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Rotation period of electron orbiting a proton

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    An electron is rotating around a proton (at rest) in a perfect circular orbit. If the radius of the orbit is r=10^-10 m, how long is the rotation period [hint: the radial acceleration is entirely due to the electric force]

    k=9*10^9
    q=1.6*10^-19

    2. Relevant equations
    1. F=(k*q^2)/r^2
    2. arad=(angular velocity)^2*r


    3. The attempt at a solution
    I found the force by equaiton 1., and I got 2.3*10^-8 N ((9*10^9)(1.6*10^-19)^2))/((10^-10)^2)
    I tried equation 2. to get (angular velocity)=sqrt(F/r) and got 15.2 rad/s
    This means that it is 2.4 rev/s (by dividing by 2pi) and 0.41 seconds per orbital period.
    That is like the world's slowest electron. Where did I go wrong?
     
  2. jcsd
  3. Feb 17, 2015 #2

    lightgrav

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    Is Force equal to acceleration? ... or is there something missing?
    always use units, to avoid slip-ups like this.
     
  4. Feb 17, 2015 #3
    Maybe not. It just says that the radial acceleration is entirely due to the electric force. To find the radial acceleration do I just need to say a=F/m?
     
  5. Feb 17, 2015 #4

    lightgrav

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    well, the radial acceleration is caused entirely by the electric force, but it is mitigated via (spread thru-out) the electron mass
    = .911×10-30 kg .
     
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