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We took up the topic of particle nature of light recently.As the light is quantised when an electron jumps from higher to lower orbit it releases energy but as the energy is quantised it will release them in distinct packets.So while the electron is coming down:

  • It's potential energy decreases as smooth fxn

  • It releases a distinct amount of energy(as seen from its optical spectra)

So the question arises what happens in between the jump.As the energy cant be released until it's sufficient to form a photon of corresponding energy,where does the energy go.

Upon asking my teacher he said it's released as radiation as electron is accelerating.But then again radiation is released as photons.

P.S.-I know elementary b.sc level maths and physics,so answers of appropriate level will be appreciated.

Rahul Narwar
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  • Does this help? https://physics.stackexchange.com/a/369093/123208 – PM 2Ring Oct 26 '18 at 06:41
  • The discussion is on its movement but I am more confused by the the process of photon formation and energy conversion of p.e. into radiation – Rahul Narwar Oct 26 '18 at 10:10
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    As John said in his last paragraph, the electron is never measured in some intermediate-energy state. When we measure it, we either see the electron in the high energy orbital, or we see it in the low energy orbital plus a fully-formed photon. – PM 2Ring Oct 26 '18 at 10:37
  • Ok I get it a bit.thanks what more is known on the process of generation of photon – Rahul Narwar Oct 26 '18 at 15:51
  • Dan's answer below is good, but also see https://physics.stackexchange.com/questions/94049/does-a-photon-instantaneously-gain-c-speed-when-emitted-from-an-electron in particular, the answer by John Rennie. – PM 2Ring Oct 27 '18 at 02:08

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Here's one way to think about it.

Consider a simplified model that can be in either of two kinds of configuration. In configuration $A$, the electron is in the higher-energy orbital and no photon is present. In a configuration of type $B$, the electron is in the lower-energy orbital and one photon has been released, flying away from the atom. Both configurations are defined to have the same total energy.

In this simplified model, the transition would be described something like this. The state starts as pure $A$. As time passes, it becomes a quantum superposition of $A$ and $B$, with the contribution of $A$ smoothly becoming smaller and the contribution of $B$ smoothly becoming larger. Eventually (very quickly by human standards), the contribution of $A$ is practically zero, and the state is practically pure $B$.

The total energy is always the same, because if $A$ and $B$ have the same energy, then so does any quantum superposition of $A$ and $B$.

The process is smooth, even though we never have just part of a photon. If we measure the number of photons, we always get an integer. But when we're not measuring the number of photons, we can have a quantum superposition of different numbers of photons. I can't define these statements in a way that is both concise and satisfying, but hopefully this concise answer at least gives you an interesting avenue to investigate.

Chiral Anomaly
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