What is the effect of photon-electron collisions in X-ray imaging?

[genefalk]

My understanding is:
A) If a photon strikes an atom and the photon has exactly the correct amount of energy to move the electron to the next (another) energy level, the electron will absorb the photon, the photon will cease to exist, and the electron will move.

B) If a photon strikes an atom and the photon contains some other amount of energy, then the atom will gain some kinetic energy.

My question:
In case B) what happens to the photon?

 

[avarmaavarma]

B) is incorrect. The only transfer of energy occurs either wholly (one quantum = 1 photon) or not at all.
It seems hard to visualize since intuitively, the photon should transfer some K.E. to the electron. In reality, no energy transfer happens unless a full photon is absorbed. Energy transfers are always quantized (all or nothing..no continuous in-between values).

 

[Bob S]

B) is incorrect. The only transfer of energy occurs either wholly (one quantum = 1 photon) or not at all.
It seems hard to visualize since intuitively, the photon should transfer some K.E. to the electron. In reality, no energy transfer happens unless a full photon is absorbed. Energy transfers are always quantized (all or nothing..no continuous in-between values).

How does Rayleigh scattering in air work? The blue skylight has roughly a smooth 1/λ⁴ spectrum.

Bob S

 

[avarmaavarma]

Rayleigh scattering does not involve absorption or emission of photons. It is an elastic process - so the final energy of the photon is the same as its initial energy. The 'blue' photons have more energy and are scattered more (than say the red). Hence the sky appears blue. So - no real 'transfer' of energy in the sense of absorption/emission actually occurs in Rayleigh scattering.

 

[Matterwave]

B) is incorrect. The only transfer of energy occurs either wholly (one quantum = 1 photon) or not at all.
It seems hard to visualize since intuitively, the photon should transfer some K.E. to the electron. In reality, no energy transfer happens unless a full photon is absorbed. Energy transfers are always quantized (all or nothing..no continuous in-between values).

This seems to be contradicted by Compton scattering wherein the photons scatter off an electron and by doing so, the wavelength of the photon changes...hence it has lost some energy...but it was not absorbed wholly.

 

[kof9595995]

This seems to be contradicted by Compton scattering wherein the photons scatter off an electron and by doing so, the wavelength of the photon changes...hence it has lost some energy...but it was not absorbed wholly.

I think that's because in Compton scattering you must knock the electron off from the atom, i.e. from a bound state to a unbound state, so the specturm is no longer quantized, in the sense that as long as your phonton has a greater than ionization energy, you can transfer energy to the electron

 

[avarmaavarma]

In Compton scattering, photons are absorbed and re-emitted. Here is a thread that somewhat discusses this. There may be more references out there. The 'change' in wavelength in Compton scattering is due to the different energy photon that is emitted.

 

[avarmaavarma]

This time with the actual thread link...

https://www.physicsforums.com/archive/index.php/t-178653.html

 

[kof9595995]

Now I'm also kinda confused:let's say if an incident photon has an energy higher than the first excited energy but lower than the second excited energy, why can't it be that the electron get excited by absorbing the photon then re-emmit another one with lower energy immediately?

 

[avarmaavarma]

That's exactly the basic concept in QM - if an energy 'quantum' isn't the EXACT right amount - nothing happens. No absorption - no part absorption. We need to move away from our 'continuous' energy absorption ideas when dealing with collisions in the microscopic world. Even the basic picture of a photon 'here' colliding with an electron 'there' is incorrect. It should say - a photon with this 'spread' (in both position and energy) collides with an electron somewhere in this region with this position spread and energy spread.
In your example above, the incident photon will not be absorbed since it isn't the exact right energy. That was why the original photoelectric experiments (Hertz et. al) found that simply increasing the energy of the incident light did nothing to 'knock out' more electrons. And lowering energy of the incident photons made no difference either - the lowest energy ultraviolet photon would still continue getting absorbed - since it was of the exact right 'quanta'.

 

[kof9595995]

Yes I know this is a fact of QM, but how do you understand it from the formalism of QM?

 

[Chopin]

Someone can correct me if this is incorrect, but I believe it has to do with resonance effects. To excite an electron that's in a bound state, you have to wiggle it back and forth at exactly the right frequency to kick it up to another energy eigenstate. If the photon has exactly that frequency, then the electron will resonate with it, and eventually become excited. If the photon has a different frequency, though, then no resonance occurs, and the electron stays where it is.

The math behind this can be found by searching for "time-dependent perturbation theory". Also, http://www.falstad.com/mathphysics.html has some really nifty little Java apps that illustrate some of these points--look for the atomic transitions apps in the Quantum section.

 

[kof9595995]

In your example above, the incident photon will not be absorbed since it isn't the exact right energy. That was why the original photoelectric experiments (Hertz et. al) found that simply increasing the energy of the incident light did nothing to 'knock out' more electrons. And lowering energy of the incident photons made no difference either - the lowest energy ultraviolet photon would still continue getting absorbed - since it was of the exact right 'quanta'.

Wait a minute, i just realize photoelectric experiment is not a good analogy to my problem, for photoelectric effect to happen you only need to provide photon with higher than work function energy, i.e. electron can absorb whatever energy higher than work function, and it says nothing about whether electron can re-emit another photon, i see no connection between my problem and photoelectric experiment.

 

[kof9595995]

@Chopin: I guess you are right, I've also been pondering if the answer lies in perturbation theory, which I haven't learn much about.

 

[avarmaavarma]

Hmm - kof9595995 - the photoelectric effect has a lot to do with the original question. The effect showed that the 'wavelike' thinking of light - that its energy could be absorbed in a continuous fashion - was incorrect. The idea of a photon having more than the exact energy that could be partly absorbed (or transferred to the electron) is in fact a remnant of the 'waveish' mode of thinking of energy transfers.

In the wave world it makes sense to assume that part of the energy would knock of the electron and the other part would continue to live on to do other useful work. That seems to be the original intent of the question that was posed (partial transfers of energy of an incoming photon to an electron)

 

[kof9595995]

I beg to differ, in particle language, photoelectric effect shows just an electron cannot absorb more than 1 photons at once, or in wave language, the amplitude of a em wave can not be arbitrarily small. But what I'm concerning is about frequency, why doesn't the electron absorb part of the photon's energy, and reduce the photon's energy by lowering its frequency?

 

[Bob S]

... But what I'm concerning is about frequency, why doesn't the electron absorb part of the photon's energy, and reduce the photon's energy by lowering its frequency?

It does. The electron (either free or bound) can "absorb" only part of the photon energy, as long as the final state is a free unbound electron plus a scattered photon. This was pointed out by Matterwave in post #5. This is called Compton scattering. See

http://en.wikipedia.org/wiki/Compton_scattering

Bob S

 

[kof9595995]

Yes, I agree with you, and I said more or less the same thing in post 6, but the thing I'm not very clear of is why can't it happen between bound state, I guess what Chopin said could be the answer to my question, anyway I have never touched time-dependent perturbation theory.

 

[loreak]

It can't happen in the bound state because electrons can't exist in between their shells, they must make a complete quantum jump to the next shell or be ionized out of the electron where they can then absorb many wavelengths because they are not in shells any longer.

The act of absorbing the right amount of energy and then re-radiating the left over amount takes time and it would require the electron to exist in a non-shell area which you can't do.

The reason electrons only exist in their shells is due to interference, just as in the double slit experiment. In an atom only certain frequencies will allow the electron to exist because inbetween the layers the electron will interfere with itself and it makes the probability of it existing there 0% so it never does.

 

[avarmaavarma]

There is no 'part' of the photon's energy! The smallest 'part' is the photon. There is no smaller part. If an electron's next energy level happens to be the exact same energy as this 'part', then it will readily absorb the photon. Otherwise it will not. It does not have an option to take part of the photon - just like a square peg does not have an option to fit into a round hole. It can only fit into a hole of the exact same size. For energy absorption purposes, the electron can actually be viewed as a hole of a certain size - and the photon can either exactly fit the whole or not.
A smaller frequency photon will not fit the hole, neither will a higher frequency photon. A higher frequency photon cannot fit part of the hole and continue on to fill some other hole.

 

[kof9595995]

@loreak

The act of absorbing the right amount of energy and then re-radiating the left over amount takes time

Are you sure of this? It seems like a hypothesis to me now.

@avarmaavarma: you are basically repeating what you said, I just don't agree what you said is relevant to my question, in compton scattering electron can absorb and re-emit photon with lower frequency, what can't it happen between bound states?

Again, I think Chopin is right, I just checked my textbook for time-dependent perturbation for absorption, it seems transition probability is a function of frequency sharply peaked at the "right" frequency.

 

[fred1234]

@loreak
@avarmaavarma: you are basically repeating what you said, I just don't agree what you said is relevant to my question, in compton scattering electron can absorb and re-emit photon with lower frequency, what can't it happen between bound states?

It's not compton scattering per se that has the property that a absorbed photon has a different frequency from the emitted photon, it's a property of the atom/molecule shells. The absorption and emission are separate events. To be absorbed the frequency/energy of the photon has to match the difference between the current energy state/shell and a higher empty energy state/shell. The probability of emission, i.e. the time between absorption and subsequent emission, is dependent on the configuration and current condition of the atom/molecule. In atoms/molecules that support lower probability of emissions may allow for another photon to be absorbed before the previous photon has been emitted. Or a photon with the frequency/energy that is able to transition two energy levels at once. At some time after the absorption, either seemingly instantaneously or a noticeable delay, the transition to a lower empty energy state/shell can occur with the emission of a photon with the frequency equal to the difference in energy states that were transitioned. (Note empty shell should be read as empty spot in the shell since a shell can contain two, one up spin and one down spin)

[you can not park your car in parking spot that already has a car in it. :-]

Are situations that are commonly described by compton scattering, e.g. X ray, more prone to the aforementioned transitions?

@loreak
Again, I think Chopin is right, I just checked my textbook for time-dependent perturbation for absorption, it seems transition probability is a function of frequency sharply peaked at the "right" frequency.

The difference between energy states, and thus the frequency of the photon, depends on the electric and/or magnetic field experienced by electron.

 

[Bob S]

It's not compton scattering per se that has the property that a absorbed photon has a different frequency from the emitted photon, it's a property of the atom/molecule shells.

Compton scattering, when viewed in the center-of-mass frame, is a completely elastic photon-electron collision. The scattered photon has the same energy as the incident photon. So Compton scattering is elastic scattering.

The absorption and emission [of a photon] are separate events. To be absorbed the frequency/energy of the photon has to match the difference between the current energy state/shell and a higher empty energy state/shell. The probability of emission, i.e. the time between absorption and subsequent emission, is dependent on the configuration and current condition of the atom/molecule. .

Exactly. A good example is the resonant absorption of the 5890-Angstrom yellow (doublet) line in sodium. The emission is delayed by the natural transition lifetime.

Bob S

 

[kof9595995]

It's not compton scattering per se that has the property that a absorbed photon has a different frequency from the emitted photon, it's a property of the atom/molecule shells. The absorption and emission are separate events.

Actually I don't quite understand what you mean by "separate events", do you mean there's always some lag of emission after absorption? But for compton scattering, if we want to describe it in a "absorption-emission" way, there's must be no lag, or there would be some time energy and momentum are not conserved simultaneously(I vaguely remember there exists such an issue, correct me if I'm wrong). So how can we be sure there is a lag bound-state absorption & emission?

 

[fred1234]

Compton scattering, when viewed in the center-of-mass frame, is a completely elastic photon-electron collision. The scattered photon has the same energy as the incident photon. So Compton scattering is elastic scattering.
Bob S

Yes, I was thinking 'the explanation the effects seen during X-Ray imaging' instead of the specific process of 'Compton scattering'. The answer to the question I asked then is, we are only talking about compton scattering.