Simple Energy Levels confusion

[alexgmcm]

In the famous diagrams of the energy levels of a hydrogen atom it seems that the energy levels get closer together as they increase such that the difference in energy between higher energy levels is less than between lower ones. But working from the electron in a box approach we get the equation:

E_{n} = \left(\frac{h^2}{8mL^2} \right) n^2
where n = 1,2,3...

So that would suggest that the energy levels would go E, 4E, 9E, 16E etc. which would mean that the gaps between the energy levels would increase? But that's different to most of the diagrams I've seen like http://www.avogadro.co.uk/light/bohr/atomspec.gif" which seem to show the energy levels getting closer together.

I found http://imgur.com/OYSV4.png" in my textbook which also suggests that the energy levels should get further apart not closer together. I know that is only considering rotational and not vibrational energies, but the vibrational energies are equally spaced and so would not lead to closer energy levels in combination.

I'm sure I've just missed something quite basic here which is so basic the books fail to mention what it is, so can someone please help?

 

[Vanadium 50]

The potential for a hydrogen atom is different than the potential for particle in a box. Different potentials give you different energy levels.

 

[alexgmcm]

So in general for atoms and molecules, do the energy levels get closer together as you move farther apart from the atom. That is, do the differences in energy between the energy levels decrease at higher energy levels?

 

[Fightfish]

So in general for atoms and molecules, do the energy levels get closer together as you move farther apart from the atom. That is, do the differences in energy between the energy levels decrease at higher energy levels?

Yup. You can perform a preliminary derivation of the relation between the energy levels and 'n' using Bohr's semiclassical model. That should be available in any standard text. For the hydrogen atom, the relation is
E_{n} = - \frac{13.6}{n^{2}} eV