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Electron potential/momentum problem

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data
    To produce x-ray radiation, electrons are accelerated in vacuum and aimed so as to collide with a target. Assume that all of the electron kinetic energy is converted into the energy of the x-ray photon on collision and that the wavelength of the resulting radiation is 1.24 angstroms.



    2. Relevant equations
    h^2/(2*(e)(m)(λ)^2)

    3. The attempt at a solution
    (6.63*10^-34)^2/(2(1.6*10^-19)(9.11*10^-31)(1.24*10^-9)^2)

    =.979 V
    /1000 for kV = .000979 kV. I entered this answer and it told me it was incorrect.
     
  2. jcsd
  3. Sep 29, 2015 #2

    andrevdh

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    That equation is unknown to me. Try starting out with the basic equation for the energy of the x-ray photon.
     
  4. Sep 29, 2015 #3

    ehild

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    What is the question?
    You did not write any equation. What does the formula in 2 mean?
     
  5. Sep 29, 2015 #4
    The question is "What potential difference must the electron be accelerated? (answer in kV) "
    Enter the electron momentum in kg*m/s

    My equation I found using KE=1/2mv^2, λ = h / p, and I guess I messed something up

    so I have the λ=1.26, the mass of an electron, the fact it is a vaccum, the charge of an electron, planks constant. What kind of formula should I be using?

    h/ λ =p
    (6.63*10^-34)/(1.24*10^-10) = 5.34*10^-24 (momentum?)
     
  6. Sep 29, 2015 #5

    ehild

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    You need a formula that contains the accelerating voltage.
    The problem asks the momentum of the electron. h/ λ =p is right if you use the wavelength of the electron, but you substituted the wavelength of the photon, so you got the momentum of the photon.
     
  7. Sep 30, 2015 #6

    andrevdh

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    The kinetic energy of the electrons are used to produce the x-ray photons. So I suggest starting out with the energy of the x-ray photons. Do you know what the equation is for the energy of a photon (according to Planck and as used by Einstein to explain the photoelectric effect)?
     
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