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Bohr Model: Atomic Structure from the Perspective of Quantum Physics

Important Formula

Formula: Bohr Model

$$L_{\class{red}{n}} ~=~ \class{red}{n} \, \hbar$$ $$L_{\class{red}{n}} ~=~ \class{red}{n} \, \hbar$$ $$\class{red}{n} ~=~ \frac{L_\class{red}{n}}{\hbar}$$

Rearrange formula

What do the formula symbols mean?

Orbital angular momentum

$$ L_{\class{red}{n}} $$ Unit $$ \mathrm{Js} $$

Orbital angular momentum of the electron in the $\class{red}{n}$-th state. It can be - according to the Bohr model - only a multiple of the reduced Planck's constant $ \hbar $, so that the electron orbits the nucleus in a stable orbit.

Principal quantum number

$$ \class{red}{n} $$ Unit $$ - $$

Principal quantum number is a natural number $ \class{red}{n} ~\in~ \{1,2,3...\} $ and quantizes angular momentum in the Bohr model as a multiple of $\hbar$.

Reduced Planck's constant

$$ \hbar $$ Unit $$ \mathrm{Js} = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{s} } $$

Reduced Planck's constant is a physical constant (of quantum mechanics) and has the value: $ \hbar ~=~ \frac{h}{2\pi} ~=~ 1.054 \,\cdot\, 10^{-34} \, \text{Js} $.

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} $.

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} \):

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

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:

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

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...

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:

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

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.

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise: Classical Angular Frequency of the Electron in the H-Atom

Determine the classical angular frequency $ \omega_n $ of an electron - depending on the distance from the hydrogen nucleus, by incorporating the quantum number $ n $ and the quantization of angular momentum.

Hint: Use the Coulomb's law, as well as the centripetal force for this exercise. Understand how angular frequency is defined. Then, determine the quantized radius $ r_n $ using the formula for quantized angular momentum, with which the electron (in the state with quantum number $ n $) orbits the nucleus.

Solution to the Exercise

To determine the angular frequency $ \omega_n $ of a classical electron in the H-atom, you first need to find out what condition the orbiting electron satisfies. In the Bohr model, it always remains at a certain distance $ r_n $ from the nucleus of the hydrogen atom - depending on the $ n $-th energy level.

For the same distance to be maintained, the centripetal force $ F_{\text{C}} $ must correspond to the attractive force $ F_{\text{C}} $ between the negative electron and the positive nucleus - which is given by Coulomb's law. So, equate the electrical Coulomb force and the centripetal force: 1 \[ \frac{Q_1 \, Q_2}{4\pi \, \varepsilon_0 \, r_{n}^2} ~=~ \frac{m \, v_{n}^2}{r_n} \]

In the hydrogen atom, there is a positively charged proton with charge $ Q_1 ~=~ +e $ in the nucleus, which is orbited by a single negatively charged electron with charge $ Q_2 ~=~ -e $. Here, $ e $ corresponds to the elementary charge with the value: $ e ~=~ 1.602 \cdot 10^{-19} \, \text{C} $. We only consider the magnitudes, so $ Q_1 = Q_2 = e $.

The mass $ m $ in equation 1 is the mass of the orbiting particle, i.e., the mass of the electron $ m := m_{\text{e}} $. And $ \varepsilon_0 $ is a physical constant, which you can also find in this Physics Equations Book.

If you substitute the two charges and the mass (but not as numbers) into equation 1, you get: 2 \[ \frac{e^2}{4\pi \, \varepsilon_0 \, r_{n}^2} ~=~ \frac{m_{\text{e}} \, v_{n}^2}{r_n} \]

You need to somehow obtain the angular frequency $ \omega_n $. It's best to express it only with physical quantities already present in equation 2. The angular frequency is related to the orbital period $ T_n $ as follows: $ \omega_n = 2 \pi \frac{1}{T_n} $. So, first, you need to determine the orbital period.

The orbital period $ T_n $ is - in this case - the circumference of the circle $ 2\pi \, r_n $ divided by the velocity $ v_n $. (You know: distance per velocity equals the required time). So, the period is given by: 3 \[ T_n ~=~ \frac{2\pi \, r_n}{v_n} \]

The frequency $ f_n $ is defined as $ \frac{1}{T_n} $. So, you have the following expression for the frequency when using 3: 4 \[ f_n ~=~ \frac{1}{T_n} ~=~ \frac{v_n}{2\pi \, r_n} \]

The angular frequency is defined as $ \omega_n = 2 \pi \frac{1}{T_n} = 2\pi \, f_n $. Thus, you need to multiply Eq. 4 on both sides by $ 2\pi $ to obtain the angular frequency: 5 \[ \omega_n ~=~ \frac{v_n}{r_n} \]

Now you have an equation for angular frequency that contains only quantities present in equation 2; namely the velocity $ v_n $ and the radius of the orbit $ r_n $.

You can rearrange the angular frequency equation 5 to isolate the velocity: 6 \[ v_n ~=~ \omega_n \, r_n \]

Substitute it into Eq. 2. Here, one factor $ r_n $ cancels out in the denominator: 7 \[ \frac{e^2}{4\pi \, \varepsilon_0 \, r_{n}^2} ~=~ m_{\text{e}} \, \omega_{n}^2 \, r_n \]

Since you're looking for the formula for angular frequency, rearrange the equation for angular momentum 7 in terms of angular frequency: 8 \[ \omega_n ~=~ \sqrt{ \frac{e^2}{4\pi \, \varepsilon_0 \, m_{\text{e}} \, r_{n}^3} } \]

Unfortunately, you don't know the radius $ r_n $ and the quantum number $ n $ is not yet present in the equation. Therefore, you use the quantized angular momentum: 9 \[ L_n ~=~ m_{\text{e}} \, v_n \, r_n ~=~ n \, \hbar \]

This eliminates the radius and introduces the quantum number into your equation. Here, $ \hbar $ is the reduced Planck constant $ \hbar ~=~ \frac{h}{2\pi} $.

Rearrange the angular momentum equation 9 for velocity: 10 \[ v_n ~=~ \frac{n \, \hbar}{m_e \, r_n} \] and substitute it into equation 2: 11 \[ \frac{e^2}{4\pi \, \varepsilon_0 \, r_{n}^2} ~=~ \frac{n^2 \, \hbar^2}{r_{n}^3 \, m_e} \] where one $ m_e $ term cancels out.

Rearrange Eq. 11 for the radius $ r_n $: 12 \[ r_n ~=~ \frac{4\pi \, \varepsilon_0 \, n^2 \, \hbar^2}{e^2 \, m_e} \]

By the way, this is the Bohr radius. Substitute it into equation 8 for angular frequency: 13 \[ \omega_n ~=~ \sqrt{ \frac{e^8 \, m_{e}^3 }{4\pi \, \varepsilon_0 \, n^6 \, \hbar^6 \, 4^3 \, \pi^3 \varepsilon_{0}^3 m_e} } \]

Simplify and take the square root. Then we get a formula for angular frequency, which depends only on constants and the known quantum number $ n $: 14 \[ \omega_n ~=~ \frac{e^4 \, m_{e} }{16\pi^2 \, \varepsilon_{0}^2 \, \hbar^3} \, \frac{1}{n^3} \]

Example: In the ground state $ n = 1 $, the electron has the following classical angular frequency: 15 \[ \omega_1 ~=~ \frac{e^4 \, m_{e} }{16\pi^2 \, \varepsilon_{0}^2 \, \hbar^3} \]

Exercise #2: Muonic Hydrogen Atom

Muonic hydrogen consists of a proton in the nucleus and a muon instead of an electron. A muon has the same charge as an electron, but it is 207 times heavier.

Calculate the binding energy of the muon in the lowest shell $ n=1$.
Calculate the Bohr radius without considering the motion of the proton.
Calculate the Bohr radius with consideration of the motion of the proton.

Tips: Use the formula for the binding energy, which consists of the kinetic and potential energy of the muon. This can be easily derived using the Bohr quantization condition: $ L = n \, \hbar $.

"Motion of the proton" means - the proton is not assumed to be stationary, find the formula for reduced mass.

Solution to Exercise #2.1

In the H-atom, the muon with charge $-e$ orbits the proton with charge $e$. The binding energy of the muon is the sum of its kinetic and potential energy in the H-atom: 1 \[ E ~=~ \frac{1}{2} \, m_{\mu} \, v^2 ~-~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{e^2}{r} \]

For the muon to move stably on a circular path, the centripetal force and the Coulomb force must be in equilibrium: 2 \[ \frac{m_{\mu} \, v^2}{r} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{e^2}{r^2} \]

Using the Bohr quantization condition, which states that the angular momentum exists only as a multiple $n$ of $\hbar$: $ L = m_{\mu} \, v \, r = n \, \hbar $, the velocity $v$ in 2 can be eliminated with the Bohr quantization condition. Rearranging for the radius $r:=r_n$ yields the Bohr radius for different states $n$ of the muon: 3 \[ r_n ~=~ \frac{4\pi \, \varepsilon_0 \, \hbar^2}{e^2 \, m_{\mu}} \, n^2 \]

First, substitute the velocity in 1 with the Bohr quantization condition. Then, substitute the Bohr radius 3 into 1. Using $ \hbar = h /2\pi $ gives the quantized energy of the muon in the H-atom: 4 \[ E_n ~=~ \frac{m_{\mu} \, e^4}{8 \varepsilon_{0}^2 \, h^2} \, \frac{1}{n^2} \]

Substitute the mass of the muon $m_{\mu} = 207 m_{\text{e}}$: 5 \[ E_n ~=~ - \frac{207 m_{\text e} \, e^4}{8 \varepsilon_{0}^2 \, h^2} \, \frac{1}{n^2} \]

Since the task requires the binding energy in the lowest shell, $ n ~=~ 1 $: 6 \[ E_1 ~=~ - \frac{207 m_{\text e} \, e^4}{8 \varepsilon_{0}^2 \, h^2} \]

Substitute the constants: 7 \[ E_n ~=~ - \frac{207 ~\cdot~ 9.109 \cdot 10^{-31} \, \text{kg} ~\cdot~ (1.602 \cdot 10^{-19} \, \text{C})^4}{8 \cdot (8.854 \cdot 10^{-12}\frac{\text{As}}{\text{Vm}})^2 ~\cdot~ (6.626 \cdot 10^{-34} \, \text{Js})^2} = -0.45 \, \text{fJ} \]

Solution to Exercise #2.2

Without considering the motion of the proton means the proton is at rest and is orbited by the muon on a fixed path. The condition for a fixed path of the muon is given by the equilibrium between the centripetal force and Coulomb force: 8 \[ \frac{m_{\mu} \, v^2}{r} ~=~ \frac{1}{4 \pi \varepsilon_0} \, \frac{e^2}{r^2} \]

Additionally, you need the condition that the radii $ r_n $ are quantized: 9 \[ 2\pi r_n ~=~ n \, \lambda \]

And the de Broglie relationship: 10 \[ \lambda ~=~ \frac{h}{p} ~=~ \frac{h}{m_{\mu} \, v} \]

Rearrange 10 for velocity $ v $ and substitute it into 9: 11 \[ v ~=~ \frac{h}{m_{\mu} \, \lambda} \]

Rearrange 9 for wavelength and substitute it into 11: 12 \[ v ~=~ \frac{n \, h}{ 2\pi \, r_n \, m_{\mu} } \]

You can now substitute the velocity from Eq. 12 into Eq. 1: 13 \[ \frac{m_{\mu} \, n^2 \, h^2}{4\pi^2 \, r^3 \, m_{\mu}^2 } ~=~ \frac{1}{4 \pi \varepsilon_0} \, \frac{e^2}{r^2} \]

Rearrange Eq. 13 with respect to the radius: 14 \[ r_n = \frac{h^2 \, \varepsilon_0 }{\pi \, e^2 \, m_{\mu} } \, n^2 \]

Solution to Exercise #2.3

Considering the motion of the proton means: The proton moves together with the muon around their common center of mass. For this, in Eq. 14, you just need to substitute the reduced mass for the mass of the muon. This includes both the mass of the muon and the mass of the proton.