quantum mechanics - 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?

Answer

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.

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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.