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I Electron in orbit of around a single proton

  1. Jan 13, 2017 #1
    Today I was doing some reading and I came across this topic. If we have a stationary hydrogen atom with a single electron in orbit around the nucleus and want to calculate the kinetic energy of the electron we would take the following approach.

    1) Using Newton's second law:
    F = ma ⇒ FE = mac ⇒ k(q1q2)/r2 = mv2/r​
    2) We know:
    charge of the electron = -e = -1.602×10-19 and the charge of the proton = +e = 1.602×10-19
    3) Now using equation in step 1:
    -k(e2)/r = mv2 ⇒ -k(e2)/(2r)=(1/2)mv2

    We know kinetic energy is suppose to be positive so why is it that it comes out negative in this case?
     
  2. jcsd
  3. Jan 13, 2017 #2

    BvU

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  4. Jan 13, 2017 #3
    Towards the nucleus of the atom.
     
  5. Jan 13, 2017 #4

    BvU

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    So FE provides the centripetal force. It's the only force around. ##mv^2\over r## also points towards the center, so it needs a minus sign too.
     
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