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Compton Effect, frequency question

  1. Dec 31, 2012 #1
    1. The problem statement, all variables and given/known data
    An X-ray with a frequency of 3.74 x 10^20 Hz is incident on a thin piece of metal. The lower frequency X ray on the other side is observed deflected at 48°. What is the frequency fo the deflected X ray?

    fi = 3.74 x 10^20 Hz
    Deflected at 48°


    2. Relevant equations
    I'm not sure... I would guess..

    Δλ= (h/mc)(1-cosθ)
    E = hf
    E = hc/λ
    (hc/λ)=hf

    3. The attempt at a solution
    I really have no idea where to begin.... If someone could just start me off that would be awesome :) This is my attempt anyway..

    Ei = hf
    = (6.63 x 10^-34 Js)(3.74 x 10^20 Hz)
    = 2.47962 x 10^-13 J

    E = hc/λ
    λ= hc/E
    = (6.63 x 10^-34 js)(3.00 x 10^8 m/s) / (2.47962 x 10^-13 J)
    = 8.02139037 x 10^-13 m

    Δλ = λf - λi
    λf = Δλ + λi
    = ((6.63 x 10^-34 Js) / (9.11 x 10^-31 kg)(3.00 x 10^8 m/s))(1- cos48) + 8.02139037 x 10^-13 m
    =4.78097437 x 10^-12 m

    Ef = hc/λ
    =(6.63 x 10^-34 Js)(3.00 x 10^8 m/s) / (4.78097437 x 10^-12 m)
    =4.16023983 x 10^-14 J

    E = hf
    f = E/h
    = (4.16023983 x 10^-14 J) / (6.63 x 10^-34 Js)
    = 6.274871536 x 10^-19 Hz

    Yeah.... I don't think this is right..

    Please help!!
     
  2. jcsd
  3. Dec 31, 2012 #2

    TSny

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    Homework Helper
    Gold Member

    x-rays are electromagnetic radiation. So, the easiest way to get back and forth between λ and f is to use the basic wave relation c = λf. Thus, λ = c/f. You can check that this will give the same result for λi.
    Looks set up correctly, but I don't get the same value for λf.
    Again, you can just use f = c/λ.
     
  4. Jan 1, 2013 #3
    Wait... was your λf value 1.06771397 x 10^-12 ??

    My calculator was in radian mode...
     
  5. Jan 1, 2013 #4

    TSny

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    Homework Helper
    Gold Member

    I get about 1.605 x 10^-12 m
     
  6. Jan 1, 2013 #5
    Hmm.... I'm not sure what I'm doing wrong then...
     
  7. Jan 1, 2013 #6
    Nevermind!! I got it now! :D Thanks :)
     
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