# Determine frequency of emitted photon

1. Aug 18, 2011

### liquidFuzz

Still some days of holidays left.

Back in school, I've must have been 14 years old or something. Our teacher showed that different substances emits different coloured light. From the top of my head I remember copper - green and sodium - yellow. He showed more, but these will do for now.

Sodium has a valence electron in 3s. Lets say I want to check whether it is the jump from 3s to 3p and back that creates photons of this yellow wave length. Now I get puzzled. from my understandings the jump between energy levels are the source of the light.

$\displaystyle E_n = -\frac{13.6}{n^2}$

But this doesn't explain why the 3p orbital is at lower energy than 3s. Since $\displaystyle -\frac{13.6}{3^2} = -\frac{13.6}{3^2}$

I don't know whether I should be embarrassed by missing the obvious or flinch back because it's so complicated. How do I calculate the frequency of a jump from 3p to 3s..?

2. Aug 19, 2011

### Zarqon

There's a couple of reasons why the formula you write is not applicable in this case. First, the value 13.6 is the ionization energy for Hydrogen; other species will have other values here, so you can't take this straight off. Secondly, the formula you use only deal with changes to the principal quantum number, the "n", and can thus not be used at all to see what happens to changes of the orbital quantum number, i.e. between the s and p.

In general, calculating the energy difference between two sets of arbitrary quantum numbers is a very difficult task, computationally, and cannot be written down in a simple formula like the one for hydrogen. Some special cases exist which are still simple, and it's possible that sodium is on the simpler side, since it's a hydrogen-like atom, however, I don't know such a formula off the top of my head.

3. Aug 19, 2011

### liquidFuzz

Mhmm... I'm still curious. Do you have any links or suggested words for a google search?

4. Aug 19, 2011

### JeffKoch

The 13.6/n^2 equation for Hydrogen comes from the simplest approximate solution to Schroedinger's equation for hydrogen, which in particular neglects spin-orbit coupling and relativistic effects - when you do a more correct treatment, the degeneracy between the different "l" levels is removed, and these energy level shifts become more pronounced as some power of Z (I forget what the power is). Dig around with words like "fine structure", "spin orbit coupling", and the like.

But yes, in general calculating these things correctly can turn into one's life work, and people devote entire professional careers to the effort.