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As known, the hyperfine splitting $Δν = 2M_eB/h$ (where $h$ is the Planck’s constant) of all $S$ spectral terms and $P$, $D$, $F$ fine subterms in hydrogen line spectrum appears because in all allowed states of hydrogen atoms, the atomic electrons have always their magnetic moment $M_e$ aligned exclusively parallel or antiparallel to the local magnetic induction $B$ originated by nucleus-proton, and these opposite orientations mean two different energy levels of the electrons.

Therefore, as both hyperfine splits of $1S$ and $2S$ terms in hydrogen spectrum have very exact values measured experimentally, $Δν_{1S} = 1.420406 GHz$ and $Δν_{2S} = 0.177556 GHz$, it results similarly exact values for nuclear magnetic induction $B = hΔν/2M_e$ experienced any time by atomic electrons both in $1s$ and in $2$s atomic state of hydrogen atom. Or, such extremely exact and constant values of nuclear magnetic induction experienced by atomic electrons is possible only if both in $1s$ and $2s$ atomic states these atomic electrons moves on perfectly circular orbits (around the central proton) placed in the symmetry plane $Oxy$ of the dipole magnetic field with axial $Oz$ symmetry originated by proton. This is an elementary and evident condition. But perfectly circular orbits of the electron are physically impossible in the Bohr’s hydrogen atom, formed through proton-electron electric attraction acting as a central force, and they are all the more improper for the hydrogen atom of quantum physics, where even the concept of orbital radius is “void of meaning”.

More, as the hyperfine split of $1S$ term is exactly eight times larger than that of $2S$ term, $Δν_{1S}/ Δν_{2S} = 8$, and the dipole magnetic field originated by proton observes a law in $1/r^3$, it results an orbital radius of atomic electron in $2s$ state exactly twice larger than its orbital radius in $1s$ state, ${r_{2}/r_{1} = \sqrt[3]8 = 2}$, and because the linear velocities of atomic electron in the two states are in the inverse ratio $v_{1}/v_{2} = 2$, it results the same orbital angular momentum of atomic electron in the two states, $mv_{1}r_{1}= mv_{2}r_{2} = ћ$, which contravenes the Bohr’s postulate $mv_{n}r_{n}= nћ$, where $n =1,2,\ldots$ . Or, this experimentally disproved Bohr’s postulate was taken over as such by quantum mechanics of hydrogen atom, and the quantified orbital angular momentum of atomic electron in hydrogen atom is a concept still valid even in modern atomic theories.

In these circumstances my question is: how should these discrepancies between experimental hyperfine splits and atomic theories be interpreted?

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closed as unclear what you're asking by Emilio Pisanty, Michael Seifert, Daniel Griscom, Mitchell, Asher Nov 12 '17 at 3:24

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    $\begingroup$ "Orbital radius", "v" and "r" are void of meaning in quantum mechanics. Atomic theories are built upon the general quantization rules which lead to Schroedinger, Pauli or Dirac equations. $\endgroup$ – DanielC Oct 11 '17 at 8:11
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    $\begingroup$ If you're trying to interpret this in the context of the Bohr model I doubt that will be successful. $\endgroup$ – John Rennie Oct 11 '17 at 10:23
  • $\begingroup$ @EmilioPisanty Now I hope that my question is entirely understandable, and consequently you will reopen it in order to allow all the interesteds to answer it. Sorin Vlaicu $\endgroup$ – Sorin Vlaicu Nov 16 '17 at 7:22
  • $\begingroup$ You've just added a second incorrect premise on top of the initial one. There are no discrepancies between experiment and theory on hyperfine structure - hydrogen spectroscopy is one of the most precise branches of physical science, with experiment and quantum mechanics agreeing to a full fifteen significant figures or more, requiring much finer corrections than just hyperfine structure. Moreover, you're still using orbital radii after acknowledging that they are in fact a useless concept in this context - I don't see how you can think of that as consistent. $\endgroup$ – Emilio Pisanty Nov 16 '17 at 7:48
  • $\begingroup$ As for reopening, any five site members with 3k rep can reopen this thread, and it is currently under review. I personally see no reason why this should be reopened (it's worse than it was before) but it's the community vote that decides. $\endgroup$ – Emilio Pisanty Nov 16 '17 at 7:53
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Any discrepancies that you get from using the Bohr model have a simple explanation:

  • The Bohr model is incomplete, unable to fully explain experiment, and should not be used to explain any aspect of atomic structure beyond the ones where it explicitly agrees with full quantum mechanical calculations.

From a quantum mechanical perspective, your argument mostly reduces to gibberish: as a simple example, you just can't say things like "the orbital radius of atomic electrons in 2s state is exactly twice larger than their orbital radius in 1s state" because there are no well-defined orbital radii for electrons in any energy eigenstates.

Instead, all atomic orbitals have a spread of possible orbital radii (in fact, all orbital radii $r\in(0,\infty)$ are possible) with some probability distribution $|\psi(r)|^2$ over them. You can indeed define an expected or average radius, $$ \langle r\rangle = \int_0^\infty r|\psi(r)|^2\mathrm dr, $$ but you can't use that average radius to compute other expectation values: $$ \langle f(r) \rangle = \int_0^\infty f(r)|\psi(r)|^2\mathrm dr \neq f\left(\int_0^\infty f(r)|\psi(r)|^2\mathrm dr \right) = f(\langle r \rangle ). $$

If you want to calculate hyperfine splittings correctly, you need to do it within full quantum mechanics.

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  • $\begingroup$ Still, the very precise values of the hyperfine splits measured for spectral terms in hydrogen spectrum (up to seven significant digits and with no spectral width) prove the very precise and constant nuclear magnetic induction experienced by atomic electrons in each atomic state of hydrogen atoms. Or, this is possible only if inside the nuclear dipole magnetic field the electrons move exclusively on perfectly circular orbits placed in the Oxy plane of this field, as long as the radiating electrons can emit their transition photons from any point on their orbits. $\endgroup$ – Sorin Vlaicu Oct 15 '17 at 3:18
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    $\begingroup$ @SorinVlaicu Electrons don't move in orbits. You're using naive, classical intuition to approach a fully quantum mechanical problem. $\endgroup$ – J. Murray Oct 15 '17 at 4:15
  • $\begingroup$ @Sorin There's no such thing as electron orbits in an atom. Explanations based on them are bound to fail, period. Use full wavefunction-based QM instead. $\endgroup$ – Emilio Pisanty Oct 15 '17 at 10:16
  • $\begingroup$ @EmilioPisanty Please read R. A. Ferrell, Amer. J. Phys, 28, (1960) 484. I quote: “It is pointed out that the so-called “hyperfine interaction” between the magnetic moment of a nucleus and the spin s of the surrounding electrons is completely classical in origin. Quantum mechanics is irrelevant and not necessary to the understanding of this coupling.”, “Alternatively, if the nuclear magnetic moment has already been determined in some other way, then it can be used as a tool, via the hyperfine interaction, for studying the structure of the electron cloud in an atom, molecule, ... “, etc. $\endgroup$ – Sorin Vlaicu Oct 16 '17 at 6:12
  • $\begingroup$ @SorinVlaicu A few words to clarify the Ferrell 1960 paper you quoted: the arguments in that paper provide (at best) a classical justification for the operators used to describe the coupling. The notion that hyperfine splittings can be described purely with classical methods is laughable ─ without quantum mechanics, there is no quantized energy level to split to begin with. If you really wanted to be unscientific, then you could possibly patch Bohr's model to include it, but you'd get a useless contraption. $\endgroup$ – Emilio Pisanty Nov 16 '17 at 10:19

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