In 1911, Ernst Rutherford discovered the atomic nucleus and posited that atoms consist of negatively charged electrons orbiting a positively charged nucleus under the electrostatic interaction, in a similar way to how planets orbit the sun under the gravitational interaction.

This is not compatible with classical physics because the orbiting electrons would be accelerating and accelerating charges generate electromagnetic radiation. The radiating electromagnetic waves would carry energy away from the electrons, causing them to lose kinetic energy and spiral into the nucleus. This would happen in about $10^{-8}\,\text{s}$, so matter would not be stable. If the classical Rutherford model were correct, everything would disintegrate into electromagnetic radiation, and since you are here reading these notes, presumably on a non-disintegrating computer while sitting on a non-disintegrating chair, the Rutherford model cannot possibly be correct.

Ignoring the catastrophic disintegration of everything in the universe for a moment (hard to do, I know), there is another problem with the classical Rutherford atom. The electromagnetic radiation emitted by an electron orbiting the nucleus ought to have the same frequency as the frequency with which the electron orbits the nucleus. Since the electron can be at any distance from the nucleus, its orbiting frequency can take any possible value, so we would expect to see a continuous distribution of frequencies of electromagnetic radiation emitted by an atom.

In fact, we do not. Atoms only emit and absorb electromagnetic radiation at a discrete set of frequencies, or equivalently wavelengths. The set of wavelengths at which radiation is emitted or absorbed by an atom is called its spectrum. Putting the light emitted by atoms through a prism is one way of observing the spectrum. The figures below show the spectrum of hydrogen in the visible light range, followed by the spectrum in a wider range of wavelengths plotted on a log-scale.

The discrete atomic spectra suggest that there is some discreteness in the fundamental structure of atoms, so perhaps some kind of quantum postulate is at work.

In 1913, Neils Bohr proposed a model for the hydrogen in which an electron can only occupy a discrete set of orbits around the proton with energies $E_1$, $E_2$, $E_3$, $\cdots$. For reasons that will become clear when we study full-blown quantum mechanics, these orbits are called stationary states. For simplicity, he initially only considered circular orbits. Attempts to generalize to elliptical orbits were made but, suffice to say, they did not work out very well, so we will stick to circular orbits here.

The allowed orbits are those for which the angular momentum is an integer multiple of $\hbar = h/2\pi$. \[L = n\hbar,\qquad\qquad n=1,2,3,\cdots,\] so Bohr's main postulate is really that angular momentum is quantized.

Now, accelerating charges are supposed to radiate energy, but if only a discrete set of energies are allowed for the orbits then there is no way that the electron can continuously transition from one orbit to another. So Bohr proposed that electrons transition between orbits by jumping discontinuously and instantaneously.

It has to be said that Bohr's model was extremely bold, by which I mean that it would have seemed completely crazy to physicists at the time. Forget blackbody radiation, the photoelectric effect, Compton scattering and de Broglie matter waves, which only require us to do a bit of fancy footwork about particles sometimes being waves and vice versa. Bohr is asking us to completely abandon Newtons laws and the laws of electromagnetism when it comes to electrons orbiting the nuclei of atoms. If there is anything that indicated that a completely new theory of physics was needed it was the Bohr model of hydrogen.

Bohr suggested that, as long as the electron stays in one of the stationary orbits then it does not emit or absorb any electromagnetic radiation. However, the electron may jump from a lower energy orbit $E_n$ to a higher energy orbit $E_n$, with $E_n > E_m$, by absorbing radiation with frequency $\nu$ that satisfies \[h\nu = E_n - E_m.\] In other words, by absorbing a photon with energy $h\nu$.

Also, the electron may jump from a higher energy orbit $E_n$ to a lower energy orbit $E_m$ by emitting radiation of frequency $\nu$, where \[h\nu = E_n - E_m,\] i.e. by emitting a photon with energy $h\nu$.

Let's now work out the consequences of Bohr's postulate $L = n\hbar$ for the possible energies of the electron orbits and the possible frequencies of radiation that can be emitted or absorbed by the hydrogen atom.

We will assume that the proton is stationary because its mass is much larger than that of the electron and only consider the motion of the electron. We need to recall a few key formulas about circular motion.

First, the centripetal acceleration of the electron (directed towards the proton) is \[a_r = \frac{v^2}{r},\] where $v$ is the speed of the electron and $r$ is the distance between the proton and the electron. Neglecting gravity, the only force acting on the electron is the electrostatic force due to the proton. This is directed towards the proton and has magnitude \[F_r = \frac{e^2}{4\pi\epsilon_0 r^2}.\] The angular momentum of the circular orbit is \[L = m_ev r,\] which is also equal to $n\hbar$ by Bohr's assumption.

In an in-class activity, you will be asked to take these four formulas an eliminate variables to show that the radius of the $n^{\text{th}}$ orbit is \[r_n = \frac{4\pi\epsilon_0\hbar^2}{me^2} = a_0 n^2,\] where \[a_0 = \frac{4\pi\epsilon_0\hbar^2}{me^2},\] is the Bohr radius. It is the radius of the lowest energy $n=1$ orbit and has value $a_0 = 0.053\,\text{nm}$.

In the in-class activity, you will also show that the electron's speed in the $n^{\text{th}}$ orbit is \[v_n = \left ( \frac{e^2}{4\pi\epsilon_0} \right ) \frac{1}{n\hbar}.\]

The energy of the $n^{\text{th}}$ orbit is then \[E_n = \frac{1}{2}m_ev_n^2 - \frac{1}{4\pi\epsilon_0} \frac{e^2}{r_n},\] where the first term is the kinetic energy of the electron and the second term is the electrostatic potential energy of the electron in the electric field of the proton. Note: here we are using the convention that the zero of electrostatic potential energy is at $r=+\infty$, so all electrostatic potential energies will be negative.

Substituting the expressions for $r_n$ and $v_n$ into the energy $E_n$ and simplifying (which I will not subject you to doing in an in-class activity) gives \[\boxed{E_n = -\frac{m_e}{2\hbar^2} \left ( \frac{e^2}{4\pi\epsilon_0} right )^2 \frac{1}{n^2} = -\frac{\mathcal{R}}{n^2}},\] which is known as the Bohr Energy.

The constant $\mathcal{R}$ is the magnitude of the energy of the $n=1$ orbit and is known as the Rydberg Energy. Its value is \[\boxed{\mathcal{R} = \frac{m_e}{2\hbar^2} \left ( \frac{e^2}{4\pi\epsilon_0}\right )^2 \approx 13.6\,\text{eV}.}\]

Note: the fact that there is a lowest possible energy for the Bohr atom and that this corresponds to a finite radius and finite speed means that the electron does not spiral into the nucleus radiating ever higher frequency radiation as it does so. It has to stop at the lowest energy state, so matter is stable and the universe is saved from disintegration. Hooray!

The set of allowed energies for a hydrogen atom is depicted below.

• $n=1$ is the lowest energy state with $E_1 = -13.6\,\text{eV}$. It is called the ground state of hydrogen.
• States with $n>1$ are called excited states.
• For $E < 0$, the kinetic energy of the electron is smaller than the magnitude of the electric potential energy. The electron is bound to the proton and it has a discrete set of possible energies.
• As $n\rightarrow\infty$, the energy levels become more and more closely spaced. If $E>0$ then the kinetic energy of the electron is larger than the magnitude of the electric potential energy. There is a continuous set of energies and the electron behaves like a free particle.

  • OK, you got me mathematical pedants, $E > 0$ cannot happen because $E \rightarrow 0$ as $n\rightarrow \infty$, so positive energy would require an $n$ larger than infinity! What this really means is that the electron behaves approximately like a free particle for very large $n$ because the energy levels are so closely spaced. Besides, there are more than two particles in the universe, so when the electron is very far away from the proton, the electric field of the proton is effectively screened off by all the other charged particles floating around, so the electron will actually be free and have positive energy for a finite, but large, value of $n$.

Right, now that we have the hydrogen atom down, let's do helium and the rest of the periodic table! Sadly, things are not so simple. As soon as we have two or more electrons, we have to take into account the electrostatic repulsion between the electrons. The Bohr model does not work because we can no longer reasonably expect the orbits to be circular. Even in full-blown quantum mechanics, we cannot even solve the helium atom without approximations. So helium, and the rest of the periodic table, will have to wait until PHYS 452.

Your chemistry professors may wonder why us physicists spend an entire course attempting to solve the hydrogen atom, only obtaining the full quantum mechanical model at the very end, when there are so many other interesting atoms and molecules out there. Us physicists like to understand the simplest system in full detail before moving on, and the hydrogen atom is our “quantum spherical cow”.

So, what can we do that is more general than hydrogen at this point. Well, we can certainly deal with ions that have heavier nuclei than hydrogen, but have lost all of their electrons except one. These are known as hydrogen-like ions. The only effect this has on our calculations is that the charge on the proton $e$ is replaced by the charge on the heavier nucleus $Ze$ (where $Z$ is the atomic number, i.e. the number of protons in the nucleus). This means that one of the factors of $e$ in all our formulas will be replaced by $Ze$. The other factor of $e$ does not change because that comes from the charge of the electron.

So, for a hydrogen-like ion, the Bohr energy becomes \[\boxed{E_n = -\frac{m_e}{2\hbar^2} \left ( \frac{Ze^2}{4\pi\epsilon_0} right )^2 \frac{1}{n^2} = -\frac{\mathcal{R}Z^2}{n^2}},\]

The other thing we could do is remove the approximation we made that the nucleus remains stationary. It is, in fact, a really good approximation because the mass of the proton is $m_pc^2 = 938\,\text{MeV}$, so the ratio of the electron mass to the proton mass is $m_e/m_p = 0.511/938 \approx 5\times 10^{-4}$. However, it is possible to make atom-like systems by replacing the proton by a positively charged particle with much smaller mass that is closer to the mass of an electron. If you go on to study physics in graduate school you will learn that every particle has an anti-particle that has the same mass but opposite charge. The anti-particle of the electron is called the positron and physicists have managed to make a bound system of an electron and positron, which is called positronium.

If we want to deal with a system like positronium where the mass $M$ of the “nucleus” is of the same order of magnitude as the mass $m_e$ of the electron then we cannot assume that the “nucleus” remains stationary. Instead we have to move to a center of mass coordinate system and replace the electron's mass in our calculations with the reduced mass \[\mu = \frac{m_e M}{m_e + M},\] which you should be familiar with from your classical mechanics courses.

In full generality, if we have a hydrogen-like ion where the “nucleus” is made of $Z$ particles each with charge $e$ and the total mass of the nucleus is $M$, then the radius of the $n^{\text{th}}$ orbit becomes \[r_n = \frac{4\pi\epsilon_0 \hbar^2}{\mu Ze^2}n^2 = \left ( 1 - \frac{m_e}{M}\right )\frac{a_0}{Z}n^2,\] and the energy of the $n^{\text{th}}$ orbit becomes \[E_n = -\frac{\mu}{2\hbar^2} \left ( \frac{Ze^2}{4\pi\epsilon_0}\right )^2 \frac{1}{n^2} = -\frac{Z^2M}{M-m_e}\frac{\mathcal{R}}{n^2}.\]

We can now explain the observed spectrum of the hydrogen atom. When dropping from level $n$ to level $m$ with $n>m$, the hydrogen atom will emit radiation with frequency given by \[h\nu = \mathcal{R} \left ( \frac{1}{m^2} - \frac{1}{n^2} \right ).\] This corresponds to what is observed in experiments.

In a sensible world, this is all we should need to say about the subject. However, since the spectral lines were discovered before the Bohr model of the hydrogen atom, sets of spectral lines that are clustered in specific frequency ranges were given fancy names, named after the person who first observed each particular set of lines.

• For $m=1$, the atom drops into its ground state and emits ultraviolet radiation. This is called the Lyman series and has $h\nu_L = \mathcal{R}\left ( 1 - \frac{1}{n^2} \right )$.
• For $m=2$, the atom drops into its first excited state and emits visible light. This is called the Balmer Series and has $h\nu_B = \mathcal{R} \left ( \frac{1}{4} - \frac{1}{n^2}\right )$.
• For $m=3$, the atom drops into its second excited state and emits infra-red light. This is called the Paschen Series and has $h\nu_B = \mathcal{R} \left ( \frac{1}{9} - \frac{1}{n^2}\right )$.

Are you getting bored of this yet? Some of the other series with higher values of $m$ also have fancy names. I promise I will never make you remember these names for an exam. In an ideal world, we would just call them the “$m=1$ series”, the “$m=2$ series”, etc. I blame chemists for the fact that the fancy names have stuck around. They are the people who most often do experimental spectroscopy and they need to have complicated jargon to make it look like their subject is difficult. I often wonder if they actually understand the simple physics underlying the experiments they are doing. I am not posting this in public am I? Please don't tell your chemistry professors that I said any of that.

1. The energy of an electron orbiting a proton at radius $r$ is \[E = \frac{p^2}{2m_e} - \frac{e^2}{4\pi\epsilon_0 r}.\] Use the uncertainty principle in the form $\Delta r \Delta p \sim \hbar$ to estimate the radius and energy of the minimum energy state in terms of $m_e$, $e$ and $\epsilon_0$. Hint: Interpret $\Delta r$ and $\Delta p$ as standard deviations. If we assume that the average $r$ and $p$ for the minimum energy state are very small then the root mean square values of $r$ and $p$ are approximately $\Delta r$ and $\Delta p$. You can therefore use $rp \approx \hbar$ to estimate the minimum energy state.
2. Use \[a_r = \frac{v^2}{r},\] \[F_r = \frac{e^2}{4\pi\epsilon_0 r^2},\] \[L = m_evr = n\hbar,\] to derive expressions for the electron speed $v_n$ and radius $r_n$ of the $n^{\text{th}}$ stationary state of the Bohr atom in terms of $n$, $\hbar$, $\epsilon_0$ and $m_e$.
3. The Bohr Atom and de Broglie Waves: Imagine that the electron orbiting the proton in a hydrogen atom is described by a standing wave that wraps around the circular orbit as illustrated below. To be a standing wave there must be an integer number of wavelengths around the circumference of the circle. Use this, together with the de Broglie relation $p = h/\lambda$ to derive Bohr's postulate $L=n\hbar$.