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What is the kinetic energy of the electron?

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data
    In the Bohr Model of a hydrogen atom, a single electron revolves around a single proton in a circle of radius r. Assume that the proton remains at rest.
    (a) what is the kinetic energy of the electron?
    (b) what is the electrical potential energy?
    (c) show that the electron's kinetic energy is equal to half of the electric potential energy.
    (give answers in terms of e, Me, Mp, and r)

    2. Relevant equations
    KE = 1/2mv^2
    F = Ma(centripetal accel.)
    a(centripetal accel.) = v^2/r
    F = (mv)^2/r = (kqq)/r^2

    KE = -1/2U
    U = GMm/r^2 = (kqq)/r
    v(orbit) = [tex]\sqrt{}GM/r[/tex]
    1/2mv^2 = GMm/2r
    F = q|E|
    |E| = F/q = kq/r^2


    3. The attempt at a solution
    a)KE = ke^2/2r

    b) I am having trouble finding a way to say that KE = -1/2U because I keep getting that...
    KE = ke^2/2r
    and that
    U = GMm/2r even though U should be equal to something like...
    U = -GMm/4r

    c)cannot find a way to relate them...
     
  2. jcsd
  3. Sep 24, 2007 #2

    Hootenanny

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    Correct
    Look up the equation for electic potentional energy :wink:
     
  4. Sep 24, 2007 #3
    ah, so for ....
    b) U=kqq/r

    and

    c) KE = ke^2/2r
    (1/2)(kqq/r) = KE = kee/2r

    do I need to relate
    KE = GMm/2r = kqq/2r any further or do the constants pretty much switch out since it is dealing with electric potential and not gravitation?
     
  5. Sep 24, 2007 #4

    Hootenanny

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    The gravitational field is negligable on the atomic scale, you only need to consider the electromagnetic field.
     
  6. Sep 24, 2007 #5
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