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If when you shine a photon into an atom for example, and this excites an electron to a higher energy level, do the electron(s) keep going higher the more light you shine, and is there an energy limit, if so why?

Secondly, if you then stop shining light, why will the electrons fall back to a lower level? Will they at all? And why? it seems arbitrary that they will unless acted on by something else.

If they do fall back, how long does it take till they do?

I am taking QM but have not reached this understanding yet.

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  • $\begingroup$ About which energy limit are you writing? The limit of the strength of the light you shine on the atom? Or maybe the limit of the energy to ionize the atom (for one electron), which depends on which electron is excited? $\endgroup$ – descheleschilder Dec 14 '17 at 10:02
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    $\begingroup$ The snarky answer is: because they can. $\endgroup$ – Derrell Durrett Dec 14 '17 at 18:07
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    $\begingroup$ @DerrellDurrett That's actually the best answer by far. Quantum mechanics gets a lot easier once you realise that everything that can happen will happen. $\endgroup$ – Dawood says reinstate Monica Dec 14 '17 at 19:57
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    $\begingroup$ Thanks, @DawoodibnKareem. I completely agree that absorbing this lesson makes QM far easier to understand. $\endgroup$ – Derrell Durrett Dec 14 '17 at 20:00
  • $\begingroup$ Rem it's not about how much light but the frequency of the light. $\endgroup$ – John ClearZ Dec 14 '17 at 20:46
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The answer is thermodynamics, and the assumption that you're working in a colder environment than the temperature corresponding to a Planck distribution where your photons would be "on average" fairly present. In other words, inside a star, where it is hotter, the atoms are NOT in their ground state most of the time - in fact, if it is hot enough, they are in their "highest state" which is an ionised state: you have a plasma. It is simply because most atomic matter has energy levels with differences that are much larger than the average photon energies at "room temperature" (about 26 milli-eV) that we tend to say that atoms and molecules are in their ground states. It is because at these low temperatures, it is statistically favorable to have energy spread out more than in concentrated excited states.

BTW, you can see that with rotational states of molecules: at room temperature, these are usually NOT in their ground state and excited rotational states don't "decay to ground state". It is because their energy levels are below 26 meV.

So when you "shine light on an atom" in a cold environment, you put it out of thermodynamic equilibrium, and it will tend back to equilibrium which is its ground state. When you "shine light on an atom" in a hot environment, it will not fall back to its ground state, because that's not its equilibrium state.

An atom in a cold environment will decay to ground state through spontaneous emission, which has an exponential time decay that is depending on the specific state and is quite difficult to calculate.

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    $\begingroup$ Note that as you're shining light on an atom, you're changing the environment, and thus which energy levels are most statistically favourable (with all those photons fying around, an atom with no excitation energy sticks out as a statistical anomaly just like an excited atom in a cold environment). This is why you are able to consistently use that light to excite atoms. $\endgroup$ – Arthur Dec 14 '17 at 11:38
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    $\begingroup$ @anna: I wasn't talking about hot atoms, but about hot environments, essentially meaning the temperature of the vacuum (at reasonable temperatures this only affects the EM field). Think of a single atom in a cavity filled with black body radiation at temperature T. In an environment where k T is of the order of the excitation energy, spontaneous emission is mostly negligible, because you will get an equilibrium with stimulated emission (by the present thermal radiation) and absorption (of the thermal radiation) so as to populate the energy levels according to statistical mechanics. $\endgroup$ – entrop-x Dec 14 '17 at 12:47
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    $\begingroup$ @arthur: yes. I should have said: after you stop shining light on it, it will... $\endgroup$ – entrop-x Dec 14 '17 at 12:51
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    $\begingroup$ @anna: I'm trying to give a better view than the standard initial pedagogical view, which somehow, tells you that "excited atoms at rest decay to the ground state by some magical rule". Excited atoms only "decay out of stationary states" because there's a coupling to another system not taken into account in the stationary states, and then it depends on the thermodynamic state of the system one couples with (here, the EM field). If you couple to a COLD EM field, yes, you go to the ground state. If you couple to a hot EM field, not at all. $\endgroup$ – entrop-x Dec 14 '17 at 16:03
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    $\begingroup$ @anna: but without QED, there wouldn't be any spontaneous decay. Without QED, all you can have is electrostatics, and then stationary states are, eh, stationary. No decay. Stable excited states. Hell, they are stationary. No time evolution. Decay is a phenomenon of coupling with QFT. You get spontaneous decay if you couple to a cold QED field. $\endgroup$ – entrop-x Dec 14 '17 at 16:19
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I want to add about spontaneous emission. Excited states of atoms are not stationary states because of atoms are not isolated QM systems. There always is interaction with electromagnetic field. Schrodinger equation for atoms in simplest form takes into account only Coulomb interaction between electrons and nucleus. In this simplified approach excited states are stationary and spontaneous emission has no place.

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    $\begingroup$ This is entirely correct. If there were no coupling with the EM field at all, atoms wouldn't be sensitive to photons and both could have different temperatures. $\endgroup$ – entrop-x Dec 14 '17 at 9:35
  • $\begingroup$ What about the vacuum quantum fluctuations? Are they not capable to induce transitions from higher to lower energy states in an atom? $\endgroup$ – descheleschilder Dec 14 '17 at 9:55
  • $\begingroup$ @descheleschilder: that's exactly what Gec was talking about. $\endgroup$ – entrop-x Dec 14 '17 at 10:27
  • $\begingroup$ I can't see it explicitly written, but I assume you mean There always is interaction with electromagnetic field? $\endgroup$ – descheleschilder Dec 14 '17 at 10:42
  • $\begingroup$ Yes, there is always interaction with the EM field. The vacuum fluctuations are the "something else" in "acted on by something else". $\endgroup$ – garyp Dec 15 '17 at 18:32
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  • do the electron(s) keep going higher the more light you shine (...)?

No, because energy levels are quantized. This means that no matter how many photons you throw at the electron (i.e. the intensity of the light source), it won't jump to an higher energy level unless the frequency ($\nu$) of the photons is right, i.e. if

$$ \nu = \frac{\Delta E} h $$

where $\Delta E$ is the energy difference between the energy levels and $h$ is Planck's constant.

  • Secondly, if you then stop shining light, why will the electrons fall back to a lower level? Will they at all? And why? it seems arbitrary that they will unless acted on by something else.

Yes, they do fall back, and the reasons are two:

  1. An atom is never truly isolated, and it will interact with the external electromagnetic field.
  2. Even if we assume that the atom is in free space, far from any source of EM field, it will still be subject to vacuum fluctuations of the EM field and thus eventually decay to a lower energy level. This process, which is called spontaneous emission, cannot be explained if the EM field is treated as a classical object, and its description requires the formalism of quantum field theory. For a more detailed discussion, see for example the Wikipedia page:

Spontaneous transitions were not explainable within the framework of the Schroedinger equation, in which the electronic energy levels were quantized, but the electromagnetic field was not. Given that the eigenstates of an atom are properly diagonalized, the overlap of the wavefunctions between the excited state and the ground state of the atom is zero. Thus, in the absence of a quantized electromagnetic field, the excited state atom cannot decay to the ground state. In order to explain spontaneous transitions, quantum mechanics must be extended to a quantum field theory, wherein the electromagnetic field is quantized at every point in space. The quantum field theory of electrons and electromagnetic fields is known as quantum electrodynamics.

In quantum electrodynamics (or QED), the electromagnetic field has a ground state, the QED vacuum, which can mix with the excited stationary states of the atom. As a result of this interaction, the "stationary state" of the atom is no longer a true eigenstate of the combined system of the atom plus electromagnetic field. In particular, the electron transition from the excited state to the electronic ground state mixes with the transition of the electromagnetic field from the ground state to an excited state, a field state with one photon in it. Spontaneous emission in free space depends upon vacuum fluctuations to get started.

  • If they do fall back, how long does it take till they do?

The probability that the transition has not happened at time $t$ is $1-p$, where $p$ is the probability that it has happened. To calculate the transition probability per unit time, you can use the Einstein coefficients.

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Classical electromagnetic theory could not explain the spectral lines seen in light spectra. It predicted that an electron caught around a proton would radiate continuously (due to the radial acceleration) and fall on the proton by emitting a continuous spectrum of light. No stable hydrogen atoms would exist.

Here is the hydrogen atom:

hydrogen energy

The fixed orbits explained the observed spectra, the Balmer and Lyman series and this was one of the main pillars for inventing quantum mechanics.

balmer

copied

Notice that the lines have a width

If when you shine a photon into an atom for example, and this excites an electron to a higher energy level,

Let us take the hydrogen atom as an example.

Yes, the atom will be excited if the photon has the frequency/energy of the difference in the energy levels

do the electron(s) keep going higher the more light you shine, and is there an energy limit, if so why?

It is not a matter of quantity but of the appropriate energy differences

Look at the energy levels . For a photon to hit an electron and ionize the atom it has to have a frequency h*nu=13.6eV. To transit to an intermediate level, the photon has to have the difference in energies, otherwise it just scatters off the field of the atom and leaves it intact.

If you give a photon the correct steps in energy , then the electron can step up until ionization. Again, it is the energy or the photon that has to match the energy level differences.

If you have a hydrogen gas, and a source with the appropriate energy levels to excite the atom then the more photons, the more excited states. General frequencies will ony by chance fit the correct difference.

Secondly, if you then stop shining light, why will the electrons fall back to a lower level?

There is a calculable quantum mechanical probability for the electrons to fall to an empty lower level, because it is a law of nature, quantum mechanical and classical, that everything goes to the lowest allowed energy level where stability exists.

Will they at all? And why? it seems arbitrary that they will unless acted on by something else.

The electron can either cascade down the levels releasing photons with the appropriate frequency/energy, or go in one step to the lowest energy level. All these are calculable probabilities in the quantum mechanical frame. Your "acted on by something else" translates in saying that the Schrodinger solutions are not enough to give widths to the predicted lines, which is true, one needs quantum electrodynamics.

A single atom in space, if in an excited state has a calculable probability to fall to the ground state ,which is modeled by using interactions with the QED vacuum.

If they do fall back, how long does it take till they do?

When you study further you will see that this is connected with the width of the line, which is not a strict energy line but has a Δ(E) again calculable. These calculations need quantum field theory , not the simple Schrodinger model, as valerio92 quotes in his answer. The Heisenberg uncertainty principle then ties the decay times with this energy width.

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  • $\begingroup$ Excellent explanation. Thank you very much for you time!! $\endgroup$ – Paulo Dec 15 '17 at 9:44
  • $\begingroup$ @anna v "everything goes to the lowest allowed energy level where stability exists". Does lowest allowed energy mean lowest potential or it takes into consideration also the kinetic energy? $\endgroup$ – ado sar Sep 19 at 12:53
  • $\begingroup$ @adosar In quantum mechanics there are energy levels in which the electrons can exist, the potential is taken into account in the solution to find those energy levels, there is no "ptential and kinetic energy". $\endgroup$ – anna v Sep 19 at 18:08
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In this article you can find the reason for spontaneous emission.

The excitation of an electron depends of course on the frequency of the photons you shine on the atom. In practice, these photons are wave packets. If we assume the outer electron is excited the photons must have energies that are equal to the difference (of which there are many) of the energy levels of the electron. If we take one frequency (or better said, a wave packet around a mean value for the frequency), associated with the energy difference between the first energy level of the electron and the second (below that energy the photon can't excite the electron), the more photons you shine on the atom, the higher the energy state of the electron (you have to shine them of course at a faster rate than the rate at which they fall back to lower energy levels). So by shining a lot of photons with the just amount of energy, the electron will eventually be knocked off the atom (the spreading of energies becomes less the higher the energy states).

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