# A Little More About the Bohr Atom Model (Additional Article)

In the eighth part of the series of articles on the development of quantum physics, I tried to explain the Bohr model of the hydrogen atom (the “planetary model”) in plain terms. However, it isn’t overly challenging to understand some simple, but quite neat, mathematics behind it. Therefore, I’d like to introduce the interested reader to a different and probably more deepening view of the Bohr model.
This article is more mathematical and technical than the vast majority of my other ones. If you don’t like equations and stuff, you’ll presumably like to read some other articles of my blog first. Anyway, I’ll try to keep this article as simple as possible (and as you’ll notice, the mathematics isn’t that hard to comprehend)!

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In the Bohr model (Niels Bohr, 1885-1962), an electron orbits the nucleus, which is just one proton in case of the hydrogen atom.

Here are the quantities which we’ll use below:

Mass of an electron ….. $m_e$
Speed of the electron ….. $v$
Radius of the electron’s orbit ….. $r$
Mass of the nucleus (generally) ….. $m_K$
Nuclear charge (generally) ….. $+Z\cdot e$ (Atomic number $Z$ times the elementary charge $e$)

Instead of describing the realistic situation in which both the electron and the nucleus circle around a common barycenter, we simplify the circumstances by making use of a common approximation: We assume that the nucleus is standing still while an electron particle of a reduced mass $\mu=m_e\cdot m_K / (m_e+m_K)\approx m_e$ is orbiting the center of the Coulomb potential in the nucleus (at $r=0$). Therefore, we have to describe a system in which only one component is moving, instead of two. This approximation can indeed be justified by pointing out that the nucleus is much heavier than the electron.

The force acting radially towards the center of the atom (the centripetal force) is the Coulomb force. Thus, centripetal force equals Coulomb force. Or mathematically,

$-\frac{\mu v^2}{r}\hat{e}_r = -\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2}\hat{e}_r$

(“$\hat{e}_r$” is the unit vector in the radial direction.)

By rearranging the equation, we get an expression for the radius of the circular orbit (the “orbital”):

$r=\frac{Ze^2}{4\pi\varepsilon_0\mu v^2}$

This radius $r$ can attain arbitrary values since there are no constraints for the (kinetic) energy $\mu v^2 / 2$ of the electron …yet.

However, let’s describe the electron in terms of its matter wave $\psi$. In doing so, we additionally impose requirements on it: We demand that the states of the atom are stable over time (or “stationary”), which means that in our model, the electron is neither allowed to leave the atom nor fall into the nucleus. Such stationary states can only be realized by standing waves. (See the main article on the Bohr model!)
Thus, we need the circumference of our circular orbit to be an integral multiple of the de Broglie wavelength $\lambda_D$.

Written in mathematical terms, our demand looks as follows:

$2\pi\cdot r = n\cdot \lambda_D$

The $n$ stands for “integral multiple” and can only be a natural number ($n=1,2,3, \dots$).

The de Broglie wavelength is $\lambda_D = h / (\mu\cdot v)$. After plugging this into the formula above, we rearrange the equation in order to get an expression for the speed $v$:

$v=n\cdot\frac{h}{2\pi\mu r}$

Now, let’s insert the speed $v$ in the above expression of the radius $r = (Ze^2) / (4\pi\varepsilon_0\mu v^2)$. After repeated conversions, we obtain a new kind of radius:

$r=\frac{n^2 h^2 \cdot\varepsilon_0}{\pi \cdot\mu \cdot Z \cdot e^2}=a_0 \cdot\frac{n^2}{Z}$

We see that the radius is just some constant value $a_0$ multiplied by $n^2/Z$.
The constant value $a_0$ is called Bohr’s radius and is

$a_0=\frac{\varepsilon_0h^2}{\pi\mu e^2} = 5.2917\cdot 10^{-11}~\text{m} \approx 0.5 \mathring{A}$

The Bohr radius is the smallest possible ($n=1$) value of the electron’s orbit in the hydrogen atom ($Z=1$).

But wait – there’s a $n$ in our formula for the radius! What’s the meaning of this?
Well, what happened here is a quantization of the electron’s circular orbits. The $n$ originates from the requirement on the “length” of the electron paths to be an integral multiple of the wavelength of the electron and it persists up to this point. $n$ can attain discrete values only ($n=1,2,3,\dots$) – values “in between” aren’t possible, because otherwise the atom in our model wouldn’t be stable. This results in discrete radii $r$, which, too, can’t have “values in between”. The radii of the allowed electron paths are therefore quantized!

Not only the radii are quantized, but also are the electron’s energies!
After tinkering around with our first equation (centripetal force = Coulomb force), we are able to find an expression for the kinetic energy of the electron. All we have to do is to rearrange the equation such that $\mu v^2/2$ is on one side of the equals sign, which is just the kinetic energy. Then, on the other side we get:

$E_\text{kin}=\frac{\mu v^2}{2}=\frac{1}{2}\cdot\frac{Ze^2}{4\pi\varepsilon_0 r}=-\,\frac{1}{2}E_\text{pot}$

Thus, the kinetic energy equals half the negative potential energy. (Surely, you might need some physical-mathematical practice to immediately identify the potential energy on the right side of the equation. But after all, no magic is involved in this step. 😉 )
Now, let’s put together the different energies in order to calculate the total energy of the electron:

$E=E_\text{kin}+E_\text{pot}=-\,\frac{1}{2}\cdot\frac{Ze^2}{4\pi\varepsilon_0 r}$

We substitute the radius $r$ by

$r=\frac{n^2h^2\cdot\varepsilon_0}{\pi\cdot\mu\cdot Z\cdot e^2}$.

(We’ve already deduced this expression earlier.)
In doing so, the overall energy becomes dependend on $n$ too:

$E_n=-\,\frac{\mu e^4\cdot Z^2}{8\varepsilon_0^2h^2\cdot n^2}=-Ry^\ast\cdot\frac{Z^2}{n^2}$

We can put all the constants together and write another single constant instead. In fact, we already did that in the last step above. The new constant $Ry^\ast$ is called Rydberg constant.

$Ry^\ast=Ry\cdot h\cdot c=\frac{\mu\cdot e^4}{8\varepsilon_0^2 h^2}$

Once again, we saw something getting quantized – this time, it was the electron’s energy. All these quantizations are represented by the $n$ ocurring in the formulas. Consequently, the electron can only have discrete values of energy. Again, values in between aren’t possible.
We can simply number the energy values consecutively by plugging in a natural number for $n$ ($n=1,2,3,\dots$).
The $n$ in the above formula represents a quantum number, in fact it stands for the “energy quantum number”, a number which specify the number of periods of the standing de Broglie waves on the circle’s perimeter.

Incidentally, the boundary condition that the de Broglie wave forming around the nucleus has to be a standing wave can just as well be replaced by the condition that the angular momentum of the electron has to be quantized.
(In doing so, one has to multiply both sides of the equation

$v = n\cdot\frac{h}{2\pi\mu r}$

with $\mu\cdot r$. For the left side of the equation we thus obtain $r\cdot\mu v$ which is just the absolute value $L$ of the angular momentum. The right side yields $n\cdot h / (2\pi)$, which we can write as $n\cdot\hbar$.
Hence, the absolute value of the electron’s angular momentum is quantized too and it can only be an integral multiple of $\hbar$. $L=n\cdot\hbar$.

Well, maybe you also like to see some tangible numerical values:
The lowest energy ($n=1$) of the electron in the hydrogen atom ($Z=1$) is $E_1 = -Ry^\ast = -13.6~\text{eV} = -E^\text{ion}$. In order to move the hydrogen’s electron from the ground state to infinity (speaking in terms of the “real world”, “infinity” just means “very far away”), 13.6 eV of energy have to be spent. We call this process ionization and consequently, the corresponding energy is the ionization energy.

There are many other mathematical spin-offs (the derivation of the Balmer series for the hydrogen spectrum or the derivation of the Bohr radius from the Heisenberg Uncertainty Principle, …), but maybe here’s the point where I should come to an end.

Hopefully, you aren’t completely confused by this more or less mathematical description of the Bohr model! But if you are, don’t worry, that’s not bad. Maybe you were able to take something along even in this case. In my opinion, the point in doing physics hasn’t always to be to develop an understanding right down to the last detail. Instead, it often suffices to grasp some impressions of the theories, the models, or nature and therefore being able to ask further questions about our world. The most important thing is to stay curious and (like Ines Lalowa once told me) say things like “…I really enjoy being amazed“.

(As a last point, I probably should note again that the Bohr model of the hydrogen atom – although being a relatively intuitively accessible model – isn’t able to describe all the experimental observations. It cannot be the “real” picture of an atom. See the main article for further informations.)