# Ionization energy of nitrogen in V/m

## Homework Statement

I have an electric field that is nominally 30 kV/m, and I'm trying to figure out if nitrogen could be getting ionized in a field of that strength.

The 1st ionization energy of nitrogen is 1402.3 kJ mol-1.

What is the relationship between joules/mole and volts/meter?

## Homework Equations

[that's the question]

## The Attempt at a Solution

[don't know where to start]

Ionization energy is measured in units of -- umm -- energy (or total energy for a mass quantity) such as eV (or kJ/mol.) V/m is not a unit of energy, so there is no answer to your question; it's like asking for the distance from New Your to Chicago in pounds.

To start on the road to what you want, try this: 1) Start with the ionization energy in kJ/mole; I assume you're working in gaseous nitrogen, so you'll need the first ionization energy of the N2 molecule; 1402kJ/mol is the energy for atomic nitrogen. If you know where to find the energy for N2, please let me know. (It's what I was looking for when I stumbled across your question.) 2) Divide that by Avogadro's Number and multiply by 1000 to get J (per molecule.) 3) Divide by the magnitude of the electron charge to get eV. Since one electron needs to be removed from the molecule, the energy in eV is numerically the same as the potential in V through which the electron must be moved.

But here's the rub: Over what distance must the electron be moved? I don't know how to determine this. It probably has something to do with the size of the nitrogen atom, the size of the N2 molecule, and/or the spacing between molecules in the air. If you know the applicable distance, then you can divide the potential computed above by that length to determine the field strength that will ionize the nitrogen.

Here's an easier way. According to Wikipedia, the dielectric strength of dry air at STP is about 3.3MV/m between spherical electrodes; it may be lower for other shapes. That's the field strength needed to strike an arc. Striking an arc is done by ionizing the air so that it becomes conductive. So, the electric field needed to ionize N2 (most of air) is somewhere on the order of 3.3MV/m. At 30kV/m I can be pretty sure you're not ionizing anything. If you're not striking an arc, you're not ionizing the gas.

rolanddubey
No, I don't know the ionization energy for molecular nitrogen -- sorry. :/

If you're not striking an arc, you're not ionizing the gas.

Actually, we know that ionization is definitely possible at lower voltages. A corona discharge (which in the atmosphere would be called St. Elmo's Fire) occurs in STP air at roughly 100 kV/m, or 1/30 the potential necessary for an arc discharge. And a corona discharge itself is evidence of an existing electron avalanche that simply became robust enough to generate visible luminosity. (Any time an electron arrives at an atom, it emits a photon, but this has to occur en masse for the effect to be visible.) In other words, electrons start shifting around in much smaller voltages, and corona/arc discharges are simply thresholds (one for luminosity and the other for dielectric breakdown).

I "think" that the relevant distance through which the electron will move will be the mean free path, which is micrometers at STP. Anyway, I'll see if I can work it all of the way through. Thanks!

No, I don't know the ionization energy for molecular nitrogen -- sorry. :/
And it's probably not worth proceeding without that, so I guess we're both stuck.

Actually, we know that ionization is definitely possible at lower voltages. A corona discharge (which in the atmosphere would be called St. Elmo's Fire) occurs in STP air at roughly 100 kV/m, or 1/30 the potential necessary for an arc discharge.[Etc.]

OK, fair point; I stand corrected on that one. Still , your 30 kV/m is less than 1% of arcing potential, so I'd still bet against (but not so confidently.)

I "think" that the relevant distance through which the electron will move will be the mean free path, which is micrometers at STP. Anyway, I'll see if I can work it all of the way through. Thanks!

I have a different suspicion, but suspicion is all that it is. I need some time to spell it out coherently, and I'm supposed to be working, so let me continue this later.