# Photon energies and attenuation [closed]

I hope you can help.

I am trying to understand which attenuation effect (photoelectric, compton scatter and pair production) is dominant at which level of photon energy and why. The levels are 10keV, 30keV and 4meV. I am sure pair production happens at the latter due to its high energy needs.

Thank you

D

## closed as too broad by Carl Witthoft, John Rennie, Kyle Kanos, ACuriousMind♦, DanuSep 1 '15 at 6:56

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The answer depends in part on the Z of the material you are looking at. This is something you can easily verify by looking at the XCOM database

To generate an example, I entered "single element, Z=25" (manganese) and selected plots for different types of interaction in the range up to 10 MeV. The result looks like this:

As you can see, the photoelectric effect dominates up to energies around 100 keV and pair production comes into play about 1.022 MeV but doesn't even reach the level of incoherent (Compton) scatter until you get above 10 MeV.

If you repeat this for different elements you will get different results. For example, repeating the above for hydrogen (but focusing in on the range up to 100 keV only) I get the following plot:

where the photoelectric effect is much less important and Compton scatter dominates for almost the entire energy range.

The full calculations are not trivial. A comprehensive review of the history of these calculations can be found in Hubbell (2006). It contains reference to more than you would ever hope to know about the subject...

UPDATE

The "simple" version of the above is as follows:

Coherent scattering occurs mostly when the scattering photon does not lose energy during the interaction. This is most likely at low energies (energy small compared to the binding energy of the electrons). The electron is too busy "doing laps" to be disturbed by the photon. So the photon just changes direction. At the energies you listed, it is unlikely to be the dominant effect.

Photoelectric effect dominates for energies that are comparable to the binding energy of the electrons. You will see a "saw tooth" on the attenuation curve, where the probability of interaction goes up each time your energy exceeds another binding level (and therefore makes more electrons available for photoelectric absorption). The last of these jumps is usually the biggest: this is known as the K edge. The location of the K edge depends on the atomic number (Z) of the nucleus. For Mn (Z=25), the K edge is at 6.5 keV; for lead (Pb, Z=82) it is 88 keV. A full list is found in [this Kaye and Laby table]) http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_2/4_2_1.html)

As the energy of the photon becomes higher than the binding energy of the K electron, you enter the domain of Compton scatter - the photon packs such a punch that the electron behaves "as though it was free", and the exchange of energy and momentum is described by the Compton equation.

Pair production doesn't come into play until you hit energies above 1.022 MeV (the minimum needed to generate an electron/positron pair); as you can see in the graphs above, the probability of pair production is smaller than the probability of Compton scatter for most nuclei at 4 MeV (although at A level they might expect you to say that pair production dominates there).

• Thank you for your help. This is an extremely confusing subject for me, I do not really have much of a physics background. The questions is for an high grade A level question which I am struggling to understand. The question is as follows - Describe and explain which of the 4 attenuation effects described (scatter, photoelectric, Compton and pair production) predominates at photon energies of 10keV, 30keV and 4MeV. Could you help further? I assume only a fairly basic answer is required. Thank you very much – D. Grey Aug 28 '15 at 18:36
• Hmmm. What do they call "scatter" in high grade A level? Rayleigh scatter? – Floris Aug 28 '15 at 18:46
• Unfortunately they do not specify, it is just 'scatter'in the notes - 'photons simply bounce of the atoms, neither the atom nor the photon are changed, the photon just changes direction of travel' – D. Grey Aug 28 '15 at 18:52
• OK - that is "coherent scatter". – Floris Aug 28 '15 at 18:52