The Bohr's radius

MonsieurWise

I know that the Bohr theory is like "Santa Claus" in Physics, but is his way of calculating radius right? Because I've made some calculations for ground state of neutral Helium and singly ionized Lithium using the Bohr's radius. What I did for Helium was to take the energy level of singly ionized Helium (negative sign) add the Coulomb repulsive energy between 2 electron (because Helium neutral has 2 electrons) in the same shell of Bohr radius. I did almost a similar thing to singly ionized Lithium. The ground state energy I've got was surprisingly close to NIST database.
So can I say Bohr was right at least at the radius of the s-orbital?

alxm

The Bohr radius is considered 'correct' for Hydrogen, in that his radius corresponds to the radius of maximal electron density in quantum theory (which inspired Bohr's 'theory of equivalence' between (semi-)classical and quantum results). The Bohr radius is also very nearly correct (using that definition) for Helium.

MonsieurWise

Wow...this makes me so happy :D. I was surprised with such a random idea I've got. thank you ^^.

Matterwave

I think the point is there is no "radius" at which the electron "orbits". The electron doesn't really make circular orbits around the nucleus, there is only a wavefunction. But the Bohr's radius does correspond to the maximum of the probability density in a Hydrogen atom as alxm suggests (though not the expectation value). (IIRC)

MonsieurWise

yeah...base on my result, I would say that Bohr was right at least at the radius of the probability density in a Hydrogen atom, but for higher state of n, using the same approach, the Bohr radius failed miserably: adding the repulsion energy of 2 electrons to the Bohr's energy level, I got positive answer (when it supposed to be negative...). So I say electron in higher n does not travel in circular orbit, which mean the Bohr's assumption was wrong.
Does the Bohr radius have the same value with the maximum of the probability density in an excited atom, at least for the s orbital (the s because the wavefunction look kinda circular)? My guest is no, but I'm not that sure...
Thanks!

edguy99

For this type of calculation, Mosley's law may be of interest to you.

MonsieurWise

Oh...thanks!!!