# Electrons in atoms

1. Oct 24, 2004

### Cheman

Right, as we know from the uncertainty principle we dont know exactly where an electron is, just where it most probably is. This implies that electrons don't follow circular paths like those of planets round the sun (as in the rutherford model) - they must take a more random path.

My question is if electrons are moving forward, like a planet, around the nucleus which is exerting a force on them, like the sun, why don't they take circular or eliptical orbits? Is it, quite simply, that electrons experience much more forces than the planets, interacting with forces from other electrons, atoms, etc?

Furthermore, does this suggest then that if there was no other matter in the universe (ie - no charges) a hydrogen atom would indeed look like one from the rutherford model - like a planet making a circular orbit of the sun?

2. Oct 24, 2004

### mathman

The model you are describing (electrons in orbit around nucleus) is an old model, which is considered quite out of date. Current models assign "states" to electrons, which are defined by quantum numbers and associated energies. If you want to picture it, think of a diffuse cloud (not too accurate a picture either) around the nucleus.

3. Oct 24, 2004

### ZapperZ

Staff Emeritus
First of all, it is not nice to do multiple posting of the same question.

Secondly, I think you seem to not know the existence of the solution to the Schrodinger equation for the hydrogen atom. Please look it up and look at the orbital solutions. There is no need to "guess" or speculate what they look like anymore nowadays.

Zz.

4. Oct 24, 2004

### Tom Mattson

Staff Emeritus
Your question presupposes that the electron can have a well-defined initial velocity in the first place. Already, this thinking is doomed to be illusory, because we have no reason to think that this classical notion applies, and every reason to think that it does not.

No, it isn't. Even in the simple case of the 1-electron atom, the electron is (in the first approximation) subject to a central force, not unlike the central force that the Earth is subject to from the Sun (again, in the first approximation). The problem is not extra forces, the problem is that at the atomic level matter behaves very differently from macroscopic bodies. The wavelike characteristics of macroscopic objects is hidden by the fact that the wavelength associated with the wave is so small. At smaller scales, the wave nature of matter becomes more apparent.

It doesn't. Even if you take away the atom, and go to the simplest possible example of a beam of electrons passing through a single-slit diffraction apparatus, the wave nature of the beam would manifest itself in the form of a diffraction pattern, just like the one exhibited by light.

5. Oct 25, 2004

### Pieter Kuiper

Electron motion can be described by circular or elliptical optics when the principle quantum number n is large. Then the correspondense principle applies: the electron's orbital period is the same as the radiation emitted in a quantum jump from n to n-1.

Nowadays laser experiments can be done on such Rydberg atoms. One can also do quantum mechanical calculations of the time evolution of small wave packets. Look at the animations of for example http://www.optics.rochester.edu:8080/users/stroud/ .

Last edited by a moderator: Apr 21, 2017
6. Oct 25, 2004

### Cheman

So let me get this straight, please correct me if I'm wrong - are you saying that an electron would do what I said bout ONLY IF it was like a planet round the sun. However, an electron is very different to this, and by its very nature is unlike things on a macromolecular level. Therefore, despite the fact that the electron is only experiencing one centrepetal force like a planets it behaves very differently to how a planet would.

Is that correct? :rofl:

7. Oct 25, 2004

### Tom Mattson

Staff Emeritus
I don't want to give the wrong impression, so let me clarify. I'm not saying that there are 2 kinds of matter, one "microscopic" and one "macroscopic". I'm saying that quantum effects aren't apparent at the macroscopic level. I'll further state that if an electron had sufficiently high energy (say, as much energy as a planet in orbit), the expectation value of its position would approach the Newtonian result, as per the correspondence principle.

Not quite. Again, it has to do with the wave effects being unapparent at the planetary level. The deBroglie wavelength of a quantum particle is &lambda;=h/p. For planets, p is much, much larger than it is for your typical atomic electron, so the wavelength is much, much shorter; many orders of magnitude shorter in fact.

8. Oct 26, 2004

### Pieter Kuiper

You do not need to go that far. Allready when the principal quantum number n=100, you get an atom with diameter $$2 n^2 a_0 \approx 1\mu{\rm m}$$. These Rydberg atoms are about the size of bacteria, and the periods of revolution are well calculated using Newtons laws and the Coulomb force.

(But of course these states are coherent superpositions of different quantum states, and behave non-classical because of that.)

9. Oct 26, 2004

### Tom Mattson

Staff Emeritus
Thank you. I didn't feel like doing any calculations, so I posted a "sufficient" condition, not a "necessary" one.