I have been reading up on multi-photon ionization recently, and I can't seem to find an explanation for this.

My understanding so far is that if you need x amount of energy to excite an electron, and if you intend to do it with n amount of photons, then you need each photon to have an energy of $\left(\frac{x}{n}\right)$.

Then when each photon collides with the electron it will temporarily "push" the electron $\left(\frac{1}{n}*100\right)$% of the way needed, and if they do it successfully, the atom will be ionized.

My question is why do you need such high intensities (1.00e12 W/cm^2) in order for this to happen? Won't it still occur even if the power is low?


1 Answer 1


You need two photons to hit the same electron very shortly after each other. The simplest way to make this happen is to increase the amount of photons you're shooting at the target which is exactly what you do when you increase intensity.

Ionisation can still occur at lower intensities, but it will happen much less frequently because the chance of two photons hitting an electron within the short time period will be smaller.

  • $\begingroup$ Interesting! Is there a formula that can calculate the probability given power? Or is it just a linear relationship? $\endgroup$
    – Username
    Commented Jun 4, 2018 at 19:45
  • $\begingroup$ The probability of one photon hitting an atom is related to the cross-section of the material. I'm pretty sure probability will increase more or less linearly with intensity, but don't quote me on that. $\endgroup$ Commented Jun 4, 2018 at 19:53
  • $\begingroup$ Alright, I will do some more research on that. I think it might be exponential though, based on other posts I have seen here. Thanks for your help! $\endgroup$
    – Username
    Commented Jun 4, 2018 at 19:55

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