 K and a pressure of log Pe = 2.7.

My attempt:
log(N+/N) = log(u+/u) + (5/2)logT - log(Pe) - χionӨ - 0.18
log(N+/N) = log(2/1) + (5/2)log6000 - 2.7 - (13.6)x(5040/6000) - 0.18
log(N+/N) = 0.693 + 21.748 -2.7 - 11.42 - 0.18
log(N+/N) = 8.14

I chose to input 13.6 for χion since the ionisation energy for the hydrogen atom in the ground state is χion = 13.6eV.

But apparently the correct solution should be 5x10^(-7). Where have I gone wrong?

Is it because the ionisation energy is 13.6 eV for the transition H <-> H+, but lower for H- <-> H?

Smythe

I discovered one mistake, (5/2)log6000 should of course equal 9.44 (not 21.748)!

Smythe

I think the log(2/1) is possibly to the base e whereas elsewhere base 10 is used. So I think the log(2/1) needs converted to base 10 to remain consistent.

Would anyone know how to do this?