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In a classical model of the hydrogen atom, the electron orbits the proton in a circular...

In a classical model of the hydrogen atom, the electron orbits the proton in a circular orbit of radius 0.053 nm. What is the orbital frequency? The proton is so much more massive than the electron that you can assume the proton is at rest. Answer is in hertz.
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Answer #1
   Electrostatic attraction between electronand proton provides the centripetal force, hence
   (1/4π0) * e * e /r2   =   m * v2/ r
   =>   9.0 * 109* (1.6 * 10-19)2 / 0.053 *10-9   =   9.1 *10-31 * v2
   v2   =   4.35* 10-18 / 9.1 *10-31   =   4.78 *1012   
   velocity ofelectron   v   =   √( 4.78 * 1012)   =   2.19* 106   m/s
   Frequency   f   =   v/ 2πr   =   2.19 * 106/ 2 * 3.14 * 0.053 * 10-9
   f   =   6.58* 1015   Hz
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