### The Rydberg equation

In 1888 the Swedish physicist Johannes Rydberg, through truly astounding mathematical detective work, discovered that electron's only bond to atoms at discrete energies.  His discovery was the precursor to quantum physics.

Recognizing that emission spectral lines from atoms represented the energy difference between discrete bonding energy levels, Rydberg was able to determine, not only that bonding energies increase with the square of the atomic number, Z, they also decrease with the square of a positive integer, n. (In the ground state n is equal to one.)

The Rydberg energy level of a single-electron ion is given by:

$E_{\infty }\approx -13.6 \frac{Z^{2}}{n^{2}}eV$

The constant, 13.6 electron volts, is often referred to as simply the Rydberg.  Follow the above link and you will discover that its official definition is quite cryptic.  It turns out, however, that the Rydberg can be derived intuitively from just a few physical constants.

To this end, I’ve put together the following introduction to those few fundamental physical constants -- presented in the order that would have been most helpful to me in making subtle connections; connections often obscured by teachers and texts. This obscuration is not intentional or conspiratorial. I think it just evolves out of some weird natural preference for esoteric and complex explanations.

There are, of course, complicated concepts in physics. That notwithstanding, simple things should be kept simple, and you should continually search for new opportunities to apply Ockham's razor.

#### Classical electron radius

Let’s start with a hydrogen atom; one electron bonded to one proton; the simplest atom in nature. But the hydrogen atom I want you to consider is special - It has collapsed in on itself until the electron is so close to the proton that the energy required to escape the proton’s electric field is equal to the electron's entire rest mass energy, (mec2); roughly half a million electron-volts. (511,000 eV is more precise and will give you the right answer on any physics or chemistry test.)

The distance between the proton and electron in this imaginary atom is a well known physical constant - the classical electron radius.  This distance is so ridiculously small, it’s meaningless to express it in meters, or any other length we can perceive - so I wont.

#### Compton wavelength

The interesting thing about the classical electron radius is its relationship to the wavelength of light that has enough energy to remove an electron from that close proximity to a proton; i.e., the wavelength of a photon with an energy of 511,000 eV.

This wavelength is also a well known physical constant: the Compton wavelength of the electron.

#### The fine structure constant

Now let’s bend the Compton wavelength into a perfect circle. The radius of this circle is called the reduced Compton wavelength and is approximately 137 times larger than the classical electron radius.

The precise ratio between the classical electron radius and the reduced Compton wavelength is another recognized constant of nature: the fine structure constant, usually denoted by the Greek letter alpha.

$\alpha \approx \frac{1}{137}$

While the above approximation will be used throughout this discussion, it should be recognized that the factor 137 is actually a substitution for 1/alpha.

Now let’s enlarge the circle another 137 times. As you might have anticipated from the emerging pattern, this new radius is also a famous length: the Bohr radius. The electrical potential energy of an electron at one Bohr radius from a proton is easily computed.

Since the electric field energy varies inversely with distance, and the Bohr radius was obtained from the classical electron radius by multiplying it times 137 twice, the potential energy at the Bohr radius is the electron’s rest mass energy divided by 137 twice:

$PE = -\alpha ^{2}m_{e}c^{2} \approx \frac{511,000}{137^{2}}\approx -27.2 eV$

This is the potential energy of a hydrogen electron at the Bohr radius.

But wait!  We already know that's twice the measured ionization energy for the hydrogen atom.  What's with that?

The simplest explanation is that the electron has a positive kinetic energy equal to half the absolute value of the potential energy.

The kinetic energy in this case is:

$KE = \frac{1}{2}m_{e}v^{2}=\frac{1}{2}\alpha ^{2}m_{e}c^{2} \approx 13.6 eV$

$\therefore \frac{v^{2}}{c^{2}}=\alpha ^{2}$

$\therefore \frac{v}{c}=\alpha \approx \frac{1}{137}$

Being exactly half the absolute value of the potential energy, this positive kinetic energy brings the net bonding energy to the observed value for the ground state hydrogen electron, -13.6 eV.

#### The de Broglie wavelength

Here’s another interesting consequence of our guess: If the electron’s velocity is 1/137th the speed of light, its de Broglie wavelength is 137 times longer than the Compton wavelength. Since the Bohr radius is 137 times larger than the reduced Compton wavelength, the circumference of a Bohr radius shell is equal to the electron’s de Broglie wavelength. I’ll take this observation to be a necessary condition for all single electron ions, i.e. the ground state shell circumference must equal the electron’s de Broglie wavelength.

$\lambda =\frac{h}{m_{e}v}=\frac{m_{e}c\lambda _{C}}{m_{e}v}=\frac{c}{v}\lambda_{C}$

#### The Rydberg beyond hydrogen

Now let’s extend our guess to see to calculate the bonding energy of single electron ions of other elements. The atomic number, Z, is the number of protons in the nucleus, and the new guess is that the ground state vibration frequency will be the Zth harmonic of the ground state hydrogen electron, i.e. Z times faster.  The general statement for ground state velocity becomes:

$v=Z\alpha c$

Since kinetic energy is a velocity squared function, you can take Z2 times the hydrogen electron’s kinetic energy for the ground state kinetic energy of any single-electron ion.

$KE=13.6Z^{2}eV$

What about potential energy? Since the electron velocity increases by a factor of Z, the de Broglie wavelength, and therefore the shell circumference (and radius), must shrink by a factor of 1/Z. Shrinking the radius by 1/Z increases the potential by the factor Z.  The electrical potential energy also increases with the number of protons, so the general potential energy also turns out to be the hydrogen potential energy times Z2.

$PE=-27.2Z^{2}eV$

(If an electron, one Bohr radius from one proton has a potential of -27.2 eV; then an electron one half a Bohr radius from two protons (He+1) has a potential of -27.2 x 4 eV.)

Adding the potential energy and kinetic energy gives the net bonding energy for the ground state of single-electron ions:

$E_{\infty }\approx -13.6Z^{2}eV$

This is the Rydberg equation for all single electron ground states.  The complete Rydberg equation can be derived from the above analysis if it is assumed that for exited states the electron ground state velocity is reduced by the factor 1/n. (The nth fundamental of the ground state frequency.)  It also requires that the number of de Broglie wavelengths in a shell circumference is always equal to the same n.  In my opinion the jury is still out on this last requirement. More about that later.