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

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Here’s another example. Let’s determine the number of Compton interactions and the number of electrons that are set into motion with energies between 0.15 MeV and 0.25 MeV when we have a slab of bone 6 mm thick, and bombard the bone with 10 4 photons of energy 0.5 MeV. We are going to do some accounting now. We are going to pretend we are accountants and we want to follow the energy and figure out where these Compton electrons go and what kind of energy these Compton electrons have. 30
First, let’s recall that the attenuation coefficient is the fraction attenuated per unit absorber b b thickness. hi k How many photons h are attenuated? d? We could ld use the h exponential approximation but when we’re estimating let’s use the thin absorber approximation. The number of photons attenuated is the number of incident photons times the attenuation coefficient times the absorber thickness. How many incident photons do we have? We have been told that the number of incident photons is 104 cdetp otoss 0. . WWhat at is s the t e Compton Co pto attenuation atte uat o coefficient coe c e t for o ½ MeV eV photons. We look that up in Table A2a in Johns & Cunningham and find that to be 0.2892 10-28 m2 per electron. What’s the thickness of the absorber? The thickness of the absorber is the thickness of the bone times the density of bone. We multiply the thickness of bone, 0.6 cm, by the density of bone, 1.650 kg/m3 , which we get from the Table 5.3. Next we multiply by 10-2 multiply by 10 to convert the bone thickness from meters to cm 2 to convert the bone thickness from meters to cm. Let’s convert that into centimeters times the electron density. Remember we have to be able to calculate the electron density to be Z over A times Avogadro’s number. Z over A is about a half. So electron density is always going to be roughly half of Avogadro’s number. That’s a very good approximation for low to moderate Z material, such as bone. Electron density comes out to be 3.19 10 23 electrons per kilogram. So the electron areal density is 3.16 10 27 electrons per meter squared. 31
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: Now we also look at the graph and s
Page 33 and 34: What we now want to find out is how
Page 35 and 36: So far, our approximations have bee
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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