What is the energy of the first excited state of hydrogen atom? How high is it above the ground state if the ground state is at −13.61 eV?
Answer:
Well, the ground state is the \(1 s^{1}\) configuration… the first excited state (search \(\text { “H I” }\)) is the \(1 s^{0} 2 p^{1}\) configuration.
The energy for this hydrogen-like atom is given by:
\(E_{n}=-Z^{2} \cdot \frac{13.61 \mathrm{eV}}{n^{2}}\)
where \(Z\) is the atomic number and \(n\) is the energy level of the orbital the electron is in.
\(E_{2}=-1^{2} \cdot \frac{13.61 \mathrm{eV}}{2^{2}}\)
\(=-3.40 \mathrm{eV}\)
[You could have wrongly chosen the \(1 s^{0} 2 s^{1}\) configuration and coincidentally gotten the “right” answer, because to a first approximation \(E_{2 p}=E_{2 s}\) in hydrogen atom.]
How many electrons does hydrogen have? What is its atomic number?
From the reference above,
which says that the first excited state lies \(10.20 \mathrm{eV} \text { above }\) the ground state. And in fact,
