Locating Li (2+) Energy Levels - eV

AI Thread Summary
The discussion focuses on finding the energy levels for Li (2+) in electron volts (eV). It highlights that Li (2+) behaves like a hydrogen-like ion, with energy levels approximated by the formula E=13.6(Z^2/n^2) eV. Participants mention that while quantum mechanics can provide detailed solutions, the energy levels are well-tabulated and readily available. The importance of filling electron shells according to quantum numbers and considering excitation energy levels is also noted. Overall, the consensus is that the energy levels can be calculated using a simplified formula, which is sufficient for most applications.
Haftred
Messages
53
Reaction score
0
Where can I find the different energy levels (in eV) for Li (2+). I can't seem to find it in my textbook or on the internet.
 
Physics news on Phys.org
You could try quantum mechanic solutions ... not for the faint hearted though...
If you know the number of electrons in Li you have to begin filling shells using the quantum numbers n,l and Ml .. which each have restrictions in their values. You keep filling the electrons in the shells (dont forget spins!) until you are left with an unfilled shell. Excitation energy levels occur when the last electron is placed in the higher energy levels and not the next available level.
Have fun
 
that's way too much work considering that the levels are well tabulated.
 
Li2+ is just a hydrogen-like ion, I believe, so its energy levels are given simply by:

E=13.6\frac{Z^2}{n^2}\ eV=\frac{122.4}{n^2}\ eV

This isn't exact because it neglects fine structure and hyperfine structure, but for most purposes, the above should suffice.
 
Last edited:
comparing a flat solar panel of area 2π r² and a hemisphere of the same area, the hemispherical solar panel would only occupy the area π r² of while the flat panel would occupy an entire 2π r² of land. wouldn't the hemispherical version have the same area of panel exposed to the sun, occupy less land space and can therefore increase the number of panels one land can have fitted? this would increase the power output proportionally as well. when I searched it up I wasn't satisfied with...
Back
Top