Asked by malina
A Li2+ ion had its electron in an excited state. When the electron relaxed to the ground state (n = 1) a photon of light was emitted with an energy of 1.884×10−17 J. What energy level was the electron in before it transitioned to the ground state?
Answers
Answered by
DrBob222
(1/wavelength) = RZ^2(1/n^2 - 1/n^2)
(1/wavelength) = RZ^2*(1/1^2 - 1/x^2) and solve for x.
R = Rydberg constant = 1.0973732
The first n is n = 1 and the second n is n = 2; that's where the 1/1^2 and 1/x^2 come from.
Z is the atomic number of the element; in this case that is 3 so Z^2 = 9.
You will need to convert energy to wavelength with E = hc/wavelength. Post your work if you get stuck.
(1/wavelength) = RZ^2*(1/1^2 - 1/x^2) and solve for x.
R = Rydberg constant = 1.0973732
The first n is n = 1 and the second n is n = 2; that's where the 1/1^2 and 1/x^2 come from.
Z is the atomic number of the element; in this case that is 3 so Z^2 = 9.
You will need to convert energy to wavelength with E = hc/wavelength. Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.