Question

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 1

Well, the ground state is the `1s^1` configuration... the first excited state (search "`"H I"`") is the `1s^0 2p^1` configuration.

The energy for this hydrogen-like atom is given by:

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

where `Z` is the atomic number and `n` is the energy level of the orbital the electron is in.

`color(blue)(E_2) = -1^2 cdot "13.61 eV"/(2^2)`

`= color(blue)ul(-"3.40 eV")`

[You could have wrongly chosen the `1s^0 2s^1` configuration and coincidentally gotten the "right" answer, because to a first approximation, `E_(2p) = E_(2s)` in hydrogen atom.]

How many electrons does hydrogen have? What is its atomic number?

From the reference above,

`DeltaE_(1->2) = 82259.158 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")`

`=` `ul"10.20 eV"`

which says that the first excited state lies `"10.20 eV"` above the ground state. And in fact,

`DeltaE_(1->2) = E_2 - E_1`

`= overbrace(-"3.40 eV")^"first excited state" - overbrace((-"13.61 eV"))^"ground state" ~~ ul"10.20 eV"`

as required.