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Uncalculable v

  1. Oct 8, 2005 #1
    for the following question:
    find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV.

    my problem:
    but when i try to calculate the v in the gamma factor, it's impossible to calculate on the calculator because the number is too small...
    any suggestions?
  2. jcsd
  3. Oct 8, 2005 #2


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    I'm not sure I see where your difficulty is. It appears that [itex]\gamma = 2[/itex] so [itex]\beta = \frac {\sqrt 3}{2}[/itex]. Did I miss something?
  4. Oct 8, 2005 #3
    i think i'm the one who's missing something... why's gamma=2? i thought that you were supposed to try the "v" in gamma (gamma= [1-v^2/(c^2)]^(-1)) so isn't gamma unknown? may i also ask what does beta represent?
  5. Oct 8, 2005 #4

    Physics Monkey

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    The energy of the electron is [tex] E = \gamma m_0 c^2 = K + m_0 c^2 [/tex] where [tex] K [/tex] is the kinetic energy of the electron. You are told that [tex] K = m_0 c^2 [/tex], right? Solve for [tex] \gamma [/tex].
  6. Oct 8, 2005 #5


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    You have [itex]\gamma - 1 = 1[/itex] so that [itex]\gamma = 2[/itex]. Also, [itex]\beta[/itex] is just v/c.
  7. Oct 9, 2005 #6
    wow... i hadn't thought of that way to solve the problem...
    so because my way is almost impossible to solve, so you guys changed your method to using the formula
    [tex] E = \gamma m_0 c^2 = K + m_0 c^2 [/tex]?
  8. Oct 9, 2005 #7


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    Doing it your way, use m=0.911e-30 kg, c=3e8 m/s
    to get gamma = 2.0 ... what did *you* use for m?

    beta is v/c (comes before gamma, inside the gamma formula)
  9. Oct 10, 2005 #8
    i used m=9.1*10^(-31) kg
  10. Oct 11, 2005 #9
    is that method right?
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