Why doesn't the atom absorb heat energy when it is low?

[sophiecentaur]

True, but what if the atom/molecule goes into a virtual state (more probable than into an excited state, when the material is transparent ...)? In this case it is not a real absorption but a failed one, always and promptly followed by the re-emission ... It would be almost like with the Huygens sources ...
.

You are assuming that the propagation is based on single atom interactions. That only happens in low density gases, in which photons are actually absorbed by specific atomic energy transitions. When that happens, the re-emitted light goes in all directions and the result is a dark 'absorption line'. Why does it go in all directions? Because there is no longer phase coherence in the original propagation direction.
PS is all this stuff strange to you? It's pretty basic. Your above argument contains a "What if" and an "almost", as if there were some doubt about the validity of what we get in basic textbooks.

 

[DanMP]

You are assuming that the propagation is based on single atom interactions. That only happens in low density gases, in which ...

And what happens (with the photons) in air and water? How they propagate? They don't interact with the atoms/molecules? [Please exclude the absorption due to atomic/molecular energy transitions, because I'm talking about transparent materials with no (or very few) absorption lines.]

 

[sophiecentaur]

And what happens (with the photons) in air and water? How they propagate? They don't interact with the atoms/molecules? [Please exclude the absorption due to atomic/molecular energy transitions, because I'm talking about transparent materials with no (or very few) absorption lines.]

They interact with the whole lattice of atoms by Elastic Scattering and not by Absorption.
Personally, I find the wave model much easier to cope with here. If you want to talk in terms of individual photons then, if a photon isn't absorbed by a particular atom (= measured / observed) then it has no location and then you have to consider all the possible paths it could take through the medium (being elastically scattered off all the atoms in the material). This boils down to the same calculation as if you started off considering waves in the first place.

 

[Mister T]

And what happens (with the photons) in air and water? How they propagate? They don't interact with the atoms/molecules? [Please exclude the absorption due to atomic/molecular energy transitions, because I'm talking about transparent materials with no (or very few) absorption lines.]

They very much do interact with the molecules.

As an aside, can you explain why the sky is blue?

 

[DanMP]

They interact with the whole lattice of atoms by ...

What "lattice of atoms" in air? Can you provide a link about this?

They very much do interact with the molecules.

I know. My job is to record [IR &] Raman spectra ... That's why I suggested virtual states ...

As an aside, can you explain why the sky is blue?

Yes, blue light is scattered more than red light. It is also the reason for the red sun at the horizon.

 

[sophiecentaur]

What "lattice of atoms" in air? Can you provide a link about this?

I was being sloppy and referring to transparent solids and liquids. Gases are transparent and very low density, nonetheless, the scattering is still elastic and involves many atoms and not just one atom per photon (that's a sort of definition of the word elastic, I think you could say.)

My job is to record [IR &] Raman spectra

And Raman scattering is Inelastic, so it does involve interaction with individual molecules. It involves absorption and re-emission so that would introduce random phase shifting which would reduce the coherence of the idealised plane wave. The fact that you are looking at Spectral detail tells you that things are different from Rayleigh scattering, which is just dependent on particle size, with a small 'tilt' over the optical band.

 

[DanMP]

And Raman scattering is Inelastic, so it does involve interaction with individual molecules. It involves absorption and re-emission so that would introduce random phase shifting which would reduce the coherence of the idealised plane wave. The fact that you are looking at Spectral detail tells you that things are different from Rayleigh scattering, which is just dependent on particle size, with a small 'tilt' over the optical band.

In wikipedia there is a diagram:

showing/suggesting that Rayleigh scattering (the most dominant of the 3 scatterings above) involves also a short "transition" to a virtual energy state, meaning that there is a very short-lived "absorption" immediately and always followed by re-emission. That's why I wrote this.

 

[sophiecentaur]

In wikipedia there is a diagram:

showing/suggesting that Rayleigh scattering (the most dominant of the 3 scatterings above) involves also a short "transition" to a virtual energy state, meaning that there is a very short-lived "absorption" immediately and always followed by re-emission. That's why I wrote this.

I guess the proof of this would be in the shape of the emerging beam pattern which would have to be affected by even a small phase uncertainty. It sounds an interesting phenomenon.

 

[DrClaude]

showing/suggesting that Rayleigh scattering (the most dominant of the 3 scatterings above) involves also a short "transition" to a virtual energy state, meaning that there is a very short-lived "absorption" immediately and always followed by re-emission.

Be careful here. Virtual transitions are virtual, meaning they are not real. It is a coherent process where "absorption" and "emission" take place at the same time. It comes from a perturbative approach, just as virtual particles do (a subject that has been discussed endlessly on PF).

 

[Mister T]

Virtual transitions are virtual, meaning they are not real. It is a coherent process where "absorption" and "emission" take place at the same time. It comes from a perturbative approach, just as virtual particles do (a subject that has been discussed endlessly on PF).

But it provides the quantum mechanical explanation of why the sky is blue?

 

[sophiecentaur]

But it provides the quantum mechanical explanation of why the sky is blue?

There is a perfectly good explanation that doesn't involve QM so why introduce something like virtual particles if they are not necessary? There is the temptation to ask 'but what is really happening?' but that's not Physics. Sometimes you can get a good prediction from calculations using virtual particles so it clearly works but are you any nearer to 'the truth'?
"The TRUTH? You can't handle the TRUTH" - brilliant film.

 

[DanMP]

I guess the proof of this would be in the shape of the emerging beam pattern which would have to be affected by even a small phase uncertainty. It sounds an interesting phenomenon.

As far as I know, there is no problem with the phase/coherence through air :)

If you find the idea/phenomenon interesting and want/need a proof for it, I suggest to try/test it for the Fizeau experiment and the Sagnac effect. [I did it and I was amazed ... and very happy with the results]

why introduce something like virtual particles if they are not necessary?

What virtual particles? It was about virtual states ... And they are both already introduced and useful.

There is the temptation to ask 'but what is really happening?' but that's not Physics.

Well, for me it is. I'm interested to find "what is really happening".

Be careful here. Virtual transitions are virtual, meaning they are not real. It is a coherent process where "absorption" and "emission" take place at the same time. It comes from a perturbative approach, just as virtual particles do

Can you elaborate on this and/or provide a link? In wikipedia I found:

a virtual state is a very short-lived, unobservable quantum state

they still have lifetimes derived from uncertainty relations

 

[DrClaude]

I looked through many sources, but I didn't find any I could cite that would give a clear justification of what I said. But it is clear that these virtual states (or virtual transitions, see below) appear only through perturbation theory (hence to comparison with virtual particles). That said, I will try to cite different sources that I hope will clarify things.

(I will emphasise in red. All other emphasis is in the original.)

[Section 11.6 Raman spectroscopy] In the energy level scheme of Fig. 11.69 the intermediate state $E_v = E_k + \hbar \omega_0$ during the scattering process is often formally described as a virtual state, which, however, is not a real stationary eigenstate of the molecule. If this virtual state coincides with one of the molecular eigenstates one speaks of the "resonance Raman effect".

[Appendix E: Raman and two-photon transitions] A Raman transition involves two laser beams with frequencies $\omega_{\mathrm{L}1}$ and $\omega_{\mathrm{L}2}$ [...] A Raman transition between two atomic levels, labelled 1 and 2, involves a third atomic level, as shown in Fig. 9.20. This third level is labelled $i$ for intermediate, but it is very important to appreciate that atoms are not really excited to level $i$.
[...]
It is vital to realize that the Raman transition has a quite distinct nature from a transition in two steps, i.e. a single-photon transition from level 1 to $i$ and then a second step from $i$ to 2. The two-step process would be described by rate equations and have spontaneous emission from the real intermediate state. This process is more important than the coherent Raman process when the frequency detuning $\Delta$ is small so that $\omega_{\mathrm{L}1}$ matches the frequency of the transition between $|1 \rangle$ and $| i \rangle$.⁶ The distinction between a coherent Raman process (involving simultaneous absorption and stimulated emission) and two single-photon transitions can be seen in the following example.

[Section 5.7 Time-Dependent Perturbation Theory] We visualise that the transition due to the second-order term takes place in two steps. First $|i\rangle$ makes an energy non-conserving transition to $|m\rangle$; subsequently, $|m\rangle$ makes an energy-nonconserving transition to $|n\rangle$, where between $|n\rangle$ and $|i\rangle$ there is overall energy conservation. Such energy non-conserving transitions are often called virtual transitions. Energy need not be conserved for those virtual transitions into (or from) virtual intermediate states. In contrast, the first-order term $V_{ni}$ is often said to represent a direct energy conserving "real" transition.

[Section 6.2 Many-level systems, (c) The first-order correction to the wavefunction]
The last equation echoes the expression derived for the two-level system in the limit of a weak perturbation and widely separated energy levels (eon 6.18). As in that case, perturbation theory guides us towards the form of the perturbed state of the system. In this case, the procedure simulates the distortion of the state by mixing into it the other states of the system. This mixing is expressed by saying that the perturbation induces virtual transitions to these other states of the model system. However, that is only a pictorial way of speaking: in fact, the distorted state is being simulated as a linear superposition if the unperturbed states of the system.

I have also found a source that eschews virtual states and transitions:

[Section 4.3 Extending the perturbation approximation] It is very important to emphasise that the language of "intermediate states" or "virtual states" has no place in this discussion. Any of these non-linear processes occur while the driving radiation is not resonant with any of the states $|j\rangle$.

Finally, I found one source that tackles both Raman and Ramsey scattering at the same time. Note that explanation is semi-classical.

[Complement A_V 1.c.β The Raman effect]
Imagine that an optical wave of frequency $\Omega/2\pi$ strikes this molecule. This frequency, much higher than those considered previously, is able to excite the electrons of the molecule; under the effect of the optical wave, the electrons will undergo forced oscillation and re-emit radiation of the same frequency in all directions. This is the well-known phenomenon of the molecular scattering of light (Rayleigh scattering). What new phenomena are produced by the vibration of the molecule?

What happens can be explained qualitatively in the following way. The electronic susceptibility of the molecule is generally a function of the distance $r$ between the two nuclei. When $r$ varies (recall that this variation is slow compared to the motion of the electrons), the amplitude of the induced electric dipole moment, which vibrates at a frequency of $\Omega/2\pi$, varies. The time dependence of the dipole moment is therefore that of a sinusoid of frequency $\Omega/2\pi$ whose amplitude is modulated at the frequency of the molecular vibration $\omega/2\pi$, which is much smaller (fig. 3). The frequency distribution of the light emitted by the molecule is given by Fourier transform of the motion of the electric dipole shown in figure 3. It is easy to see (fig. 4) that there exists a central line of frequency $(\Omega - \omega) /2\pi$ (Raman-Stokes scattering) and frequency $(\Omega + \omega) /2\pi$ (Raman-anti- Stokes scattering).

It is very simple to interpret thesis lines in terms of photons. Consider an optical photon of energy $\hbar \Omega$ which strikes the molecule when it is in the state $|\varphi_v\rangle$ (fig. 5-a). If the molecule does not change vibrational state during the scattering process, the scattering is elastic. Because of conservation of energy, the scattered photon has the same energy as the incident photon (fig. 5-b; Rayleigh line). However, the molecule, during the scattering process, can make a transition from the state $|\varphi_v\rangle$ to the state $|\varphi_{v+1}\rangle$. The molecule acquires an energy $\hbar \omega$ at the expense of the scattered photon, whose energy therefore is $\hbar(\Omega - \omega)$ (fig. 5-c): the scattering is inelastic (Raman-Stokes line). Finally, the molecule may move from the state $|\varphi_v\rangle$ to the state $|\varphi_{v-1}\rangle$, in which case the scattered photon will have an energy of $\hbar(\Omega + \omega)$ (fig. 5-d; Raman-ani-Stokes line).

(note the description in terms of the scattering of a photon, not absorption-emission)

 

[DanMP]

I looked through many sources, but I didn't find any I could cite that would give a clear justification of what I said. But it is clear that these virtual states (or virtual transitions, see below) appear only through perturbation theory (hence to comparison with virtual particles). That said, I will try to cite different sources that I hope will clarify things.

Thank you very much for your interest and effort.

After reading all the material you offered, I'm not yet convinced that

"absorption" and "emission" take place at the same time

(note the description in terms of the scattering of a photon, not absorption-emission)

True, but this is probably because we usually say absorption/emission when we deal with transitions to/from a real energy state, not a virtual one. That's why I/we wrote "absorption"/"emission" (and "transition").In your last quote I found:

the molecule, during the scattering process, can make a transition from the state ...

This suggests that there is a process that takes time, probably very close to zero, but still different from zero, as my quote from wikipedia (in my previous post) also suggested: "they still have lifetimes derived from uncertainty relations".

By the way, when a photon interacts with a molecule or with an atom, there is the possibility, if the energy is right, to be absorbed, but most of the time the energy is not right and the photon is not absorbed. The question is: there is a trial and error process (that may take time), or the "decision" requires no time at all? Keep in mind that atoms and molecules are systems, made of smaller parts ...I couldn't ignore (again, in you last quote):

Imagine that an optical wave of frequency Ω/2π strikes this molecule. This frequency, much higher than those considered previously, is able to excite the electrons of the molecule; under the effect of the optical wave, the electrons will undergo forced oscillation and re-emit radiation of the same frequency in all directions.

If the 'optical wave' is "caused" by one photon and 'the electrons will undergo forced oscillation and re-emit radiation [photons?] of the same frequency', this process, without the "absorption" of the incident photon, would produce many similar photons, and this is true only for lasers (with a different mechanism).

 

[DrClaude]

This suggests that there is a process that takes time, probably very close to zero, but still different from zero, as my quote from wikipedia (in my previous post) also suggested: "they still have lifetimes derived from uncertainty relations".

I think Wikipedia is completely wrong here. If there would be any instant, however short, where the system could be found in the virtual state, then the state would be real (since we could measure the system in that state), not virtual.

It is important to note that the time-energy uncertainty principle is not Heisenberg's uncertainty principle, since time is not an observable in QM. And that idea of employing the uncertainty principle to justify what is happening is again reminiscent of the problem with virtual particles, where some will say that the energy needed to create the particle can be "borrowed" from the vacuum, so long as the lifetime of the particles is short, which is a very questionable statement.

By the way, when a photon interacts with a molecule or with an atom, there is the possibility, if the energy is right, to be absorbed, but most of the time the energy is not right and the photon is not absorbed. The question is: there is a trial and error process (that may take time), or the "decision" requires no time at all? Keep in mind that atoms and molecules are systems, made of smaller parts ...

QM doesn't work that way. The interaction of the photon with the atom (or molecule) will result in the atom and the electromagnetic field being in a superposition of non-excited atom + one photon and excited atom + no photon. Whether the photon was absorbed or not can only be discovered by means of a measurement. That measurement can be done at any time, and in that sense the transition is instantaneous (there is no intermediate state, so the atom can not be at any time "in the process" of absorbing a photon).

If the 'optical wave' is "caused" by one photon and 'the electrons will undergo forced oscillation and re-emit radiation [photons?] of the same frequency', this process, without the "absorption" of the incident photon, would produce many similar photons, and this is true only for lasers (with a different mechanism).

I did point out that this is a semi-classical explanation. The optical wave is here a classical electromagnetic wave, which is being scattered by molecules. There are no "new" photons coming out.

 

[DanMP]

I think Wikipedia is completely wrong here.

So, why it is still unchanged?

It is important to note that the time-energy uncertainty principle is not Heisenberg's uncertainty principle, since time is not an observable in QM.

It is not Heisenberg's uncertainty principle, but it is something worth considering, as you may see in this article.

QM doesn't work that way. The interaction of the photon with the atom (or molecule) will result in the atom and the electromagnetic field being in a superposition of non-excited atom + one photon and excited atom + no photon.

In my opinion QM is not quite complete, at least in explaining things. I found something to support this in Nature:

What Is Real?: The Unfinished Quest for the Meaning of Quantum Physics Adam Becker Basic: 2018.

All hell broke loose in physics some 90 years ago. Quantum theory emerged — partly in heated clashes between Albert Einstein and Niels Bohr. It posed a challenge to the very nature of science, and arguably continues to do so, by severely straining the relationship between theory and the nature of reality. Adam Becker, a science writer and astrophysicist, explores this tangled tale in What Is Real?.
...
What Is Real? is an argument for keeping an open mind. Becker reminds us that we need humility as we investigate the myriad interpretations and narratives that explain the same data.

So why not keeping an open mind and investigate the Fizeau experiment and the Sagnac effect as I suggested?

I did point out that this is a semi-classical explanation. The optical wave is here a classical electromagnetic wave, which is being scattered by molecules. There are no "new" photons coming out.

So how is the "non-classical" explanation? How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

 

[Mister T]

How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

New photons are produced in the medium.

 

[Nugatory]

So, why is [wikipedia] still unchanged?

For the same reason that Sisyphus's rock is still at the bottom of the hill
And seriously, kidding aside, there's a reason why Wikipedia is not in general an acceptable source under the forum rules.
(Although to be fair the anonymous Wikipedian who wrote that section may not have intended to make the suggestion that you took away, that "there is a process that takes time...").

In my opinion QM is not quite complete, at least in explaining things. I found something to support this in Nature:

That is a fairly non-controversial position, as quantum mechanics has never claimed to explain things in the sense that you're using the term here. Nonetheless, you would be better served by Becker's book than by Skibba's review of it.

So why not keeping an open mind and investigate the Fizeau experiment and the Sagnac effect as I suggested?

The first step in the open-mind review is to understand the best current theory. It's difficult to improve on something when you don't know what it does and does not do well.

 

[DanMP]

The first step in the open-mind review is to understand the best current theory. It's difficult to improve on something when you don't know what it does and does not do well.

That's why I asked the question:

how is the "non-classical" explanation? How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

New photons are produced in the medium.

Ok, but from one photon entering the medium (air, water), we don't usually get more, similar photons at the other end, as wikipedia suggests ...

 

[Nugatory]

Ok, but from one photon entering the medium (air, water), we don't usually get more, similar photons at the other end, as wikipedia suggests ...

I'm not seeing that suggestion in that Wikipedia article (which only mentions "photons" in one place, unrelated to this discussion).

 

[Lord Jestocost]

So how is the "non-classical" explanation? How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

As Peter W. Milonni writes in “Answer to Question #21 [‘‘Snell’s law in quantum mechanics,’’ Steve Blau and Brad Halfpap, Am. J. Phys. 63 (7), 583 (1995)]” (American Journal of Physics 64, 842 (1996)):

“In the quantum-mechanical description of a plane wave incident on a dielectric medium, each photon has a probability amplitude to be scattered by anyone atom. The complete probability amplitude for a photon to be found at any point inside or outside the medium is the amplitude for it to get there without any scattering, plus the sum over all the possible paths by which it can get there via single- and multiple-atom scattering. The result of this superposition of all possible probability amplitudes is an amplitude that is nonzero both inside and outside the medium.”

 

[DanMP]

I'm not seeing that suggestion in that Wikipedia article (which only mentions "photons" in one place, unrelated to this discussion).

This is from Wikipedia (the underline and the comment in red are mine):

At the atomic scale, an electromagnetic wave's phase velocity is slowed in a material because the electric field creates a disturbance in the charges of each atom (primarily the electrons) proportional to the electric susceptibility of the medium. (Similarly, the magnetic field creates a disturbance proportional to the magnetic susceptibility.) As the electromagnetic fields oscillate in the wave, the charges in the material will be "shaken" back and forth at the same frequency.[1]:67 The charges thus radiate their own electromagnetic wave [photons, right?] that is at the same frequency, but usually with a phase delay, as the charges may move out of phase with the force driving them (see sinusoidally driven harmonic oscillator). The light wave traveling in the medium is the macroscopic superposition (sum) of all such contributions in the material: the original wave plus the waves radiated by all the moving charges. This wave is typically a wave with the same frequency but shorter wavelength than the original, leading to a slowing of the wave's phase velocity. Most of the radiation from oscillating material charges will modify the incoming wave, changing its velocity. However, some net energy will be radiated in other directions or even at other frequencies (see scattering).

 

[DanMP]

“In the quantum-mechanical description of a plane wave incident on a dielectric medium, each photon has a probability amplitude to ...

Ok, but how is this related to my question? It doesn't say anything about the speed of light in the medium.

 

[Lord Jestocost]

It doesn't say anything about the speed of light in the medium.

In case one wants to understand how the apparent speed of light in media comes about, I recommend to thoroughly read chapter 31 “The Origin of the Refractive Index” in “The Feynman Lectures on Physics, Volume I “(http://www.feynmanlectures.caltech.edu/I_31.html).

 

[Nugatory]

This is from Wikipedia (the underline and the comment in red are mine):

[1]:67 The charges thus radiate their own electromagnetic wave [photons, right?]

Not right. A light wave is not made up of photons the way you're thinking.

 

[Mister T]

This is from Wikipedia (the underline and the comment in red are mine):
[...]The charges thus radiate their own electromagnetic wave [photons, right?]

You need large collections of photons for the electromagnetic wave approximation to be valid.

 

[DanMP]

In case one wants to understand how the apparent speed of light in media comes about, I recommend to thoroughly read chapter 31 “The Origin of the Refractive Index” in “The Feynman Lectures on Physics, Volume I “(http://www.feynmanlectures.caltech.edu/I_31.html).

Thank you!

I just read it. It was interesting, but still a classical explanations ... with a peculiar way of seeing the electrons in the atoms:

Each of the electrons in the atoms of the plate will feel this electric field and will be driven up and down (we assume the direction of E0 is vertical) by the electric force qE. To find what motion we expect for the electrons, we will assume that the atoms are little oscillators, that is, that the electrons are fastened elastically to the atoms, which means that if a force is applied to an electron its displacement from its normal position will be proportional to the force.
You may think that this is a funny model of an atom if you have heard about electrons whirling around in orbits. But that is just an oversimplified picture. The correct picture of an atom, which is given by the theory of wave mechanics, says that, so far as problems involving light are concerned, the electrons behave as though they were held by springs. So we shall suppose that the electrons have a linear restoring force which, together with their mass m, makes them behave like little oscillators, with a resonant frequency ω0.

This is not the modern view of the atom, and apparently not consistent with the views expressed here.

 

[Nugatory]

Thank you!

I just read it. It was interesting, but still a classical explanations ... with a peculiar way of seeing the electrons in the atoms:

Yes of course it is - you entered this thread in #33 looking for a classical explanation. Much of the subsequent discussion has been about trying to stop you from confusing yourself by unnecessarily introducing quantum misconceptions into the discussion.

This is not the modern view of the atom

Of course not. It is an accurate and thorough classical analysis of a problem that can be treated classically and for which the methods of qantum electrodynamics are overkill. You don't have to be satisfied by it, but if you aren't willing to learn quantum electrodynamics, there's nothing better on offer except this and other semiclassical descriptions.

 

[Mister T]

... with a peculiar way of seeing the electrons in the atoms

Not peculiar to people who work in that area. In fact, the casual learner is better served by this simple model of matter than the more complex models presented everywhere from chemistry textbooks to popular science books.

 

[Lord Jestocost]

This is not the modern view of the atom, and apparently not consistent with the views expressed here.

This is a complete different story and refers to the acceleration of an atom as a whole.

 

[cmb]

. If the photon energy doesn't match this energy difference, then the atom's energy level doesn't change. We describe that by saying that the photon is not absorbed.

There is something about this thread I don't like!

So ... if this is the whole story ...
firstly, photons from a black body radiator should never be able to excite an atom, because the 'analogue' energy from a BB emitter will never exactly match the energy levels between electron states, there will always be a little bit left over.
secondly, hydrogen electrons have, as a maximum, an energy state of ~13 eV , so any photons with more than 14 eV cannot ionise hydrogen!

I don't accept those corollaries, please can we look at this proposition that incoming energy has to be exactly the electron band levels, but also if an atom does absorb energy higher than its band energy, what happens to the rest?

 

[phinds]

Thanks to everyone for your opinion.

Just so you are clear, no one here has given you an opinion, they have given you facts. Science does not work on opinions.

 

[sophiecentaur]

because the 'analogue' energy from a BB emitter will never exactly match the energy levels between electron states,

You seem to be wanting a classical and a quantum argument to apply at the same time "Exactly" does not apply. The "analogue" energy you are describing can be regarded as consisting of many different photons and not a continuum. There is a finite spread of energy levels amongst the H atoms and there are a finite number of photons with energy that fall within this spread of atomic energy levels. So the absorption is not of a single energy but in a restricted band of energies. Different substances have different widths of absorption lines and you can look at this as different bandwidths of receiver.

if an atom does absorb energy higher than its band energy, what happens to the rest?

The surplus energy goes into Kinetic Energy of the released electron.

 

[cmb]

You seem to be wanting a classical and a quantum argument to apply at the same time "Exactly" does not apply. The "analogue" energy you are describing can be regarded as consisting of many different photons and not a continuum. There is a finite spread of energy levels amongst the H atoms and there are a finite number of photons with energy that fall within this spread of atomic energy levels. So the absorption is not of a single energy but in a restricted band of energies. Different substances have different widths of absorption lines and you can look at this as different bandwidths of receiver.

The surplus energy goes into Kinetic Energy of the released electron.

I don't think quantum theory allows for a 'band' of energies for electron states. What is the range of this 'band'?

I am aware that the energy bands can change depending on electromagnetic effects (Zeeman/Stark), is that what you mean?

In regards the idea that a freed electron has any 'spare energy' manifest as kinetic energy I understand the "mechanical logic" of it but I don't find this convincing for the following 'philosophical' reason; if the state of the freed electron was defined by both the atom from which it came and the energy that impinged on that atom, then it naturally implies that the electron and the atom's nucleus were like separate components of a two-component system, but that turns my understanding of quantum physics around because surely it is the 'atom' that defines the energy levels of the electron, not the nucleus of the atom as a distinct object from the electron?

 

[Dale]

What is the range of this 'band'?

The range of the band for an individual transition depends on the characteristic time of the transition. For a transition that happens quickly the linewidth is broad, and for a transition that is very unlikely the linewidth is narrow.

When there are many transitions involved it gets more complicated with overlapping transitions forming broad bands, as described.