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Relativistic Kinetic energy, momentum, speed.

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data

    An Electron (rest mass=9.11*10^-31kg) is accelerated to an energy (mass energy+kinetic energy) of 30*10^6 eV (30 MeV). What is its kinetic energy? Its momentum? Its speed?
    (Note: 1 eV = 1.602*10^-19 Joules; c=2.998 * 10^8 ms^-1)
    2. Relevant equations


    3. The attempt at a solution
    Okay, so this problem was on a test that I missed and I have been trying to understand it/figure it out since then. I solved for Ek (kinetic energy) I believe. (Em= Mass energy; Ek = Kinetic Energy)

    Em + Ek = (3*10^6 eV)* ((1.602*10^-19 J)/(1eV))
    =4.806 * 10^-13 J here I converted to Joules.

    Then I solved for Em
    Em = mc^2
    =(9.11*10^-31 kg)(2.998 * 10^8 m*s^-1)^2
    =8.188*10^-14 J

    So, plugging into Em + Ek = 4.806*10^-13 J and rearranging to solve for Ek i get
    Ek= 4.806*10^-13 J - Em
    Ek= 4.806*10^-13 J - 8.188*10^-14 J
    Ek= 3.9872 * 10^-13 J
    I'm not too sure if I found Ek the accurate way.
    Now, if I did, I'm a little lost in how to find my momentum and speed.
    I know momentum is p = ymv = mv/(sqrt(1-(v/v)^2)) with y being gamma
    and I believe after using that to find my momentum I can find velocity by rearranging the Ek equation of
    Ek = (1/2)mv^2 to v= sqrt(2Ek/m)

    Im just not sure if i'm heading in the right direction. A little guidance would be truly appreciated. Also, I wasn't sure where to post this question.
     
  2. jcsd
  3. Feb 10, 2015 #2

    Simon Bridge

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    Not a lot of point converting to Joules: you are just making work for yourself and adding a way to make mistakes.
    What is wrong with all energies in keV and speeds as a fraction of the speed of light.

    Thus total energy E is given by: ##E = \gamma E_0## ... E0 is the rest mass energy of an electron - which is 0.511 keV or 0.000511MeV and gamma is given by ##\gamma=(1-v^2)^{-1/2}##, with v as the fraction of the speed of light.

    Kinetic energy T is then: ##T=E-E_0## and the energy-momentum relation is: ##E^2-E_0^2+p^2##

    But otherwise you did OK: KE is the difference between total and rest energy.
    Momentum is found from the energy-momentum relation.
    You can use the equation above for unified units, for SI units it becomes:
    $$E^2=E_0^2+p^2c^2$$
     
  4. Feb 11, 2015 #3
    Agh, yes I see. Thank you so much!!!!
     
  5. Feb 11, 2015 #4

    Simon Bridge

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    NO worries, welcome to PF.
     
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