Posted by **Lauren** on Sunday, November 4, 2007 at 11:22am.

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

- Chemistry -
**drwls**, Sunday, November 4, 2007 at 11:48am
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**, Sunday, November 4, 2007 at 11:55am
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**, Sunday, November 4, 2007 at 12:22pm
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**, Sunday, November 4, 2007 at 12:59pm
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**, Sunday, November 4, 2007 at 1:16pm
so i got 1/4797297.498

- Chemistry -
**Lauren**, Sunday, November 4, 2007 at 2:07pm
so I did 1 divided by 4797297.498

and got 2.085 X 10^-7

- Chemistry -
**DrBob222**, Sunday, November 4, 2007 at 2:17pm
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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