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Kinetic energy of electrons/protons

  1. Feb 15, 2005 #1
    Hi, I'm working on the following:

    An electron starting from rest acquires 6.3 keV of KE in moving from point A to point B. (a) How much KE would a proton acquire, starting from rest and moving to point A? (b) Determine the ratio of their speeds at the end of their respective trajectories.

    So I understand that the proton will also have a KE of 6.3 keV because it has an equal but opposite charge to the electron (part A). But could someone give me a hint about the whole ratio of speeds part? Would I use some equation for velocity, and if so, how would I use the keV in it? Or do I use the respective masses of electrons and protons? Thx.
     
  2. jcsd
  3. Feb 15, 2005 #2
    Use the relativistic equation

    [tex] E = \gamma m c^2[/tex]

    where m is the rest mass to find the velocities of e and p.
     
  4. Feb 15, 2005 #3
    Think about energy conservation.

    There talking about kinetic energy,
    The equation for kinetic energy is

    [tex] KE = \frac{1}{2} m v^2 [/tex]

    using conservation of energy, how would the KE of the electron compare to the KE of the proton?

    edit: This is assuming your looking at it from the "classical" sense. If not, look to the post above
     
  5. Feb 15, 2005 #4

    dextercioby

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    There's no need for relativistic formula,the ratio is approx 1/100 for the electron and 1/180000 for the proton (if the latter has the same 6.3 KeV).

    Daniel.
     
  6. Feb 15, 2005 #5
    I'm sorry Dexter, I don't think I understand what you're saying?? The answer, according to my friend, is something like 1:43. So I am looking for the ratio of their two velocities. This is dependent on mass, correct? Thanks.
     
  7. Feb 15, 2005 #6

    Andrew Mason

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    Are we to assume that it acquires this KE by passing through an electric field? There are other ways it can acquire kinetic energy.

    AM
     
  8. Feb 15, 2005 #7

    dextercioby

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    I was making the ratio:
    [tex] (\frac{KE}{Rest \ mass \times c^{2}})_{electron} \sim \frac{6.3KeV}{511KeV}\sim \frac{1}{85} [/tex]

    Well,i approximated (not too accurately,though) to 1/100 and called the use of relativistic energy formula a mere complication...

    Daniel.
     
  9. Feb 15, 2005 #8

    xanthym

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    SUMMARY OF ESTABLISHED PROBLEM ITEMS:

    [tex] :(1): \ \ \ \ (ElectronKineticEnergy) = (ProtonKineticEnergy) = (6.3 keV) [/tex]

    [tex] :(2): \ \ \ \ \frac { M_{electron} V^{2}_{electron} } {2} = \frac { M_{proton} V^{2}_{proton} } {2} = (6.3 keV) [/tex]

    {Electron Mass} = (9.1093897e-31 kg)
    {Proton Mass} = (1.6726231e-27 kg)


    ~~
     
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