Why helium is easier to ionize than N2?

[BlackPowder]

Helium has a higher 1st ionization energy (24.58eV) than N₂ (15.6eV) and O₂ (12.06eV). For an atmospheric room-temperature helium, why it is easier to get ionized than the daily life air under a same discharge setup? For example, for the Paschen curves, N₂ locates at the left of He which means that for the same voltage and pressure, N₂ requires us to move electrodes closer.

The most common answer I heard of this is that N₂ is diatomic molecule which has more degrees of freedom than helium has. However, I don't think this can explain the following fact that N₂ has a higher rate coefficient of electron impact ionization than He has, based on the calculation of the Boltzmann solver "BOLSIG+", even for a mean electron temperature (Te) much lower than the ionization thresholds of N₂. Considering the electron temperature distribution function (EEDF) is near Maxwellian, if the mean Te is very low, the rate of N₂ should higher.

 

[TeethWhitener]

Maybe I’m misreading these graphs, and this isn’t my area of expertise, but it looks like helium has the higher rate coefficient. Whether that’s enough to explain the breakdown, I don’t know. Dielectric breakdown is a complex process that depends on the ability for a gas to sustain a cascading ionization. There are a lot of moving parts.

 

[BlackPowder]

Maybe I’m misreading these graphs, and this isn’t my area of expertise, but it looks like helium has the higher rate coefficient. Whether that’s enough to explain the breakdown, I don’t know. Dielectric breakdown is a complex process that depends on the ability for a gas to sustain a cascading ionization. There are a lot of moving parts.

Thank you for the reply. I have moved a step forward about this problem. Here is something funny. The attached figure shows the results of the Boltzmann solver solving for Townsend coefficient and e-impact ionization rate coefficient of He and N₂ under different mix ratio.

For the Townsend coefficient, there is a clear trend that more helium making higher value. Also, for the e-impact ionization rate, pure helium has a higher rate than pure nitrogen. However, for their mixture, the rate coefficient of N₂ is always higher than He, although both of them decrease when the He-N₂ density ratio decreases. For the 90%-10% case, the rate coefficient of N₂ is even higher than the value of pure He at high E/N.

This means that He can somehow make N₂ very easy to be ionized. But still have no idea why this happens.

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[TeethWhitener]

This is quite interesting, and again, I'm not an expert in this area. It seems pretty clear that the ionization process induces some interaction (either energy or charge transfer) between He and N₂ which doesn't happen in the pure species. This might also happen with pure helium: He₂⁺ is a stable bound species, and helium dimers form stable excimers, whereas I'm not sure that N₂ forms excimers (and N₂⁺ is less stable than neutral dinitrogen). Again, dielectric breakdown is a complex process that probably involves a number of species, and the ease or difficulty of it is likely determined by the relative stabilities of all of these species.

 

[mfb]

Could excited states play a role in the ionization process? Electron collisions with helium are used in HeNe lasers for example, the excited helium from electron collisions is used to create population inversion in neon in these lasers.

 

[TeethWhitener]

Probably. That’s why I mentioned the He₂ excimer. It’s a long-lived state with a decently high binding energy. It’s also probably easier to ionize than an individual helium atom.

 

[BlackPowder]

There is a type of ionization called "Penning Ionization" that an excited species which at an energy level higher than the ionization threshold of another species can ionize that species during collision. In the mixture of He and N₂, the Penning ionization is He* + N₂ => He + N₂ + e. This is very common in actual experiment since He has two metastable states: He(2¹S) at 20.6eV, and He(2³S) at 19.8eV. Both of them are higher than the ionization threshold of N₂, which is 15.58eV. However, my results above are solutions of a Boltzmann solver which includes electron impact collisions only. Penning ionization was not involved when I got the figure. This is why I am so confused.