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Electron-positron creation

  1. Dec 16, 2008 #1
    an electron-positron pair, each with a kinetic energy of 220 Kev, is produced by a photon. find the
    energy and wavelength of the photon.

    started with E=hf=pc , and the fact that p=mv, but his not working
     
  2. jcsd
  3. Dec 16, 2008 #2
    never mind I solved this using
    total E = rest E + KE
    E = Mc2 + 220Kev double this since we have e- and e+ and you have the protons E
    then use E=hf to find wavelength.
     
  4. Dec 19, 2008 #3

    turin

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    Homework Helper

    Be careful. How do you find wavelength from this? It may seem like a silly question, but actually, you can't even solve this problem without more information or assumptions (i.e. the relative directions of the electron and positron) that have no reason to be true in general. BTW, did you mention a proton, or was that a typo?
     
  5. Dec 20, 2008 #4
    hi there.
    ok let me step you through my thinking
    1.the energy of the photon will be the energy of the positron plus the energy of the electron

    energy of e- = rest energy + kinetic energy
    = Mc2 + 3.52x10-14 (220 Kev)
    = 8.1791x10-14J + 3.52x10-14J
     
  6. Dec 20, 2008 #5
    hi there.
    ok let me step you through my thinking
    1.the energy of the photon will be the energy of the positron plus the energy of the electron

    energy of e- = rest energy + kinetic energy
    = Mc2 + 3.52x10-14 (220 Kev)
    = 8.1791x10-14J + 3.52x10-14J
    X2 since we have e- and p+
    = 2.33982x10-13J = 1.46239 Ev
    is the total energy that the proton must of had in order to create this pair


    the E = hc x wavelength-1 (dont know how to do fractions here)
    substitute values and
    wavelength is is 8.494x10-13 M

    what do you think?
     
  7. Dec 21, 2008 #6

    turin

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    Yes. Momentum must also be conserved.



    What do you think c is in this case. The actual equations that you used here were
    [tex]
    E=h\nu
    [/tex]
    and
    [tex]
    u=\nu\lambda
    [/tex]
    where you have assumed that u=c. However, the correct equation to use for the wavelength is
    [tex]
    p=h\lambda
    [/tex]
    Unfortunately, your problem statement does not give enough information to determine the momentum of the photon, so the best you can do is to come up with an upper limit on the momentum (and thus a lower limit on the wavelength).

    For example, if the electron and positron are produced so that they travel in opposite directions, what must be the momentum of the photon? And, therefore, what must be its wavelength?
     
  8. Dec 21, 2008 #7

    Avodyne

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    The problem here is that this process is impossible; a real photon cannot convert itself to an electron-positron pair. If the problem refers to a virtual photon, then turin is correct that there is not enough information.
     
  9. Dec 21, 2008 #8

    turin

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    This process is not impossible. The problem is that the OP is using the wrong physical principle. The OP did not specify that this photon was "real", and, as you point out, it cannot even be real anyway. But at any rate I personally hate that terminology. Every photon is just as real as every other photon, regardless of how far from its mass-shell it is.
     
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