In photoelectric effect, why does photon prefer K shell electron?

[amylase]

Homework Statement

In photoelectric effect, why does photon prefer K shell electron?

Asked differently:
In photoelectric effect, why does photon prefer electron of closest binding energy, rather than going for another electron of much lower binding energy?

Homework Equations

http://www.sprawls.org/ppmi2/INTERACT/

The Attempt at a Solution

This is what I understand about photoelectric interaction: If a photon's energy matches or is higher than the binding energy of an electron, the photon will be absorbed by that electron. This allows the electron to overcome its binding energy and be liberated from the atom. So for example a photon can be absorbed by a K shell electron which then escapes.

I am cool with above.

My question is: if that photon had enough energy to break free K shell electron, that photon could have broken free an L shell, or M shell electron instead (they are of lower binding energies hence easier to liberate). Why would the photon preferentially choose to free the one that's hardest to free (K electron)? That photon could easily have chosen to free the M shell electron and given it more residual energy to run away with.

Hope my question is clear: why does photon prefer K shell electron in photoelectric effect when it could easily choose an easier target (eg. M shell electron)?

Thank you.

 

[Simon Bridge]

Check: does the photoelectric effect prefer the K shell? How do you know this? Does your own reference support this idea?

eg. in a solid metal, wouldn't the ejected electrons come from one of the crystal structure bands rather than from atomic shells?