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I am thinking of the high power LEDs at around 250nm wavelength and an output power of some 30mW optical, in continuous mode either divergent beam or focused.

Qmechanic
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    Be really careful of those unless you want cataracts at an early age: even very low power UV is very nasty to the sight, see here, particularly my answers to comments at the end of the question. – Selene Routley Feb 06 '15 at 11:24
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    Since I am looking for something to replace Polonium, this looks like the lesser of two evils –  Feb 06 '15 at 12:10

2 Answers2

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Let's look at the primary constituent of air, nitrogen. The ionization energy of nitrogen is approximately $1400~\text{kJ/mol}$. This works out to be approximately $2.25\times10^{-18}~\text{J/atom}$.

The energy content of a laser beam is $E = h c /\lambda$. This means we can solve for a wavelength, $\lambda$ that provides enough energy to ionize a single atom of nitrogen.

This wavelength works out to be roughly $88~\text{nm}$.

So no, it would appear that $250~\text{nm}$ is not high enough energy to ionize the components of air.

While I certainly appreciate the upvotes (and maybe they are a sign this is an okay way to approach the problem), the are some questions raised in the comments about some discrepancies with other approaches. So before upvoting, please read the comments and see if there is a way to reconcile things.

And if this is okay and you still want to upvote, then please do so but help out and leave a comment saying there is no problem!

tpg2114
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  • Did you mean $10^{-18}$, not $\exp(-18)$? – Ruslan Feb 06 '15 at 11:28
  • @Ruslan Yes, I am used to dealing with floating point outputs which use $e$ rather than $\times 10$. I will correct it if it is confusing. – tpg2114 Feb 06 '15 at 11:30
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    Of course it is. Either you use float-style 2.25e-18, or you use $2.25\times10^{-18}$, but not mix them. – Ruslan Feb 06 '15 at 11:31
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    @Ruslan Oh, yeah, I see the problem. Adding the exponent. Cut me some slack, it's way too early in the morning and I haven't had any coffee yet. – tpg2114 Feb 06 '15 at 11:32
  • Do you have a reference for this? I mean, it sounds right, but I can't upvote it because I don't know. But I was going to calculate the electric field at a focus and compare it with the breakdown potential of air. If the focus is $(10{\rm \mu m})^2$ then the intensity is $3\times 10^8{\rm W,m^{-2}}$, implying an electric field of $340{\rm kV,m^{-1}}$, or about a ninth of the breakdown potential. Maybe the ionisation energy is the main part of a calculation of the breakdown potential. Experimentally, which calculation would be right? – Selene Routley Feb 06 '15 at 11:40
  • @WetSavannaAnimalakaRodVance Reference for which part? I just considered it from an energy balance perspective, and so the only number that needs a reference would be the ionization energy of nitrogen. Which I should add is the ionization energy of $N$ and not $N_2$, which means the laser would have to first break apart the $N_2$ before it could ionize it. So it might actually take an even smaller wavelength. – tpg2114 Feb 06 '15 at 11:44
  • And that's a very good question about which one would be correct in real life. Can you post the analysis you were going to do and see how the end results compare? Maybe they are close enough, maybe they are orders apart? – tpg2114 Feb 06 '15 at 11:45
  • I'm not disagreeing with you. Your calculation just got me thinking: it looks perfectly sound and I began wondering how it would compare with a breakdown calculation - and then of course the question arises exactly how would calculate breakdown theoretically, and I think that beginning with a method like yours would be a good start. I've upvoted, because there's bound to be more than a grain of truth to it. – Selene Routley Feb 06 '15 at 11:48
  • Details: I've just used $I = \sqrt{\frac{\epsilon_0}{\mu_0}} E^2$ for the intensity, and I've assumed that at best you can focus an LED down to a $(10{\rm \mu,m})^2$ spot. The latter is probably going to be optimistic because the LED chip is quite extended and you have a fairly high etendue source. My method, however, would be independent of wavelength, which doesn't agree with yours. – Selene Routley Feb 06 '15 at 11:49
  • @WetSavannaAnimalakaRodVance I'm actually finding many references that use far-infrared lasers to induce ionization. So maybe the problem is my analysis is looking at when a single photon hits a single atom, in which case it needs to be a very short wavelength. But if you can focus a great many, lower energy, photons onto a single atom it can still transfer enough energy to ionize it? Perhaps that is the distinction between our two methods. Is your approach really independent of wavelength, isn't $E \propto \lambda$? – tpg2114 Feb 06 '15 at 11:54
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    NIST (http://webbook.nist.gov/cgi/cbook.cgi?ID=C7727379&Mask=20#Ion-Energetics) lists the ionization energy of N2 as 15.581 eV, so it looks like you did pretty well. – Jon Custer Feb 06 '15 at 14:04
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    Note that the IE of oxygen (O2) is lower than nitrogen (triple bond trumps, in this case) - about 12eV. This lowers the required wavelength to something in the ~100nm range - still too low a WL for a LED, but ozone generation (due to O2 ionization) is usually the first thing one will notice when UV starts getting hard enough. In practice this is usually around 130nm and shorter. – J... Feb 06 '15 at 15:22
  • Reading all this comments made me remember of the discussions in my physics classes about classical and quantum calculations of kinetic energy of electrons by photoelectric effect: where classical one leads to wavelength independence and light intensity dependence, and quantum one leads to wavelength dependence and intensity independence. Therefore, lets remind ourselves that classical EM is good only in low frequencies (say... infrared or lower). At higher ones (say... ultraviolet) the classical theory of electromagnetism begins to breakdown and can't be trusted. – Physicist137 Nov 03 '15 at 03:03
  • @WetSavannaAnimalakaRodVance Although your calculation seems classical instead of quantum, I was curious. How have you calculated $10\mu m^2$ spot? – Physicist137 Nov 03 '15 at 03:15
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    I believe this is the right approach. You need a single photon to have enough energy to knock off an electron, unless you are in the nonlinear regime where two photons arrive in very close succession (and effectively double-bump an electron). That happens at very high power densities... not achieved with an LED. – Floris Feb 10 '17 at 22:52
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well. There is a paper that can clearly answer your question what you can expect since it is the same wavelength.

http://aip.scitation.org/doi/figure/10.1063/1.3692090

Figure #4 gives that from small intensities < 10^9W/cm^2 you'll have linear ionization of organic atmospheric impurities (250 nm= 5ev quanta energy which is capable to provide direct photoelectric effect from most materials except Platinum). Though the actual yield will be low (<10^12 electrons/cm^-3) it will actually be air ionization.

met
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