Bohr's atomic model and Bohr and Rydberg equations

[Nick Jackson]

Hello,
well, I am totally new to this section of physics so my question may sound ridiculous, but here it is:
When I was reading about the Bohr's atomic model, I learned about the Bohr and Rydberg equations (E=-2,18*10^18*Z^2/n^2 J and 1/λ=RZ^2(1/n1^2-1/n2^2) as well as their proofs. Then I read about the "shaking down" of this atomic model (please excuse my terrible english, I am greek) which I understand but, when I asked a couple of physicists, they told me that the equations remain and just show the largest possibility of an electron to be in that place. Now I get that too. What I don't get is HOW these equations remain intact. I mean the proof uses the assumption that the electron does angular motion and makes use of the formulae Fc=mv^2/r and L=Iω. However, we know now for sure that for l>0 (the azimuthal quantum number) the shape of the orbital discards the theory of the circle orbits.
Every suggestion is welcome!
Thank you!

 

[DrDu]

I can't give you a completely satisfactory answer to your question, but only two remarks:
1. There is an extension of the Bohr model, the Bohr Sommerfeld model where orbits are no longer circular but elliptical depending on angular momentum.
2. The hydrogen atom (or more generally the problem of Keplerian orbits) has a high but somewhat hidden symmetry, SO(4) which is related to the fact that the Runge Lenz vector is a constant of motion. This symmetry dictates most of both the classical and quantum mechanical behaviour of the system. Hence the two lead to remarkably similar conclusions.
You may have a look at this
http://math.ucr.edu/home/baez/classical/runge_pro.pdf

 

[Nick Jackson]

I can't give you a completely satisfactory answer to your question, but only two remarks:
1. There is an extension of the Bohr model, the Bohr Sommerfeld model where orbits are no longer circular but elliptical depending on angular momentum.
2. The hydrogen atom (or more generally the problem of Keplerian orbits) has a high but somewhat hidden symmetry, SO(4) which is related to the fact that the Runge Lenz vector is a constant of motion. This symmetry dictates most of both the classical and quantum mechanical behaviour of the system. Hence the two lead to remarkably similar conclusions.
You may have a look at this
http://math.ucr.edu/home/baez/classical/runge_pro.pdf

Thank you very much for your answer, it has been very helpful and the expansion of the Bohr's model answers many of my questions in general. Unfortunately, even with the resource you provided me with, I can't conclude why the mathematical statement stays intact...
Thanks very much anyway! :)

 

[DrDu]

Mathematically, the Bohr Sommerfeld quantization rule can be derived using the asymptotic WKB approximation to the Schroedinger equation. From this one would expect the energy levels to come out right for high principal quantum numbers n. That this quantization is in fact exact for all n is quite a peculiarity of the hygrogen problem. In the quantization of other systems, the Bohr Sommerfeld quantization is usually not exact.