# Why does the effectiveness of low-Z shielding increase with photon energy?

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Concrete for example - you need less concrete to obtain the same lead equivalent for photon energy 500 keV than for 200 keV. What is the reason for this?

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Ok, I did some thinking ;) Is it because for low Z materials, photons with higher energy interacts with absorber almost by Compton Effect only? And Compton is independant of atomic number.

Astronuc
Staff Emeritus
Can one show some calculations or data to support the assertion that "you need less concrete to obtain the same lead equivalent for photon energy 500 keV than for 200 keV." The mass attenuation coefficient continually decreases as a function of gamma energy, although the mass energy-absorption coefficient increases slightly between 0.2 to 0.5 MeV. But this is misleading, since the 500 keV gamma will scatter to a lower energy, and that photon will scatter, and so on.

https://physics.nist.gov/PhysRefData/XrayMassCoef/ComTab/concrete.html

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https://archive.org/details/jresv38n6p665

I asked this question as I was reading about shieldings in PET departments, where they suggest lead/concrete ratio 12-15, while ratio for 150 keV X-Ray is 80... (These are example regulations from my country).

https://archive.org/details/jresv38n6p665

I asked this question as I was reading about shieldings in PET departments, where they suggest lead/concrete ratio 12-15, while ratio for 150 keV X-Ray is 80... (These are example regulations from my country).
It's a property of lead rather than a property of the concrete.
Have a look at the cross-section per unit mass of lead below, and compare it to the same for concrete:

Lead has an elemental absorption edge at around 88keV, so the cross-section for lead is massively increased when the photon energy, as the photon energy is further increased, the photo-ionisation cross-section due to that particular energy-level decreases, becoming more comparable to that of the concrete.