Question

How can I calculate the excited state energy level?

Answer 1

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

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For hydrogen-like atoms, i.e. `"H"`, `"He"^(+)`, `"Li"^(2+)`, etc., the energy levels are given by:

`E_n = -Z^2 cdot "13.61 eV"/n^2`

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

So for `"He"^(+)`, the first excited state energy level would be the `1s^0 2p^1` configuration:

`color(blue)(E_2) = -2^2 cdot "13.61 eV"/2^2`

`= color(blue)(-"13.61 eV")`

And its ground state energy would be:

`E_1 = -2^2 cdot "13.61 eV"/1^2`

`= -"54.44 eV"`

So, its first excited state lies `"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")`