Amount of photons released during decrease of energy of vibration

[mcfaker]

An atom has the following energy of vibration: E= nhv. n can only be in the range of whole numbers.

So if an vibrating atom changed energy, say from 6hv to 2hv, it means that 4hv of energy is released in light/photons

Does that mean that 4 photons are "released" by the atom (?), because The energy of one photon is equal to hv ?


Thanks for the help!

 

[Borek]

No, single photon is emitted. It just has different ν.

 

[mcfaker]

But an atom can have several electrons, each electron emitts just one photon. So if 4hv energy is emitted, there must be 4 photons emitted because the energy of 1 photon is hv.
It can't be that an atom that emitts 2hv, or 4hv of energy, only emitts 1 photon?

 

[Borek]

You are apparently mixing things and TBH I have no idea what where to start, as it is not clear what misconceptions are behind your reasoning.

Atoms don't vibrate - vibrations are characteristic of molecules, at least diatomic. Quantum oscillator energy is given as $E_n=(n+\frac 1 2)\hbar \omega$.

Single electron transition (which has nothing to do with oscillations) emits a single photon of some characteristic energy.

You can convert this energy to the photon frequency, but it is not that all photons in the whole universe have the same frequency ν and the same energy hν.

Just because atom has several electrons doesn't mean they all get excited at the same time.

When the oscillator falls from an excited stated to the more basic one it can do it in one large step (emitting $k\hbar\omega$ energy) or in k steps, emitting $\hbar\omega$ in each. But it has nothing to do with electrons.

 

[mcfaker]

Thanks, does oscillation energy of an atom represent the total energy of an atom? or is it just partial energy?
I know that atoms have internal potential energy& kinetic energy.