8 posts • Page 1 of 1
I just think of the equation as v=R(1/nf^2 - 1/ni^2) but i've seen that it only works for emission problems. However, I noticed that making the Rydberg constant negative in this same formula for absorption problems works as well. It just depends if it's an emission or absorption problem :)
It depends on whether the problem tells you if it is an emission or absorption of energy, and what which values of n it gives you for the transition of energy. Like the above post says, it depends on the context of the problem.
In v=R(1/(n2)^-1/(n1)^2), n2 is the lower energy level and n1 is the higher energy level. I like to use v=R(1/(nF)^-1/(nI)^2), where nF is the final energy level and nI is the initial energy level. In my opinion, using nF and nI instead of n2 and n1 is "better" since I have seen others switch around n2 and n1, and it also makes it clear which energy level is the final one and which is the initial one.
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