# Energy levels, de Broglie

1. Mar 31, 2008

### dalarev

The Formula for computing E(n) = -(Z^2)/(n^2) * Ei

is fairly straight forward. Am I right in thinking this formula refers to the energy level of the atom with Z number of protons? This formula, what it yields at least, doesn't depend on the number of electrons the atom has?

Also, when trying to find the wavelength of the emitted/absorbed photon, should we use hc/E or h/p ? What makes a wave a de Broglie wave? I have deeper doubts but I don't even know where to start, I hope with more replies it'll become clearer.

2. Mar 31, 2008

### Staff: Mentor

This formula works only for hydrogen and for "hydrogen-like" ions, that is, ions with only one electron, e.g. He+, Li++, Be+++, etc. If there's more than one electron, the electrons interact with each other and this affects the energy levels.

It doesn't make any difference, because for a photon, E = pc. This follows from the general formula for energy, momentum and rest mass:

$$E^2 = (pc)^2 + (m_0 c^2)^2$$

with $m_0 = 0$ for a photon.

3. Mar 31, 2008

### dalarev

I suppose for the purpose of my class (and my exam today) this general formula will be enough. Actually, they love to use the energy 13.6 eV in the book, I believe this is the same case that only applies for single electron-atoms.

That's exactly right..I didn't think it through very well.

What about when they ask for the energy of the photon that is emitted? The wavelength of the emitted/absorbed photon will always equal the Energy needed for the electron to move from one energy level to the next, correct?

4. Apr 1, 2008

### G01

The wavelength of an emitted/absorbed photon will be related to the energy decrease/increase of the electron that absorbed it by:

$$\Delta E=hc/\lambda$$

Last edited: Apr 1, 2008