# In the Bohr model, how do you derive that the angular momentum of an electron in a stationary state is a quantized integral value of $h/2\pi$?

I understand that Bohr postulated that electrons can only occupy orbits at certain radii, and that in order to move from one orbit (or stationary state) to another, it would have to absorb or emit a quantum of energy from a photon. This would lead to the fact that the change in energy between stationary states is $$\Delta E = E_2 - E_1 = hf,$$ where $h$ is Planck's constant and $f$ is the frequency.

And though I conceptually understand how this would also imply that the angular momentum of and electron is also quantized, I can't quite derive how the angular momentum is an integer multiple of $h/2\pi$.

It made sense to me, but I did not have the reputation points in order to comment and ask a follow up question. That is why I am posting here. In the top ranked answer by Kenshin, it reads:

$$\Delta E = h\nu = \frac{hv}{2\pi r} \tag 1$$

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

$$\frac{(L_2 - L_1)v}{r} = \frac{hv}{2\pi r}.$$

My question is why is potential energy, $U$, used to find the $ΔE$? Why not kinetic energy, or why not total energy?

• Notice that the electrical force is attractive here and the potential energy is actually negative, so if we use the magnitude of the physical quantities, for circular trajectories we actually have $U=-\frac{Lv}{r}$. This sign is ultimately not too relevant for the conclusion because of the relation $E=U+K=-2K+K=-K$. Hence, you might just get a sign shift but still reproduce the result you were looking for. Feb 1 '18 at 10:55
• And a factor of 2, with respect to that answer. Still quantization of $L$ follows. Feb 1 '18 at 11:06

As I mentioned in the comments, the energy in circular trajectories for electrostatic Coulomb attraction satisfies $E = U + K = - 2 K + K = -K$. More explicitly the energy takes the form \begin{equation} E = \frac{m v^2}{2} - \frac{k}{r} \, , \end{equation} and the condition $U=-2K$ can be seen as the constrain $k = mr v^2$. Now if we plug this value of $k$ and follow the same reasoning that is suggested in the question, that involves assuming average velocities and positions between the states, we get quantization of $L$ but in the form $\Delta L = 2 \hbar$. Whether we can consider those assumptions valid is up for debate but frankly I will also make somehow similar assumptions of my own so pick your poison.
I propose the alternative of using $k = mr v^2$ a posteriori. Consider a jump that will manifest as $r \rightarrow r + \epsilon \Delta r$ and $v \rightarrow v +\epsilon \Delta v$. Then we find at first order in $\epsilon$, \begin{equation} L = m r v \rightarrow L'= mrv + \epsilon \, m \left(v \Delta r + r \Delta v\right) \, . \end{equation}
Let's do the same with the energy. We have a first order in $\epsilon$, \begin{equation} E \rightarrow E' = \frac{m v^2}{2} - \frac{k}{r} + \epsilon \left(\frac{ k}{r^2} \Delta r + m v \Delta v\right) \, . \end{equation} Now if we use $k = mr v^2$, we have \begin{eqnarray} E \rightarrow E' &=& \frac{m v^2}{2} - \frac{k}{r} + \epsilon \frac{1}{r} \left(m v^2 \Delta r + m v r \Delta v\right) \\ &=& \frac{m v^2}{2} - \frac{k}{r} + \epsilon \frac{v}{r} \, m \left( v \Delta r + r \Delta v\right) \, . \end{eqnarray} By comparing $\Delta L = L' - L$ and $\Delta E = E'- E \,$ we find the desired $\Delta E = \frac{v}{r} \Delta L$ , which using Bohr quantization rule leads to $\Delta L = \hbar$.