Kinetic energy of electrons/protons

  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. 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. 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. dextercioby

    dextercioby 12,328
    Science Advisor
    Homework Helper

    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).

  6. 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. Andrew Mason

    Andrew Mason 6,965
    Science Advisor
    Homework Helper

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

  8. dextercioby

    dextercioby 12,328
    Science Advisor
    Homework Helper

    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...

  9. xanthym

    xanthym 412
    Science Advisor


    [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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