# Why Do Stark Manifold Graphs All Have Negative Energy?

I have been studying Rydberg-Stark State Atoms and their Stark Manifolds (like the one on Wikipedia: http://en.wikipedia.org/wiki/File:Hfspec1.jpg) and I was wondering, Why does the y-axis (of Energy of Rydberg Atom) only show Negative Energies? I thought it was impossible for something to have a negative energy? - How is this possible?

Thanks in advanced for any help!

BTW, I understand the concept of a Stark Manifold (which shows the Stark Shifting of Ryderg atoms in Different Electric Field Strengths), and the concept of High and Low-Field-Seeking States of these atoms.

Take for example the $n = 13$ line on the graph, which intercepts the $y$ axis at $650\text{cm}^{-1}$. Converting $650\text{cm}^{-1}$ to a wavelength by taking the reciprocal and dividing by $100$ gives the corresponding wavelength as $\lambda = 15.38\mu\text{m}$, and converting to an energy using $E = hc/\lambda$ gives $E = 1.292 \times 10^{-20}\text{J}$ or $0.080\text{eV}$.

Now, there's a convenient way to calculate the energy of the $n$th energy level of hydrogen:

$$E_n = - \frac{13.6}{n^2} \text{eV} \tag{1}$$

where the energy is negative because we take the energy of the unbound electron to be zero, and binding the electron into the atom decreases its energy. Obviously decreasing from zero gives a negative energy.

Anyhow, if we use equation (1) to get the energy of the $n = 13$ energy level we get:

$$E_{13} = - \frac{13.6}{13^2} = -0.080\text{eV}$$

which is exactly what we worked out corresponds to $650\text{cm}^{-1}$ on the graph.

So the graph is just showing the energy of the $n$th levels of hydrogen, and they're negative because the energy of the unbound electron is being taken as the zero.