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

Question

okay I saw the rydberg formula but how am i or what numbers am i supposed to plug where

Answer 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

Answer 2

DrWLS showed you what to do.
1/lambda = R_HZ²*[(1/N₁²)-(1/N₂²].

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

Answer 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

Answer 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?

Answer 5

so i got 1/4797297.498

Answer 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.

Answer 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.

Answer 8

so I did 1 divided by 4797297.498

and got 2.085 X 10^-7

Answer 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 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.