Question

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

Answer 1

The Rydberg formula has terms 1/n1^2 and 1/n2^2 in it, wwhere n1 and n2 are the quantum numbers of the lower and upper electron enery levels of the transition. In your case, n1=2 and n2=3. There should also be a Z^2 tem, and in your case Z=3. The wavelength should be t 1/9 of the wavelength of the Balmer red line of hydrogen, which has the same quantum numbers. This puts the corresponding Lithium line into the so-called "vacuum ultraviolet".

Answer 2

so am I solving for Rh

This is what the equation should look like right:

1/9= Rh (1/2)-(1/3)

Answer 3

No. You are supposed to be solving for the wavelength, not Rh. The Rydberg constant for Li+++ is nine times the Rydberg for H+, because Z=3. That makes the wavelengths 1/9 as large, for the same values of n1 and n2.

Your formula is also wrong. The quantum numbers get squared.

Take a look at "Ryberg constant for hydrogen" at
(Broken Link Removed)