# Relation of Hydrogen orbitals to its spectral series?

I'm looking for the link between the Rydberg formula for hydrogen spetcral series

$$\frac{1}{\lambda_{\mathrm{vac}}} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

and this image. Is it right to say that the Balmer Series is created by all transitions from (x, _, _) to (2, _, _) where x>2 and _ are "don't cares"?

Are the three (2, _, _) states all of the same energy?

• The Rydberg formula you have there is for determining the wavelengths of the transitions between different energy levels of the hydrogen atom, which are characterized merely by the principal quantum number $n$ (that establishes the energy of the state). In the image, each image represents a state with the form $|n,l,m\rangle$, so you can see that each row represents the same value of the energy (same $n$) and it's different degeneracies by adding the possible values of $l$ and $m$. – Charlie Nov 14 '18 at 17:38
• The Balmer series definition is correct, and it's just a particular case of the general formula for transitions. Also, all the three (2,-,-) you see there correspond to the same energy (states with the same degeneracy), and the same applies for the other rows. – Charlie Nov 14 '18 at 17:39
• There are different n in the first three rows,\ but I get your point. – Jasper Nov 14 '18 at 17:40
• My bad, yeah you're right. I meant only the ones with the same (n,-,-) have the same energy. – Charlie Nov 14 '18 at 17:43

The 'link' is, in short, the Schrödinger equation.

The orbitals plotted in the image ─ the wavefunctions $$\psi_{n,l,m}(r,\theta,\phi)$$ ─ are the hydrogenic solutions to the hydrogen Schrödinger equation, $$\left[ -\frac{\hbar^2}{2m}\nabla^2 - \frac{e^2}{4\pi\epsilon_0} \frac{1}{r} \right]\psi_{n,l,m}(r,\theta,\phi) = E_{n,l,m} \:\psi_{n,l,m}(r,\theta,\phi),$$ for which you require the energy to be $$E_{n,l,m} = E_{n} = -\frac{m e^{4}}{(4\pi\epsilon_0)^{2}\hbar ^{2}}{\frac {1}{2n^{2}}},$$ independently of $$l$$ and $$m$$.

Thus,

Are the three (2, _, _) states all of the same energy?

Yes.

Is it right to say that the Balmer Series is created by all transitions from (x, _, _) to (2, _, _) where x>2 and _ are "don't cares"?

Yes, that is correct (though it's important to note that selection rules generally apply, and not all the transitions within that set actually contribute any significant signal).

For further details, see any introductory textbook on quantum mechanics.