# How can I calculate the excited state energy level?

May 1, 2015

This is only known exactly for the hydrogen-like atoms. Otherwise, it is done experimentally via photoelectron spectroscopy.

For hydrogen-like atoms, i.e. $\text{H}$, ${\text{He}}^{+}$, ${\text{Li}}^{2 +}$, etc., the energy levels are given by:

${E}_{n} = - {Z}^{2} \cdot \frac{\text{13.61 eV}}{n} ^ 2$

where $Z$ is the atomic number and $n$ is the quantum level.

So for ${\text{He}}^{+}$, the first excited state energy level would be the $1 {s}^{0} 2 {p}^{1}$ configuration:

$\textcolor{b l u e}{{E}_{2}} = - {2}^{2} \cdot \frac{\text{13.61 eV}}{2} ^ 2$

$= \textcolor{b l u e}{- \text{13.61 eV}}$

And its ground state energy would be:

${E}_{1} = - {2}^{2} \cdot \frac{\text{13.61 eV}}{1} ^ 2$

$= - \text{54.44 eV}$

So, its first excited state lies $\text{40.83 eV}$ above its ground state. That matches the electronic energy level difference here from NIST:

329179 cancel("cm"^(-1)) xx 2.998 xx 10^10 cancel"cm"/cancel"s"

xx 6.626 xx 10^(-34) cancel"J"cdotcancel"s" xx ("1 eV")/(1.602 xx 10^(-19) cancel"J")

= "40.82 eV" ~~ ul("40.83 eV")