# Geometry of the atomic structure

1. Dec 24, 2013

### res3210

Hey guys,

I was looking at both the time-dependent and time-independent schrodinger equations, and I notice that we often choose to solve these in spherical coordinates. I understand that we do this because they are convenient for problems with azimuthal symmetry. However, how do we know that this geometric model is actually accurate? Is it because Bohr made the assumption that the hydrogen atom can be modeled as a spherical system? I understand that QM has been tested extensively and we have seen that it is a very accurate model, is it because of this that we assume that our geometric assumptions are correct?

Thanks for the help and information in advance,

Ryan

2. Dec 24, 2013

### ZapperZ

Staff Emeritus
Hum... forget about QM. When you were doing your E&M problem, and you were given a central, spherically symmetric potential or charge distribution, how did you know that choosing spherical coordinate was the best choice?

Once you have answered that, look at the potential term in the Schrodinger equation for the H atom, and compare.

Zz.

3. Dec 24, 2013

### Maui

There are images of atoms taken by scanning tunneling microscopes. They are easy to find online.

4. Dec 24, 2013

### res3210

Of course, it is because the electric field points radially outward, which is experimentally verified. So I suppose we assume the same thing for H. If we see that it behaves electrically equivalent to an electrically charged sphere, then that should lead us to the spherical geometry conclusion.

5. Dec 24, 2013

### Bill_K

We write down the Schrodinger equation for a Coulomb potential, V = Ze2/r. At that point the physical assumptions are done with, and we can solve that equation any way we like. Spherical coordinates are an obvious choice but not the only one. In fact the hydrogen atom can be solved in parabolic coordinates also. See "www.ejournal.unam.mx/rmf/no546/RMF005400609.pdf‎" [Broken]. Also here.

Last edited by a moderator: May 6, 2017