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.
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ 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. $\endgroup$– Selene RoutleyCommented Feb 6, 2015 at 11:24
-
1$\begingroup$ Since I am looking for something to replace Polonium, this looks like the lesser of two evils $\endgroup$– user56903Commented Feb 6, 2015 at 12:10
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
15
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!
-
2$\begingroup$ Of course it is. Either you use float-style
2.25e-18, or you use $2.25\times10^{-18}$, but not mix them. $\endgroup$– RuslanCommented Feb 6, 2015 at 11:31 -
1$\begingroup$ @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. $\endgroup$– tpg2114Commented Feb 6, 2015 at 11:32
-
1$\begingroup$ NIST (webbook.nist.gov/cgi/…) lists the ionization energy of N2 as 15.581 eV, so it looks like you did pretty well. $\endgroup$ Commented Feb 6, 2015 at 14:04
-
3$\begingroup$ 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. $\endgroup$– J...Commented Feb 6, 2015 at 15:22
-
1$\begingroup$ 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. $\endgroup$– FlorisCommented Feb 10, 2017 at 22:52
$\begingroup$
$\endgroup$
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.
