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I've studied electrostatics and we always use Coulomb's law to calculate the force on a particle due to another particle. In an atom there are many electrons and many protons, and if we "take a picture" of an atom, at any given moment, we should see something like this (which is possibly wrong, since matter and waves behaves in the same ways under some conditions. https://en.wikipedia.org/wiki/Matter_wave and because Heisenberg uncertainty principle exists)

enter image description here
(these are just the first two orbitals).

Now, we could draw for every single particle the net force acting on that particle. The protons and the neutrons are held together by the strong nuclear force (of which I don't really know anything about..) but what about the electrons? If they all repel each other, and at the same time they are attracted to the nucleus, arrived at a certain position (wherever they may be) they should have a net force acting on them equal to zero and therefore stop. I just know some very basic details of quantum physics, and some electromagnetism and chemistry so I don't know what I'm missing. I'm just thinking about electrons as simple "balls" and in an intuitive way, but could you please tell me what is wrong with my reasoning?

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  • $\begingroup$ Zero force does not imply zero motion. It implies that the acceleration is zero, and therefore constant velocity. $\endgroup$
    – mike stone
    Apr 23 '21 at 17:14
  • $\begingroup$ @mikestone Yeah yeah correct, but it's pretty much like an equilibrium point (or it is?), if a particle moves a little from that point it would be pushed back in that position no? $\endgroup$ Apr 23 '21 at 17:33
  • $\begingroup$ Thinking about "simple balls" is completely wrong, because an accelerating charged particle emits electromagnetic radiation, and all the electrons in the atom would lose their energy and stop orbiting the nucleus is a tiny fraction of a second. $\endgroup$
    – alephzero
    Apr 23 '21 at 18:29
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    $\begingroup$ How many electrons are there in this atom? $\endgroup$
    – my2cts
    Apr 23 '21 at 19:01
  • $\begingroup$ "I'm just thinking about electrons as simple "balls" and in an intuitive way, but could you please tell me what is wrong with my reasoning?" -- What's wrong with your reasoning is that you are just thinking of electrons as simple balls and in an intuitive way. $\endgroup$
    – WillO
    Apr 23 '21 at 22:52
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The "ordinary physics" picture of atoms with twirling electrons orbiting the nucleus like little satellites is inapplicable, because it ignores quantum effects which dominate at the size scale of an atom.

An electron in motion carries with it a certain amount of kinetic energy as it zooms about, simply because it is above absolute zero. If the electron is freely moving through space, it can increase or decrease its kinetic energy by almost any amount by colliding with and recoiling off of other things (like electrons) that happen to be moving a bit faster or slower, but if you confine the motion of an electron by (for example) putting a convenient proton nearby so coulomb's law pulls on it, then you discover that the electron cannot possess any energy amount it wishes, and it cannot shift from one energy level to any other arbitrary energy level, once it becomes trapped in the proton's electrostatic grip. Then it can possess only certain discretely-separated energy levels called orbitals.

In the lowest-energy orbital, the electron zooms around as a probability cloud in 3-dimensional space surrounding the proton and has no way to decrease its kinetic energy, so there it stays.

Adding another proton to the nucleus allows for another electron to take up residence around it in similar fashion, but not only do the two electrons repel one another, but the energy level rules imposed by quantum mechanics force the electrons to line themselves up in their probability clouds in a nonrandom manner which gets more and more complicated as more and more electrons join in at progressively higher and higher orbital energy levels.

Wikipedia has some nice graphics of these orbitals, that have names like 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, and so on, each with a different size and shape- and all the consequence of quantum mechanics.

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Now, we could draw for every single particle the net force acting on that particle.

Not at the level of atoms, electrons protons and neutrons, you cannot. It was found necessary to invent the theory of quantum mechanics to describe molecules and atoms and their content.

The protons and the neutrons are held together by the strong nuclear force (of which I don't really know anything about..) but what about the electrons?

Force is the same concept at the quantum level and classically only when defined as a dp/dt, a momentum transfer interaction. The classical Coulomb potential is used in the quantum mechanical equation and defines the energy levels which the electron can occupy in the specific atom, but as in your picture, it is only orbitals that can be predicted , probability loci where the electron can found when measured, orbitals , not orbits. The electrons in different orbitals do not interact, that is why atoms are stable, they have to stay in their given energy level. They do not repel as classical particles, because all the Coulombic interaction has been taken into account in solving the quantum mechanica equation with the correct potential to get the energy levels and the orbitals. See the example of the hydrogen atom.

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The electostatic force is $\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$. It depends on the distance between the charges. If you have an atom with $N$ protons and $N$ electrons, and you were to look at the forces on 1 particular electron, this would entail summing the forces from all $N$ protons on it, as well as the $N-1$ other electrons. The distances between those particles and the one electron you're looking at matter greatly, and will almost certainly not sum exactly to $0$.

There's a pretty rough picture which sometimes gets invoked. In this rough picture, the electrons are all moving so fast that they are basically "smeared" in a spherical way at a constant radius from the nucleus. As you know, from Newton's shell theorem, an electron will not feel any net force when inside one of these spherical shells. So if there are $N$ electrons, and we look at the $n^{th}$ outermost, it will feel the attractive force from the $N$ protons, as well as the repulsive force from the $n-1$ other electrons which are in a smaller shell than it. So roughly, it'll feel an attractive force of $(N - n + 1) \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2}$ towards the center of the atom.

Because all the electrons will roughly feel a net inward force, they have to keep moving in order to not crash into the nucleus, much like how the planets must have some orbital velocity in order not to crash into the sun.

Obviously to get an actually correct picture of what's going on you need to treat the electrons as wave functions and use Schrodinger's equation.

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