link to lecture transcript - UT-H GSBS Medical Physics Class Site

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We now have the differential scatter cross section. The energy interval we’re looking at from 0.15 to 0.25 MeV is 0.1 MeV. To obtain the cross section we integrate the differential cross section over the energy interval. Assuming the differential cross section is constant over that energy interval, a reasonable assumption, we find the cross section is 7 10-30 meters squared per electron. To calculate the number of electrons generated in this energy range we multiply the number of incident photons, 104 , times the cross section, 7 10-30 meters squared per electron times the absorber thickness 3 16 1027 per electron, times the absorber thickness, 3.16 10 electrons per meter squared squared, giving us an answer of 220 electrons. In general when we want to calculate the number of electrons that are liberated, we take the differential energy cross section, multiply it by the energy interval, multiply it by the number of incident photons, multiply that by the electron density. The cross section section, then then, is the probability of an interaction per unit incident photon fluence. fluence Is everyone comfortable with this kind of a calculation? The problem sets will give you more practice in doing calculations of this nature. Our basic goal in doing these problems is to be able to know where the energy is going. Remember, we need to follow the energy. 34
So far, our approximations have been based on the model of a free electron gas. For a free electron gas we have no binding energy. There is no force attracting the electrons to the atom, but that’s not completely correct. So now we are going to make a minor refinement and account for binding energy. To do this, we have to use quantum mechanics. We need the wave functions for atoms. You can only do quantum mechanics relatively accurately on fairly simple atoms. The effects of binding energy are important only for low energy incident photons for which the binding energy is comparable or at least not very much less than the energy of the incident photon. 35
Page 1 and 2: Today, we continue our talk about C
Page 3 and 4: We previously derived an equation f
Page 5 and 6: So σ 0 is the classical cross sect
Page 7 and 8: Let’s look at this graph. Notice
Page 9 and 10: What about the dependence of the Co
Page 11 and 12: The energy transferred to the elect
Page 13 and 14: If we plot the energy transfer coef
Page 15 and 16: The energy transfer coefficient has
Page 17 and 18: What about the scatter coefficient?
Page 19 and 20: Using this equation we can now plot
Page 21 and 22: This probability is represented by
Page 23 and 24: This figure is a plot taken from Jo
Page 25 and 26: We can always calculate what the ma
Page 27 and 28: Remember dσ/dE is relatively const
Page 29 and 30: Now we also look at the graph and s
Page 31 and 32: First, let’s recall that the atte
Page 33: What we now want to find out is how
Page 37 and 38: These values were tabulated by Hubb
Page 39 and 40: So what is the effect of binding en
Page 41 and 42: At higher energies there is no diff
Page 43 and 44: Note that this graph is for very lo
Page 45: The fraction of energy transferred

link to lecture transcript - UT-H GSBS Medical Physics Class Site

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