Possible inconsistency in Bohr atom model 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? 
 A: 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.
A: 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.
