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We are taught that in photoelectric effect if the frequency of light is lower than the threshold, then no matter how long a metal is exposed to it there won't be any ejection of electrons.

This made me consider a thought experiment. If we shoot two photons at an isolated atom, having energies equal to half of the ionisation enthalpy of the atom and hence a sum which is equal to the ionisation enthalpy of the atom. Now, in case they hit the atom at the same exact instant, would there be any photoelectric effect?

If no, then why? Why no Photo E Effect when some low frequencies are exposured for more than 20-30 hrs?

I felt like the explanation is that since electrons take energies at discrete amounts, taking some energy from one photon and the rest from the others won't work. However, hitting the atom at the same instant should actually let it happen.

My little thought experiment might be a little vague. Thus, My direct question is why PEE starts only at a particle frequency? Why increasing the intensity of light of lower wavelengths don't work?

uddhav saikia
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1 Answers1

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The answer is 'yes, sort of.' The approach isn't necessarily to get two identical wavelength photons to be absorbed simultaneously, but rather to have one photon "bump up" an electron to a higher orbital, then the second photon kicks the electron free before it can reradiate back to ground state.

In fact, 2-photon absorption is an effect used for passive Q-switching in some laser designs. In this setup, rather than ionization occurring, the material changes its absorptivity quite dramatically when the electron is pushed up to a certain high orbital. In the ground state, the optical transmissivity is low, but once the incident photon density reaches a critical level, 2-photon absorption happens, the material's transmissivity goes close to 1.0 .

Carl Witthoft
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