Radiological physics  linear attenuation coefficientby stacey_1995 Tags: attenuation, coefficient, linear, physics, radiological 

#1
Feb314, 09:48 AM

P: 1

I am completing a radiological physics experiment about attenuation and half value thickness.
I'm struggling to understand linear attenuation coefficient and what it means when it says its and exponential relationship. I understand that attenuation is the reduction in intensity of an xray beam when it passes through a material but that is as far as i can get my head around. any insight into this would be helpful. Thanks 



#2
Feb1214, 02:01 PM

Sci Advisor
P: 2,568

Think of an xray beam as a large number of photons (N) travelling through space.
Now put a thin piece of some material in its path. Some of those photons will interact with the material, while others will pass right through. Say that n photons interact. The number that interact (n) depends on the number that are incident on the thin material. It also depends on the thickness of the material. If you double the thickness, you'll get twice as many interactions, for example. So if we have a material of thickness dx, the number of interactions you get is n = N dx u. That "u" is in there as a constant of proportionality,which is a property of the material. Lead, for example will generally give rise more interactions than air. That factor "u" is the linear attenuation coefficient. If you rearrage the equation, you should be able to convince yourself that u is the fraction of photons that interact in the material per distance dx. If you think about though, n depends on N. So what happens when I put a whole lot of that material in the way? I know  I already told you that if I double the thickness, I double the number of interactions, but that was an approximation that relies on the material being really thin. To consider a "thick" case  you get into the calculus of it all. A thick case is really just a whole bunch of thin cases stacked up together. Each time you go through a thin piece of material, you have u N dx interactions. But the N in each case keeps changing because the number of photons that start out on any given piece depend on how many interactions occured in the piece before. The change in the number of photons each time we pass through a small piece of material is dn =  u N dx. The minus sign is because we lose photons. Rearranging, this gives us a differential equation: dn/N = u dx Once you learn to integrate, you'll learn that this has a solution: Nf = Ni exp(u x), where Ni is the initial number of photons and Nf is the final number after you've passed through your entire material. And there's your answer about where the exponential comes in. It's a direct result of the fact that the change in the number of photons that occurs through a small piece of material is directly related to the number of photons that were incident upon that small piece, which depended on the number of photons that were incident on the previous piece, etc. I know this is a little late, but I hope that helps. 


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