1.Ionization energy equals minimal energy electron needs to overcome atom’s Coulombs and gravitational fields.

Ionization energy depends on which state atom currently is. For ground state ionization energy highest. For every next states it is lowering.

So, if atom is not at ground state it needs less energy to be ionized.

2.Lifetime of excited states very small, so for some monochromatic photon stream, falling onto some atoms, the probability, that atom, being already in the excited state receive new photon is tiny. But not zero, as understand.

3.Besides excited states, which atom can occupy for a short time, there are metastable states, where atom can stay long.

4.Important thing, I assume, do not know for sure, that in some atoms, there is metastable state, that atom occupies by losing some energy after being excited to state $E_n$ from state $E_{n-1}$, via photon with energy $E_n - E_{n-1}$, which ionization energy is lower than $E_n - E_{n-1}$.

For example: some atom with energy states: $E_0 = 30eV$, $E_1 = 10 eV$ …,and with metastable level between first and ground $E_{0.5} = 20eV$, being at ground state receives photon with $E = 20eV$, being excited to first level for a short time, and jumps to metastable level $E_{0.5}$, staying there.

Not leaving $E_{0.5}$ state, atom receives next same $E = 20eV$ photon, and should be ionized.


According to third photoeffect law, ionization does not depends on stream intensity, and can be achieved only starting from specific wavelength of single photons.

So, actually, it is not true?


1 Answer 1


Physics is not mathematics. Models work in their domains, but may need modification other circumstances. At high electromagnetic intensities, various "nonlinear" optical phenomena are evident. Multiphoton ionization is a known effect.

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    – SuperCiocia
    Mar 27, 2023 at 18:00

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