# Why are electron energy levels negative?

##### 1 Answer

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_(oo) = lim_(n->oo) E_n#

#= lim_(n->oo) {-"13.60569253 eV" cdot Z^2/n^2}#

#=# #"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 = -(Ze^2)/(4pi epsilon_0 r)# where

#e# is theelementary chargeof 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 = -(mu Z^2 e^4)/(8epsilon_0^2 h^2 n^2)#

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

#E_n = -"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:

#DeltaE = -"13.60569253 eV" cdot Z^2 cdot (1/n_f^2 - 1/n_i^2)#