The impact of electron density on photoelectric effect

Question

I'm a second year radtion therapy student and i'm looking at the attentuation of KV energy beams. I'm wondering why for the photoelectric effect to occur does there need to be a high amount of 'inner'/ tightly bound electrons. Is it because this increases the probability of an incident photon actually hitting a tightly bound electron with enough force to knock it from it's orbit? If so why does the probability of PE increase with higher z numbers, surely lower Z numbers will have more loosely bound electrons which are easier to knock from their orbits?

Answer 1

It's not that simple. Binding energies depends on $\mathcal Z$ number and also on particular electron configuration, as per chart :

Also, if we take gamma photons as a reference,- full photoelectric cross-section is given by Sauter formula :

$$ \sigma _{ph}={\frac {3}{2}}\phi _{0}\alpha ^{4}{\biggl (}Z{\frac {E_{e}}{E_{\gamma }}}{\biggr )}^{5}(\gamma ^{2}-1)^{3/2}{\Biggl [}{\frac {4}{3}}+{\frac {\gamma (\gamma -2)}{\gamma +1}}{\Biggl (}1-{\frac {1}{2\gamma (\gamma ^{2}-1)^{1/2}}}\ln {\frac {\gamma +(\gamma ^{2}-1)^{1/2}}{\gamma -(\gamma ^{2}-1)^{1/2}}}{\Biggr )}{\Biggr ]} \tag 1$$

Where, $$\gamma ={\frac {E_{\gamma }-E_{B}}{E_{e}}} + 1 \tag 2$$

So, photoelectric effect cross-section depends on atomic number $Z$ and on electron binding energy $E_B$.