Penetration of X Rays in body tissue


by sophiecentaur
Tags: body, penetration, rays, tissue
sophiecentaur
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Feb20-14, 08:54 AM
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I have been looking for information about the way Radiotherapy works and, needless to say, there is much more involved than just local frying of tissue. I have been told and also read, that the dose received in tissue is not necessarily maximum at the surface. For instance, the dose seems to be a maximum at 0.4cm for 2MV photons and 2.3cm for 10MV photons. I cannot think why, if the absorption is exponential with depth (or so I would expect) the maximum effect (energy from the beam) is not right at the surface. 2.3cms is very deep under the surface. What happens between the surface and that depth to make the dose less?

I have looked in all the places on the web that I can think of but, as usual, what's available is either the chatty home-medecine advice or the latest papers which all assume the reader knows this sort of basic thing. Any ideas? I think this is probably the right forum as it's basically Physics.
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phyzguy
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Feb20-14, 09:58 AM
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For the photon energies used, the rate of energy loss dE/dx is approximately proportional to 1/E. This means they deposit most of their energy near the end of their track. It is not an exponential attenuation like scattering where a single scattering event causes a photon to lose all of its energy and to be lost from the beam. There is continuous energy loss along the track due to the interaction of the photons with the electrons in the matter, and this energy loss increases as the photon loses energy, so most of the energy gets deposited at the end of the track.
sophiecentaur
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Feb20-14, 10:03 AM
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Great answer. Makes perfect sense. Thanks a lot.

256bits
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Feb20-14, 10:06 AM
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Penetration of X Rays in body tissue


See
http://en.wikipedia.org/wiki/Bragg_peak
256bits
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#5
Feb20-14, 10:22 AM
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Quote Quote by phyzguy View Post
For the photon energies used, the rate of energy loss dE/dx is approximately proportional to 1/E. This means they deposit most of their energy near the end of their track. It is not an exponential attenuation like scattering where a single scattering event causes a photon to lose all of its energy and to be lost from the beam. There is continuous energy loss along the track due to the interaction of the photons with the electrons in the matter, and this energy loss increases as the photon loses energy, so most of the energy gets deposited at the end of the track.
Does that work for photons?
They ionize along their path and the buildup of ionization peaks at a certain depth and then falls exponentially. For that reason, several beams of ionization beams are to used to obtain a maximum of ionization at a certain depth, and lessen tissue damage ouside the peak.
sophiecentaur
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Feb20-14, 05:27 PM
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Ah yes, the Bragg Peak!! It's only taken 110 years since it's invention for me to have taken it on board! We are not dealing with a linear system here and, as a Radio Engineer, I didn't take that into account.


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