Question

Compton ScatteringExercise 10:The equation for Compton scattering of a photon off of an electron is.?′=?+(ℎ??)(1―cos?)If using nm for wavelength, the quantity (h/mc) is 0.00242631nm (to a ridiculous number of sig figs, but you can round off later.) Notice how λ’ is always bigger than λ, because the scattered photon always has less energy – unless the angle is zero, which means nothing happened.A) Suppose a photon of energy 248eV scatters off of an electron in such a way as to maximize the change in wavelength (180. degrees). Show that the wavelength has negligible change, even with the optimal angle. (I get about 0.1% change.) [Such a photon actually has very little chance to do a Compton scatter. It CAN ionize any atom, so it is still a radiation risk for exposed tissue.] FYI, you need the equation from Exercise 1.B) Suppose the photon energy is 24,800eV and it scatters optimally (180. degrees). Find the wavelength and energy of the scattered (x-ray) and therefore how much energy was lost to the electron. I get about 2000eV.C) Repeat B) for a photon of energy 124,000eV

D) Repeat B) for a photon of energy 511,000eV (and note the value of λ.) This one has a LOT of energy lost from scatter of 180. degrees - about 2/3 of the energy.E) The 511,000eV photons of part D are (gamma rays) produced when positrons and electrons annihilate. This energy is the rest energy of an electron (and why λ is so familiar in the Compton equation). PET images are Positron-Emission-Tomography pictures, based upon the principle that where concentrations of radioactive Fluorine-18 accumulate in the body, positrons will be emitted and collide (essentially in the same region) with a valence electron, producing a pair of these gamma rays in opposite directions (momentum conservation.) The detector system traces pairs of gammas backward and, system wide, creates a map of the radioisotope concentration.The whole thing is thrown off when one of these gamma rays does a Compton scatter because it goes off in some random direction. But if the angle is big enough, the gamma ray loses enough energy that the detector can see that it scattered, and therefore DISCOUNT that particular ray. OK, then. Here’s what you do. Suppose the detector accepts photons of energy 511 ± 20 keV. Note the units of keV. Show that a scatter angle of 15.0 degrees will result in a scattered photon that IS accepted, but that a 25.0 degree angle will result in a photon that is disregarded. These numbers are typical for PET imaging systems. Scatter rejection is merely kinda good.

Answer 1