Verify that the ground state energy $E_0$ is $13.6\text{ eV}$ by using $E_0=\frac{2{\pi }^2{q}_e^4{m}_e{k}^2}{h^2}$.

Hi Ryan, thanks for the question. $E\_0$ is the ground state only for hydrogen, whereas other "hydrogen-like" atoms have ground states in terms of $E\_0$ of $E = -Z^2E\_0$, where I substituted $n = 1$ for the ground state in equation [\[30.26\]](https://openstax.org/books/college-physics/pages/30-3-bohrs-theory-of-the-hydrogen-atom#eip-673) $E\_n = -\\dfrac{Z^2}{n^2}E\_0$, and $Z$ is the number of protons in the atom. The electron is more tightly bound to nuclei with more protons. Equation [\[30.26\]](https://openstax.org/books/college-physics/pages/30-3-bohrs-theory-of-the-hydrogen-atom#eip-673) is pretty limited, however, since it's only for "hydrogen-like" atoms, which means there is **only one** electron. A better term for them would be "hydrogen-like ions", rather than atoms, since, with the exception of hydrogen, they would all be charged. When additional electrons surround the nucleus (it's no longer a hydrogen-like ion in other words), things get more complicated and there doesn't exist a formula for calculating the ground state energy of the inner-most election in that circumstance. You might find the Wikipedia discussion of [ionization energy](https://en.wikipedia.org/wiki/Ionization_energy) interesting. The "ground state energy of the hydrogen atom" is the same as the "ionization energy of hydrogen". All the best, Shaun