# Ionization temperature of an atom

1. Apr 28, 2012

### Boltzmann2012

Hi,
How do we calculate the temperature required to ionize an atom, say, helium. Also how can we find the number of atoms ionized or unionized at a given temperature?

Boltzmann

2. Apr 28, 2012

### AliAhmed

You should check out the Saha Ionization Equation(s) (or Saha-Langmuir Equation). I must say, wikipedia actually has a good page on the equation(s) and its use.

3. Apr 28, 2012

### Boltzmann2012

Thanks. I have read it and it is nice.

But what should we do when the ionization fraction tends to 1, I.e the Saha ionization eqn is valid only for thermal equillibrium.

Regards

Boltzmann

4. May 1, 2012

### AliAhmed

- Firstly, sorry it took me so long to reply. I have to say I was stumped by your question and was looking for my notes when I learned about the equation, but unfortunately I couldn't find them.

- I'm just going to tell you what I know. When I learned it, my professor wrote the equation differently (in the equation below, I modified the variables so it has the same variable names as that for Wikipedia):
$\frac{g_{i}}{g_{i+1}}$$\frac{n_{i+1}}{n_{i}}$$\frac{n_{e}\Lambda^{3}}{2}$ = e$^{-\frac{\Delta\epsilon}{k_{B}T}}$

- As the entire term on the left tends to 1, then it will take an infinitely large temperature and thus an infinitely large amount of energy to further ionize atoms (in most cases we consider the first ionization state (so n0 and n1 for example). Therefore, this acts as a limiting scenario. I remember my professor ascribing a name to this scenario (as though someone had discovered it and had their ascribed to it). But to be honest I don't remember all that well so this is my educated guess.

- If I didn't answer your question I'm sorry, but that's all I've got (good question)!

5. May 3, 2012

### Boltzmann2012

Thank you for the reply. It was most useful.

Regards

L.Boltzmann

6. May 3, 2012

### K^2

The irony here is that the equation for fraction of atoms ionized at a given temperature is derived from Boltzmann distribution.

7. May 4, 2012

### Boltzmann2012

Ah, Thank you for reminding.

Ludwig Boltzmann