Chemistry
posted by Lauren on .
What is the wavelength of the transition from n=4 to n=3 for Li2+? In what region of the spectrum does this emission occur? Li2+ is a hydrogenlike ion. Such an ion has a nucleus of charge +Ze and a single electron outside this nucleus. The energy levels of the ion are Z^2RH/n^, where Z is the atomic number
okay I saw the rydberg formula but how am i or what numbers am i supposed to plug where

The energy levels are
En = Z^2RH/n^2
You left out the 2.
In your case, calculate En for Z = 3, with n = 4 and again for Z = 3 and n=3. Call those energy numbers E4 and E3.
E4  E3 is the wave number (1/wavelength) of the photon emitted.
Invert that to get the wavelength 
DrWLS showed you what to do.
1/lambda = R_{H}Z^{2}*[(1/N_{1}^{2})(1/N_{2}^{2}].
R_{H} = 1.0967758341 x 10^7
Z = 3
N_{1} = 3
N_{2} = 4
Solve for lambda. 
okay so it should look like this
1/lambda= 1.0967758341 x 10^7(3^2)*[(1/3^2)(1/4^2)]
1.0967758341 x 10^7(9)(.0486)
i got 4798394.274 
I don't get that. Your work LOOKS ok to me except your answer is 1/lambda. You need to take the reciprocal, right?

so i got 1/4797297.498

Plug that into your calculator; i.e., divide 1 by that large number and the answer will be the wavelength. That's what you are trying to find.

One note.
Since you are carrying all the other numbers to such high precision, I would suggest you carry the 0.0486 you obtained to the same number of significant figures. 
so I did 1 divided by 4797297.498
and got 2.085 X 10^7 
That's what I have. Usually the answer is expressed either in Angstrom units or in nanometers. Your answer is 208.5 nm or 2085 Angstroms. Again, if you want to carry that many places for R_{H}, then you are justified in more placed in the answer. I don't know how your professor wants it done. Many use 1.097 x 10^7 for R_{H}.