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Centripetal force for an electron

  1. Jun 30, 2008 #1
    In a few chemistry books/resources I've seen the Bohr model of the atom is developed equating the electrostatic force to a centripetal force. Yet these sources also depict the centripetal force as being directed outward, away from the nucleus. (a single proton for the H atom) One even called it the centrifugal force. Another looked at centripetal force as a "separate" force, that is:

    (Ze^2)/(4*pi*e0*r^2) - (mv^2)/r = 0 e0 = epsilon naut

    The equation is correct of course, but the author stated this as an "equilibrium" condition, almost as if the net force should be 0 for a "stable orbit." Some of these derivations seem quite wrong. Isn't centripetal force a "requirement" for circular motion, in this example? If Fnet = F1 + F2 + ... Fn then centripetal force should simply be Fnet, not one of the component forces, like electrostatic attraction/repulsion, right? The centripetal force being directed outward is probably the most discomforting. Am I missing something here, or am I misinterpreting the authors?

    Gray's "Chemical Bonds" gives the "stable orbit" condition and directed the centripetal force outward.

    http://chem1.che.caltech.edu/chem1a/LectureNotes/Series01AtomsAndBohrModel.pdf [Broken] also draws the force outward and calls it centrifugal, if this helps any.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 30, 2008 #2


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    Welcome to PF tachyontensor,

    Let us first clear up the matter of centripetal vs. centrifugal force. The presence of the centrifugal force depends on which frame of reference we are using. When viewed from an inertial reference frame (i.e. that of the nucleus), a body in circular motion only requires a centripetal force (i.e. a force directed towards the centre of the circle). However, to apply the inertial laws of motion from the electron's point of view (i.e. a non-inertial reference frame) we must include an 'imaginary' pseudo-force, which is the centrifugal force.

    If you have a look through this forum, you'll find plenty of threads discussing the differences between the centripetal and centrifugal forces. You can also take a look at the related articles in the PF library.

    With respect to the equilibrium comment, I agree with you, the centripetal force is not a component but rather a requirement for circular motion. However, I think perhaps that you are misreading what the author means by equilibrium condition and I certainly don't think that he/she would have [intentionally] presented the centripetal force as a component. Do you agree that for an object of mass [itex]m[/itex], moving at velocity [itex]v[/itex] in a circle of radius [itex]r[/itex] the net force [itex]F_\text{net}[/itex] must obey the following equality?

    [tex]F_\text{net} = \frac{mv^2}{r}[/tex]

    Assuming that you agree, then this is indeed an equilibrium condition. If the net force was greater or less than this then a orbit would be unstable. If you consider the above a valid equilibrium condition then it logically follows that one could also consider the following an equilibrium condition:

    [tex]F_\text{net} - \frac{mv^2}{r} = 0[/tex]

    I hope this helps.
    Last edited by a moderator: May 3, 2017
  4. Jun 30, 2008 #3
    Thanks for your help, but I still don't understand why the centripetal force would be drawn away from the nucleus. From what I understood, both authors were always referring to inertial reference frames, which is why I found the presence of the word 'centrifugal' to be odd for what looked like a centripetal force. My memory failed me for the equilibrium condition. Let me quote Gray from Chemical Bonds:

    "For a stable orbit to exist the outward force exerted by the moving electron trying to escape its circular orbit must be opposed exactly by the forces of attraction between the electron and the nucleus. The outward force, F0, is expressed as

    F0 = [m(sub e)*v^2]/r

    This force is opposed exactly by the sum of the two attractive forces that keep the electron in orbit - the electrostatic force of attraction between the proton and the electron, plus the gravitational force of attraction. The electrostatic force is much stronger than the gravitational force, thus we may neglect the gravitational force. The electrostatic attractive force, F(sub e), between an electron of charge -e and a proton of charge +e, is

    F(sub e) = -(e^2)/(r^2)

    The condition for a stable orbit is that F0 + F(sub e) equals zero:

    [m(sub e)*v^2]/r - (e^2)/(r^2) = 0 or [m(sub e)*v^2]/r = (e^2)/(r^2) "

    Since this is written out in scalar form, the electrostatic and centripetal forces tend to disagree in sign, and apparently direction. Since the electrostatic attraction is directed towards the proton, like a central force, and the centripetal force is (as far as I've seen) is also towards the center, shouldn't the centripetal force change sign if written out in vectors? That is, if the problem employed polar coordinates, both forces would have the same sign and one component each under the vector r? (no angular components?)
    Last edited: Jun 30, 2008
  5. Jun 30, 2008 #4


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    From what you have quoted below, the author makes no comment regarding inertial reference frames, nor does he mention any centripetal force. Therefore it is safe to assume that any force directed radially outwards is in fact the centrifugal pseudo-force. The force is drawn away from the nucleus because it is the centrifugal pseudo-force and not the centripetal force.
    The outward force makes it quite clear that we are dealing with a centrifugal pseudo-force rather than the centripetal force. The comment regarding the electron trying to escape it's orbit further confirms that we are dealing with the centrifugal pseudo-force and non-inertial reference frames.
    since we are dealing with a centrifugal pseudo-force, which always acts away from the axis of rotation, it makes sense that the two forces differ in direction: the coulomb force towards the nucleus and the centrifugal pseudo-force radially outwards.
  6. Jul 1, 2008 #5
    I'm such an idiot. That's so obvious too. Thanks for all your help, I really should have noticed that earlier.
  7. Jul 1, 2008 #6


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    I wouldn't say that you were an idiot at all. The quoted text would do well in my opinion to add a little more detail to the derivation and explain that non-inertial reference frames are in use.
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