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

taffer33
Messages
4
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
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
 
Last edited:
https://archive.org/details/jresv38n6p665
it's in this article for example

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).
 
taffer33 said:
https://archive.org/details/jresv38n6p665
it's in this article for example

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:

z82.gif


concrete.gif


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.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
11
Views
2K
Replies
17
Views
4K
Replies
21
Views
4K
Replies
3
Views
2K
Replies
15
Views
3K
Back
Top