I feel like I'm missing something really fundamental regarding the cross section of the photoelectric effect. I'm looking at this chart, and it seems that the lower the energy of the incident photon, the more likely a photoelectric interaction is to happen. Higher energy interactions get less likely, until a cut-off point, at which they no longer occur.   If the energy of the ejected electron is equal to the incident photon energy minus the binding energy, I understand why a lower limit exists (since there must be enough energy to eject the e-), but why should there be an upper limit? Why does the cross section reduce with energy at all?
90011
Consider the diagram below:  (Author: Joshua Hykes )  From this diagram we can see that the absorption coefficient for the photoelectric effect generically decreases with the increase in energy of the photon. What is the physical reason for this decrease. (i.e. why does the probability of photoelectric absorption decreases with increasing photon energy).  Note: I am aware that the large spin is due to more inner electrons been able to be lifted.  Edit For Clarity  I will here rephrase the question: in the above diagram there is a general decrease in the absorption coefficient due to the photoelectric effect (i.e. the absorption of a photon by an electron and subsequent emission of that electron) with energy (ignoring the occasional absorption edge). This is of course linked to a decrease in the probability of the photoelectric effect occurring (and a corresponding decrease in the cross section). My question is: what is causing this decrease with increasing energy. (For the purpose of this question I am ignoring the contribution due to the Compton effect and pair production and simply focusing on the cross section due to the photoelectric effect.)
48919
How does quantum electrodynamics actually explain HOW reflection occurs on a microscopic scale?   Note that Feynman's QED lecture series/book is not sufficient, as he only assumes that light DOES reflect ('turn around and go back') in order to expound his path integral theory. My question is why does light have the propensity to turn around in the first place.  Is it just the absorption and re-emission of photons, and if so, why does it happen so uniformly (i.e. on a shiny thing, entire scenes are reflected near-perfectly). In essence, why are flat things shiny? Are all the molecules arranged at exactly the same angle?
85057
Is it an accurate statement to say that free electrons in a metal experience NO restoring force when they interact with electromagnetic waves? I understand that the electrons exist in a space filled with ions, and doesn't the cumulative potential that is present due to the presence of the ions exert an electric field on the electrons. Even in the case of simple metals, where you can say that the nucleus is shielded by the valence electrons, so called Coulomb shielding, how significant is the shielding. Naively, it seems to me that because the charge in the nucleus is not balanced by the charge in the bound electrons, there should be some net potential that the free electron sees.
23410
It's a common understanding that atoms emit or absorb light when the energy of the photons is equal to the the difference in the energy levels in the atom. What I don't understand is how does an electron absorb light inside an atom? Is it that the atom as a whole absorbs light or the individual electrons absorb the light packets? Also I would like to know, what does one mean when you say that an electron absorbs photon. How does an electron absorb a photon??
86455
Consider the atomic spectrum (absorption) of hydrogen.     The Bohr's model postulates that there are only certain fixed orbits allowed in the atom. An atom will only be excited to a higher orbit, if it is supplied with light that precisely matches the difference in energies between the two orbits.   But how precise does 'precisely' mean. Of course, if we need energy $E$ to excite the electron to a higher energy level, and I supply a photon with just $E/2$ I would expect nothing to happen (since the electron cannot occupy an orbit between the allowed ones). But what if I supplied a photon with energy $0.99E$, or $1.0001E$ or some such number. What will happen then?  I think that the electron should still undergo excitation precisely because the lines we observe in the line spectrum have some thickness. Which means that for a given transition, the atom absorbs frequencies in a certain range.   Is my reasoning correct? If not, why? How does Bohr's model explain this? How about modern theory?   If I'm right, what is the range of values that an atom can 'accept' for a given transition?
105537
My question may be very basic, but I can't think of a reasonable explanation for this.  Consider a solid charged sphere. Now, we have an electric field inside the solid sphere, but at any particular point there are charges infinitesimally close to the point. As I see it, these charges should contribute to an infinite value of the electric field.  One might possibly argue that at any given point, infinitely-close charges are distributed uniformly everywhere and hence their effects cancel out. But at the surface of the sphere the uniformity of infinitely-close charges is not present and hence here we must have infinite field, but that is not the case as well! Where are we going wrong?  Also, if we think of electric field at that point in continuous charge distribution, the contribution of electric field might exist finitely from nearby charges with both charge and distance between them tending to zero, but for the very small charge present exactly there, the electric field contribution from itself must be infinite with charge being finitely small and distance being exactly 0. What's wrong here?  I have strong suspicions, is it sensible and correct to define electric fields at such a close vicinity of other continuous charges?
110807
OK, lets formulate it differently and say water works as a blue passing / red restricting filter.  It is actually observable. Just do a dive in a swimming pool with white light (maybe even at night) and a pool with a metal casing or white tiles. The further you look, the more blue it is.  So why is this so? Why is water not transparent?  And how does it happen on a molecular, subatomar or electron shell level?  Do 'red' waves become absorbed and retransmitted as 'blue' light waves or do only the 'red' waves get absorbed and the 'blue' can pass?  Can Maxwell help?
22694
So I've been looking into particle-antiparticle pair production from a gamma ray and don't understand one thing.  Let's say I have a 1,1 MeV photon and it hits a nucleus - electron-positron pair with some momentum will be created and the nucleus will probably get some momentum as well because of the impact.  But why does the photon need the nucleus at all? Why can't it just fly through space and suddenly, with some probability, change into a electron-positron pair with momentum? I see that the momentum of the system wouldn't be conserved but I don't really understand how the nucleus helps it.
61421
I'm still struggling a bit with some ideas around double slit experiments. One that keeps cropping up for me is the role of observers.   Imagine a classic double slit experiment with a hidden observer who has arranged an apparatus to detect which slit the electrons are passing through. This person and their measurements are hidden to you and you have no interaction with them.  So the question is, do you see an interference pattern or not?  Additionally: And if the answer is Not, then is the reason because they "disturbed" the electron (by say firing photons at them) or is it for another reason? And if it is because they "disturbed" the electron, then how is it that unobserved electron's are not disturbed since they certainly interact with other objects, for example other atoms in the matter around the slit(s) will feel a slight gravitational tug as it passes through.
77472
Essentially this question boils down to "why does the energy of light depend on frequency?".    The analogy my textbook (pg. 272 principles of chemistry A molecular Approach 3rd edition by Nivaldo J. Tro) gave was that photons with higher frequency are like baseballs, whereas photons with lower frequency are like ping pong balls. Meaning that each photon has higher energy.  This StackExchange page () seems to say the opposite, that the energy of each photon is the same and what changes is the amount of photons per area.
64776
I understand that the photon needs to have enough energy to produce a lepton and it's antimatter partner, and that all of the properties are conserved, but why does the photon do this in the first place? What's going on "behind the scenes" to transform a single, neutral particle (the photon) into two charged particles (i.e. electrons)? Why is the nucleus necessary?
81639
I am currently reading about the interactions between light and matter, but I keep coming across conflicting explanations.  My initial understanding (using classical electrodynamics) was that light (EM) waves do not actually travel through glass. Instead, they are absorbed by the atoms and/or electrons within the glass, which then emit new EM waves. The new EM waves emitted by the energised particles make up the forward-radiation, which is the light observed from the opposite side of the glass. So it appears as if light is travelling through the glass, but it's actually being absorbed and then re-emitted with (mostly) the same properties.  The article  by Chad Orzel, based on a related SE question, confirms the above:      To understand the propagation of a wave through a medium, you can   think of each component of the medium– atoms, in the case of a glass   block– as being set into motion by the incoming wave, and then acting   as a point source of its own waves.   But an answer to a different but related question,  conflicts with the above:     the electrons in glass are tightly bound to atoms so they are not free   to move like the electrons in a metal and therefore they do not absorb   the photons.   I know the above is referring to the quantum electrodynamics understanding, but I think it still conflicts as it claims that EM waves are not absorbed by the electrons in glass because they are too tightly bound to their atoms.  Also, the  provides a small explanation of the mechanics behind refraction within glass, which kind of conflicts with both of the above:     In the case of dielectrics such as glass, the electric field of the   light acts on the electrons in the material, and the moving electrons   generate fields and become new radiators. The refracted light in the   glass is the combination of the forward radiation of the electrons and   the incident light. The reflected light is the combination of the   backward radiation of all of the electrons.   From my understanding, the above claims that some of the light is absorbed and re-emitted by the electrons (which conflicts with the linked answer), and that the rest of the incident light travels through the glass (makes up the rest of the refracted light) (which I think conflicts with the linked article, claiming the light is absorbed (although the article doesn't technically state whether all of the light is absorbed and re-emitted, so maybe that's where I'm going wrong(?))  I'm sure this is my misunderstanding, so I would be very grateful if you could clear it up for me:  When a light wave is incident on a sheet of glass, do the particles that make up the glass absorb the waves and re-emit them in the forward direction? Or do the light waves manage to travel through the glass, without being absorbed and then exit the glass?
42056
On a systems level, I understand that as electrons are pushed into a wire, there is a net field and a net electron velocity.  And I've read that the net electron drift is slow.  But electricity travels through the wire, essentially at c, and I want to understand that mechanism.  My apologies if my question is poorly stated, I know bare bones undergraduate quantum theory circa 1990s but it doesn't explain the motion of electricity in detail.  Here's my conception, I'm hoping someone will fill in the holes, ha ha.  An electron moves into the wire.  It's got a kinetic energy.  After travelling a short distance, it spontaneously emits a photon, which hits another electron in a valence shell.  That electron then presumably does the same.  If this conception is simply wrong, please enlighten me.  The questions that arise:   Presumably at this level, electrons are acting more like waves and less like particles, but is there any classical component in the picture, ie are electrons coming in imparting other electrons with kinetic energy through repulsion, or does it not work that way? If electrons momentarily have energy, then pass it on by a photon, what determines when that photon is emitted, and what frequency it will be?  I assume that electrons in this cloud are not limited by any kind of exclusion principal, and that any frequencies are possible? Why should a photon emitted by an electron be in the direction of travel?  Conservation of momentum tells me that if an electron is moving, the photon should be emitted in that direction, slowing the electron, but could an electron emit a photon in the opposite direction?  If it did, I assume it would somehow have had to absorb energy from elsewhere? That sounds possible by analogy with quantum tunneling. What is the mechanism by which electrons propagating increase the temperature of the material?  Are they transmitting energy to the electrons in the valence shell, which tug at the nucleus, do some photons hit the nuclei directly, or is there some other way?  Presumably, electricity travels slower than light, because there is some time in each exchange, and some time when electrons are moving at sublight speeds before emitting a photon.  By how much is this slower than light, and what is the speed of each interaction?
27474
and they are on opposite ends of the electromagnetic spectrum, then why can't light travel through walls which is right in the middle of the spectrum?    This question has already been asked . However, I am not entirely satisfied by the answer given on that page which relies on fanciful analogies and metaphors of ants, elephants etc. I am looking for a better explanation.    I think the crux of the matter, and my dilemma, relates to formula for . This is a well known formula used to explain the fact that low frequency waves have more penetration than high frequency waves.     But then how come gamma waves have such high penetration?     Are there some assumptions behind derivation of this formula which break when we consider very high frequency waves?    Or, are there some new factors that need to be taken into account as we move into the high frequency regime?     If given a radio source and a gamma source of equal intensity, then will the radio source have more penetrability than the gamma source per formula for ? If not, why not?  Thanks for all those who posted answers to this question. The answer being proposed is that light waves have the right energy to interact with atoms and electrons in the matter and thus get absorbed. This is a quantum mechanical explanation. The skin depth, on the other hand, is derived purely on basis of classical electrodynamics. So I see there are two mechanisms at work here? Does anyone agree with this? If so, the net absorption will be the sum of absorption due to skin effect + absorption due to atomic physics. Now if we take gamma rays - agreed there will be no absorption due to atomic physics but there should be absorption in accordance with skin effect. And so I come back to original problem.
75540
Pardon me for my stubborn classical/semiclassical brain. But I bet I am not the only one finding such description confusing.  If EM force is caused by the exchange of photons, does that mean only when there are photons exchanged shall there be a force? To my knowledge, once charged particles are placed, the electromagnetic force is always there, uninterruptedly. According to such logic, there has to be a stream of infinite photons to build EM force, and there has to be no interval between one "exchange event" to another. A free light source from an EM field? The scenario is really hard to imagine.  For nuclei the scenario becomes even odder. The strong interaction between protons is caused by the exchange of massive pions. It sounds like the protons toss a stream of balls to one another to build an attractive force - and the balls should come from nothing.  Please correct me if I am wrong: the excitations of photons and pions all come from nothing. So there should be EM force and strong force everywhere, no matter what type of particles out there. Say, even electrical neutral, dipole-free particles can build EM force in-between. And I find no reason such exchanges of particles cannot happen in vacuum.   Hope there will be some decent firmware to refresh my classical brain with newer field language codes.
11990
Followup to "".  It is argued that even a tiny residual charge would result in huge amounts of electricity in bulk matter, everything would be different, etc. I do not find that a convincing answer: suppose $n$ protons plus $n+1$ electrons are neutral. Why wouldn't we also expect there to be $n$ protons to every $n+1$ electrons? That is, there is no bulk matter problem if every $n$'th atom is a negative ion (for hydrogen).  Now, what empirical lower bound can we give for $n$ in that kind of scenario?
114737
In the following video clip at 2:10,    ,   Professor Walter Lewin talks about a misconception people have that the energy going through a wire to a resistor is in the form of kinetic energy of electrons. He proves this cannot be so as follows. The current density is J = I/A = Vne where V is the drift velocity (or average velocity), n is the number of electrons per volume, and e is the charge of an electron. A is a cross-sectional area of the wire.   We can make A as large as we want (keeping  the current constant), and therefore V will have to become very small, and the electrons will have very little kinetic energy. Yet the resistor (say a light bulb) dissipates the same amount of power P=(I^2)*R. Therefore, it must be that the form of energy is not the kinetic energy of electrons.  My first question is, if we make A larger why does it have to be that V goes down? Perhaps n goes down - we increased the volume (by increasing the cross-sectional area), so there should be fewer electrons per volume?  My second question, my main question is, if the energy is not the kinetic energy of the electrons, what does in fact bring energy to the resistor and how does it heat up?