How do I find the endpoint energy for photons when an .75 MeV electron beams stops inside a tungsten target? I just don't understand how to solve it without the scattering angle.
No scattering angles means the angle is zero? The scattering angle varies from 0 to 180° (0 to [itex]\pi[/itex]). http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compeq.html#c1 Here is a characteristic X-ray for tungsten at 100 kV. http://www.bh.rmit.edu.au/mrs/subject/mr100/prodxray.html#Fig4 Is the energy supposed to be 0.75 MeV!? or 0.075 MeV (75 keV)? Perhaps I am not understanding the question.
Here is the exact question, from a notoriously vague teacher: A .75 MeV electron beam strikes and stops inside a tungsten target. a. Determine the endpoint energy for the photons. b. Determine the momentum of the hiest energy photons. c. Determine the wavelength of the end-point photons. Wouldn't an angle of 0 make the equation (lamda2-lamda1=(h/mc)(1-cos(0))) go to 0?
That's not the Compton effect. It is bremsstrahlung (stopping radiation). The max photon energy would be .75 MeV.
Yeah, I realized he was just trying to get that point across, that the max energy would be .75 MeV. I'm just used to more complicated questions with this guy. Thanks for all the prompt help!