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Finding the Kinetic Energy of Compton Scattering

  • #1
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Homework Statement


X-ray photons of wavelength 0.02480 nm are incident on a target and the compton-scattered photons are observed at 90 degrees.(a) What is the wavelength of the scattered photons?(b) What is the momentum of the incident photon and scattered photon? (c) What is the kinetic energy of the scattered electrons?(d) what is the momentum (direction and magnitude) of the scattered electrons?


Homework Equations


1/E'-1/E = 1/mc^2(1-cos(theta))
E=pc


The Attempt at a Solution


(lamda)'/hc-(lamda)/hc = 1/mc^2(1-cos(90)) (where mc^2 equals .511)
the (lamda)'=hc/.511(1-cos(90))+.02480 = .2426e-11 nm
then for b. E=pc so p=E/c=h/(.02480)=2.67e-23 and p'=E'/c=h/(lamda)'=6.626e-34/2.46200e-21=2.69e-13.
c then is E=pc=8.01e-15 = Rest Energy and E'=p'c=(2.69e-13)(3e8)=8.07e-5
then K=E-rest energy = 8.07e-5-8.01e-15=8.07e-5. Finally d is P=sqrt(2.69e-23^2+2.67e-23)=7.3212e-12

Is any of this correct if not where did i go wrong I think my methods are correct I am not sure about my units?
 

Answers and Replies

  • #2
672
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ok I screwed up the first part should be .02722nm. P'=E/c = h/(lamda)' = (4.136e-15)/(2.722e-11)=.000151947 and P=E/c=h/(lamda)=(4.136e-15)/(2.480e-11)=.000166774. For part (c) E'=p'c=(.000151947)(3e8)=45584.1 and E=pc=(.00016674)(3e8)=50032.2. K = E-rest energy = 50032.2-45584.1=4448.1. (d) p=sqrt(.000166774^2+.000151947^2)=.000225
 
  • #3
672
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Ok so once again I screwed up the first part should be .02722nm or 2.722e-11. then P'=E/c=h/(lamda)'=(6.626e-34)/(2.722e-11)=2.43423953e-23 and therefore p = E/c=h/(lamda)=(6.626e-34)/(2.480e-11)=2.671774194e-23. Which I then use to find the energies E'=p'c=7.302718589e-15 and E=pc=8.015322581e-15. Thus K=E'-(rest energy)= 7.302718589e-15-8.015322581e-15=-7.12603992e-16 which is IMPOSSIBLE coreect ugggg what did I do. p=sqrt(2.671774194e-23^2+2.43423953e-23^2)=3.614e-23
 

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