# Why are electron energy levels negative?

Dec 28, 2017

It is because the electron is trapped in the atom. It must overcome a negative, stabilizing potential energy (through input of energy from outside sources) for the atom to be ionized.

The zero of the potential energy is then defined to be when the electron has escaped the atom; that is, when the atom has been ionized of that electron.

For a hydrogenic atom:

${E}_{\infty} = {\lim}_{n \to \infty} {E}_{n}$

$= {\lim}_{n \to \infty} \left\{- \text{13.60569253 eV} \cdot {Z}^{2} / {n}^{2}\right\}$

$=$ $\text{0 eV}$:

A hydrogen-like atom is the simplest example of what quantum physicists would call an "infinite spherically-symmetric well":

The proton(s) and electron interact via a coulomb potential $V$ (the potential energy of charge-charge interaction):

$V = - \frac{Z {e}^{2}}{4 \pi {\epsilon}_{0} r}$

where $e$ is the elementary charge of a proton and the negative sign accounts for the negative charge of the electron.

• When $V$ is negative, two charges are attracted.
• When $V$ is positive, two charges are repelled.

This is how it works out, regardless of sign conventions.

When solving the radial part of the hydrogen-like atom Schrodinger equation, one must rationalize that the presence of the proton stabilizes the electron motion, and that naturally correlates with negatively-valued energy levels.

When all is said and done, the energy expression becomes:

${E}_{n} = - \frac{\mu {Z}^{2} {e}^{4}}{8 {\epsilon}_{0}^{2} {h}^{2} {n}^{2}}$

When all the units are worked out and the energy is converted to electron-volts,

${E}_{n} = - \text{13.60569253 eV} \cdot {Z}^{2} / {n}^{2}$

And just to bring up a familiar form, one can show that this is really the Rydberg equation in a different form by subtracting the final and initial states:

$\Delta E = - \text{13.60569253 eV} \cdot {Z}^{2} \cdot \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$