 Dear students, so far we have discussed the interaction of heavy charged particles and large electrons with matter. Today we will discuss the interaction of electromagnetic radiations with matter. So, electromagnetic radiations essentially comprise of gamma rays and x-rays. So, gamma rays as you know already appear from the de-excitation of the excited nuclei, whereas the x-rays come from de-excitation of excited atoms. In addition to this discrete gamma rays from the decay of excited nuclei, we also have Brimstallion which we discussed during the interaction of beta particles with matter and this Brimstallion essentially I have a continuous spectrum starting from 0 to the n-point unit 0 beta or even for that matter if a positive. Now, the Brimstallion you know already that when the accelerated particles moving, see if it is accelerated in the vicinity of a new PS, then it emits a Brimstallion radiation. So, all of them, all interactions, all the types of radiation gamma rays, x-rays and Brimstallion, they are electromagnetic radiations and the interactions with matter are same. So, this we will call them mostly in terms of gamma rays. So, gamma rays predominantly interact by three modes, the photoelectric effect, the contents scattering and para production. So, the major deposition of energy in the gamma ray detector which essentially we will see how they manifest in the gamma ray spectrum comprises of these three processes. And since the gamma rays are photons, they will not cause the ionization directly in the medium, but by these three processes, photoelectric effect, contents scattering and para production, ultimately the net result is ionization and excitation. By secondary means, they will produce electrons which in turn will cause ionizations. So, there is no primary ionization in the interaction of this gamma photons with the matter. It is mostly by secondary ionization, whatever the electrons are produced as we will discuss very soon in these three processes, they will cause ionization and therefore, we get the secondary ionization. Another important point is that since the gamma rays are photons, their specific ionization is very small compared to that of electrons. So, their ranges as I discussed, their ranges are very high. So, about one tenth to one hundredth of the electrons, they will cause very less specific and means the per unit thickness in the medium, how many ion pairs they can produce that is called as their specific ionization, number of ion pairs produced per unit thickness. So, that is very small and that is why the gamma rays or even the put the x-ray, the install and they all travel a long distance in any material. But then we know, we will discuss very soon that if the i-z materials which can stop the gamma rays effectively and so, they are the ones which are used to stop these gamma rays. So, let us discuss the main three processes, what is the mechanism, what are the manifestations in the gamma ray spectrum. The photoelectric effect, all you all know from the Einstein's theory of photoelectric effect, then when the photon interacts with an atom as a whole. So, it is not with the electron, but it is there with the absorbed atom as a whole. Then in the vicinity of the atom, the photon disappears, it is completely absorbed. So, it can also be called as photoelectric absorption. And so, the atom gets excited and the excited atom gets loses its energy by emission of electron. In the case of photoelectric effect, probability is maximum for emission of the highest bound electron. So, that means the k-cell electron has highest probability of emission provided. The energy of photon is more than the binding energy of that particular electron. So, essentially the effect of photoelectric effect is to release one electron from the atom and the kinetic energy of that electron will be the initial energy of the photon minus the binding energy of the electron in that shell. So, this is the schematically shown here, the incident photon comes, it is absorbed by the atom as a whole. So, the important point is that the momentum of the photon, say photon has a momentum e by c that is balanced, that is it has to be conserved and that is conserved by the atom as a whole. So, all the momentum is taken up by the atom as a whole. And so, the excited atom that emits the electron from the k-cell or l-cell depending on its energy. So, it is important for the conservation of momentum that the electron which is closest to the nucleus is participating in the photoelectric effect. And the photon, the electron that is produced, we will be calling as the photoelectron. The photoelectron energy as you can see if we have a 1 MeV gamma ray and binding energy of electron may be about few tens of electron to kEV, then the electron will be coming out with a very high energy. And so, that electron can cause subsequent ionization in the medium. So, this process of photoelectric effect essentially leads to a full energy deposition of the gamma ray is lost. So, all energy of gamma ray photon is deposited in the material in form of electron and the electron will stop now in the medium. So, essentially it will cause ionization. The photoelectric effect is the predominant mode of interaction at the low energy of photon as we will very soon see the dependence of the cross section for this process, the energy of the photon and the jet of the absorptivity. And as I mentioned that for this momentum conservation, the cell closest to the nucleus is excited. So, that electron coming from the k-cell has the highest probability. So, the highest probability of absorbing the photon is for most tightly bound electron that is k-cell. So, if the energy of the photon is more than the k-cell binding energy, then it is the predominantly k-cell electron that is imitated. Now, so what essentially has happened there is a hole in the k-cell electron has been imitated. So, the post photoelectric effect what happens? It is a hole in the k-cell, then the electrons in the higher cells will jump down to the lower shell and in the process you will see the x-rays being imitated and subsequently this when it comes to the balanced electrons even this instead of x-ray emission the OZ electrons may also be imitated. So, as a result of the photoelectric effect we have photoelectrons, x-rays and OZ electrons being imitated and all these x-rays again will cause photoelectric effect their energies are much smaller. So, they will be having much smaller ranges and so all of them ultimately deposit energy in the medium and you get the full energy deposition in the substance. So, the probability of the photoelectric effect depends heavily upon the z of the absorber z to the power n where a z is the absorber atomic number and this n value is quite high 4 to 5. Whereas, it is depends upon the energy of the gamma ray in inverse way at e to the power 3.5. So, these dependencies have come from the electromagnetic theory and so we will not go into details how these relationships have been arrived at. But the point to note is that the the probability of photoelectric effect is very dependent upon the z higher the z higher the section for the photoelectric effect and lower the energy higher the probability for photoelectric effect. So, if you what I have shown here is in the this in as a function of energy of the photon at the the projection or the what we call as the attenuation coefficient for the photoelectric effect it is coming down with energy you can see here. So, higher the energy lower the probability of photoelectric effect for a particular z for a particular atomic number you can see here. So, this is the k h this is the l h and so on. So, what happens that in general that cross section should come down with energy, but whenever there is a the photon energy matches with the k-cell binding energy there is a jump in the cross section for photoelectric effect. Similarly, whenever the energy of the photon is equal close to the l-cell binding energy then there is a jump in the cross section and so the cross section for photoelectric effect will have a z at the respective cell binding energies. So, the photoelectric effect is the one process that leads to complete position of the energy of the photon in the detector system and so the typical spectrum you can see here. So, what happens support this is the count versus energy of the gamma ray then if the energy of the photon was e 0 then it will lead to a sharp peak ideally it should be a delta function sharp line, but the detector has its own resolution and so it becomes a Gaussian peak. So, there is a broadening in the peak of the gamma ray so, but this is what. So, there is no other effect of manifestation of the photoelectric effect in the gamma spectrum except that suppose this is the high jet material the detector is high jet material. So, the x-rays are emitted and x-rays might escape it. So, there will be x-ray escape peaks particularly it happens in sodium iodide which we will discuss subsequent lectures. The iodine is a one of the components of sodium iodide and the iodine x-ray might escape from the detector system and so there will be a peak full energy minus iodine kx-ray that is called the iodine x-ray escape. Another process of the photon interaction is content scattering and as the name itself implies it is a scattering. So, it is like an elastic scattering between the photon and the loosely bound electron. So, here the electron that is participating in content scattering is either a valence cell almost loosely bound electron. So, there is a collision between a photon, a photon comes strikes the electron and the photon is emitted at theta degree the required electron goes at 5. So, it is like an elastic scattering between the photon and the electron and the energy of the incident photon is shared by the scattered photon and the required electron. So, you can set up the in fact the equations for the conservation of energy and momentum of the photon before and after the collision and then we will not go into because this is a very simple derivation you can set up. So, finally, we want to know what is the energy of the photon that depends upon the angle at which the photon is indicated and so the h nu dash the energy of the content scattered photon is given by h nu incident energy upon 1 plus h nu upon m0 c square m0 c square m0 is the response of electron c is the speed of light and 1 to 1 minus 4 theta. And so, accordingly the energy of the electron will be h nu minus h nu dash. So, the energy of the incident photon is shared between the electron and the scattered photon. So, the manifestation of this you will see in this when we see what happens at cos theta equal to 0 when theta equal to 0 cos theta equal to 1 and then so this term becomes 0. So, you have h nu dash equal to h nu h nu dash equal to h nu. So, that means at 0 degree the photon goes undeflected there is no electron energy is 0 when the electron when the photon is scattered at 0 degree. But when the photon is scattered at 180 degree then cos theta is equal to minus 1 and you put here the value of cos 1 minus minus 1 become 2. So, h nu dash is h nu upon 1 plus h nu upon 2 m0 c square. So, it should be 2 h nu upon m0 c square 2 h nu upon m0 c square. So, at 180 degree this is the minimum value of the photon scattered photon energy. Now, let us do some simplification if the gamma ray energy is much higher than the rest mass energy of the electron m0 c square means 0.511 mev in that case you can actually substitute this term you can neglect. So, it becomes h nu dash would be m0 c square by because this will get cancelled with this m0 c square would go in the numerator and it becomes m0 c square by 2 m0 c square is nothing but 0.511 mev or 511 kv half of it will be 256 kv. So, at 180 degree the gamma ray photon energy is very high then you will find there is something called as a compliant edge because the energy of the electron will be h nu minus 256 and so at a energy full energy of the gamma ray minus 256 there is a peak that is called the it is not a peak it is a hump so, that is called the compliant edge. So, it is a typical now feature of content scattering which we will see in the subsequent slide you can see here the the content edge this is so, at let us this is 256 if the photon energy is much higher than the rest mass energy of electron that is 511 kv. So, you can see here depending upon the scattering of the photon at 0 degree or 180 degree you get a continuum in the gamma spectrum as a result of content scattering this is the full energy. So, content scattering does not lead to the full energy deposition, but what happens if your detector is very big in size then the content scattered of photon can further undergo photoelectric effect or content scattering and then you can have a full energy deposition because the even the content scattered for here is assumed that the photon is scattered out of the system, but if the photon can further interact with the medium then you can lead to energy deposition. And the probability of the content scattering has a mild dependence on the z and the energy of the photon. So, here what is shown here is the energy dependence of the photon. So, as the as the energy of the gamma ray increases the probability the content scattering cross section decreases with the energy, but it is a very mild dependence. Now, the third process is pair production. So, essentially in this process also the photon is completely lost the photon is like absorbed in the Coulomb field of the nucleus and the energy of the photon is used in producing a pair of electron and positron. So, this process is seen schematically here that the incident photon comes in the vicinity of the nucleus and the energy of the photon is taken up by the nucleus and instead of that the internal photon it is like the universe of the annihilation of the gamma ray, the pair annihilation of the positron with the electron. So, a pair of electron and positron is produced and the energy of the photon is now going into the kinetic energy of the positron and electron, but to create the pair of electron and positron you require the rest mass energy of the pair that is 1.02 MVV and so that is the threshold for this pair production process. The remaining energy h nu minus 1.02 is going as the kinetic energy of the electron positron pair. This positron will get thermalized in the medium and it can meet another electron in the medium and annihilate the electron when it is thermalized to give you 2 511 KeV gamma rays which we discussed in the interaction of positron heat matter. The net result of pair production as you can see here you will have a high energy electron and 2 511 KeV gamma rays and then these 2 511 KeV gamma rays can further interact in the medium either they can get deposited in the system or they can get escape or so. So, if then either there can be full energy deposition the detector is large enough or one of the gamma ray one of the 511 KeV may escape this called single escape peak or both the gamma ray this 511 KeV gamma ray photons may escape from detector that is called the double escape peaks. So, what you will have here that from the suppose this is the gamma ray e gamma counts. So, if this is the full energy then you will find and this is the Compton this is the photopic you will see the single escape and double escape peaks on the gamma spectrum because of this 511 and this is 1.2 corresponding to single escape and out escape. So, this is what happens that when the pair production takes place. So, you can immediately make out that since the gamma energy is higher than 1.02 pair production will be dominant process. So, the probability of pair production is varying with the energy of the gamma ray and the jet of the absorber in this fashion jet square into ln of e gamma here the e dependence is shown. So, there is a threshold for the pair production and as you go and increase the energy of the photon the pair production probability increases. So, let us say 2 MeV 3 MeV you will find pair production will dominate the interaction of gamma rays with matter. So, the net projections for all the processes the photoelectric the Compton scattering and the pair production are shown in this slide here. So, you can see this is the photoelectric effect you can see the k h and it is going down by the after 1 MeV or so actually this is for the sodium hydride. So, it is impacting the sodium and iodine in the material then the Compton scattering is you can see here Compton scattering this is a photoelectric effect. So, it is coming down as 1 by e and the pair production is going up beyond 1.02 immediately. The net result of these processes is the projection for the whole process interaction will be you come down and then again start going up. So, this is the way the gamma ray photons will interact with the media and the when the gamma ray is traveling through a absorber. So, how does the intensity of the gamma ray change with the thickness of the absorber material that I have tried to illustrate here is the incident gamma ray photon of in an intensity i 0 is passing through a thin slab of the absorber material and the transmitted intensity is i then the the different the the decrease in the intensity of the gamma ray minus d i by b x in thickness proportional to i. Why this has come because in this case of gamma ray where the gamma ray photon is interacting in one step either it will undergo photoelectric effect or Compton scattering or pair production in one event itself the photon is removed from the initiating flux and so that process has got a like you know radioactive decay radioactive decay atom decays and so you have an exponential decay with time. In the case of this interaction of gamma ray in one interaction itself the photon is disappearing from this beam intensity and so you have the similarly the first order rate law and the solution of the differential equation is i equal to i 0 a raise to minus mu x that is what I have tried to show here that the ratio of the transmitted to the incident intensity is exponentially decaying thickness of the absorber material as the thickness is increasing the gamma ray since the transmitted intensity is decreasing. And this so you can see this mu you can get from here by fitting this equation and that mu is the total so we are not trying to break up them but this mu whatever we get is the sum of the three attenuation coefficients mu is called the attenuation coefficient how does the gamma ray get attenuated. So mu photoelectric effect mu Compton scattering and mu pair production the sum total of this is the total attenuation coefficient. Now see the attenuation coefficient we define two types of attenuation coefficient one is the linear attenuation coefficient mind you the quantity in exponential e raise to minus mu x the mu x has to be dimensionless quantity anything in exponential dimensionless quantity. So if x is in thickness in centimeter mu will be in centimeter inverse so that is what is obviously this is x is linear in thickness in centimeter mu is called the linear attenuation coefficient. But in the interaction as I discussed earlier also suppose you can have a one inch thick material with pores inside and same thing you can compress the thickness becomes one centimeter but the density is different. So the density normalized thicknesses are called mass dependent thicknesses and so that where the density independent mass attenuation coefficient can be obtained by dividing the linear attenuation coefficient by the density. So if you multiply this by it should be rho density of the material mu by rho so it will be centimeter inverse on gram into centimeter 2. So this is the density 1 by rho this is the mu. So it becomes mu m that is centimeter square per gram this is called the mass attenuation coefficient. So if you write the thickness in gram per centimeter square so you multiply the thickness by density x into density is centimeter into centimeter cube this is called the mass dependent density dependent thickness. So normally in this interaction so we will always talk about density dependent thicknesses and so we have that. So then this becomes independent of density of the material so what about value you get mu m it is independent of the density whereas this attenuation coefficient linear one depends on the density of the. Now I will just touch upon the interaction of neutrons with matter neutrons are also neutral particles they are neutral particles so they do not have any pulomic interaction with the material but they will undergo a scattering elastic scattering or nuclear reactions and that is how we make use of these interactions to detect the neutrons. In the case of neutrons so we can classify neutrons into slow neutrons or harsh neutrons. In fact there are other classification like epithermal neutrons I mean there are some regions of energy of neutron where there are high resonances like here I have shown here the cross-section for the neutron induced reactions fall as we increase the energy of neutron in fact this is called one by V law velocity of the neutron is increasing so the cross-section is decreasing. So thermal neutrons the neutrons of low energy are more reactive than high energy and somewhere in the middle around 1 to 2 electron volt so there are resonances in the cross-sections there are whenever the nucleus is matching with its energy that is energy so what we want to show is that thermal neutrons are more reactive and so many times in to protect neutrons we need to thumb like the neutrons. So we will discuss the interaction of slow neutrons and harsh neutrons little bit separately slow neutrons are highly reactive and why it is called cadmium cutoff the energy of neutron less than 0.5 electron volt is called cadmium cutoff because at 0.5 electron volt cadmium has got this kind of ordinances and so if you put cadmium the gamma rays are the neutrons are totally you know absorbed by this cadmium so that is called the cadmium cutoff means the energy which is absorbed by the cadmium. So if you wrap a sample by cadmium that then low energy neutrons are all absorbed by the cadmium you will see only the fast neutrons. So the thermal neutrons though it will undergo elastic scattering but that is not important from the point of view of detecting the neutrons the slow neutrons will predominantly undergo nuclear reactions it can be neutron captured and gamma it can be neutron induced proton emission followed by preceded by neutron capture and alpha or fissure neutron induced fissure. So you can see detectors based on boron bf3 gas where boron 10 interacts with the neutron to give alpha and lithium 7 so these are the charged particles and they will be depositing energy in the medium similarly helium 3 gas and p reaction tritium so the tritium and proton they are the charged particles similarly the 235 uranium neutron induced fission the fission fragments are charged. So the net result of the interact nuclear reactions by neutrons with different detector materials will be the portion of charged particles which will interact with the like heavy charged particles and deposit their energy in the substance. The fast neutrons can also undergo inelastic collisions apart from elastic collisions and they can excite the nuclei or they can undergo nuclear reactions. So when it is undergoing elastic scattering we have required NPI like a neutron can interact with the proton and knock it off from the place so we can have the heavy charged particles as a required nuclei or the fast neutrons can also undergo nuclear reactions like NP and alpha NF neutron diffusion and N2M types of reactions. So this neutrons detection can be done using the nuclear reaction in fact mostly the neutrons are detected using the nuclear reactions and they what is the range that neutron the how much distance they can travel like they are neutral particles so the the energy loss mechanism is first initially only by elastic scattering or inelastic scattering. So if a neutron is colliding with a nucleus of mass number A coming number Z then it can just give a kinematics you can set up and that we did in the case of heavy charged particles and electrons the energy received by the required nucleus will be 4 mm upon upon m plus m square into energy of the incident neutron this is the same formula we derived for the heavy charged particles so same thing applies here. Now let us assume that for protons the protons and neutrons are relatively same mass so m equal to small m equal to 1 and so if you put the value here then you will find the energy of the required will be energy of the neutron. So that means what with the proton when the neutron is interacting neutron can give all energy in one single collision like in the case of us electrons us electron will give all energy in one collision so the when you want to reduce the energy of neutron using a moderator the proton or the hydrogen material is the best water. So that is why you will find water is used as a moderator in the reactors there are other problems with the water so we use any water. Now finally the cross sections this is called the microscopic cross section when a neutron is interacting with the nucleus we call cross section as sigma but when we say neutron beam is passing through a material we call capital sigma is called macroscopic where n is the density so nuclei per cc number of nuclei per cc or number of atoms also you can say cc in the material and so the sigma essentially is like you know decay constant of a neutron like when we have the beam at innovation of the neutron you can see here that the neutron beam is going then i0 to i the neutrons also will follow the similar type of expression i equal to i0 is to minus sigma x where this sigma is the microscopic cross section and many a times we associate a mean free path that means what if the distance traveled by the neutron without attracting the material that is one over sigma so these are in fact huge in the reactor physics when the neutron neutron economy is being discussed how the neutron is diffusing in the detector system these concepts become important in that. So I will stop here and subsequently discuss the detection of relations by different means. Thank you.