A double covalent bond of nitrogen ($\mathrm{N}_2$) has an ionization potential of 15.58eV. Now lets go into an example. If I have a laser of wavelength $\lambda = 477\:\mathrm{nm}$ and I calculate the photon energy as $E = 1240\:\rm{eV\: nm}/\lambda$ then I get $2.6\:\mathrm{eV}$ per photon.
Now if I do $15.58\:\mathrm{eV}/2.6\:\mathrm{eV}$ it equals ${\sim}6$ photons.
Now, to figure out what pulse energy is required for 1 photon I do $$ E =\frac{hc}{\lambda} = \frac{6.626\times 10^{-34}\:\mathrm{J\: s} \times 3\times 10^{17}\:\mathrm{nm/s}}{477\:\mathrm{nm}} \approx 4.167\times 10^{-19}\:\mathrm J, $$ and for 6 photons that's $2.500\times 10^{-18}\:\mathrm J$.
Now if I want to get 6 photons in a pulse then I would need a pulse energy of $2.500\times 10^{-18}\:\mathrm J$, right? All I want to know is why does it require such low amounts of energy to do this, when all I see is that you need a lot of watts to ionize. Now as I was typing this I thought that maybe that equation shows how many joules of energy are in each photon, but yet this video says "Joules required per photon". But if this equation is not right then what equation should I use?
Answer
There really isn't much more to say that isn't well summed up by Jon Custer's initial comment,
Perhaps you should step back a bit and dig into what is happening physically in multi-photon ionization. Basically, you need in a short time to have happen to be there at pretty much the exact same time. All the others? Well, the timing just wasn't right.
In quantum mechanics, the totalitarian principle tells you, basically, that if something can happen, it must happen. However, it doesn't tell you how often it will happen.
For the multiphoton ionization you're asking about, your calculations show that there is a six-photon ionization process that's possible for the nitrogen molecule subjected to a 477nm laser. However, what you don't learn from that calculation is how likely that process is to happen: if your laser is weak, you're likely to have one or two electrons get ionized on occasion; that might be once per million ages of the universe, but it's still a positive rate so the process is happening.
To get the pretty sparks of laser-induced optical breakdown in air, however, you don't just need a nonzero rate, you need enough electrons to kick-start the avalanche process of dielectric breakdown. That is, you need the ionization rate be macroscopically meaningful, not just nonzero.
So, what is this ionization rate, and how does one calculate it? Well, it calls on a completely different set of concepts than just the photon-energy layer that you've used to compute whether the processes are possible. More specifically, what you want is the multiphoton ionization cross section, which depends critically on the intensity of the laser (that is, the concentration of the laser energy, not just the total energy of the pulse).
I won't go hunting on a wild-goose chase of moving goalposts for the specific parameters you've posted, but I'll start you off with this paper, which measures the cross-section of $(9\pm1)$-photon ionization of molecular nitrogen under the influence of a $1.06\:\mu\rm m$ neodymium-glass laser. (Note that ionizing from the ground state at this wavelength would take some 14 photons; here there's probably nontrivial additional physics going on.) Generally, the ionization probability through an $n$-photon process scales (by definition of the latter) as $$ \mathrm{ionization\ rate} =\mathrm{constant}\times\mathrm{intensity}^n, $$ i.e., as the $n^\rm{th}$ power of the intensity. This scaling is extremely important, but you still require at least one data point to fix that constant, and this is pegged by the single measurement that at intensity $I_0=2.44\times10^{12}\:\rm{W/cm^2}$ the ionization rate is about $r(I_0)=3\times 10^6/\rm s$, i.e. about three million per second. (It's important to note that these numbers are not transferable, and they do not apply to other molecular species or other driver wavelengths. If you want numbers for other configurations, do your own literature search, like everyone else.) That data point allows us to estimate the ionization rate at arbitrary intensities $I,$ as $$ r(I) = \left(\frac{I}{2.44\times10^{12}\:\rm{W/cm^2}}\right)^9\times 3\times 10^6/\rm s. $$
This is why you need high-power lasers: because if the intensity is, say, a factor of ten smaller than the reference intensity, at $I=2.44\times10^{11} \:\rm{W/cm^2}$, then the ionization rate will drop by nine orders of magnitude. That's a billion times. There's other factors involved, but when you have a multiphoton process with this kind of scaling, that stuff-to-the-ninth-power dependence trumps most everything else.
Since you do seem to need some help with the actual numbers, let me plug in a few for you. Suppose that we take the pulse energy $E$ as in your question, i.e. just nine photons in the pulse, with pulse energy $E=9h\nu = 10.5 \:\mathrm{eV} = 1.7\times 10^{-18}\:\rm J$, and let's use pulse of one-picosecond duration $\tau$ with beam waist $w$ of a hundred microns. Then the peak intensity is roughly $$ I=\frac{E}{\pi w^2\tau}=53\:\rm{W/cm^2}, $$ and we expect it to produce, on average, $$ r(I)\times\tau = \tau\times\left(\frac{E/\pi w^2\tau}{2.44\times10^{12}\:\rm{W/cm^2}}\right)^9\times 3\times 10^6/\rm s \approx 10^{-132} $$ electrons per atom per shot - i.e. nothing anywhere near remotely measurable. On the other hand, if we put in something more macroscopically meaningful as our pulse energy - say, a per-pulse energy of $E=1\:\rm mJ$, corresponding to one watt of average power at a 1kHz repetition rate, then that increases to about $10^{-5}$ electrons per atom per shot, on average; when you add in the number of atoms in your focal volume, that can give you enough electrons to seed the avalanche process. But once you start playing with those numbers, it will hopefully become obvious just how crucial is the dependence on both the beam waist, the pulse energy, and the pulse duration.
To put things a bit more informally, the $n^\rm{th}$-power dependence is exactly what you would expect if the process required the photons to be present not just in the pulse somewhere, but directly on the molecule all at the same time. (That's not the actual origin, but the analogy is useful.) You can't just say "my pulse has nine photons, I can ionize one electron" - you need all those photons to lie on top of each other both temporally (thus requiring a short pulse, and changing the dependence from pulse energy to the peak power, a.k.a. energy flow rate) and spatially across the transverse directions of the pulse (thus requiring a tight focus, and changing the dependence from power to intensity (more precisely, irradiance) and adding the $/\rm{cm}^2$ to the units of the quantity of interest).
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