# Maximum number of Spectral Lines "A better quantum model shows that there will be n^2 transitions"?

I was parsing the following post What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state? and came across with the reply from @porphyrin.

cite : "lease note that these answers are only true in the Bohr model, a better quantum model shows that there will be $$๐^2$$ transitions. The orbitals are now labelled with increasing orbital angular momentum, โ=0,1,2,3..."

Please, does someone can point me out to references or more extended explanations, from where does the $$๐^2$$ come from?

Update The general formula for calculating number of spectral lines is (n2โโn1โ)(n2โ โn1โ+1)โ/2; which "merely" relate to triangular numbers n(n+1)/2. I wonder about $$n^2$$ since the sum of two consecutive triangular numbers give a square number; what does the (n-1)n/2 constituent part describe? Is this due to Pauli exclusion Principle; Bose/Einstein Vs Fermi/Dirac distribution?

Really appreciate.

PS: I don't have 50 reputation to comment directly

• Phys. chem. textbooks cover this as well as specialist books on spectroscopy and, of course, books on quantum mechanics. Mar 23 at 12:51
• Dear @porphyrin thanks for the reply, I have updated my question. Please could you provide me with references or articles explaining the idea; I'm lost with precise keywords.. Mar 23 at 13:11
• @Xavier These type of questions can easily become "rabbit holes" unless you are comfortable with the basic model systems of quantum mechanics. Do you know the quantum mechanical treatment of angular momentum and the treatment of the hydrogen atom and selection rules ? Mar 23 at 15:40
• @HansWurst, thanks :) - I'm used to live in rabbit hole questions. I feel confortable. Yesterday night I finally end up in the Zeeman Effect and start to read about selection rules.. does it have something to do with $n^2$ Mar 24 at 12:22