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The Energy of a Multiparticle System

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider an electron (charge -e and mass m) in a circular orbit of radius r around a fixed proton (charge+e). Remembering that the inward Coulomb force ( ke[itex]^{2}[/itex]/r[itex]^{2}[/itex]) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to -[itex]\frac{1}{2}[/itex] times it's PE; that is, T = -[itex]\frac{1}{2}[/itex]U and hence E = [itex]\frac{1}{2}[/itex]U. (This result is a consequence of the so called virial theorem. Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius r around a fixed proton. Electron 2 approaches from afar with kinetic energy T[itex]_{2}[/itex]. When the second electron hits the atom, the first electron is knocked free and the second is captured in a circular orbit of radius r'.

    2. Relevant equations

    coulomb force : ke[itex]^{2}[/itex]/r[itex]^{2}[/itex]

    virial theorem T = nU/2




    3. The attempt at a solution

    I'm not really worried about the second part of this problem quite yet. Right now I'm not really sure how to go about proving that the kinetic energy is -[itex]\frac{1}{2}[/itex] times the potential energy... can someone get me started on the right track?
     
  2. jcsd
  3. Feb 26, 2012 #2
    You could use the virial theorem! :p

    It's been a while since I've done this but I'm pretty sure it's just a case of writing down the kinetic and potential energies and then stating the correlation.
     
  4. Feb 26, 2012 #3
    haha sorry I know it probably seems obvious but we haven't even had the virial theorem mentioned in class. So I'm not really sure how to use it? :/ but I'll try to figure it out
     
  5. Feb 26, 2012 #4
    Don't worry, the virial theorm is part of action based lagrangian mechanics, you probably haven't encountered it.

    You should use the fact that the orbit is circular, that is to say that r doesn't change.
    So try playing about with the equation for the inwards force and the outwards centripital (centrifugal? I can't never remember which is the outwards pushing one)
    [itex]a_{outwards} = \frac{v^2}{r}[/itex]
    [itex]a_{outwards} + a_{inwards} = 0[/itex]

    See if you can work out the kinetic energy from that then see what happens when you compare it to the potential energy
     
  6. Feb 27, 2012 #5
    I still can't get this. I'm sorry I'm just slightly frustrated with this and now it's late haha can I just get an explanation for this :/ I'm struggling in this class right now
     
  7. Feb 28, 2012 #6
    I've been given warnings for being too overzealous with my hinting so I'll try my best to help without giving you the answer


    Mod note: removed overzealous answer
     
    Last edited by a moderator: Feb 28, 2012
  8. Feb 28, 2012 #7
    Thank you so much! This really helps a lot :)
     
  9. Feb 28, 2012 #8
    no problem buddy!
     
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