# How do molecules bubble off a photon?

After being excited by a photon, an electron of a photoactive molecule jumps to a higher electronic state. When it relaxes, the molecule emits a photon (in simple terms). How is this photon "generated"? Photons are particle/waves, right? So somehow this particle has to form. I picture it in a way similar to when you blow through a ring of soap-water and a bubble forms.

How does the photon really form?

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When Particle meets it's antiparticle (for example electron meets positron) they will annihilate and 'generate' two photons: Example that shows electron-positron annihilation: $e^+e^-\rightarrow \gamma\gamma$ And Energy of that photon will be $E=2m_ec^2$ ($2m_e$ because we have two particles(electron an positron) which have same mass) and the frequency of that photon is: $f = \frac{2m_ec^2}{h}$ – Gigi Butbaia May 2 '14 at 14:05
And when electron goes from higher energy state to lower energy state it emits light, because at higher energy state it has more energy than in lower energy state, so when it goes down it has to 'lose' energy in order to go down so it emits light(photon) – Gigi Butbaia May 2 '14 at 14:11
I appreciate your response but you're not saying anything that I havn't said in my post. – TMOTTM May 4 '14 at 18:57