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It is well known that Bohr model is not totally right. But I recently discovered a very curious inconsistency (if I am right) which I haven't seen explained anywhere.

The first postulate of Bohr theory is that the Orbital momentum of the electron is quantized $L = mvr=nh$ (where $h$ mean the Dirac constant). This means that if there is a transition between level $n=5$ to $n=1$ (Balmer series) the orbital momentum changes by $4h$!!! Based on this rule and his second postulate Bohr finds the right energy for this transition (and all others as well). But this transition is a release of just one photon and a photon has spin $1h$. $(h,0,-h)$. So it can change the orbital momentum with $1,0$ or $-1$ and not by $4$. Maybe the spin of electron can also change with $1$ (from $\frac{1}{2}$ to -$\frac{1}{2}$) which combined changes orbital momentum by most 2. I'm very surprised to make such conclusion.

I am wrong here?

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closed as unclear what you're asking by ACuriousMind, Kyle Kanos, Norbert Schuch, Daniel Griscom, Mark Mitchison Jan 4 '16 at 1:26

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  • $\begingroup$ 2 days and no refutation. If you can not refute the statement I appreciate if someone admits there is something wrong in Bohr model. $\endgroup$ – Mercury Nov 28 '15 at 21:02
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    $\begingroup$ What is commonly called the "Bohr model" doesn't even include spin - it just has the angular momentum of supposedly orbiting electrons discretized. Why are you surprised to get some strange results when you combine reasoning about spin with a model that doesn't include spin? $\endgroup$ – ACuriousMind Jan 3 '16 at 22:07
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    $\begingroup$ Perhaps I'm misunderstanding something here, but isn't it known that the Bohr model is faulty? So what's the big deal about finding out that it is in fact faulty? $\endgroup$ – Kyle Kanos Jan 3 '16 at 22:45
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Be careful with the formulation of your question (and comment), see here why: http://math.ucr.edu/home/baez/crackpot.html.

Now for your question. Electronic transitions in atomic hydrogen such as the Lyman series that you mention (Balmer involves $n=2$ as lowest state) are associated with a change in the principal quantum number $n$. The principal quantum number is somehow associated with the size of your electron orbital. You apply the quantization of orbital angular momentum to an orbital with fixed radius (particle on a ring). For the orbital angular momentum, typically denoted as $\ell$, the selection rules with $\Delta\ell=\pm 1$ that you mention do hold. However $n$ does not represent an angular momentum, so the selection rules are different. So basically you got confused by using the symbol $n$ to represent orbital angular momentum.

Note in addition that Bohr considered selection rules by expressing the classical motion of the electron in a Fourier series and allowing only transitions between levels that have identical Fourier components.

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    $\begingroup$ Yes n does not represent angular momentum in QM. There it is l=n-1,....,0. But not so in the Bohr model. In his model Bohr definitely derives the energy of the electron $E=R/n^2$ from angular momentum $L=n\hbar$ of THAT same electron. So we have an electron with momentum n and energy $E=R/n^2$. Is this momentum $L=n\hbar$ which starts the calculation of E coming out from nowhere? $\endgroup$ – Mercury Dec 2 '15 at 22:42
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    $\begingroup$ Well, since you invoke the angular momentum of the photon that was discovered 20 years after Bohr published his model, I hope you also allow me to make use of what we know now. Of course the principal quantum number and maximum value for the orbital angular momentum are strongly connected, so that from the quantization of the orbital angular momentum by Bohr you can determine the corresponding principal quantum number. $\endgroup$ – Paul Dec 3 '15 at 0:55
  • $\begingroup$ Nevertheless this an inconsistency of the model. QM proves the model is not exact. But this a more obvious prove anyway. $\endgroup$ – Mercury Dec 4 '15 at 15:20
  • $\begingroup$ QM proves of course that the Bohr model is not exact. But from todays viewpoint this here is more obvious and simple. By the way Bohr could easily notice that his radiation can take any angular momentum k*hbar. k any number. $\endgroup$ – Mercury Dec 4 '15 at 15:34
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As far as Bohr was concerned, or could possibly know at the time, there was no such thing as angular momentum selection rules for radiative transitions, so the model isn't "wrong" as such - you're comparing it to a much higher standard than it can possibly accommodate. The model is not inconsistent, because it does not contain any contradictory statement within it - it says nothing of the angular momentum of radiation.

I should also point out at this point that the selection rule $\Delta l=1$ is true only for dipole transitions, but that this breaks down when you include the effect of the spatial variation of the wave. One can have quadrupole transitions for which $\Delta l=2$ or $0$, and even octupole transitions with $\Delta l=3$. (For $\Delta l=4$ you need a hexadecapole transition, which is still possible in principle, though I'm unsure if it's been observed in atoms.) These lines come from very small couplings, which means that they're very weak and very narrow, so you need to be actively looking for them to find them, but that doesn't mean they're not there. For some examples see e.g. this paper.

Finally, you're taking the most restrictive version of the Bohr model, without even trying to take it in its full generality. More specifically, you're restricting yourself to circular orbits, which corresponds in the full quantum theory to the case $|m|=l$, without allowing for the states with $|m|<l$ - or, in other words, for elliptical orbits. As it happens, the Bohr model can be expanded to include elliptical orbits, via the use of Bohr-Sommerfeld quantization. Nowadays that's more of a historical curiosity than anything else, because the full quantum theory followed it within ten years or so, but it's the correct generalization. Once you include it, it becomes perfectly possible to make a transition from $n=1$ to $n=5$ while only changing angular momentum by $\Delta l=1$.

Taking the Bohr model, confronting it with angular momentum selection rules, and not giving it full rein in how it treats its own internal angular momentum could be called intellectually dishonest in some places. So no, you haven't proved Bohr wrong, and you should take it easy with comments like this one.

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  • $\begingroup$ In your efforts to defend Bohr model (or Bohr itself?) you in fact confirm my observation. For more info look at the following answer. $\endgroup$ – Mercury Jan 3 '16 at 19:36
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    $\begingroup$ I'd recommend you read the John Baez article that Paul linked to. As it is, you have an opinion and you have given all the indications that you're not prepared to change your mind so I'm not really going to argue with you. As I said already, criticising the Bohr model over its angular momentum behaviour without extending it to Bohr-Sommerfeld quantization is like bashing the restricted three-body problem for not conserving energy - it's just short of dishonest. $\endgroup$ – Emilio Pisanty Jan 3 '16 at 23:24
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    $\begingroup$ And if you're proud you've discovered that the Bohr model is not the whole truth, there's this Schrödinger guy you should read about. In any case, good luck with your opinions, and getting them taken seriously by the scientific community. $\endgroup$ – Emilio Pisanty Jan 3 '16 at 23:24

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