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Biomedical Imaging by Sebastian Aeffner, Timo Aspelmeier, Tim Salditt

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Inserting p = γm0 and E=γm0c2=p2c2+m02c4gives

ΔEmax=2mec2β2γ2[ 1+(mem0)2+2γmem0 ]1.(5.29)

The maximum energy transfer depends on the ratio of the rest masses of the electron and the incoming particle mem0, as well as on the ratio of the initial particle velocity and the speed of light β=vc, which is contained in γ. We distinguish four cases:

nonrelativistic caseextreme relativistic case
β ≪ 1, γ ≈ 1β → 1, γ ≫ 1
incoming proton = m0 mpΔEmax ≈ 2meυ2ΔEmax = E
incoming electron m0 = meΔEmax = EΔEmax = E

In the case of nonrelativistic protons, the relative energy loss in a single interaction with a resting electron is limited by

ΔEmaxE2mev212mpv2=4memp1500.(5.30)

Since collisions are usually ...

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