 Hello and welcome back to my channel and the lecture series on quantum mechanics. In today's video, we are going to do a very detailed discussion on the Compton effect. Now in my previous lectures, we have done some discussions on similar experiments like blackbody radiation and photoelectric effect that gave us some very interesting conclusions about the nature of light. The blackbody radiation experiment suggested that the blackbody walls emit electromagnetic radiation in discrete values of h nu, h being the Planck constant and nu being the frequency. And in photoelectric effect, we found that the electrons absorb electromagnetic energy in the form of h nu or packets of energy called photons. So this gave us the photon picture of radiation. In the photon picture of radiation, light or electromagnetic radiation is composed of these discrete packets or particles having energy E is equal to h nu. So today we are going to talk about the Compton effect and Compton effect played a very important role in providing an irrefutable evidence to this picture of light where light is behaving like a packet, like a particle, like a photon. First let's talk about the experimental observations and then we will talk about the theoretical predictions or explanations provided by Compton. So we have a source here which emits a beam of monochromatic x-rays of wavelength lambda and this beam passes through a pair of collimeters and falls on some kind of a material like graphite material and then it gets scattered in various directions. So here you can see that there is a detector, an x-ray spectrometer that is capable of detecting the incident radiation or the scattered radiation at different angles. So let's suppose that this is an angle with respect to the incident radiation direction. So the detector is capable of measuring the intensity of radiation or the scattered radiation at different angles. When we make observations as to what kind of a wavelength is visible or detected by this detector, we get this kind of a graph. So in the x-axis we have wavelength and in the y-axis we have an intensity and majority of these graphs are associated with two distinct peaks. One of the peak corresponds to the wavelength of the incident radiation and then there is an additional peak corresponding to a wavelength which is greater than that of the incident radiation. So essentially what is happening is that in the scattered x-rays there is a wavelength which is larger than that of the source x-ray wavelength. Essentially the nature of the scattering is such that there is some sort of a wavelength shift that happens between the scattered radiation and the incident radiation. So when we expose a material with some kind of an incident x-ray having wavelength lambda, in the scattered radiation you end up finding an additional wavelength lambda dash which is greater than lambda by some kind of an amount. What is interesting is that this wavelength shift that happens between the scattered radiation and the incident radiation is dependent on the angle. So if you look at the angle for small angles, the shift is very small. For large angles the shift is very large. And this is very interesting and peculiar observation about this particular effect that when you have small angles the wavelength shift is very less and for larger angles the wavelength shift is more. So this is a very important observation that if theta is small, del lambda is small that means the shift is small, shift in the wavelength between the scattered and the incident radiation if theta is large del lambda is large. This shift in the wavelength between the scattered radiation and the incident radiation is known as Compton shift. Now the question is why is this shift happening in the first place? Why is the scattered radiation having wavelength greater than the wavelength of the incident radiation? Now this is something that cannot be explained by the classical idea of electromagnetic waves. In the classical picture of electromagnetic waves when an electromagnetic wave interacts with matter. We say interacts with matter we are mostly going to talk about interaction with an electron. So therefore what happens here in the material is that you have electrons which are stuck to different kinds of atoms. The incident radiation comes and provides some sort of an energy to the electron and then the electron behaves accordingly based on that nature of the interaction. So when an incident electromagnetic wave interacts with an electron because the electromagnetic wave in the classical picture is essentially oscillating electric and magnetic field lines and the electron is a charged particle. So because of this oscillating electric magnetic field lines the electron being the negatively charged particle will also start oscillating back and forth. And because of this oscillating back and forth the frequency of these oscillations is exactly the same as a frequency of the incident radiation. So after some time when this oscillating electron releases the energy in the form of electromagnetic radiation in all directions then the scattered radiation must also have the same frequency and wavelength as that of the incident radiation. So this is the classical picture of what happens in some sort of an interaction between radiation and matter. So how can we explain the existence of a wavelength shift in this particular experiment? Why is it that there is a wavelength shift that happens in the scattered radiation and this wavelength shift is proportional to in some fashion to that of the scattering angle. Well here Compton postulated that the incident radiation is not composed of electromagnetic waves in the classical picture but instead is composed of photons just like what Einstein assumed in the photoelectric effect experiment. Remember the photoelectric effect experiment and the photoelectric effect Einstein suggested that the incident radiation is composed of these photon particles these photons being discrete packets of energy and every single electron that is released from the surface absorbs the energy of the photon one on one. So in the similar fashion Compton also assumed that when light interacts with an electron let's suppose then the incident radiation is composed of these photon particles and this photon particle interacts with the electron just like two particles interact with each other in various kinds of collisions. So for example if you have watched a billiards game you see these balls colliding with other balls and they are exchanging energy and momentum in that kind of a process this particular phenomena of collision of billiards balls can be replicated here between the interaction between a photon and an electron. You see the shift in the perspective of nature of the interaction in the classical picture the electrons absorb the energy by becoming oscillators and then releasing that energy in the form of electromagnetic radiation but in the quantum picture the incident radiation is composed of photon particles and these photon particles are capable of colliding with the electrons and exchanging energy and momentum just like two particles collide with each other. Once we make this switch from the wave picture to the particle picture then everything starts making sense. So we can explain the Compton effect and make the predictions of all these observations based on that particular idea that photon is colliding inelastically with an electron just like two particles collide with each other. So let me first draw a diagram. So this is the picture of the inelastic collision between the photon and the electron. So essentially when the photon or the incident radiation interacts with the electron in the material the electron is probably bound to some atom and if the photon has sufficient energy then the electron is dislodged from that atom and now that electron behaves like a free electron and from which the photon can get scattered. So let us assume that the photon is a particle that is interacting with the target electron and you end up getting some kind of an inelastic collision where a particle collides with another particle and both of them go in different directions, they exchange momentum, they exchange energy. So let us represent the energy expressions and the momentum expressions and try to apply conservation of momentum and energy just like we do in the case of collisions between particles and then see whether the results that we achieve is able to explain this kind of a wavelength shift that is dependent on the angle of scattering. So first of all we have an incident photon and we know from our discussions with black body radiation and photoelectric effect the incident photon has how much energy? The incident has energy of h nu. This is the Planck postulate right that a photon essentially composed of energy h times frequency where h is a Planck's constant. Now the photon also has some kind of a momentum. Here I am going to invoke another expression which is basically the momentum of the photon is equal to the energy upon c, e upon c so e is h nu so h nu upon c. Now some of you may be familiar with this expression, some of you may not be familiar with this expression. This expression essentially comes from the relativistic expression for energy of massless particles right. So we had some discussions about relativity previously in my channel and relativity you are probably familiar with the expression that the energy of any particle is given by m naught square c to the power 4 plus p square c square whole root over where m naught is the rest mass of the particle p is the relativistic momentum of the particle right. Now particles that move at the speed of light. So photon is a particle that is moving at the speed of light has rest mass equal to 0. For particles that are moving at the speed of light or massless particles m naught is equal to 0. In that situation we end up getting e is equal to p c. So this is an expression that is valid for a photon also because ultimately photon is a massless particle it's rest mass is 0 but it has momentum it has energy and their relationship is this. So p is equal to e upon c that's it. Now what about the target electron? We assume that the target electron is initially at rest so if it is initially at rest it only has rest mass energy e naught is equal to m naught c square and its momentum is equal to let's suppose 0 right p I am just writing p naught here is equal to 0 because it was initially at rest. Now this whole interaction takes place and the photon gets scattered in a particular direction at an angle theta with respect to the incident direction in that situation the photon loses energy so its frequency changes right. So now the photon of the scattered photon has an energy h nu dash you see initially it had a frequency of nu but now it loses energy because it provides some of its energy to the target electron. So it loses energy its frequency decreases and its momentum also becomes p dash let's suppose we call it. So this is e dash and this is p dash is equal to so p dash is equal to h nu upon c but now the scattered electron has both rest mass energy as well as some kind of a kinetic energy. So let us write the most general expression which is the expression for relativistic total energy so let me write this. So the energy of the scattered electron let's suppose e e is equal to m naught square c to the power 4 plus p e square c square root over where p e is the momentum of that scattered electron. So this is the picture incident photon with some energy gets scattered from a target electron it exchanges momentum energy the scattered photon goes in a certain direction with less frequency and the scattered electron goes in a certain direction with gained energy. So the loss in the energy of the photon is equal to the gain in the energy of the electron. So we can apply the conservation laws of momentum and energy in this particular picture. So first let us apply the conservation law of momentum. So conservation of linear momentum. So the conservation of linear momentum simply says that the momentum of the photon initially plus what is the momentum of the target electron 0 so I can just write this in the left hand side is equal to the momentum of the scattered photon p dash plus the momentum of the electron let me make it a vectorial expression. You see momentum is conserved independently in two perpendicular directions so along the x axis momentum is conserved along the y axis momentum is conserved to make the calculations simpler I have written the expression in vector notation that means the overall momentum so the vector quantity contains the x axis component and the y axis component is conserved. So essentially the momentum is conserved along the x axis and the y axis separately. So let me rewrite this expression in terms of the magnitude of the momentum of the electron you see I want to sort of remove the P e term because essentially I am more interested in the wavelength of the scattered photon with respect to the theta here. So I want to remove these terms that are unknown to do that what I am going to do is essentially I am going to write that P e is equal to p minus p dash these are vector expressions ok. Now what is this? This is the difference between two vector quantities p and p dash now you are probably familiar with vector notation right so if I may write down a vector notation let's suppose you have a vector here a all right and then you have another vector here b and then you have this vector here which is c. So if you are familiar with vector notation then you know that c is equal to a minus b right and let's suppose the angle between a and b is given by theta in that situation if I take the dot product of c with itself so let's suppose I say c dot c then because the angle between c and c itself is 0 so this comes out to be c magnitude square and this is equal to nothing but a minus b dot product of a minus b. So if I separate out all these terms what I end up getting is a times a I end up getting magnitude of a square plus b times b I end up getting magnitude of b square and then I have minus a b and minus b a. So I can write this as 2 magnitude of a magnitude of b and then we have cos theta because essentially this is a dot product and the angle between a and b is theta so minus 2 a b cos theta. So this is something that you probably are already familiar with but anyways I derived it that if you have this kind of a vector expression c is equal to a minus b then the magnitude of c square is equal to the magnitude of a square plus b square minus 2 a b cos theta. This is something that I can apply here so if I apply this here then I end up getting the magnitude of p e ok I am just going to remove the arrows and all the stuff because we are dealing with magnitude now. So the magnitude of p e square is equal to the magnitude of p square plus the magnitude of p dash square minus the magnitude of p p dash cos theta what is the direction or the angle between p and p dash theta. So therefore this is the expression that we have now. So if I want to simplify this let's simplify this p e square is equal to what is p square p is equal to h nu upon c so I end up getting h nu upon c whole square what is p dash is equal to h nu dash upon c so h nu dash upon c whole square minus 2 again h nu upon c h nu dash upon c cos theta. So if I take h upon c square common then this expression ends up becoming p e square is equal to h upon c I take this as common and in the bracket you have nu square plus nu dash square minus 2 h upon c whole square is taken out you end up having h nu nu dash cos theta alright. So this way I can substitute p e somewhere in another expression to remove p e completely from these expressions. So this is p e square is equal to h upon c whole square times nu square plus nu dash square minus 2 nu nu dash cos theta let me say that this is equation number 1. So based on conservation of linear momentum I come up with this particular expression now I want to apply conservation of total energy and let's see if we can use this expression there to simplify everything alright. So let me first rub this alright okay I have erased this part of the board because I wanted to create some space to do further calculations but the essential expression that we just now derive moments back is written here p e square is equal to whatever the expression is so this is point number 1. Now let us apply conservation of energy principle here the conservation of energy says that the energy of the incident photon plus the energy of the target electron so the target electron also has some energy right because the mass has energy according to relativity so the target electron has a rest mass energy. So this is equal to the energy of the scattered photon which is e dash plus the energy of the scattered electron here. So let us quickly plug in all the values e is simply equal to h nu e naught is equal to m naught c square e dash is equal to h nu dash and ee is equal to root over m naught square c to the power 4 plus p square c square m naught here is the rest mass of an electron. So if I take this h nu dash to the left hand side I end up getting h nu minus nu dash plus m naught c square and we have this left if I square the whole expression then this is square is equal to m naught square c to the power 4 plus p e this is the momentum of the electron right p e square c square. So if I simplify the terms in the left hand side what should we get alright. So first of all we should get h square nu minus nu dash square plus m naught square c to the power 4 plus 2 times this and this. So twice h nu minus nu dash m naught c square in the left hand side right and in the right hand side what do we have m naught square c to the power 4 plus p e square c square. So here we will end up cancelling this particular term m naught square c to the power 4 same in the left hand side and the right hand side. So again let me simplify by taking out all the terms here. So I end up getting h square nu square minus h or plus h square nu dash square minus 2 h square nu nu dash right when we take out all the terms from this expression a minus b square and I am left with plus 2 h nu minus nu dash m naught c square right. Now let us look at the right hand side expression the right hand side expression contains p e square c square. What is p e square? p e square is this expression right that we just now obtained some moments back using conservation of linear momentum. So if I substitute p e square here the c square and the c square gets cancelled and we are left with the remaining expression which is h square nu square plus h square nu dash square minus 2 h square nu nu dash cos theta. Here can I cancel a few terms? Yes I can this term gets cancelled h square nu square and we can also I think cancel out h square nu dash square. So the remaining terms are the remaining terms essentially become 2 h nu minus nu dash m naught c square is equal to if I take this to the right hand side 2 h square nu nu dash minus 2 h square nu nu dash cos theta. So if I simplify this further this is 2 h nu minus nu dash m naught c square is equal to so if I take the 2 h square common and nu nu dash common I end up getting 1 minus cos theta. So finally it is simplifying now. So here 2 2 gets cancelled h gets cancelled and what else gets cancelled nothing else right. So let me write the expression so nu minus nu dash nu dash upon nu nu dash is equal to h upon m naught c square 1 minus cos theta. Let me rewrite this expression in terms of wavelength because we are essentially dealing with the wavelength in term in the experimental observations. So what is nu nu is nothing but c upon lambda right. So if I write here c upon lambda what is nu dash this c upon lambda dash. So lambda dash is the scattered wavelength and lambda is the incident wavelength and in a denominator we have c upon lambda and c upon lambda dash. I hope you can see this essentially this is equal to h upon m naught c square 1 minus cos theta. So here the c gets cancelled right in the numerator you have the c gets cancelled with the denominator and there is another c in the denominator which gets cancelled with this. So finally I can come up with an expression let me rewrite the expression here okay. Let me rewrite this expression here. I come up with an expression basically lambda lambda dash this is going to the numerator and then this is going to become lambda lambda dash times 1 upon lambda minus 1 upon lambda dash is equal to h upon m naught c 1 minus cos theta. So here again lambda lambda dash will get cancelled out and essentially what you will get here lambda lambda dash lambda lambda dash lambda dash minus lambda right. So lambda lambda dash will get cancelled out and we will be left with lambda dash minus lambda. So I can just write it straight away as lambda dash minus lambda which is equal to what lambda dash minus lambda is the wavelength shift or the countant shift right which is just del lambda okay. So del lambda is equal to how much h upon m m naught c 1 minus cos theta. So essentially we finally have this expression now for the wavelength shift that the wavelength shift is equal to h upon m naught c 1 minus cos theta where h is a plan constant m naught is the breast mass of the electron c is the speed of light and theta is the scattering angle. So we have derived an expression for the wavelength shift or the countant shift based on the particle picture of radiation by assuming that the incident radiation is composed of photons which are behaving like particles and these photons are interacting with the electrons in the manner that two particles collide with each other and by applying conservation of momentum and energy we come up with this expression for the wavelength of the scattered photon. So what does this equation predict? First of all it predicts that there is going to be a wavelength shift right. Remember when we talked about the classical picture of how an electromagnetic wave interacts with an electron and the electron scatters that radiation in different directions where no such wavelength shift takes place. Well by assuming the particle picture of radiation the photon picture of radiation there is in fact a wavelength shift that means the scattered photon has a wavelength greater than the incident photon or the frequency lesser than the incident photon because of loss of energy due to this kind of a collision kind of an interaction right. So first of all the particle picture the photon picture does predict a wavelength shift when this sort of an interaction takes place and not only a wavelength shift but also this wavelength shift depends upon the scattering angle. So as this scattering angle increases the wavelength shift also increases which is what we saw in this particular experimental observation. Therefore Arthur Compton who derived this expression has provided an irrefutable evidence that light is indeed composed of these photon particles that are capable of colliding with traditional particles like electrons. If you look at the nature of this kind of a wavelength shift you see that with increase in angle the wavelength will be increasing therefore I can draw another graph here let me just draw a graph where we have in the x axis we have let's suppose theta and in the y axis we have let's suppose the wavelength shift then because this is 1 minus cos theta so if I represent 1 minus cos theta as a graph I should end up getting something like something like this you see there you have it. So at zero angle you end up getting zero wavelength shift at angle of 180 degrees that means completely backward that means if the photon collides with the electron and goes completely backward that is where the maximum loss of energy takes place which is where the maximum wavelength shift also is going to happen. Now it's very interesting that the wavelength shift does not depend upon the wavelength of the incident photon whatever the wavelength of the incident x-ray is the wavelength shift is independent of that it is only dependent on the scattering angle so let me rub the board I hope that you have absorbed everything I've discussed let me rub the board again and give you a little bit more information about this whole thing all right all right so as I said the maximum Compton shift happens at the backwards direction it happens when theta is equal to pi because for that kind of a interaction where the particle is rebounded in the backward direction that is which corresponds to maximum loss of energy as it is evident from here right because cos theta comes out to be minus 1 right so del lambda is equal to h upon m0 c 1 minus minus 1 so 1 plus 1 so this comes out to be 2h upon m0 c so this is the value for the maximum value of Compton shift that is actually possible in any kind of a direction in fact this term h upon m0 c has a very special sort of a name h upon m0 c is called Compton wavelength because h here is the plant constant m0 is the rest mass of the electron c is the speed of light this is a constant number that has a value of 2.43 into 10 to the power minus 12 meters or 0.0243 angstrom so essentially the maximum wavelength shift that is possible is twice the Compton wavelength in fact I can rewrite this Compton shift expression as the general expression for the wavelength shift del lambda is equal to lambda c 1 minus cos theta as simple as that this is the expression for the Compton shift which essentially just depends upon the angle at which the measurement is being made so not only the photon picture is capable of demonstrating the wavelength of the scattered radiation is greater but it is also giving us a relationship by what amount the wavelength will change based on the particular angle now it is not necessary that all kinds of wavelengths are going to lose equal amounts of energy because this kind of an expression really makes us a sort of a gives an idea that because the wavelength shift is independent of the incident wavelength therefore maybe the incident wavelength how does it relate to the loss in energy well there is a another additional relationship which is that of the fractional loss of energy okay so let me also give you some idea about that fractional loss of energy I would say or fractional increase in wavelength so essentially what is happening is that the loss of the energy of the incident photon is what translates into the gain in energy of the scattered electron right the loss in energy of the photon is a gain in the energy of the electron but if I change the incident photon wavelength how does that affect what amount of energy is lost or gained for that let me just say that the relativistic energy of the electron what is the relativistic energy of the electron the relativistic energy of the electron is essentially composed of two parts you could say one is a rest mass energy of the electron and the other is a relativistic kinetic energy of the electron right so this is basically equal to the rest mass energy of the electron plus I would say ke but it simply means the relativistic kinetic energy now usually the electrons are not really traveling at very high speeds so the relativistic kinetic energy would approximate to the kinetic energy expression that we use in our day-to-day lives so if I use this expression the kinetic energy essentially represents the energy gained by the electron right so this is equal to E minus E naught now if I apply this in our original expression for conservation of energy so in our original expression for conservation of energy what we had E plus E naught is equal to E dash plus EE right so here if I say that E minus E dash is equal to EE minus E naught this is essentially equal to ke it simply represents the idea that the loss in energy of the photon is the gain in the energy of the electron so for relativistic scenarios you have relativistic kinetic energy but for day-to-day lives where electrons are not really traveling at that high speeds it is essentially normal kinetic energy right so if I want to come up with an expression for what is the ratio of the kinetic energy that is gained by the electron or the amount of energy that is lost by the photon with respect to the incident photon energy then I can come up with this particular expression so ke upon so I say ke upon let's suppose the initial energy of the photon is equal to how much so ke upon the initial energy of the photon so ke is equal to E minus E dash right E minus E dash upon E so what is this what is E that is h nu minus h nu dash upon h nu so what is this so h essentially gets cancelled and we are left with c upon lambda minus c upon lambda dash upon c upon lambda so again this comes out to be how much the c gets cancelled in the numerator and the denominator and the lambda goes to the numerator and we are in up with lambda and lambda lambda dash lambda dash minus lambda so essentially again the lambda lambda gets cancelled and finally we come up with another expression which is the kinetic energy gained by the electron or the energy lost by the photon upon the energy of the incident photon is equal to what is lambda dash minus lambda that is the Compton shift so del lambda upon what is lambda dash that is lambda plus lambda or del lambda right lambda dash is equal to incident photon plus the Compton shift and this comes up or this provides us with an additional relationship that is useful for us how is it useful this is useful because this gives us an idea about the amount of loss of energy that takes place with relationship to wavelength right so what is the wavelength lambda here lambda is the wavelength of the incident photon so if lambda is large the fractional loss of energy is less and if lambda is less the fractional loss of energy is larger right so for a low energetic photon lambda is greater so fractional loss of energy is less and for a high energetic photon lambda is smaller and fractional loss of energy is greater that means for low energetic photons or greater wavelength the fractional loss of energy is less and for high energetic photons or smaller wavelengths the fractional loss of energy is very high. That means for visible radiation where the radiation is somewhere around 5000 or 6000 angstrom the fractional loss of energy due to Compton effect would be very less maybe less than 1% but for gamma photons or x-rays having wavelengths of the order of let's suppose 0.01 angstrom the fractional loss of energy would be very high maybe around 50% loss of energy. So you see that for low energetic photons the loss of energy is very less for high energetic photons the loss of energy is very high. So clearly Compton effect dominates for high energetic photons which brings me to another question that might arise in your mind why is it that you have end up essentially getting two peaks right. So one is a peak associated with the incident photon wavelength and the other peak is associated with the Compton shifted wavelength photon right. The reasoning that there are two peaks in the experimental observations is because whenever the incident radiation interacts with matter some of the photons are capable of dislodging the electrons from their atoms right and some are unable to do so because the electrons are essentially bound to the atom right and when a photon interacts with that electron either it would release the electron from the atom or it would not be able to do so. So in those circumstances when the photon is incapable of releasing the electron from its bound condition with the atom in those situations the photon is essentially interacting with the whole atom itself. So because the mass of the atoms is many many times larger than the mass of the electron in that situation we end up getting the scattering in the classical picture which is basically known as Rayleigh scattering where the photon is scattered in various directions without any increase in wavelength. So for low energetic photons which are incapable of dislodging the electron from the atom we end up getting the normal scattering so therefore we end up getting lambda right. But for those situations where the wavelength is capable of dislodging the electron from the atom so now the electron is not bound anymore it's free there we end up getting this kind of an inelastic collision. So for those situations where the electrons are still bound to the atom we end up getting normal classical Rayleigh scattering so we end up getting the original incident radiation and for those cases where the photon is capable or photon is successful in dislodging the electron from its atom so that now the electron is a free particle in that situation the scattering that happens represents this kind of a increase in the wavelength or loss of energy according to the quantum mechanical picture here. So you see that both the phenomena is essentially happening in this particular picture but with increase in frequency as I just now mentioned the Compton effect phenomena dominates. So the Compton effect phenomena is usually not seen for low energetic radiation like visible light or infrared or you know radio waves but it is very much dominant whenever x-rays or gamma rays interact with matter. Alright so I think I have explained as much as I could about this particular experiment. Essentially this is an experiment where high energetic photons when they interact with matter they end up getting scattered in different directions where the scattered radiation has a wavelength greater than the incident radiation. This phenomena cannot be explained by the classical theory of radiation but it can be explained by the photon picture of radiation where the incident photon is composed of these particles called photons and these photons are interacting with the electrons just like two particles will collide with each other and by applying conservation of momentum conservation of energy we can come up with an expression for the shift in the wavelength and also we can come up with the amount of loss of energy that takes place. So therefore Compton effect finally hit the nail on its head I would say because we have been following a series of experiments like right in blackbody radiation the energy of electromagnetic oscillations were emitted in the form of discrete values. In photoelectric effect the energy of the incident photon was absorbed by the electrons in discrete packets. In Compton effect these discrete packets or photons are actually behaving like particles they're colliding with other particles just like particles with collide with each other thereby providing irrefutable evidence that the photon picture of light is here to stay. For whatever reason light does behave like a particle in these experiments in blackbody radiation in photoelectric effect in Compton effect light is behaving like a traditional particle. So this makes way for the dual nature of radiation light has dual nature in certain experiments like diffraction interference it behaves like a wave it demonstrates wave properties in certain experiments like blackbody radiation photoelectric effect Compton effect it behaves like a particle it has particle nature. So light has split personality sometimes it behaves like a particle sometimes it behaves like a wave so we will carry forward this discussion in the next lecture for today this is all this is the Compton effect I am Divya Shruti Das this is for the love of physics thank you very much have a nice day take care