Rydberg Equation

[Dan_Jin_1K]

Hey guys, I'm a little confused on when we should be using the Rydberg Equation that the textbook provided (V = R (1/n1^2) -(1/n2^2)) vs the one that we learned about in lecture (E= -hR/n^2).

Can someone explain to me which one we should use and why?

Thank you!

[Jessica Huang 1M]

You can use either one. I personally use the E=-hR/n^2 because I understand what I am looking for. That equation would tell me the energy of the specific energy level and I can follow that. I don't like using the Rydberg equation because I don't understand how I can get from energy levels to wavelength. It's up to you on which equation to use. :)

[Edgar Khachatryan 3G]

Both can be used: the latter is just a simplified version.

[Myra_Zhan_2N]

Essentially, both can be used, but I personally prefer to use the latter one. When you have a question asking about the change from one energy level to another, I find it easier to plug the values in individually into the second equation. Afterwards, you can find the change in E by subtracting E (final)- Energy (initial).

[Chem_Mod]

An analogy to help you better understand this would be the thermometer example. The energy calculated from equation E=-hR/n^2 would be the increments on the thermometer (or energy level of the potential well in this case). And if I want to know the temperature change (energy difference) between to increments (levels), the Rydberg Equation can be used. But this is not to say that E=-hR/n^2 cannot be used to calculate energy difference between 2 energy levels, you still can. I encourage you to derive the Rydberg Equation from E=-hR/n^2 by setting two energy level (one being n_initial and other being n_final). It is not necessary, but learning the derivation should answer your question fully.

[Sebastian]

If you use e=hv and plug in -hr/n^2 for e, the h cancels, yielding the v=r(n^2-n^2)