Energy Levels Revisited

We met $E_n = -13.6\,\text{eV}/n^2$ earlier as the Bohr result. Now that we have set up the radial equation properly, we can see why the spectrum is discrete and pin down exactly where the $n^2$ comes from.

Quantization from the boundary conditions

Recall the effective 1D radial equation for $u = rR$ ,

-\frac{\hbar^2}{2\mu}\frac{d^2 u}{dr^2} + \left[-\frac{e^2}{4\pi\epsilon_0\,r} + \frac{\hbar^2\,\ell(\ell+1)}{2\mu r^2}\right]u = E\,u,

with $u(0) = 0$ and $u(r\to\infty)\to 0$ . For $E < 0$ the solution behaves like $e^{-\kappa r}$ at large $r$ . Demanding that the polynomial part terminate (so the wavefunction stays normalizable) forces a discrete condition: the series must cut off after a finite number of terms. That happens only when

n \equiv n_r + \ell + 1

is a positive integer, where $n_r$ is the degree of the polynomial. This integer $n$ is the principal quantum number.

The energy formula

Working through the termination condition gives the energies in terms of fundamental constants:

E_n = -\frac{\mu}{2\hbar^2}\!\left(\frac{e^2}{4\pi\epsilon_0}\right)^{\!2}\frac{1}{n^2} = -\frac{E_R}{n^2}, \qquad E_R = \frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2} \approx 13.6\,\text{eV}.

The constant $E_R \approx 13.6\,\text{eV}$ is the Rydberg energy. Notice the energy depends only on $n$ , not on $\ell$ or $m$ — a special "accidental" degeneracy of the pure Coulomb potential that we examine in the next lesson.

The Rydberg formula for spectral lines

When the electron falls from level $n_i$ to a lower level $n_f$ , it emits a photon carrying the energy difference:

E_\gamma = E_{n_i} - E_{n_f} = 13.6\,\text{eV}\!\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right).

This is the Rydberg formula. Transitions ending on $n_f = 2$ form the Balmer series, which falls in the visible band. The first Balmer line, $n=3 \to n=2$ , is the deep-red H $\alpha$ line.

The takeaway

Requiring a normalizable radial solution quantizes the energy as $E_n = -13.6\,\text{eV}/n^2$ , with $n = n_r + \ell + 1$ . Emission lines follow the Rydberg formula $E_\gamma = 13.6\,\text{eV}(1/n_f^2 - 1/n_i^2)$ .

Try it

This is a numerical exercise — return a number. Compute the energy (in eV) of the photon emitted when hydrogen relaxes from $n = 3$ to $n = 2$ , the H $\alpha$ Balmer line.

Run your code to see the quantum state.