Energy to completely ionize gas in star

[blkqi]

Please look this simple problem over.

A star is composed of 91% H and 9% He. What energy is required to completely ionize this gas mixture?

Energies of ionization are 13.6eV for H and 54.4eV and 24.5eV for the 1st and 2nd ionizations of He.

My method:

Let n be the total number of atoms, then n(H)=.91n and n(He)=.09n.
The energy required to completely ionize the gas is

(13.6 eV)(.91n)+(24.5+54.4 eV)(.09n)=(19.477 eV)n

Equating this with the average kinetic energy from rms speed (kinetic theory of gases),

(3/2)kT*n=(19.477 eV)*n

we find that T=151,000 K. So at 151,000 K the gas mixture is a pure plasma..

I'm unsure of the validity of my first step, where I find that the average energy of each atom would be 19.477 eV/atom. This energy is enough to break the ion potential on hydrogen, but not helium. Perhaps it is the remainder of this energy (kinetic energy of the free electrons, almost 6 eV) that ionizes the He? Also, in the completely ionized gas we could say that H+ is two particles (electron, proton) and He++ is three particles (2 electron, alpha) so perhaps n is multiplied at ionization; should this be taken into account?

Note I didn't use the Saha equation. As far as I know the Saha equation is not fit to predict complete ionization since the number density of neutral atoms would be 0...

 

[gtoguy97]

It's difficult to know for sure without more context about the problem, but it looks like your approach is generally correct. The 19.477 eV per atom you found is the energy required to ionize the atoms in the mixture, so this is the average energy needed to completely ionize the gas mixture. You're right that some additional energy will be required to break the ion potentials of the He and provide kinetic energy for the free electrons. However, since the total energy to completely ionize the gas is simply the sum of the energies required to ionize each component (H and He), you don't need to take into account any additional energy requirements.