Does Bohr's Model of Hydrogen Still Hold Up Against Quantum Mechanics?

[PeteGt]

I searched the forums to find my answer and didn't find anything within the first couple searches.

Here's my question:

Bohr developed his model of the hydrogen atom. Then Quantum mechnics came along.

Does Bohr's model of Hydrogen work for itself? I always thought even though wave mechanics describes everything else appropriately, that hydrogen's model like bohr described was totally okay. I was about to try to work through the calculus and figure it out, but I was wondering if anyone had a quick answer, or maybe even better got a mathcad document I could look at?

Does hydrogen's electron orbit in a discreet and exact orbit at Bohr's radius? Or does it exsist in this s orbital like everything else does?

Thanks in advance.

Peter

 

[theFuture]

The energy levels work out, but there are key physical flaws in the Bohr model.

Schrodinger's equation allows us to look at the electron as probability density, and nothing else. It does not orbit classically as the Bohr model might suggest.

 

[Claude Bile]

The two major failings of Bohr's model was its failure to predict fine spectral structure and the models inability to predict the behaviour of more complex atoms such as Helium.

Sommerfield proposed an alternative model to Bohr's which could explain fine structure, however it too could not be applied to more complex atoms.

In Schrodinger's model, electrons are not localised and thus thus do not orbit in a classical sense. Bohr's radius still retains physical signifigance, for while the time averaged displacement from the nucleus is zero, the R.M.S displacement is the Bohr radius.

Claude.

 

[quantumdude]

Amazingly, Bohr and Schrodinger agree on the n-dependence of the energy levels of the hydrogen atom. But as has been noted, Bohr doesn't go any further than that. Also, the Bohr model gets the orbital angular momentum totally wrong.

 

[zefram_c]

In Schrodinger's model, electrons are not localised and thus thus do not orbit in a classical sense. Bohr's radius still retains physical signifigance, for while the time averaged displacement from the nucleus is zero, the R.M.S displacement is the Bohr radius.

One can expand on this and say that the expectation of the radial displacement from the nucleus is the Bohr radius. As pointed out, the time averaged displacement is nil due to the radial symmetry. But the probability that the electron is inside the actual nucleus is very small.

 

[PeteGt]

Gotcha,

So you're saying that <r> is Bohr's Radius, but doing a square of the Radial * Spherical WAvefunctions would lend a possiblity of finding the electron outside of <r> but never at the nucleus. Am I reading this correctly?

Peter

 

[Claude Bile]

The highest probability density is a the nucleus, however because the volume of the nucleus is small compared to the volume of the s orbital, the chance of finding the electron in the nucleus is vanishingly small (nearly zero, but not quite).

Claude.