Recent content by sindre00

E = hc/lambda, hence E= (6.6*10-34)*(3.0*108) / (5.89*10-7m) So this will give me the energy difference of the ground state and excited state as the energy of the photon is equal to the energy difference?

Ok, thank you. So according to my question, which formula would be the most appropriate to use for calculation of the energy difference in ground and exited state in a sodium atom? Thank you

Ok. So does every atom have a constant number (which does not have to be calculated) for every energy level?

Ok. Thank you. So the atom emits an photon when it drops down from the excited state to a lower state, and absorb an photon when it goes from a lower to a higher energy state. I do know the different states of a hydrogen atom (calculated by the rydebergs formula) but is it possible for use that...

Homework Statement The emission spectrum of thermally excited sodium atoms practically consists of a single intensive line at 589 nm wavelength. What is the energy difference (in eV units) between the excited and ground states of the sodium atom? Homework Equations E = hc/lambda, we also know...