# Chemistry

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 hydrogen-like 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

1. 8
1. 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

posted by drwls
2. DrWLS showed you what to do.
1/lambda = RHZ2*[(1/N12)-(1/N22].

RH = 1.0967758341 x 10^7
Z = 3
N1 = 3
N2 = 4
Solve for lambda.

posted by DrBob222
3. 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

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

posted by DrBob222
5. so i got 1/4797297.498

posted by Lauren
6. 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.

posted by DrBob222
7. 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.

posted by DrBob222
8. so I did 1 divided by 4797297.498

and got 2.085 X 10^-7

posted by Lauren
9. 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 RH, 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 RH.

posted by DrBob222

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