What made Bohr quantise angular momentum and not some other quantity? Bohr's second postulate in Bohr model of hydrogen atom deals with quantisation of angular momentum. I was wondering, though: why did he quantise angular momentum instead of some other quantity?  
 A: As dmckee commented, angular momentum is barely mentioned in Bohr's revolutionary 1913 paper "On the Constitution of Atoms and Molecules" (Philos. Mag. 26 , 1).
Instead, Bohr's bases his argument on Planck's hypothesis that the radiation from a quantum harmonic oscillator "takes place in distinctly separated emissions, the amount of energy radiated out from an atomic vibrator of frequency" $\nu_o$  "in a single emission being equal to" $nh \nu_o$, with $n$ a whole number and $h$ Planck's constant.  (I've changed some of Bohr's notation to conform with modern usage.)
Now a harmonic oscillator has the same frequency no matter what its energy, but not so a hydrogen atom.  
After calculating the relation between frequency $\nu_c$ and ionization energy $W$ for a classical closed orbit of semi-major axis $a$, charge $e$, and electron mass $m$:
$$ \nu_c = \frac{1}{\pi} \sqrt{\frac{2}{m}} \frac{W^{3/2}}{e^2} \quad , \quad 2a = \frac{e^2}{W}$$
Bohr offers two approaches:


*

*Consider an electron starting at rest, far from the atom, and ending in a stable closed orbit (itself a bold assertion, since classically there are no such orbits).  Since the starting "frequency" is 0, Bohr splits the difference and postulates that the radiated emission's frequency $\nu_r$ is half the frequency of the end orbit $\nu_c$.  Setting $W=h \nu_c/2$ (the energy radiated away), Bohr derives explicit expressions for the end orbit characteristics.

*As an alternative, Bohr considers transitions between two nearly classical orbits of very low frequency, where the start and end orbit frequencies are nearly identical.  This situation is very similar to the harmonic oscillator, so the Planck hypothesis can be applied directly:  the radiation frequency equals the orbit frequency.  The results are consistent with the first approach.


Only then does Bohr note: "While there obviously can be no question of a mechanical foundation of the calculations given in this paper, it is, however, possible to give a very simple interpretation of the result", namely the stable circular orbits have quantized angular momentum $L=n \hbar$.
By 1918, in "On the Quantum Theory of Line-Spectra", Bohr has adopted the technique of adiabatic invariants to demonstrate that the action $I$ of a hydrogen atom is quantized just like the harmonic oscillator: $I=nh$.  He notes that "this condition is equivalent with the simple condition that the angular momentum of the particle round the centre of the field is equal to an entire multiplum of $h/(2 \pi)$. 
I think "multiplum" = "multiple".
It took awhile to appreciate the fundamental nature of quantized angular momentum.
A: Bohr postulated that electrons orbit the nucleus in discrete energy levels, and electrons can gain and lose energy by jumping between energy levels, giving off radiation of frequency $\nu$ according to the formula
$$\Delta E = E_2 - E_1 = h\nu$$
where $\nu = \frac{1}{T}$, where $T$ is the period of orbit, as in classical mechanics.
Now during the transition, let $r$ be the average radius and $v$ be the average velocity of the particle.  Making such a simplification allows us to calculate the period of orbit:
$$T = \frac{2\pi r}{v}$$
Therefore,
$$\Delta E = h\nu = \frac{hv}{2\pi r} \tag 1$$
Also, we know the kinetic energy at a particular energy level is given by
\begin{align}
\text{K.E.} & = \frac{mv^2}{2} = \frac{Lv}{2r}, \quad \text{so therefore} \\
-U & = 2KE = \frac{Lv}{r}
\end{align}
Again, taking $r$ and $v$ to be the average radius and velocity during the transition, we get
$$\Delta E = \frac{(L_2 - L_1)v}{r}. \tag 2$$
Equating $(1)$ and $(2)$ gives
$$\frac{(L_2 - L_1)v}{r} = \frac{hv}{2\pi r}.$$
Therefore,
$$L_2 - L_1 = \frac{h}{2\pi} = \hbar$$
Therefore, each energy level differs from the next by an angular momentum of $\hbar$.  It is therefore reasonable to postulate that if the lowest energy level has no angular momentum, then each energy level from then on has an angular momentum of $n\hbar$ where $n$ is an integer.

Below is the modern de-Broglie method:
From the definition of angular momentum,
$L = rp$, where L is angular momentum, r is radius of orbit and p is momentum.
We also know that momentum is related to wavelength of a particle from the de-broglie relation:
$$p = \frac{h}{\lambda}.$$
Combining these gives
$$L = \frac{rh}{\lambda}.$$
Ok, now let us consider an electron orbiting a nucleus.  
The circumference of the orbit is $2\pi r$, and because we want the electron to form a standing wave orbit, we require that $\frac{2\pi r}{\lambda}$ be an integer, in order for the wave not to interfere with itself.  That is,
$$\frac{2\pi r}{\lambda} = n,$$
where $n$ is some integer. Now we can substitute our definition of $L$ from above into this equation to give:
$$\frac{2\pi L}{h} = n$$
and re-arranging gives,
$$L = \frac{nh}{2\pi} = n\hbar$$
Therefore, quantising angular momentum allows for the electron wave to not interfere with itself during orbit.
A: Angular momentum is a conserved quantity (in a closed system) and this is true also for the angular momentum that is carried by the electromagnetic (EM) field.  This conservation is a manifestation of rotational symmetry and the azimuthal part of the EM field emitted must be single valued.  In other words, when rotating the EM field in the azimuthal ($\phi$) direction (perpendicular to the $z$ direction) an integer number $n$ of $2\pi$ (360 degrees) it must have the same value as before the rotation.  This necessary single-valuedness after a rotation of $2n\pi$ can only be realised if the angular momentum attains only integer values $n$, i.e. if the angular momentum is quantized (at the classical level!).
A: This is an answer for why there is a constant n involved in the angular momentum classically.
$$F_c=\frac{1}{r}mv^2=k\frac{q^2}{r^2}=F_p$$
$$L=rmv  $$
$$r=k\frac{q^2}{mv^2}=k\frac{q^2}{m(\frac{L}{mr})^2}$$
$$r=\frac{L^2}{kq^2m}$$
$$\frac{1}{2}mv^2=\frac{1}{2}m\omega^2r^2=\frac{1}{2}m\omega^2(\frac{L^2}{kq^2m})^2=\frac{1}{2}k\frac{q^2}{r}$$
$$\omega^2rL^4=mk^3q^6$$
After rearranging
$$r^3v^6=\frac{k^3q^6}{m^3}$$
$$rv^2=\frac{kq^2}{m}$$
This shows that  r must be multiplied by $n^2$ and v must be divided by n. 
This implies that L can have n as a variable if we have the mass known. So $L=mr_0v_0n$ where $r_0$ and $v_0$ is when n=1. But why n must be integers it does not say. We have that n=1 can be arbitrary so it can be set to n=1 when we are at the bohr radius.
