# Only we can calculate the energy emitted when n_x rarr n_tonly in the hydrogen atom not in any other atom. When will be the new equation which can apply to all atoms found in the future?????

Apr 27, 2017

Because the hydrogen atom has only one electron, so there are no electron repulsions to complicate the orbital energies. It is these electron repulsions that give rise to the different energies based on the angular momenta of each orbital shape.

The Rydberg equation utilizes the Rydberg constant, but the Rydberg constant, if you realize it, is actually just the ground state energy of the hydrogen atom, $- \text{13.61 eV}$.

-10973731.6 cancel("m"^(-1)) xx 2.998 xx 10^(8) cancel"m""/"cancel"s"

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

$= - {13.60}_{739}$ $\text{eV}$ $\approx - \text{13.61 eV}$

Thus, it is constructed FOR the hydrogen atom.

It would be very impractical to construct a working equation for more complicated atoms, because rather than one orbital energy per $n$, we would have $\boldsymbol{n}$ orbital energies at each $n$, and $2 l + 1$ orbitals for each $l$ within the same $n$.

We would also have to account for the spectroscopic selection rules that require $\Delta l = \pm 1$, rather than allow all possible transitions.

Instead of one electronic transition upwards, say, for $n = 2 \to 3$, we would have to, for light atoms, only take $2 s \to 3 p$, $2 p \to 3 s$, and $2 p \to 3 d$, and we cannot take $2 s \to 3 d$ for instance. Of course, you may also accidentally get $2 s \to 2 p$, which does not satisfy $n = 2 \to 3$.

This would make for a very complicated equation for general chemistry students to dissect...