# Why is there an upper energy limit on the photoelectric effect? [duplicate]

I feel like I'm missing something really fundamental regarding the cross section of the photoelectric effect. I'm looking at this chart, and it seems that the lower the energy of the incident photon, the more likely a photoelectric interaction is to happen. Higher energy interactions get less likely, until a cut-off point, at which they no longer occur.

If the energy of the ejected electron is equal to the incident photon energy minus the binding energy, I understand why a lower limit exists (since there must be enough energy to eject the e-), but why should there be an upper limit?
Why does the cross section reduce with energy at all?

## marked as duplicate by rob♦Jul 24 '17 at 13:40

• Did you read this: physics.stackexchange.com/questions/269997/…. Great user name :), ,BTW – user163104 Jul 24 '17 at 12:15
• Oh fab, that's exactly what I was after! Sorry for making a duplicate post guys. – SherlockOhms Jul 24 '17 at 12:50

The cross section reduces for higher frequencies, because our material will get more and more transparent for the incoming light.

Think about it in this way: the electrons follow the oscillating E-field of the photons. Now imagine the field oscillates faster and faster. At some point the electrons will be unable to follow it hence they won't get any energy from it and the photon just passes by.

In this regime other processes set in and dominate for the higher frequencies (pair production).

To give you a short answer for your question, in a crude approximation the photo-electric cross-section has the following expression:$$\sigma=const \cdot \frac{Z^n}{E^3}$$ Here Z is the nucleus atomic number and n is an integer (usually between 4 and 5). Of course, E is the energy of the photon. So as you can see, the probability that a photon will produce the photo-electric effect, decreases with energy.

You can read more detailed explanations here: $\textbf{Davisson, C. M. (1965). Interaction of gamma-radiation with matter}$.

Hope this helps :)

• Hi Rob, thanks for your response. I'm afraid I'm still a bit in the dark.. can you explain qualitatively why cross section is proportional to 1 / E^3? – SherlockOhms Jul 24 '17 at 12:47