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So basically I have this idea, we can't see how does an electron really move because of its small size and the wavelength of light being comparatively being much larger. What if we took two spheres (which are macroscopic that is can be seen with the naked eye), one more massive than the other and having a mass ratio of 15368.5978977 (exact ratio between the mass of a proton to that of an electron). Now we gave the massive sphere a +e charge(positive electronic charge) and the less massive sphere a -e charge(the electronic charge).

Now we suspend the macroscopic proton model (M.P.M) in a vacuum (suppose outer space) in absence of any significant gravitational field. Then we shoot the macroscopic electron model (M.E.M) towards the M.P.M with such a velocity that when it finally arrives in the M.P.M electric field, its Electrical PE + Motional KE is exactly equal to that of an electron in the hydrogen atom.

Now if the electric, gravitational forces are the same, shouldn't the M.E.M move around the M.P.M exactly like an electron does around the nucleus in the hydrogen atom?

Batwayne
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  • No. A big problem with that would be that you lose many of the desired microscopic effects if the balls are big. My understanding of any of it is basically surface level; but I don't see how you could get a large scale system to behave exactly like a proton and electron; considering they would have to be still made up of protons and electrons. The effects of the particles that make it up would alter the results. – JMac Jan 31 '17 at 18:20
  • @JMac but isnt the motion of the electron around the proton a result of just the coulombs force and gravitational force? – Batwayne Jan 31 '17 at 18:22
  • I'm pretty sure it is far more complicated than that. Electrons don't behave the same as macroscopic matter. It has wave properties. AFAIK the real interpretation of electrons around a nucleus don't even consider it to have one location; but a probability of locations. – JMac Jan 31 '17 at 18:24
  • @JMac (first comment) that should probably be an answer – David Z Jan 31 '17 at 19:07
  • @DavidZ I have a very limited understanding of atomic physics so I don't want to unknowingly mislead someone by answering with half-information. – JMac Jan 31 '17 at 19:12
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    @JMac Still, it is an answer to the question. (At least it looks like one to me.) If you're concerned that the information (or half-information, whatever) you're providing is likely to mislead someone, don't post at all. Otherwise, put it as an answer. But not as a comment. Comments are meant for suggesting improvements to the question, including by requesting clarification and linking to related resources. I will be coming back to delete these comments shortly but I thought I'd let you know so that you could turn yours into an answer if you like. – David Z Jan 31 '17 at 19:16
  • @DavidZ My comment was mostly why I was downvoting it; but I do get your point. That said I don't think I will make this into an answer. Although I somewhat addressed the question I didn't really consider it answer quality as I didn't really talk about any of the physics. – JMac Jan 31 '17 at 19:23
  • @JMac OK, no problem! I appreciate that you tried to explain why you wanted to downvote the question. I'd just ask that if you do this in the future, explain in a way that doesn't answer the question - perhaps focus more on what should be changed about the question to make it less downvote-worthy. Or if your explanation does answer the question and you really like it that way, post it as a proper answer instead. – David Z Jan 31 '17 at 19:28
  • Your question claims "15368.5978977 (exact ratio between the mass of a proton to that of an electron)" but that's 8.37 times the actual mass ratio of 1836.15267343. – Scott Centoni May 09 '23 at 19:30

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We do not have to do your experiment because it can be described by classical electrodynamics, and classical electrodynamics is so very well validated macroscopically that it is not very smart to make models when the mathematics is there.

The solution does not describe what is happening with the hydrogen atom, which is one of the reasons quantum mechanics had to be invented in place of classical mechanics. In classical mechanics and electrodynamics there is no stable system of two charges the way you describe, whereas the hydrogen atom is stable. The two charges in your classical model will attract each other and fall on each other neutralizing the charge while at the same time radiating away energy continuously.

The Bohr model was invented to stop this destructive scenario, since the hydrogen atom was stable. So to start with stability was imposed by postulates so that the series fitting the radiation spectra of the hydrogen atom could be arrived at. Quantum mechanics with its postulates fitted these spectra using the Schrodinger equation.

In quantum mechanics only the probability of the existence of a particle with a given (x,y,z) can be predicted and known. An average momentum from the energy of the hydrogen energy level may be estimated, but the electron is not in an orbit, but in an orbital, a probability locus..

anna v
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An electron doesn't move around the nucleus in a hydrogen atom, so no.

Why don't electrons crash into the nuclei they "orbit"?

Community
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OrangeDog
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You can temporarily excite one electron of an atom to a very high quantum number $n$, which is called a Rydberg atom. When an electron is in this state, it behaves much like a classical particle. A macroscopic model like you propose would behave classically.

Scott Centoni
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