$$\require{cancel}$$

The Puzzle

It was known since the time of Maxwell that light (and all EM radiation) is produced when electric charges accelerate. Rutherford also showed that atoms had a localized, hard nucleus, which meant that atoms consisted of negatively-charged electrons bound by the electric force to a positively-charged nucleus, like satellites are bound in their orbits by gravitational forces. But this posed a conundrum – electric charges in orbits are accelerating, so they should constantly be radiating, and since light carries away energy, the electrons should spiral down into the nucleus. What is more, there is no reason to believe that any light frequencies released by atoms that lose energy should be preferred over any others. But it was known for some time that viewing light emitted by various elements through a diffraction grating reveals distinct sets of spectral lines – lines that are separated according to frequency. And finally, Planck's discovery that light comes in particle-like packets called photons with energies proportional to the frequency meant that these atoms can only emit very specific amounts of energy.

Bohr's Model

The idea of the existence "matter waves" (wave functions for particles with mass) was just taking hold, and Niels Bohr came up with a simple-but-remarkably-effective model that incorporated this idea into the simple model of an orbiting particle. Bohr reasoned that a particle whose location is extended in space in the form of a wave and is orbiting a central point would have to interfere with itself when it gets all the way around. For the particle to remain in such an orbit, it would have to (as we call it now) "match boundary conditions" – its wave function would have to be in phase with itself when a full orbit is completed. This requirement (with the assumption that the orbit is circular), along with deBroglie's formula for the wavelength of a matter wave, and the coulomb potential for two point charges is all that was required to complete this model. Let's see what this semi-classical approach predicts for the simplest of atoms – hydrogen...

The coulomb force between the two equal charges (which we will call $$e$$) of a single proton and single electron is given by:

$F\left(r\right)=\dfrac{e^2}{4\pi\epsilon_o r^2}$

The assumption that the electron follows a circular orbit requires that this force causes a centripetal acceleration, so calling the mass of the electron "$$m_e$$," we have:

$m_e \dfrac{v^2}{r} = \dfrac{e^2}{4\pi\epsilon_o r^2} \;\;\; \Rightarrow \;\;\; p^2 = \left(m_e v\right)^2 = \dfrac{m_e e^2}{4\pi\epsilon_o r}$

We put this in terms of the magnitude of the electron's momentum, so that we can use the deBroglie relation. Applying Bohr's assumption that an integer number of full wavelengths fit within the circular orbit (so that it meets itself in phase) gives:

$\left. \begin{array}{l} deBroglie: & p=\dfrac{h}{\lambda}\; \\ Bohr: & 2\pi r = n\lambda \;\end{array} \right\} \;\;\; \Rightarrow \;\;\; p^2=\left(\dfrac{nh}{2\pi r}\right)^2=\dfrac{n^2\hbar^2}{r^2}$

Plugging this back into Equation 5.1.2 and solving for the radius of the orbit gives:

$r_n =n^2 \left(\dfrac{4 \pi \epsilon_o \hbar^2}{m_e e^2}\right) \equiv n^2 a_o$

The orbit radii are thus quantized! The quantity $$a_o$$, derived purely from physical constants, is called the Bohr radius, which is a useful unit of length measurement, even in the more enlightened quantum mechanical model yet to come.

Of course, quantization of the orbital radii was not what we were after – we are interested in the energy levels of the hydrogen atom, which are somehow involved in providing energy to emitted photons, and we want to know why the hydrogen atom doesn't radiate away all its energy. To move this to a discussion of the atom's energy, we first note its electrical potential energy is:

$V\left(r\right) = -\dfrac{e^2}{4\pi\epsilon_o r}$

Comparing this with the first equality (multiplied by $$\frac{r}{2}$$) in Equation 5.1.2, we see that there is a simple relationship between the kinetic and potential energy, which means we also have a simple relationship between the total energy and the potential energy:

$\frac{1}{2}m_ev^2 = \frac{r}{2}\left(\dfrac{e^2}{4\pi\epsilon_o r^2}\right) \;\;\; \Rightarrow \;\;\; KE = -\frac{1}{2}PE \;\;\; \Rightarrow \;\;\; E_{tot} =KE+PE=\frac{1}{2}PE=-\dfrac{e^2}{8\pi\epsilon_o r}$

But the radii are quantized, so plugging this in gives quantized energy levels:

$E_n = -\dfrac{e^2}{8\pi\epsilon_o r_n} = -\dfrac{1}{n^2}\left( \dfrac{m_e e^4}{2\left(4\pi\epsilon_o\right)^2\hbar^2}\right)$

The energy spectrum is also quantized. The quantity in parentheses is a unit of energy known as a Rydberg, which $$\approx 13.6eV$$.

Bohr reasoned that the hydrogen atom doesn't radiate away all of its energy, because the lowest energy level (which we would now call the ground state) corresponds to one wavelength fitting in the orbit, so $$n=1$$ is the lowest it can go.

Emission/Absorption Spectrum

The problem of only seeing certain spectral lines is also solved in this model, if one insists that the atom can only exist in one of these quantized states, and therefore the only energy transistions it can make are between the allowed energy levels. A transition that lowers the energy level of the hydrogen atom from \(n_1) to \(n_2) frees up the amount of energy equal to the difference, which then goes into an emitted photon according to Planck's relation:

$hf = \Delta E = E_{n_1} - E_{n_2} = -13.6eV\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$

This matched perfectly with experiment! So although there are many problems with this model, our more evolved version will need to agree with this energy spectrum. Note that this quantized energy change works both ways – when a hydrogen atom absorbs a photon, it must absorb an amount of energy that carries the atom from one of its quantized energy levels to another (higher) one.