How can I calculate the excited state energy level?

1 Answer
Truong-Son N.
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. #"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:

![www.physics.nist.gov)

#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")#

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