# Rydberg formula for hydrogen

I've been told that if a hydrogen atom is exposed to electromagnetic radiation of wavelength $$\lambda$$ such that Rydberg's formula

$$\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^2}-\frac{1}{n_{2}^2}\right)$$

is valid for two integers $$n_1$$ and $$n_2$$, then the electron in the hydrogen atom would go from level $$n_1$$ to level $$n_2$$.

1. If this is true, what would happen if, given a fixed $$\lambda$$ and an initial energy level $$n_1$$, $$n_2$$ turns out not to be an integer?

2. I've been reading that, actually, that equation doesn't describe what I've written above but the contrary process. Namely, that if the electron of the hydrogen atom is already excited and at an energy level $$n_2$$, then when it goes back to a lower level $$n_1$$ a photon of wavelength $$\lambda$$ is emitted. Are both processes (electron going from $$n_1$$ to $$n_2$$ and viceversa) described by that equation or just one of them?

3. I'm having trouble getting this whole level thing. I mean, a hydrogen atom has only one orbit, so how could its only electron go from one orbit to another if theres is just one of them?

• The hydrogen atom doesn't have any orbits. It has orbitals and there is an infinite number of them. In which orbital the electron "can be found" depends on the excitation state of the atom. One can "walk" atoms up and down these energy levels, although hydrogen is not a very good system for absorption spectroscopy of this kind. First of all it doesn't come as atomic hydrogen by as molecular $H_2$, secondly, the first series of excitation energies are in the ultraviolet, which is experimentally more difficult than measurements in the optical and near IR. – CuriousOne Apr 2 '16 at 4:08

1. Such a transition can't occur because $$n_1$$ and $$n_2$$ must be integers since there's no state with non-integer primary quantum number. So hydrogen atom doesn't emit a photon which has wavelength that doesn't satisfy Rydberg's formula.