Beer-Lambert relation to dosage

[David G.]

So as I understand Beer-Lambert, it describes the attenuation of intensity/flux/fluence. My question is, suppose you have:

• some set object of interest
• fixed at some far distance from a source (so the rays are ~parallel)
• a shield (e.g. layer of lead) is placed in front of the object, that reduces the radiation intensity by, say 50%

Then would this necessarily result in a 50% reduction in exposure and absorbed dose?

Thanks in advance!

 

[e.bar.goum]

Yes, definitely. If you have half the intensity hitting you, you necessarily have half the dose. It's energy conservation.

ETA: Well, no. You can have, say, x-rays produced from the scattering of the gammas on atomic electrons. So you could get a dose from the x-rays.

 

[ChrisVer]

Well I think the distance will also play a role, since the flux drops with distance squared. But if you keep the distance as fixed, and you want to look at the particular "dose" you will get from such a distance, then yes the absorber will do exactly that job...

 

[mfb]

ETA: Well, no. You can have, say, x-rays produced from the scattering of the gammas on atomic electrons. So you could get a dose from the x-rays.

That is an important point. The flux of the initial radiation of whatever type will reduce by 50% if Beer-Lambert is applicable, but you can get secondary particles. In the worst case (very high-energetic primary particles), the absorbed dose of material behind the shielding increases.

 

[David G.]

That is an important point. The flux of the initial radiation of whatever type will reduce by 50% if Beer-Lambert is applicable, but you can get secondary particles. In the worst case (very high-energetic primary particles), the absorbed dose of material behind the shielding increases.

This is the sort of thing that had me concerned. Also whether the exposure to absorbed dose is necessarily linear. Thank you all for your feedback.