- #1
skrat
- 748
- 8
Homework Statement
Two times ionized Lithium (Z=3) in excited states emits two photons, one with ##\lambda _1=72.91 nm## and another one with ##\lambda _2=13.5 nm##.
In which excited state was originally the ion?
Homework Equations
The Attempt at a Solution
Since two times ionized Lithium has only one electron I assume I can say that the energies of eigenstates are calculated as ##E_n=\frac{R_y}{n^2}##.
I somehow imagined that I have to sum the energy of photons and equal that with the expression above.
##\frac{hc}{\lambda _1}+\frac{hc}{\lambda _2}+R_y=R_y(1-\frac{1}{n^2})##
But this gives me some weird result...
What do I have to do? :/