# Nuclear charge and atomic orbitals

1. Apr 17, 2013

### kthejohnster

If I use hydrogen-like orbitals for other atoms, how are the number of nodes and probability density affected?

2. Apr 18, 2013

### Staff: Mentor

Assuming you are considering one-electron atoms (or negelcting electron-electron interactions), then the only difference is in a scaling factor for $r$ that depends on $Z$. Note that this scaling factor also depends on the mass of the nucleus (if you are not considering an infinitely heavy nucleus).

3. Apr 18, 2013

### cgk

For other atoms (more than one electron), the number of nodes of the H-like orbitals should be fine, but the probability density would be just wrong. Using hydrogen-like orbitals for anything else than hydrogen and helium is highly NOT recommended! As least not if you care about the results you get.

If you need a proper approximation of the orbitals, you can use the AO functions of a generally contracted Gaussian basis set from quantum chemistry. For example, the ANO-RCC sets should be very accurate and useful for such a purpose. Depending on your case, much smaller Gaussian sets might also do the trick. (but stay away from STO-anything sets--these are even worse than using H-orbitals).