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Speed of Free Electron Moving in Space

  1. Dec 14, 2007 #1
    The total energy of a free electron moving through empty space is E=1.5mc^2, where m is the mass of the electron and c is the speed of light. What is this electron’s speed?

    1. c (the speed of light)
    2. 0.7453c
    3. 0.8660c
    4. 0.9428c
    5. 0.9950c
    6. 0.9999c


    I thought since that Etotal = 1.5mc^2 then....

    Etotal = KE = .5mv^2

    1.5mc^2 = .5mv^2 and solve for v (masses would cancel which is what I was looking for)

    What is wrong about this approach?
     
  2. jcsd
  3. Dec 14, 2007 #2

    Doc Al

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    Staff: Mentor

    Using KE = .5mv^2 is only good when v << c.

    Hint:
    [tex]E = \gamma m c^2[/tex]
     
  4. Dec 14, 2007 #3
    I believe that the proper relativistic kinetic energy equation would be

    [tex]E_k = (\gamma - 1)mc^2[/tex]

    Or did you make that mistake on purpose, Doc Al?
     
  5. Dec 14, 2007 #4
    oh shoot...i forgot about that. Do I need to use the non-approximated version of KE? (I can never remember it.. but i'm looking it up right now)
     
  6. Dec 14, 2007 #5

    Doc Al

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    E is total energy, of course. :wink:

    (There's no need to find kinetic energy.)
     
  7. Dec 14, 2007 #6
  8. Dec 14, 2007 #7
    so mc^2 would cancel out on both sides and I would be solving 1.5 = gamma right?
     
  9. Dec 14, 2007 #8

    Doc Al

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    Yep. That's all you need to do.
     
  10. Dec 14, 2007 #9
    ahh...got it, thank you :-)
     
  11. Dec 14, 2007 #10
    Oh, I get it... the total energy of the free electron is the sum of its kinetic energy and the energy from the mass-energy equivalence? The potential energy of the electron is zero, since it's free, right?

    [tex]E_{total}= (\gamma - 1)mc^2 + mc^2[/tex]
    [tex]\ \ \ = \gamma mc^2[/tex]

    Is this so?
     
  12. Dec 14, 2007 #11

    Doc Al

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    Staff: Mentor

    Yep.
     
  13. Dec 14, 2007 #12
    Ah, thanks...
     
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