Rydberg Equation

[charlotte_jacobs_4I]

I have been looking around and I can't find a clear answer to tell me if in the Rydberg equation it is R((1/(the original value of n))-(1/(the final value of n)) or the opposite way for the final and initial values. Please let me know if you can! Thank you!

[Yvonne Du]

I think in the book they stated that it should be n(initial)-n(final). I ran into the same problem. For one of the homework questions, I got a negative answer when I put n(initial)-n(final), but the answer key says it is positive. They flipped the other to get a positive number. I think I'll go with the initial-final as long as I don't get a negative number? Don't know if this helps.

[charlotte_jacobs_4I]

I was running into the same problem. I'm not sure but is the answer ever supposed to be negative?

[Chloe Thorpe 1J]

I was wondering this as well. The textbook says it's initial-final, so I've been using that for my calculations.

[Hailey Boehm 2H]

I was also wondering about this. In the atomic spectra video module it says that the equation is v=-R[(1/n1^2) - (1/n2^2)], where n2>n1. This would mean n2 is n initial. However, Mr. Lavelle also recommends to not use this equation because it is really easy to mess up the signs, so you should just use En= -hr/n^2 and just subtract Efinal - E initial.

[Ashita Tanwar 3H]

Does this mean that the equation given in the atomic spectra module always uses the lower energy level "n" as n1, regardless of whether it is the final or initial value? For example, if an electron was dropping from n=4 to n=2, then would the equation be the same if the electron was jumping up from n=2 to n=4?

[Tony Chung 2I]

I believe I heard during lecture that it is final - initial.

[Megan_Ervin_1F]

Does the order of final and initial change if the problem is having to do absorption instead of emission?