# Energy levels, de Broglie

## Main Question or Discussion Point

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.

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jtbell
Mentor
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?
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.

Also, when trying to find the wavelength of the emitted/absorbed photon, should we use hc/E or h/p ?
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.

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.
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.

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.
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?

G01
Homework Helper
Gold Member
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?
The wavelength of an emitted/absorbed photon will be related to the energy decrease/increase of the electron that absorbed it by:

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

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