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Why helium is easier to ionize than N2?

  1. Sep 1, 2018 #1
    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. Rate Coefficients for 0.5He0.5Air.jpg
     
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  3. Sep 6, 2018 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Sep 6, 2018 #3

    TeethWhitener

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    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.
     
  5. Sep 10, 2018 #4
    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.
    Townsend and E-Impact Rate Coefficients.jpg
     
    Last edited: Sep 10, 2018
  6. Sep 11, 2018 #5

    TeethWhitener

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    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.
     
  7. Sep 11, 2018 #6

    mfb

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    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.
     
  8. Sep 11, 2018 #7

    TeethWhitener

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    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.
     
  9. Sep 13, 2018 #8
    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.
     
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