Why does the cross section continue to decease with increasing energy for the photoelectric effect? For energies higher than the binding energy of K shell, shouldn't the cross section still be high since k electrons should absorb the energy to cause ionization and rest should be simply converted to kietic energy?
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$\begingroup$ Is this for photons interacting with atoms? $\endgroup$– David ZCommented Sep 24, 2016 at 15:39
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$\begingroup$ Sorry i forgot to specify, this is for photoelectric effect. $\endgroup$– Sabeeka NazeerCommented Sep 24, 2016 at 15:42
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1 Answer
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The energy must be transferred to a single electron, a single photon can only interact with a single electron. If you have a higher order Feynman diagram where the struck electron directly emits a virtual photon to another electron, you get a suppression factor of $1/137$ in there. This makes those things improbable and we can probably ignore them, at least in the leading order.
And there is a simple dimensional argument that it has to decrease eventually: At stupendously high photon energies, the only relevant energy scale is that of the photon. The cross section has dimension of length squared, that is $\text{Energy}^{-2}$. The relevant scale is the photon energy, so from this you get that $\sigma \sim E^{-2}$ for energies beyond any resonance or something else.
Whether this argument already holds for the photoelectric effect, I don't know.
answered Oct 2, 2016 at 20:44