Chemistry
posted by Lauren .
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

Chemistry 
drwls
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 
Chemistry 
DrBob222
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. 
Chemistry 
Lauren
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 
Chemistry 
DrBob222
I don't get that. Your work LOOKS ok to me except your answer is 1/lambda. You need to take the reciprocal, right?

Chemistry 
Lauren
so i got 1/4797297.498

Chemistry 
DrBob222
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.

Chemistry 
DrBob222
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. 
Chemistry 
Lauren
so I did 1 divided by 4797297.498
and got 2.085 X 10^7 
Chemistry 
DrBob222
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}.
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