I have been wondering why only electrons revolve around protons instead of protons other way around. They have electrostatic force and I think mass factor has nothing to do here. Then why?
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32$\begingroup$ Electrons do not "revolve around" the nucleus. They have a probability to be found near the nucleus, and they have the property of angular momentum but you really should not imagine it as proper movement... $\endgroup$– StianCommented Sep 20, 2021 at 13:32
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10$\begingroup$ A good in-depth answer should probably mention positronium and muonium as interesting examples of "atoms" with no protons. $\endgroup$– Ilmari KaronenCommented Sep 20, 2021 at 14:58
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26$\begingroup$ Think again. Mass has everything to do here. $\endgroup$– my2ctsCommented Sep 20, 2021 at 16:02
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5$\begingroup$ Why is the sun at the center of the solar system, not the Earth? $\endgroup$– Charles HudginsCommented Sep 21, 2021 at 10:25
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1$\begingroup$ @Xfce4 of course the sun isn't really the center, but it is objectively much closer to the center of mass of the system and very nearly at the focus of the ellipse the earth travels in. As with the proton and the electron, this is because the sun is much more massive than the earth. $\endgroup$– Charles HudginsCommented Sep 24, 2021 at 20:15
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5 Answers
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NB: I interpreted the question to essentially mean, why do protons rather than electrons reside in nuclei?
Electrons repel each other with a Coulomb force that grows very large when they are close together. Protons also repel each other in the same way, but the difference is that protons are also attracted to each other and to neutrons by the even stronger strong nuclear force (since protons are made up of quarks that feel the strong force), which acts over short range ($\sim 10^{-15}$m) and thus can be bound into dense nuclei (when you mix them with neutrons).
Electrons are point-like particles and not made up of quarks. They do not interact via the strong nuclear force and cannot be bound into dense nuclei.
Note that you can do physics in any coordinate system you choose. If you had an "electron nucleus" surrounded by a cloud of protons, you would probably still choose to define the nucleus as the centre. But electrons don't aggregate into nuclei for the reasons I explained.
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17$\begingroup$ This is a much more useful answer than the other! Might be worth adding that this is (kind of) because electrons are not quarks, nor are they made up of quarks, while protons and neutrons are composed of quarks. $\endgroup$ Commented Sep 20, 2021 at 14:53
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5$\begingroup$ @CarlWitthoft I do not agree, it is a reasoning of why the nuclei have a large mass while electrons cannot attain such densities , but it has nothing to do with explaining why the heavy mass of the nucleus defines a tight orbital for it in the atomic center of mass. The Bohr model gives an envelope of how the full quantum mechanical calculation would give the orbitals. Take the hydrogen atom: in the center of mass system from adding the two momenta to zero, the heavy proton will have a much smaller velocity than the light electron, making a much smaller circle. Similar to $\endgroup$– anna vCommented Sep 20, 2021 at 17:49
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1$\begingroup$ This seems to answer an older version of the question before an edit was made $\endgroup$– Paul T.Commented Sep 20, 2021 at 18:09
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5$\begingroup$ Given the OP's acceptance of the other answer, I would assume they were really asking the edited version even if they did not articulate it at first. $\endgroup$– Paul T.Commented Sep 20, 2021 at 19:27
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5$\begingroup$ The point made by this answer is important in general, but for a neutral ${}^1\text{H}$ atom there is only one proton and one electron, so the strong force does not come into play. (Not at the atomic scale, anyway. One could talk about the mass of the proton being ~90% from the dynamics of its quarks, which is largely governed by the strong force, but that takes us back into the territory of the other answer IMO.) $\endgroup$– zwolCommented Sep 21, 2021 at 14:02
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Considering a classical model of two particles, they both actually revolve around the center of mass of the system. Same thing applies to the motion of the Earth-Sun system. If one of the two objects is a lot more massive than the other then the center of mass is very close to the massive object, even inside the volume occupied by the massive object. Then the motion of the lighter one is almost like revolving around the massive one. But for electrons in atoms, the "revolving around" is not a good description. The electrons are not moving around classical trajectories and some of them may have a non-zero probability to be inside the nucleus. Hovewer the configuration of the atom, even when described by quantum mechanics, depends on the ratio of the masses of the components. So the mass factor is essential.
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2$\begingroup$ Maybe studying the Bohr model would help the OP , as there the classical orbits are used :hyperphysics.phy-astr.gsu.edu/hbase/Bohr.html . Its results are consistent with the calculation of the quantum mechanical theory,. $\endgroup$– anna vCommented Sep 20, 2021 at 9:17
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$\begingroup$ @annav Maybe one should stress that only circular classical orbits are used within the Bohr model! This restriction is lifted (to some extend) in the Bohr-Sommerfeld model, which includes elliptical orbits. And even though I think it is nice to know the history of quantum mechanics to get a bigger picture, I do also think that one should be very carefull when applying classical reasoning to quantum systems! $\endgroup$ Commented Sep 20, 2021 at 16:55
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2$\begingroup$ To put in some numbers, the Sun-Jupiter mass ratio is about 1047.3. The lightest nucleus is a proton, and the proton-electron mass ratio is about 1836.2. $\endgroup$ Commented Sep 20, 2021 at 22:25
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$\begingroup$ Sorry i didn't saw yours. You explanation is also good $\endgroup$– CyberaxCommented Sep 21, 2021 at 2:22
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$\begingroup$ @AlmostClueless It's not so dangerous to make a classical analogy in this case because Schrodinger's equation for the hydrogen atom arises from quantizing the Hamiltonian for two bodies orbiting each other with an inverse square central force. $\endgroup$ Commented Sep 24, 2021 at 20:19
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The title of the question has had too many edits by various people, so I want to clearly answer the content:
why only electrons revolve around protons instead of protons other way around. They have electrostatic force and i think mass factor has nothing to do here.
The mass factor is important, because in systems bound with any type of force , either classical or quantum mechanical (and the atom is a quantum entity) have to obey the law of conservation of momentum . Momentum is the vector $p$ = $mv$, where $m$is the mass and $v$ the vector velocity, so that is how the mass comes in.
For example lets take the atom of hydrogen: The dimensions of the hydrogen atom are one angstrom=100.000fermi , the dimension of the proton is approximately one fermi. The orbital of the electron occupies a region 100.000 times larger than the proton dimension.
To see how momentum conservation affects the bound state of electrons in an atom , for hydrogen: the electron is ~0.5MeV, the proton ~1000Mev. If one measures the electron velocity in the hydrogen atom and thus measures the momentum, the proton momentum has to be equal and opposite . Because of the very large mass difference, the velocity will be very small. This means that a plot of the orbital of the electron in the center of mass system covers a large area, whereas the orbital of the proton will be located within its volume.
So it is because of the large difference between the masses of electrons and nuclei that one assumes that the center of mass is at the nucleus and the electrons have orbitals around it.
(In the comments the analogy of the orbit of the sun around the barycenter of the planetary system has been given, where the barycenter is often within the volume of the sun).
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10$\begingroup$ Did you use . as both decimal and thousand separator? $\endgroup$ Commented Sep 21, 2021 at 8:16
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2$\begingroup$ It's probably less confusing to use
,as a thousands separator for the Ångstrom and electron region. We prefer that convention on this site. $\endgroup$– PM 2RingCommented Sep 21, 2021 at 8:27 -
11$\begingroup$ @PM2Ring: Whichever convention is used, the same separator shouldn't be used for decimal and thousand.
,could also be confusing to some European readers. I'd prefer to use a en.wikipedia.org/wiki/Thin_space or not use anything at all. It might take a bit more time to read 100000, but at least it cannot be confused with 100. $\endgroup$ Commented Sep 21, 2021 at 8:32 -
2$\begingroup$ The key factor is clearly the strong interaction needed to counteract and overcome electrostatic repulsion (ProfRob's answer). You simply cannot bunch electrons up the way nucleons are in the nucleus. Granted, even if you could (and the nucleons wouldn't lose that property) the mass difference would probably create a two-entity system where a heavy nucleus is still in the center, "orbited" by a single electron "nucleus". $\endgroup$ Commented Sep 21, 2021 at 10:29
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2$\begingroup$ @Joshua "100.000 times larger" and "0.5MeV". I've never heard of Fermi units before, so I had to check if the factor is 100 or 100000. $\endgroup$ Commented Sep 21, 2021 at 21:11
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With a hydrogen-1 atom, the proton will have an orbital centered on the center of mass of the atom, but as a proton has a mass about 2000 times the mass of an electron, its motion will be much less. For larger atoms, the proton and neutrons in the nucleus are constantly exchanging virtual particles. While the nucleus is often modeled as having "protons" and "neutrons", to a great extent the nucleus is just a soup of quarks that can separate into proton-globs and neutron-globs of quarks given the right conditions. We can't separate out a "proton" wavefunction from this soup in the same way that we can separate out an electron wavefunction. The nucleus has one overall wavefunction that encompasses all the quarks that make it up, and the "protons" and "neutrons" are more fluctuations in this wavefunction than separate wavefunctions.
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Essentially the core point is $m_{proton} \approx 1836\, m_{electron}$. If you take the quantum two-body problem i.e. the Schrodinger equation for two particles with a coulombian interaction and you do the approximation $m_{proton} >> m_{electron}$ you will get the discretized energy levels which belong to the hydrogen atom.
See here for a quick review.
In addition the quantum two body problem formulation solves the problem of the stability of a hydrogen atom because, in classical mechanics, you have energy losses due to an accelerated charged particle.
