Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Uniform circular motion question

  1. Dec 14, 2003 #1
    I'm trying to solve this problem correctly, but my calculations yield a different result than the correct answer according to my professor.

    In a semi-classical model of the neutral hydrogen atom, an electron of charge -e and of mass 9.1*10^-31 undergoes uniform circular motion around the much more massive proton with charge +e. The radius of the electron's orbit is 5.3*10^-11 m. The speed of the electron in its orbit is ____ m/s.

    Code (Text):
    (Force Elec) = (m * a)

    => e^2/(4*pi*E0*radius^2) = mass*velocity^2/radius

    =>velocity = (e^2/(4*pi*E0*radius*mass))^(1/2)

    where 1/(4*pi*E0) = 8.99*10^9 Nm^2/C^2 and e = 1.602*10^-19
    When I solve the equation I get ((1.6*10^-19)^2/(8.99*10^9 * 5.3*10^-11 * 9.1*10^-31))^(1/2) = 2.43*10^-4 m/s, however they say the correct answer is 2.2*10^6 m/s

    Maybe I'm doing something wrong? I would hope the prof did everything correctly. :smile:
  2. jcsd
  3. Dec 15, 2003 #2
    I get 2.184 x 10[tex]^6[/tex] -same as your professor's, and using the same numbers as you posted. Check your arithmetic.
  4. Dec 15, 2003 #3

    What you should do is equate the electric potential energy to the kinetic energy of the electron...Thats all that you need to do and lo behold you have the answer.


    [tex](e^2)/(4\Pi \epsilon_{0}r) = 1/2*(mv^2 )[/tex]

    From the above equation you can find the velocity...

  5. Dec 15, 2003 #4
    Re: ...

    This is the equation I was using. The mistake I made was putting the value for [tex]1/(4\Pi \epsilon_{0}) = 8.99*10^9[/tex] in the denominator of my calculation, instead of the numerator where it should go.
  6. Dec 15, 2003 #5

    As u said you must substitute [tex]1/4\Pi\epsilon_{0} = 8.99 * 10^9 [/tex] in the numerator.


    [tex]8.99*10^9 * (e^{2})/r = 1/2 * mv^2[/tex]

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook