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In terms of Classical Physics, what force would be required to hold an electron in orbit around a proton, i.e., an hydrogen atom? I have done a calculation, but need verification.

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closed as off-topic by Yashas, Hritik Narayan, John Rennie, Kyle Kanos, ZeroTheHero Jun 25 '17 at 13:40

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    $\begingroup$ What's your calculation? $\endgroup$ – user154420 Jun 24 '17 at 18:53
  • $\begingroup$ Embarrassing, but I simply took the Bohr Radius, and plugged it into the Coulomb force equation! I know. $\endgroup$ – Nedward Jun 24 '17 at 18:58
  • $\begingroup$ Countt010, I am trying to get a thumbnail calculation for A Level students, such that they can appreciate the idea of producing protons for use in a collider. I know I have oversimplified, but... $\endgroup$ – Nedward Jun 24 '17 at 19:17
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I've made a numerical answer to complement the existing answer, with some sanity checks along the way. I trust that you know that this is entirely a toy universe, and none of this stuff has very much meaning, as said in the other answer.

Speaking entirely classically, an electron orbiting a proton will feel a Coulomb force of: $$F=-\frac{e^2}{4\pi \varepsilon_0 r^2}.$$ If the electron is staying at constant $r=r_B$, then this force must be balanced by another force. The magnitude of this force is found by plugging in the right constants, as: $$F=8.24\times 10^{-8}~N.$$ Sounds small right? Consider the acceleration this force would correspond to, through Newton's second law as: $$a= \frac{F}{m} = 9.05 \times 10^{22}~m s^{-2}.$$ Now that is huge! But that's of course understandable. Consider this force coming from centripetal acceleration by being in circular motion, then the velocity the electron would be moving at would be: $$v=\sqrt{\frac{F r}{m}} = 2.18 \times 10^6~ ms^{-1}.$$ That does of course mean the electron orbits ~$10^{15}$ a second, around a tiny tiny amount of space. To get an electron travelling at $2.18 \times 10^6 ~ms^{-1}$ to completely change its velocity vector $10^{15}$ a second is going to require a mighty acceleration.

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  • $\begingroup$ Yes, I do know this is not valid. However, as an exercise in thought, leading to an analysis of why it is not valid, it is a useful entree for my students. As you point out, and I also calculated this, the acceleration is, er..., quite large. But thanks for your time. $\endgroup$ – Nedward Jun 25 '17 at 7:33
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The classical model (planetary model) states that, in a similiar manner to planetary orbits, where the gravitational force is balanced by the centrifugal force, the electron is held in balance of electric force and centrifugal force. This theory fails, as we know, because accelerating charges radiate, thus losing energy and inevitably falling into the nucleus.

This problem is solved in the quantum viewpoint- the energy of an electron around the atom is quantized. First came the bohr model which was simply a quantized version of the planetary model (only certin radii are allowed). Then came the de-Broigle model, where the electron takes it's wave-like form and can only be found in radii where it can form a standing wave.

The modern conception is that of orbitals- the electron is merely a probability density around the atom, which behaves according to the different spherical harmonics.

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  • $\begingroup$ Electron isn't probability density. Density, wawefunctions etc. are only ways to describe it's behavior. $\endgroup$ – Mithoron Jun 24 '17 at 22:51
  • $\begingroup$ @Mithoron I think this is more a discussion of semantics- the probability density is as much "the electron" as the particle is- until it isn't measured, it is superposition-ly "smeared" on the entire distribution $\endgroup$ – A. Ok Jun 24 '17 at 22:56

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