Bohr Model: Atomic Structure from the Perspective of Quantum Physics

by Alexander FufaeV

Table of contents

Earlier, it was believed that an atom is structured like a dough with raisins: positively charged dough that is evenly distributed and negatively charged raisins embedded in the dough. This model is called the Thomson model (or Plum pudding model).

Since the Thomson scattering experiment, we know that this model is incorrect. Rather, the positively charged matter of an atom is concentrated in a small place: The Rutherford model takes into account this experimental result and assumes that an atom consists of a positively charged nucleus with electrons orbiting around it on arbitrary paths.

But the Rutherford model has a crucial problem:

Why don't the electrons just fall into the nucleus?

From classical electrodynamics, we know that circularly moving charges (that is, accelerated charges) radiate energy. Therefore, the electrons should gradually lose their energy and fall into the nucleus. However, this contradicts the observations as electrons remain stable in their orbits around the nucleus. The assumption of the Rutherford model, where electrons orbit the nucleus on arbitrary classical circular or elliptical paths, cannot be correct. A new theory had to be developed to explain this phenomenon.

Niels Bohr developed a novel model that integrated the insights of the emerging quantum mechanics at that time. This model is now known as the Bohr model and is based on the following three assumptions:

Electrons move on fixed circular orbits around the atomic without radiating energy.
Electrons can jump between the individual orbits, thereby absorbing or releasing discrete amounts of energy.
The radii of the electron orbits are at specific distances from each other, so that the orbital angular momentum of the orbiting electron is always a multiple of the Planck's constant.

In the following chapters, we take a closer look at these assumptions.

Postulate #1: Fixed orbits without energy radiation

An electron moving in a circular orbit constantly changes its direction of motion. It does not move in a straight line but must constantly deflect around the nucleus. The changing direction of motion of the electron, represented by a velocity vector, results in acceleration of the electron. According to classical electrodynamics, an accelerated electron must lose kinetic energy and thus decrease its velocity.

The Coulomb force exerted on the electron by the nucleus acts as a centripetal force on the orbiting electron. The decrease of the velocity leads to the fact that the orbital radius of the electron becomes smaller. As a result, the Coulomb force between the electron and the nucleus increases as the two particles come closer together. This process would continue until the electron spirals into the nucleus. According to classical electrodynamics, an atom would thus be unstable.

To artificially solve this problem, Niels Bohr assumed something contradictory to classical electrodynamics: A negatively charged electron orbits the positively charged nucleus on a stationary path without losing energy. While the electron is attracted by the charged nucleus, this attraction does not cause the electron to fall into the nucleus.

This assumption of stationary orbits is completely inconsistent with classical physics and only consistent with the emerging field of quantum physics at that time.

Postulate #2: Discrete energy emission and energy absorption

Let us number the stationary orbits of the electrons with the number $ \class{red}{n} = 1,2,3,... $. We call this number (principal) quantum number.

We assign the quantum number $ \class{red}{n} = 1 $ to an electron moving in the circular orbit closest to the nucleus of the atom. Regarding the electron moving on the $ \class{red}{n} = 1 $ circular orbit, we say: The electron is in the ground state.
We assign the quantum number $ \class{red}{n} = 2 $ to an electron moving in the next circular orbit farther from the nucleus. Regarding the electron moving in the $ \class{red}{n} > 1 $ circular orbit, we say: The electron is in an excited state.
And so on.

Now, if an electron moves around the nucleus on a stationary orbit with quantum number $ \class{red}{n} $ and does not radiate any energy, then its energy, let us call it $ W_{\class{red}{n}} $, must always remain constant as long as the electron moves on the $ \class{red}{n} $-th circular orbit. The energy $ W_{\class{red}{n}} $ cannot take on arbitrary values, but only certain values. Wir sagen dazu: The energy of the electron in an atom is quantized.

The electron can jump from the $ \class{red}{n} $-th circular orbit to a $ \class{green}{m} $-th circular orbit or vice versa. We call this change of circular orbits a quantum leap. Here $ \class{green}{m} = 2, 3, 4, ... $ is also a quantum number numbering a larger circular orbit than $ \class{red}{n} $: $ \class{green}{m} > \class{red}{n} $.

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Here $ \class{red}{n} = 1 $ was chosen as an example. Then: $ \class{green}{m} > 1 $.

And now we come to the crucial point of Niels Bohr's second assumption: The energy difference $ \Delta W = W_{\class{green}{m}} - W_{\class{red}{n}} $ is not arbitrary but exactly corresponds to the energy $ h\,\class{violet}{f} $ of a photon with the frequency ( \class{violet}{f} \):

Second Bohr assumption: energy difference of two circular orbits

$$ \begin{align} h\,\class{violet}{f} ~=~W_{\class{green}{m}} - W_{\class{red}{n}} \end{align} $$

Here, $ h = 6.626⋅10^{-34} \, \mathrm{Js} $ represents the Planck constant, introduced by Max Planck several years ago. For an electron to transition from the $ \class{red}{n} $-th orbit to the $ \class{green}{m} $th orbit, it must absorb a photon of energy $ h \, \class{violet}{f} $.

If the energy of the photon, $ h \, \class{violet}{f} $, is less than the energy difference required for the quantum leap, $ W_{\class{green}{m}} - W_{\class{red}{n}} $, then the photon will not be absorbed, and the electron will remain in the $\class{red}{n}$th orbit.
If the energy of the photon, $ h \, \class{violet}{f} $, is greater than the energy difference required for the quantum $ W_{\class{green}{m}} - W_{\class{red}{n}} $, then the electron can transition to the $\class{green}{m}$th orbit. The excess energy, $ \Delta W $, can be emitted, for example, in the form of a photon with lower frequency.
If the energy of the photon, $ h \, \class{violet}{f} $, is significantly greater than the energy difference required for the quantum leap, $ W_{\class{red}{m}} - W_{\class{red}{n}} $, it can even ionize the atom, meaning it can eject the electron from the atom.

Example: Transition of the electron in the Hydrogenium atom

The electron is in the ground state of the hydrogenium atom ($ \class{red}{n} = 1 $) with an energy of -13.6 eV. It absorbs a photon with energy sufficient to put the electron into the first excited state ($ \class{green}{m} = 2 $). The energy of the electron in the excited state is -3.4 eV. At what light frequency do we need to irradiate the H atom to put it into the first excited state?

To do this, rearrange Eq. 1 with respect to the frequency $ \class{violet}{f} $ and insert the energies and the Planck's constant:

$$ \begin{align} \class{violet}{ f } ~&=~ \frac{W_{\class{green}{2}} - W_{\class{red}{1}}}{h} \\\\
~&=~ \frac{-3.4 \, \mathrm{eV} + 13.6 \, \mathrm{eV}}{ 6.626 \cdot 10^{-34} \, \mathrm{Js} } \\\\
~&=~ \frac{ \left(-3.4 \, \mathrm{V} + 13.6 \, \mathrm{V}\right) \cdot 1.602 \cdot 10^{-19} \, \mathrm{C} }{ 6.626 \cdot 10^{-34} \, \mathrm{Js} } \\\\
~&=~ \class{violet}{ 2.47 \cdot 10^{14} \, \mathrm{Hz} } \end{align} $$

Postulate #3: Angular momentum is a multiple of Planck's constant

A stationary $ \class{red}{n} $-th circular orbit with a fixed radius $ r_{\class{red}{n}} $ and a fixed velocity $ v_{\class{red}{n}} $ of the electron on this orbit leads to a fixed orbital angular momentum $ L_{\class{red}{n}} $. The formula for angular momentum is known from classical mechanics:

$$ \begin{align} L_{\class{red}{n}} ~=~ \class{brown}{m_{\text e}} \, r_{\class{red}{n}} \, v_{\class{red}{n}} \end{align} $$

Here $ \class{brown}{m_{\text e}} $ is the mass of the electron. In the third postulate, Niels Bohr assumes that the angular momentum of the electron is a multiple of Planck's constant:

Third Bohr assumption: Quantized orbital angular momentum of the electron

$$ \begin{align} L_{\class{red}{n}} ~=~ \class{red}{n} \, \hbar \end{align} $$

The electron in the ground state thus has the smallest possible angular momentum:

$$ \begin{align} L_{\class{red}{n}} ~&=~ \class{red}{1} \cdot \hbar \\\\
~&=~ \class{red}{1} ~\cdot 6.626 \cdot 10^{-34} \, \mathrm{Js} \\\\
~&=~ 6.626 \cdot 10^{-34} \, \mathrm{Js} \end{align} $$

The electron in the first excited state has an orbital angular momentum twice as large:

$$ \begin{align} L_{\class{red}{n}} ~&=~ \class{red}{2} \cdot \hbar \\\\
~&=~ 13.25 \cdot 10^{-34} \, \mathrm{Js} \end{align} $$

And so on...

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Quantized Angular Momentum, Radius, and Velocity in the Bohr Model.

Substitute equation 3 for angular momentum into Bohr assumption 4 and bring $ \class{brown}{m_{\text e}} $ and $ v_{\class{red}{n}} $ to the other side of the equation where $ \hbar $ is. Then we can calculate the de Broglie wavelength $ \lambda_{\text{dB}} = h / \class{brown}{m_{\text e}} \, v_{\class{red}{n}} $ of the electron to make an interesting interpretation of circular orbits:

$$ \begin{align} r_{\class{red}{n}} ~&=~ \class{red}{n} \, \frac{ \hbar }{ \class{brown}{m_{\text e}} \, v_{\class{red}{n}} } \\\\
~&=~ \class{red}{n} \, \frac{ h }{ 2 \pi \, \class{brown}{m_{\text e}} \, v_{\class{red}{n}} } \\\\
~&=~ \class{red}{n} \, \frac{ 1 }{ 2 \pi \, \lambda_{\text{dB}} } \end{align} $$

If we bring $ 2 \pi $ to the other side of the equation, we see that the left side corresponds exactly to the perimeter $ U_{\class{red}{n}} = 2 \pi \, r_{\class{red}{n}} $ of a circle:

Circumference of the circular path of the electron by using its de Broglie wavelength

$$ \begin{align} U_{\class{red}{n}} ~=~ \class{red}{n} \, \lambda_{\text{dB}} \end{align} $$

So this formula states: If we interpret the electron as a matter wave with wavelength $ \lambda_{\text{dB}} $, then the /strong>circumference of the circular orbit is always a multiple of the wavelength of the electron.

So far so good. With these three assumptions, it is possible with the Bohr model to understand, for example, the discrete emission and absorption spectra of hydrogenium. In the next lesson, we will look at the concrete application of the Bohr model to the hydrogenium atom and we will see how perfectly this model works for the hydrogenium atom.