The emission of a photon and change in electron radius is a single, inextricable event.
It happens on order of Planck time, too fast for any measurement to be possible, no matter how refined the observing apparatus, because the observing apparatus would necessarily change the state of the electron by measuring it.
In other words, the events are as close to being simultaneous and instantaneous as makes no difference.
2007-08-03 07:33:08
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answer #1
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answered by lithiumdeuteride 7
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First, it is important to realize that the 'orbits' in quantum mechanics are not like classical orbits. Instead, they are fuzzier 'orbitals' that are distributed around the atom. The electrons do not have definite positions like they do in classical physics. Similarly, the photons are described both by waves and by particle descriptions.
Because of this, there is no time at which the electron is 'between' orbitals. Instead, there is a probability that it is in each one and that probability changes over time. Initially, it is most likely to be in the higher energy orbital. Then the probability that it is in that orbital decreases and the probability that it is in the lower orbital increases. Simultaneously, the probability that a photon has appeared increases from 0 to 1.
Since the orbitals of the electron are not physically separated (the overlap actually determines the likelihood of the transition), the shift is instantaneous without there having to be an infinite velocity at any point. The photon is emitted when the shift actually happens (which is probabilistic).
2007-08-03 08:35:53
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answer #2
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answered by mathematician 7
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I believe the current hypothesis/theory/ is that the electron emits the photon, causing it to jump to a lower orbit. This idea of the photon as the catalyst certainly makes sense when you consider that electrons jump up when they absorb photons, and they oughtn't jump up before they absorb more energy.
There's no problem with the electron moving down at infinite speed; the electron has mass, so it only be accelerated to a finite speed.
2007-08-03 07:29:13
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answer #3
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answered by lockedjew 5
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Like nature_boy11 and lithiumdeuteride say, the emission of the photon and the step down to a lower orbit are both simultaneous and instantaneous.
I'd like to add one more point.
Do not look at it as "going down to a lower orbit" in the classical sense like a particle (or a satellite) moving closer to the nucleus (or planet). Because this is the classical/Newtonian way of looking at it. But we are dealing with quantum energies and quantum distances here.
Besides,... the electron does not really have a fixed position or distance away from the nucleus. -- if you look at it as a wave. We only give it fixed distances to make certain calculations simpler.
But in truth, it is neither particle nor wave, but exhibits characteristics of both, depending on the conditions. And depending on what you, the observer, want to determine.
It is convenient to speak of distances and assign certain energy values to certain distances from the nucleus, but this is only for convenience's sake. The truth is not so simple.
It is better to look at it as a step down from having one particular amount of energy to a lower amount of energy. And the jump is in a discrete step, not a gradual one, hence the word "quantum." (or "quantum leap" if you like). There is no in-between value. Welcome to the world of the quantum.
2007-08-03 07:59:07
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answer #4
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answered by ╡_¥ôò.Hóö_╟ 3
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The electron drops down to the lower-energy orbit instantaneously, and the photon is emitted at this exact moment.
2007-08-03 07:31:39
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answer #5
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answered by Nature Boy 6
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