alxm said:
Bohr's model was bad experimentally because it did not reproduce the fine or hyperfine structure of electron levels.
Bohr's model was bad theoretically because it didn't work for atoms with more than one electron, and relied entirely on an ad hoc assumption about having certain 'allowed' angular momenta.
Quantum mechanics has completely replaced Bohr's model, and is in principle exact for all atoms. While relying on sounder assumptions. It also explains a huge amount of other things. (like why Bohr's model was a good approximation)
alxm, the fine structure was first named by Sommerfeld, and it meant the relativistic effect(by the difference of the relativistic mass).
Later, the fine structure was replaced by the spin-orbital interaction.
But accidentally, these values
coincided.
In the hydrogen atom, there are more coincidences when you consider the spin-orbital interactions.
In page 167 Atomic physics by Max Born
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The case of hydrogen is
peculiar in one respect. Experiment gives distinctly
fewer terms than are specified in the term scheme of fig 9; for
n=2 only two terms are found, for
n=3 only three, and so on.
The theoretical calculation shows that here (by a mathematical coincidense, so to speak) two terms sometimes coincide, the reason beeing that the relativity and spin corrections partly compensate each other. It is found that terms with the same inner quantum number j but different azimuthal quantum numbers l always
strictly coincide.
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In QM, the Dirac equation doesn't mean the probability amplitude, and the uncertainty principle(HUP) needs the probability density. This means HUP and the relativistic correction (by the Dirac Eq.) are inconsistent? I have not yet found the clear answer.
For multi-electron atoms, for example, the 3s electron of the Sodium D line comes close to the nucleus through the inner electrons.
But the Lande-g-factor does not contain the influence of the inner electrons at all. The delicate spin and orbital precession is so stable to stand the influence of the inner electrons? (See also
this thread.)
About the nucleus,
also in the Bohr model, considering the nuclear mass and movement, if we use "the reduced mass" of an electron, the better results are obtained.
In QM, The orbital angular momentum of the S state electron is zero.
This means the electron is scattered or trapped by the nucleus each time it hits the nucleus? If so, can this keep the two electrons of Helium stable?
There are some experiments using the probability density of the electron at the nucleus (\mid\psi_{1s}(0)\mid^2 = 1 / \pi r_{0}^3). r_{0} is the Bohr radius).
In the Bohr model, the similar value exists. The magnetic field at the nucleus created by the S state electron is proportional to 1 / r_{0}^3.(The Story of Spin).
The spin has the strange property such as the "two-valued" rotation.
So you mean "
What is real" is the strange thing which goes back to its original form by the 4pi rotaion (not 2pi)?
In QM, whether you measure or not,
the probability density of the hydrogen S state electron near the point at infinity is
not zero. And this electron has the ground state nergy. But why can it have the ground state energy which sign is minus? (though the potential energy is almost zero near the point at infinity.)
