Only we can calculate the energy emitted when n_x rarr n_tnxntonly 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?????

1 Answer
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, -"13.61 eV"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) "eV" ~~-"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 bbn orbital energies at each n, and 2l+1 orbitals for each l within the same n.

We would also have to account for the spectroscopic selection rules that require Deltal = pm1, rather than allow all possible transitions.

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

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