How can I calculate the excited state energy level?

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How can I calculate the excited state energy level?

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Answer 1 · 2015-05-01T06:01:27

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

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

$E_{n} = - Z^{2} \cdot \frac{13.61 eV}{n^{2}}$ $E_n = -Z^2 cdot "13.61 eV"/n^2$

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

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

$E_{2} = - 2^{2} \cdot \frac{13.61 eV}{2^{2}}$ $color(blue)(E_2) = -2^2 cdot "13.61 eV"/2^2$

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

And its ground state energy would be:

$E_{1} = - 2^{2} \cdot \frac{13.61 eV}{1^{2}}$ $E_1 = -2^2 cdot "13.61 eV"/1^2$

$= - 54.44 eV$ $= -"54.44 eV"$

So, its first excited state lies $40.83 eV$ $"40.83 eV"$ 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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How can I calculate the excited state energy level?

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