abotiz said:
Thank you for your replies.
I don't think I follow or agree with your answer fzero, I feel as you are adressing another issue.
Lets recall the facts that we know, and what could be of importance:
1) Binding energy
2) Photon energy
3) Number of electrons in each shell
We know that the K-shell electron has the highest probability. In general, cross sections is determined by the amount of "targets" (atoms or electrons). K-shell can only have 2 electrons, whereas the L-shell can have 8. So the numbers are against us. We conclude that the amount of electrons in a shell is either of no importance or not significant.
Let's simplify the question that we want to address to: what is the relative probability that a photon with energy ##\hbar\omega## ejects an electron from the K-shell versus ejection from the L-shell. If we let ##i## label the states the electrons in the atoms, then the relative probability of ejection can be roughly expressed as
$$P_i = \rho_i p_i,$$
where ##\rho_i## is the density of the state ##i## in the sample and ##p_i## is the probability amplitude for ejection from the ##i##th state. If we are dealing with a single atom with ##Z>10##, then roughly
$$ \rho_K \sim \frac{2}{Z},~~~\rho_L \sim \frac{8}{Z},$$
so the next part that needs to be addressed is the relative values of ##p_K## and ##p_L##.
Photon energy is a variable that tells us if there is 0% chance (hv<E_bind) for ionization (ejection of electron) or >0% (hv>E_bind). Its not a very important variable if written alone. If in atom X the K-shell electrons are bind with Z eV, and in atom Y the L-shell electrons are bind with Z eV, the highest propability would still be with K-shell electrons.
The photon energy is more important than that because the probability factors ##p_i## depend on the photon energy, as well as on the binding energy. In terms of the quantum mechanical amplitude, these data are encoded in the wavefunctions. The amplitude involves both the wavefunction of the initial-state bound election (whose energy eigenvalue is directly related to the binding energy) and the wavefunction of the ejected electron (which is approximated by a plane wave depending on the final state momentum, which is in-turn directly related to the energy of the initial-state photon by energy conservation).
When we put everything together we find that
$$ \frac{P_K}{P_L} \sim \frac{c}{4} \omega,$$
where ##c## is a numerical factor. The precise value of ##c## depends on the numerical factors that I did not keep track of in my computation, but will also depend on all the complications of the many-body system of a real atom, like charge screening, that should really be addressed by using more realistic wavefunctions in the computation. Nevertheless, we can conclude that, whatever the value of ##c##, for
$$\omega > 4/c,$$
the probability to eject an electron from the K-shell is larger than the probability to eject it from the L-shell, which agrees with the experimental situation.
Now the binding energy is related to the amount of electrons which in turn relates to amount of protons in the nucleus and finally to the binding energy, again (i.e. more protons = higher binding energy). We know that the cross section for photoelectric interaction increases with the atomic number, or you could say increases with amount of protons in nucleus, or increases with the amount of binding energy.
The focus should not be on the atom number or amount of protons because that only explains the difference in cross sections between different elements. If you focus on the binding energy you include both the different elements and the difference between the K,L and M-shells.
The first-order effect of ##Z## on the cross section that you are describing here is related to the probabilities that the photon interacts or does not interact with the atom. An atom with more electrons is "bigger" and therefore has a higher cross section. This is not the question that I am trying to answer, rather I am assuming that the photon does in fact lead to electron ejection.
So I believe the answer to the topic question lies in the effect or consequence of different binding energy.
As I have commented above, the binding energy is an important input, which is accounted for by using the bound-state wavefunction in the calculation of the amplitude for the process. For a real atom, this is a complicated wavefunction, but we expect that the approximation of using hydrogenic wavefunctions should still give some useful qualitative information.
I hope that this clarification makes the analysis a bit clearer.