# Electron IN Atoms

1. Aug 26, 2004

### darkar

What energy do electrons poses in atoms?

2. Aug 26, 2004

### Locrian

What atom? What orbital level? Be more specific.

For instance, the energy level of an electron in a hydrogen atom in orbital level n is

$$E_n = -(me^4)/((4*pi*e_o)^2(2h_b*n^2))$$

Gah, pardon the terrible laytex. That should give you an idea though. Just do a search on the web for a prettier written formula.

3. Aug 26, 2004

### marlon

Like Locrian pointed out an electron has different possible enery-values, each corresponding to the orbital they are on. Each capital level is denoted by a quantumnumber n that has the "values" 1,2,3,.. corresponding to the K,L,M,N,...levels. Per level there are n possible sublevels denoted as s(0),p(1),d(2),f(3)-levels (these are the orbitals, or the probability value of the angular solutions of the Schrödinger-equation.). Per orbital there are 2n+1 possible "sub-sub"levels. for example d has number 2, so there are 5 different d-orbitals that contain maximum two electrons of the same energy when no EM-field is applied on the atom. These 5 d-robitals each correspond to a different energy...

Per sub-sub-level there is also the magnetic quantumnumber that has two values for spin 1/2-particles. Each energy-level is split in two when an extern magnetic field is applied over the atom, due to the two possible values of this quantumnumber. This effect is called the Zeemann-effect.

regards
marlon

4. Aug 26, 2004

### ZapperZ

Staff Emeritus
Er... I'm going to nitpick here a bit since I see a few problems...

1. We all agree that the original question was VAGUE.

2. I have no idea what formula Locrian is quoting.

3. At the SIMPLEST level, if we adopt the Bohr model (which we all know is not correct, but is damn accurate for H atom, so-so for slightly heavier atoms, and completely whacky for very heavy atoms), then one can find the "energy" of the electronic state from usual sources.

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

4. We need to be clear about one thing here. The principle quantum number, n, typically dictates the dominant energy state of the atom. The various orbitals (the s,p,d, etc) that are within each n are DEGENERATE states. By that we mean they have the identical energy state within each n. It is only when the degeneracy is REMOVED, either via spin-orbit coupling, application of external fields, etc., that they will have separate energy states.

Zz.

5. Aug 26, 2004

### marlon

I think with the correct addendum of Zz, the question is solved...

regards
marlon

6. Aug 26, 2004

### Locrian

Well, if you plug in all the constants and convert them to units of eV, you get something like -13.5eV/n^2 ... maybe that's more familiar?

7. Aug 27, 2004

### darkar

Other than that, electrons always move in the electron cloud right? Can you say that they have kinetic energy?

8. Aug 27, 2004

### da_willem

He probably meant: $$E_n=-\left[\frac{m}{2\hbar^2} \left(\frac{e^2}{4\pi \epsilon_0} \right) ^2 \right]\frac{1}{n^2}=\frac{E_1}{n^2}=\frac{-13.6eV}{n^2}$$

Wich is the famous "Bohr formula"