Rutherford's experiments confirmed the existence of light-weight electron clouds in a mostly empty atom, and that they occupy some space around the nucleus. What made us conclude that they can move? Can't it be vice versa: nucleus moves around the electron cloud?

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    $\begingroup$ By 1911 both the charge to mass ratio as well as the charge of electrons were known, so it must have been clear that electrons were far lighter than nuclei. The remainder is momentum conservation. $\endgroup$
    – CuriousOne
    Jan 4, 2015 at 7:08
  • $\begingroup$ That's like asking why the planets move round the Sun. Couldn't the planets stay still and the Sun move around the Solar System? $\endgroup$ Jan 4, 2015 at 7:28
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    $\begingroup$ Swami, the idea of the electrons moving is based on the idea they orbit the nucleus like little planets. This is profoundly incorrect. The electrons are delocalised within an atom. They do not have a well defined position and they do not move in the naive sense of the word. $\endgroup$ Jan 4, 2015 at 7:59
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    $\begingroup$ @JohnRennie : may I add some clarification to what you said? I will delete it if you disagree with it. Movement in the "naïve sense of the word" means following a trajectory, i.e. at each time t having a position, $\vec r$ and a velocity $\vec v$. $\endgroup$
    – Sofia
    Jan 4, 2015 at 15:46
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    $\begingroup$ @Sofia: yes, that's exactly what I meant. You describe it very well :-) $\endgroup$ Jan 4, 2015 at 16:11

1 Answer 1


As John and I sais, the electrons in the atom don't move in the sense that they follow some trajectory and have at any time a position and a velocity. We don't really understand how they behave in the atom. Though, they move.

Please see a simple calculus: let's take the hydrogen atom with the electron on the lowest level, n = 1. The wave-function is very simple, spherically symmetrical, proportional to $e^{-r/a_0}$, where $a_0$ is the Bohr radius.

Let's calculate the average linear momenta $P_x$, $P_y$, and $P_z$. Since the wave-function is real,

$<P_x> = -i\hbar \int \psi(r) \frac {∂\psi(r)}{∂x} d\vec r = 0$,

$<P_y> = -i\hbar \int \psi(r) \frac {∂\psi(r)}{∂y} d\vec r = 0$,

$<P_z> = -i\hbar \int \psi(r) \frac {∂\psi(r)}{∂z} d\vec r = 0$,

where the integral is taken over all the atom. So, on average, no net movement in some direction. However,

$<P_x^2> = -\hbar^2 \int \psi(r) \frac {∂^2\psi(r)}{∂x^2} d\vec r = \frac {\hbar^2}{a_0^2} C^2 \int \frac {x^2}{r^2}e^{-2r/a_0} d\vec r$,

where $C$ is the normalization constant. A similar calculus can be done for $<P_y^2>$ and $<P_z^2>$, and adding the results,

$<P^2> = \frac {\hbar^2}{a_0^2}$.

Thus, the QM says that the electrons in the atom, move. How they do this movement? We don't exactly understand, given that they don't have trajectories.

About the vice-versa, i.e. the nucleus moving around the electrons - well, movement is relative. But the mathematics if we take the nucleus as moving around the electron would be much more complicated. Usually we take the calculus to the center-of-mass frame. Given the disproportion in mass between the nucleus and the electron, the center-of-mass practically coincides with the nucleus.

  • $\begingroup$ This is one of the many things about QM that drove me nuts. So would it be safe to say that when bound by a nucleus, electrons act more like a standing wave than a particle "zooming" around the nucleus? Whereas when an electron is free of a nucleus, it acts more like a single particle that can "move" in the simple sense? [I am not specifically dissecting the wave-particle duality thing, just trying to clarify for my own edification.] $\endgroup$ Jan 4, 2015 at 18:26
  • $\begingroup$ @honeste_vivere : Wait a bit, I am busy now, but I'll answer you soon. $\endgroup$
    – Sofia
    Jan 4, 2015 at 18:36
  • $\begingroup$ @honeste_vivere : what is "zooming"? Anyway, what I can say is that none of the picture is good. It's not encouraging, but this is the situation. People that deal already for years with QM, accept (with no pleasure) that we can't get for these particle a classical picture of a moving particle along a trajectory. In the atom they don't form standing waves with maxima and minima. The wave-function of the electron on the ground level is completely spherically symmetrical. When the electron is free, it behaves rather as a wave, it can produce interference. $\endgroup$
    – Sofia
    Jan 4, 2015 at 18:53
  • $\begingroup$ Oh sorry, zooming is careless slang for moving along at a relatively large velocity. To clarify, though, would not a spherical wave-function be similar to a standing wave represented by spherical harmonics? Isn't that how we represent electron orbits in hydrogen (there are Leguerre polynomials or something like that too, right?)? As for free electrons, I asked my question that way because we can detect them as individual particles, right? $\endgroup$ Jan 4, 2015 at 18:56
  • $\begingroup$ @honeste_vivere : About the wave-particle duality, it just means that these microscopic particles behaves both as waves and as particles. When a beam of such particles passes through two-slits (the famous double-split experiment), we obtain on a photographic plate after the slits, an interference pattern. But, each single particle leaves on the plate one single dark spot, as any ordinary particle. $\endgroup$
    – Sofia
    Jan 4, 2015 at 18:58

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