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I'm having trouble understanding the simple "planetary" model of the atom that I'm being taught in my basic chemistry course.

In particular,

  1. I can't see how a negatively charged electron can stay in "orbit" around a positively charged nucleus. Even if the electron actually orbits the nucleus, wouldn't that orbit eventually decay?
  2. I can't reconcile the rapidly moving electrons required by the planetary model with the way atoms are described as forming bonds. If electrons are zooming around in orbits, how do they suddenly "stop" to form bonds.

I understand that certain aspects of quantum mechanics were created to address these problems, and that there are other models of atoms. My question here is whether the planetary model itself addresses these concerns in some way (that I'm missing) and whether I'm right to be uncomfortable with it.

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    $\begingroup$ One more reference - Why doesn't orbital electron fall into the nucleus of Rb85, but falls into the nucleus of Rb83? $\endgroup$
    – voix
    Jan 25, 2012 at 15:40
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    $\begingroup$ to 1:They are on the lowest energy level. They can't decay to lower ones. to 2: they don't stop, the planetary model is just that, a model(and a pretty bad one). $\endgroup$
    – P3trus
    Jan 25, 2012 at 17:13
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    $\begingroup$ similar question on mathoverflow, with some detailed answers: mathoverflow.net/q/119495 $\endgroup$
    – user4552
    May 26, 2013 at 17:34
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    $\begingroup$ The planetary model is pretty bogus, don't trust it too much. $\endgroup$
    – DanielSank
    Jun 20, 2015 at 7:36
  • $\begingroup$ Because of its wave nature, the electron in its ground state is actually smeared symmetrically about the proton (ignoring spin-spin effects) and spherically symmetric charge distributions do not radiate. See also physics.stackexchange.com/q/264123 $\endgroup$
    – jim
    Jun 27, 2016 at 15:13

15 Answers 15

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You are right, the planetary model of the atom does not make sense when one considers the electromagnetic forces involved. The electron in an orbit is accelerating continuously and would thus radiate away its energy and fall into the nucleus.

One of the reasons for "inventing" quantum mechanics was exactly this conundrum.

The Bohr model was proposed to solve this, by stipulating that the orbits were closed and quantized and no energy could be lost while the electron was in orbit, thus creating the stability of the atom necessary to form solids and liquids. It also explained the lines observed in the spectra from excited atoms as transitions between orbits.

If you study further into physics you will learn about quantum mechanics and the axioms and postulates that form the equations whose solutions give exact numbers for what was the first guess at a model of the atom.

Quantum mechanics is accepted as the underlying level of all physical forces at the microscopic level, and sometimes quantum mechanics can be seen macroscopically, as with superconductivity, for example. Macroscopic forces, like those due to classical electric and magnetic fields, are limiting cases of the real forces which reign microscopically.

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    $\begingroup$ anna v: Follow up question (excuse me if it's silly): why would the electron fall into the nucleus? It'd be losing its charge but how would that affect its kinetic energy? $\endgroup$
    – Fingolfin
    Nov 15, 2016 at 12:44
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    $\begingroup$ @xci13 a rotating charge does not lose its charge, more so the electron keeps its charge . A rotating electron accelerates, and classically an accelerating or decelerating charge emits radiation losing its kinetic energy. As it loses energy it spirals in and fall on the nucleus, classically that is. $\endgroup$
    – anna v
    Nov 15, 2016 at 14:02
  • $\begingroup$ Thank you! Can you explain a bit further how the radiation affects the kinetic energy? I still don't grasp why the radiation would affect the kinetic energy at all. Again, sorry for the novice question. $\endgroup$
    – Fingolfin
    Nov 15, 2016 at 14:09
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    $\begingroup$ classically radiation takes energy with the Poynting vector, and energy conservation assures that the accelerating electron loses it (in the system where the nucleus is at rest). en.wikipedia.org/wiki/Poynting_vector#Interpretation $\endgroup$
    – anna v
    Nov 15, 2016 at 14:20
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    $\begingroup$ At a basic level, without the existence of acceleration radiation, the planetary model could work. So the OP is not right about why the model fails. Otherwise, it should be surprising the Moon doesn't crash in the Earth. $\endgroup$
    – G. Bergeron
    Dec 8, 2016 at 10:18
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I can't see how a negatively charged electron can stay in "orbit" around a positively charged nucleus. Even if the electron actually orbits the nucleus, wouldn't that orbit eventually decay?

Yes. What you've given is a proof that the classical, planetary model of the atom fails.

I can't reconcile the rapidly moving electrons required by the planetary model with the way atoms are described as forming bonds. If electrons are zooming around in orbits, how do they suddenly "stop" to form bonds.

Right. There are even simpler objections of this type. For example, the planetary model of hydrogen would be confined to a plane, but we know hydrogen atoms aren't flat.

My question here is whether the planetary model itself addresses these concerns in some way (that I'm missing)[...]

No, the planetary model is simply wrong. The Bohr model, which was an early attempt to patch up the planetary model, is also wrong (e.g., it predicts a flat hydrogen atom with nonzero angular momentum in its ground state).

The quantum-mechanical resolution of this problem can be approached at a variety of levels of mathematical and physical sophistication. For a sophisticated discussion, see this mathoverflow question and the answers and references therein: https://mathoverflow.net/questions/119495/mathematical-proof-of-the-stability-of-atoms

At the very simplest level, the resolution works like this. We have to completely abandon the idea that subatomic particles have well-defined trajectories in space. We have the de Broglie relation $|p|=h/\lambda$, where $p$ is the momentum of an electron, $h$ is Planck's constant, and $\lambda$ is the wavelength of the electron. Let's limit ourselves to one dimension. Suppose an electron is confined to a region of space with width $L$, and there are impenetrable walls on both sides, so the electron has zero probability of being outside this one-dimensional "box." This box is a simplified model of an atom. The electron is a wave, and when it's confined to a space like this, it's a standing wave. The standing-wave pattern with the longest possible wavelength has $\lambda=2L$, corresponding to a superposition of two traveling waves with momenta $p=\pm h/2L$. This maximum wavelength imposes a minimum on $|p|$, which corresponds to a minimum kinetic energy.

Although this model is wrong in detail (and, in fact, agrees with the actual description of the hydrogen atom even more poorly than the Bohr model), it has the right ingredients in it to explain why atoms don't collapse. Unlike the Bohr model, it has the right conceptual ingredients to allow it to be generalized, expanded, and made more rigorous, leading to a full mathematical description of the atom. Unlike the Bohr model, it makes clear what is fundamentally going on: when we confine a particle to a small space, we get a lower limit on its energy, and therefore once it's in the standing-wave pattern with that energy, it can't collapse; it's already in the state of lowest possible energy.

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    $\begingroup$ Just to add there's a calculation here of how long a hydrogen atom would last with the planetary model. It works out at 1.6 × 10−11 s. See page 3. That's without relatavistic corrections, which reduce the lifespan of the atom. physics.princeton.edu/~mcdonald/examples/orbitdecay.pdf $\endgroup$
    – Robert Walker
    Dec 9, 2017 at 17:46
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The treatment of electrons as waves has combined with spherical harmonics (below image) to form the foundation for a modern understanding of how electrons "orbit."

enter image description here
Tweaks to the spherical harmonic differential equations yields the Schrodinger equation, which yields the accepted models of electron orbital structures:

enter image description here

The only element for which the Schrodinger equation may be solved exactly (approximation is necessary for the rest) is Hydrogen:

enter image description here

These models predict essentially zero probability that an electron will enter the nucleus for most orbitals. In the orbitals where there is some time that an electron spends time in the nucleus it is believed to be energetically unfavourable for the electron to bind to the proton. If electrons were merely point charges this would not be possible, but the wave-nature of electrons creates phenomena such as the Pauli-exclusion principle that predict otherwise.

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    $\begingroup$ All the s-states have an anti-node at the center, and predict that the electron spends a small but non-negligible fraction of the time in the nucleus. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jun 14, 2013 at 21:27
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    $\begingroup$ When it is energetically favorable, they do. It's called "electron capture". See physics.stackexchange.com/q/4481 for a slightly longer discussion. Or physics.stackexchange.com/a/9418/520. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jun 14, 2013 at 22:03
  • $\begingroup$ Would also be interesting to compare with why positronium is unstable and magic isotope numbers. $\endgroup$
    – Ciro Santilli OurBigBook.com
    Dec 27, 2019 at 17:59
  • $\begingroup$ How often do you find a hydrogen atom by itself, with one proton and one electron and nothing connected to it? If only we had a solution for an H2 molecule.... The only thing we have an exact solution for, is vanishingly rare. $\endgroup$
    – J Thomas
    Jun 15, 2022 at 16:58
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    $\begingroup$ The top comment here is simply incorrect. The wave function amplitude is highest at the nucleus, but the probability is actually analytically 0. This is because the radial probability density function is $r^2 |\psi(r)|^2$, which means that there is a 0 probability at 0 distance from the nucleus! $\endgroup$
    – Matt Hanson
    Aug 29, 2023 at 14:00
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Briefly,

  1. The Bohr--planetary model doesn't really address these issues.

Bohr, a genius, just asserted that the phenomena at the atomic level were a combination of stationarity while being in an orbit, and discrete quantum jumps between the orbits. It was a postulate that yielded some agreement with experiment and was very helpful for the future development of quantum mechanics solely because it got people to think about stationarity and discreteness.

2 It is totally useless for discussing chemical bonds. You are quite right to be uncomfortable with it.

3 It would be stretching a point, but you could see the Quantum Mechanics of Heisenberg and Schroedinger as the only way to salvage the planetary model of Bohr, by finally coming up with an explanation for the stationarity of an electron's state around (but no longer considered as « orbiting ») the nucleus and an explanation for discrete jumps as a response to perturbations from outside. But this required seeing the electron more as a wave and hence not having any definite location along the orbit.

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    $\begingroup$ Bohr did not just assert it. Bohr created the correspondence principle to explain how to quantize. $\endgroup$
    – Ron Maimon
    Oct 7, 2012 at 14:12
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here's an answer from Dr.Richard Feynman http://www.feynmanlectures.caltech.edu/II_01.html#Ch1-S1

You know, of course, that atoms are made with positive protons in the nucleus and with electrons outside. You may ask: “If this electrical force is so terrific, why don’t the protons and electrons just get on top of each other? If they want to be in an intimate mixture, why isn’t it still more intimate?” The answer has to do with the quantum effects. If we try to confine our electrons in a region that is very close to the protons, then according to the uncertainty principle they must have some mean square momentum which is larger the more we try to confine them. It is this motion, required by the laws of quantum mechanics, that keeps the electrical attraction from bringing the charges any closer together.

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  • $\begingroup$ Why don't protons have some mean square momentum which keeps them apart and out of the nucleus? $\endgroup$
    – J Thomas
    Mar 2, 2023 at 3:35
  • $\begingroup$ The strong nuclear force, which acts over very short ranges, is far more powerful than the electromagnetic force. Thus, the nucleus, while composed of quantum mechanical particles, still compresses together far more than the electrons and nuclei. $\endgroup$
    – Matt Hanson
    Aug 2, 2023 at 2:40
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From the askers perspective, the explanatory powers of most of these answers seem pretty bad. I prefer Emilio Pisanty's answer here: Why isn't Hydrogen's electron pulled into the nucleus? because it explains exactly how the uncertanity principle dictates the facts of this atomic reality.

The summarized problem is that, if the charged and attracted electron and proton fell into each other, we would know exactly their position, and by the Heisenberg uncertainty principle our knowledge of the momentum would be immensely small, it could be anything. The chances therefore of the momentum being large enough to "escape" this essentially electrostatic attraction are very large. Therefore, the electrons recede to an average distance from the nucleus. The electron is in the position it is (or rather average position) to keep these two opposing forces in balance.

The Heisenberg uncertainty acts as a force of repulsion, in similarity with the effect of compressing a gas. More compression=more pushback.

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Sometimes electrons do "crash into the nucleus" - it's called electron capture and is a mode of decay for some unstable isotopes.

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There is no orbit around nucleus, since expectation value for angular momentum for ground state $\psi_0$ is zero; $\langle{\psi_0}\rangle=0\;.$ That is why we cannot talk about classical planet model, like Bohr did. Also Heisenberg's uncertainty principle prevents electrons from having well defined orbits. Electron is just somewhere outside nucleus.

Since proton is positively charged and electron is negatively they have attractive Coulomb's force. But tiny quantum particles, as electrons, behave as waves and they cannot be compressed into too small volume without increasing their kinetic energy. So electron on its ground state $\psi_0$ is on equilibrium state between Coulomb's force and strange quantum pressure.

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Electrons don't crash into the nucleus of an atom. The reason is deep-rooted in quantum mechanics. According to Heisenberg's uncertainty principle, the uncertainty in position and momentum are related by $$\Delta x\Delta p_x\geqslant\hbar/2$$ When the electron approaches closer to the nucleus, the electron gets confined within a smaller region of space so that the uncertainty in position $\Delta x$ of the electron decreases. Accordingly, the uncertainty in momentum $\Delta p_x$ increases. This means that the electrons have an energy higher on the average and thereby the system deviates from equilibrium. If the electron falls in the nucleus i.e., $\Delta x\rightarrow0$, then $\Delta p_x\rightarrow\infty$ which implies infinite energy. So, in order to maintain stability of the system, the electrons try to remain away from the nucleus.

However if the electron manages to get crashed into the nucleus, then it would gain an infinite amount of energy according to the uncertainty principle which is impractical to occur in nature.

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  • $\begingroup$ This is not exactly correct as the width of the nucleus is a known finite number, i.e. $\Delta x \neq 0$. $\endgroup$
    – Mathews24
    May 17, 2020 at 1:11
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    $\begingroup$ @Mathews24 Yes, the size of the nucleus is known and it can't be equal to zero according to quantum mechanics. That's the reason I had used $\Delta x\rightarrow 0$. This doesn't mean $\Delta x=0$. You can refer the first chapter of Quantum Mechanics by Landau and Lifshitz for a more rigorous explanation. $\endgroup$
    – Richard
    May 17, 2020 at 2:29
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@user56903,@dmckee, and and @user41827 all point out that electrons do smash into the nucleus, and probably fairly often. But they don't continuously radiate until they have lost so much energy that they are stuck in the nucleus. Instead, presumably they usually get spat back out with all the considerable energy they had going in. Only when the nucleus is in just the right state to absorb that energy do they get to stay.

When the planetary model of the atom failed, there were various possible ways to fix it up.

  1. Decide that there are some circumstances where an accelerated charge does not radiate, and try to characterize which circumstances those are.

  2. Look for ways that electrons could accelerate while cancelling the radiation they produce. If two electrons were in the same circular orbit, 180 degrees apart, would their radiation cancel? (Probably not.) If they were in exactly the same orbit with opposite spin, would they cancel? (At that time, who would know? I don't think they knew about electron spin then.) Etc.

  3. Imagine ways that electrons could stay away from the nucleus without accelerating. For example, there could be a force similar to the Strong Force that pushes them away if they get too close. It would follow peculiar laws that could be worked out.

  4. Imagine that electrons are not little spheres but instead little thread-like things. Any time that one of them wraps exactly around its orbit, its radiation cancels. (Not really, but I'm sure there are ways to fudge that.) Similarly if it wraps exactly twice, exactly three times, etc.

There are lots and lots of ways to imagine it, and I'm sure that many of them could have been adapted to fit the real-world data.

As I understand it, what the physicists of the time did was to find math that fit the data, and let it go at that. Sometimes someone tries to explain the math, and mostly fails. For example we can hand-wave that electrons are waves, and sometimes they're standing waves, and the behavior of standing waves blah blah blah. Or the Heisenberg Uncertainty Principle requires not just that you can't measure everything, but that everything is really and truly indeterminate and therefore electrons do not move and do not radiate. Since the math gets the right answers, it doesn't really matter what explanation we use when we try to hand-wave explanations.

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The hypothesis that an electron would radiate as it accelerates towards an atomic nucleus is based on the unmotivated assumption that the electron only has retarded potentials. If no assumption is made about whether its potentials are retarded or advanced, then it is reasonable to assume that the advanced and retarded potential are equally big. This leads to stable conditions according to Nordström 1920

G. Nordström
Note on the circumstance that an electric charge moving in accordance with quantum conditions does not radiate
Proc. Roy. Acad. Amsterdam 22, 145-149 (1920)

Leigh Page 1924 also calculates this in

Advanced Potentials and their Application to Atomic Models
Leigh Page
Phys. Rev. 24, 296 – Published 1 September 1924

Page mentions something that he 97 years ago considered a problem

Nevertheless fluctuations of energy would take place during each period, and the converging and diverging waves should combine to form standing waves. The absence of any evidence for such waves constitutes a serious objection to the theory.

Today we know that vacuum indeed contains a lot of electromagnetic waves.

Page also refers to Arthur Constant Lunn:

A path out of the dilemma has been suggested by Nordstrom who shows that if the field of a charged particle is half retarded and half advanced, instead of being wholly retarded as usually assumed, Maxwell's equations remain valid and the net energy radiated by an electron in describing a periodic orbit vanishes. The same suggestion was proposed by Lunn at the Wisconsin colloquium in 1922.

Ethan Marx also mentions a general case where radiation is avoided by not excluding the advanced potentials. It is mentioned as time symmetry (T-symmetry).

A Time Symmetric Classical Electrodynamics
E. Marx (Army Research Lab, Adelphi)
Int.J.Theor.Phys. 15 (1976) 891-899
DOI: 10.1007/BF01807711

The "certain aspects of quantum mechanics" that were invented as a solution to the problem of electron orbits are thus avoidable.

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  • $\begingroup$ Is this a comment? It doesn't seem to be an answer. $\endgroup$
    – orome
    Sep 9, 2021 at 17:48
  • $\begingroup$ @orome I summarize and give four references. The four references are answers. $\endgroup$
    – David Jonsson
    Jan 5 at 16:46
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Think a little further. When the electrons get accelerated closer to the nucleus they radiates some energy away which fills the vacuum and get scattered by other electrons, accelerating them. This finally becomes an equilibrium condition. This was calculated by Puthoff in 1987.

Ground state of hydrogen as a zero-point-fluctuation-determined state
H. E. Puthoff
Phys. Rev. D 35, 3266 – Published 15 May 1987

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  • $\begingroup$ Is this a comment? What it is it referring to? $\endgroup$
    – orome
    Sep 9, 2021 at 17:48
  • $\begingroup$ @It is an answer according to Puthoff. It refers to Puthoff's answer. $\endgroup$
    – David Jonsson
    Sep 10, 2021 at 12:56
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This is because of Heisenberg's uncertainity principle. If the electron would have lose it's kinetic energy via synchotron radiation, then it would crash into the nucleus according to the classical mechanics. But because of quantum mechanics, if the electron crashed into the nucleus the position and the velocity both would be known i.e 0, which is prohibited via Hiesenberg's uncertainity principle. Also if this happened, our Earth would not exist which is not observed via experiment and Anthropotic principle proving the principle.

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A planet orbiting a star with eccentricity smaller than unity would have to lose kinetic energy in order to spiral into the star. This could happen in the long run for a planetary system due to emission of gravitational radiation, and due to tidal forces, heating up the star or the planet followed by radiative cooling. In quantum mechanics this cannot happen.

If the planet has eccentricity equal to unity, analogous to an s orbital, it crashes straight into the star where its kinetic energy is converted into heat. Again, in quantum mechanics this cannot happen.

Whether quantum mechanics explains why, or only how, by construction, such an atomic collapse does not happen is a matter of interpretation. Note that electron capture by some nuclei as discussed in other replies requires that the weak interaction is taken into account. I interpret the original question as being about any nucleus, not just the ones susceptible to electron capture.

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    $\begingroup$ I get negatives, delete votes bit no argument. Is physics.stackexchange.com about popularity or physics. Give me some arguments ipo anonymous, emotional negativity. $\endgroup$
    – my2cts
    Jul 26, 2019 at 8:46
  • $\begingroup$ as i understand it physics.stackexchange is supposed to be about the standard interpretation of physics. This means that the standard model with QM is all that should be discussed. If you have an interpretation which is not standard, you can expect to be voted down. There is no need for discussion or argument, beyond the belief that your idea is not standard. $\endgroup$
    – J Thomas
    Jul 9, 2022 at 3:46
  • $\begingroup$ @JThomas There is no standard interpretation of physics. $\endgroup$
    – my2cts
    Jul 9, 2022 at 16:36
  • $\begingroup$ it looks to me like the standard model with QM and QED etc is the standard interpretation. There are some rough edges that nobody has a good explanation for, but it seems to be pretty clear to everybody what's standard and what isn't. If it isn't clear, you can mostly tell by what gets voted down here. If something doesn't get voted down it might be that people just haven't noticed it, but if it is a standard answer to a question it won't get voted down. $\endgroup$
    – J Thomas
    Jul 10, 2022 at 21:37
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    $\begingroup$ When I look at what you said this time, I agree that I don't see you making any claims that specifically deny anything in the standard physics. I dunno. I guess my second hypothesis is that maybe some people have gotten so used to downvoting you that they just do it when they notice you've posted something, regardless what you say. $\endgroup$
    – J Thomas
    Jul 12, 2022 at 23:06
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While all these answers are fundamentally correct, especially with regards to Schrodinger and the shell model of electrons, there is one very basic means of radioactive decay, that of electron capture, which has not yet been discussed. Yes indeed, electrons orbiting around the atom can be captured into the nucleus. (For reference, see http://en.wikipedia.org/wiki/Electron_capture) Electron capture is a process in which a proton-rich nuclide absorbs an inner atomic electron, thereby changing a nuclear proton to a neutron and simultaneously causing the emission of an electron neutrino. Various photon emissions follow, as the energy of the atom falls to the ground state of the new nuclide. Electron capture is a common decay mode for isotopes with an over-abundance of protons in the nucleus. What is interesting about the phenomenon of electron capture is that it depends not on the electrons in the electron cloud of the atom, but rather on the nucleus. Thus, one can not ignore the fact that the behavior of electron capture is dependent solely on the nucleus, not the electrons. For example, if the nucleus is, for example, Carbon-9, 100% of this isotope will decay via electron capture to 9-Boron. Yet Carbon-14, which has the same electric charge and same number of electrons in an identically configured electron cloud, never decays via electron capture. Quantum physics, especially when the answer is focusing on the electrons of the atom, has trouble explaining the behavior of Electron Capture with a sufficient credibility. So in answer to your question, electrons do indeed fall into the nucleus, via the phenomenon of electron capture, yet that behavior can not be explained by examining the quantum physics of the electrons.

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    $\begingroup$ The quantum mechanics of electron capture is very well understood. $\endgroup$
    – Brandon Enright
    Mar 5, 2014 at 1:31
  • $\begingroup$ I am very aware of the explanations offered by quantum mechanics, and no, they do not answer my questions. $\endgroup$
    – user41827
    Mar 5, 2014 at 23:04
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    $\begingroup$ That's fine but don't provide an answer saying "Quantum physics, especially when the answer is focusing on the electrons of the atom, has trouble explaining the behavior of Electron Capture with a sufficient credibility." just because you have questions about the process. $\endgroup$
    – Brandon Enright
    Mar 5, 2014 at 23:09
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    $\begingroup$ You must examine the quantum physics of the nucleus, not the electrons. The quantum physics of the electrons says the phenomenon can't happen, yet it happens all the time. That's why they had determined the existence of the electron neutrino, the particle that allows this to happen. Anyone who says that an electron can't fall into a nucleus, because quantum physics prevents it, is incorrect. The electron neutrino is the mediator of this process, and this allows it. $\endgroup$
    – user41827
    Mar 5, 2014 at 23:30
  • $\begingroup$ Do not misunderstand me. I have no questions. Let me clarify. What I am saying is that the answer will not be found by examining the quantum physics of the electrons. It is not the electrons that regulate this process. It is the quantum physics of the nucleus, which has been very much ignored in these previous answers. $\endgroup$
    – user41827
    Mar 5, 2014 at 23:31

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