 Most probably everybody has knowledge of it. But I would like to stay in the general formula, because it's the general formula for every reaction. Why is the general formula? Because this gives you the reactions per time, normalized through the projectiles per area per time times the target nuclei. And this is an absolute number. So any form of a reaction, any type of reaction, is obeying this formula in its general form. And I mean when you measure absolute. Is anybody that doesn't understand what means absolute measurement? You understand? What is an absolute measurement? An absolute measurement is a measurement that gives you a quantity that is independent of what? Of reference system? Of your device? Of what else? No, but this is your device. This is your device, end of time. So, which means when you talk about cross-section, you have to define in your paper, in your publication, in whatever, what cross-section do you mean? Is it a relative cross-section? Or is it an absolute cross-section? I'm sure somebody of you knows an example of a relative measurement, to give me. Who is doing here activation measurements? You. Why is your measurement relative and not absolute? Or when? For example, you make neutron activation, on your sample, and then you go and measure, and you measure counts and whatever. In most of the cases, you have a gold foil for which you know very well the cross-section. You activate your foil, and through the activation and the known cross-section of your gold and something, it is controlled, but in many cases, you have this kind of measurements. You know something and you go and you determine a cross-section or some other quantity relative to that. This is not an absolute measurement. You give an absolute value because there is an absolute value known, but it's not your reaction. So, something else. When you make experiments for nuclear structure, are you aware of this kind of measurements? Let me show you an example. Nuclear structure means you have a nucleus. Here is its ground state. A, Z, X, ground state. By the way, can somebody tell me an example in nature when we say ground state? What is the ground state of aluminum? Yeah. You all play with aluminum. Everywhere in your house is aluminum. Is this aluminum in ground state or not? Why not? Maybe it's not, but why not? No, you have aluminum. OK, this is a foil of aluminum. Suppose this is aluminum. Is this aluminum in ground state or not? Yes, please. Aluminum 27 is stable. So, is it in ground state or not? Yes. Why? I'm telling you that the aluminum 27 is represented by a square well where the nucleons are. And this has an energy epsilon. Is it in ground state or not? Of course it is. Because this energy has to do with its binding energy. So the zero energy for the macrocosmos is here. And this is zero. So everything you touch and is not activated is not no radioactivity. It's in ground state. OK? But what happens now if you give a little bit of energy in that system? It gets excited. OK? So take 27 aluminum. I give energy, OK? When I give energy, depending on the energy, I create the aluminum and the aluminum can be in different excited states. Clear? So this is my first excited state, my second and so on. So let's make it a bit more general now. I have an accelerator, OK? And I have a carbon beam, I have the ion beam and I'm shooting a foil. It could be aluminum, it could be palladium, it could be whatever, OK? And I make a reaction, OK? Just to give you an example, that was the reaction in my PhD thesis. What gives me that? 120 plus 13, 133 minus 3, this should be 130, OK? So if I know this, the Z, you will find that this is barium, OK? Doesn't matter. Now, I create barium. This barium is in most of the cases, OK? In an excited state here, OK? In some continuum. And what I do now is I go and have a measurement. This is my beam, OK? My 120 tin. And I have various detectors here, OK? I make an experiment at E60 MeV, an experiment at 70 MeV, and so on, and so on. And I want to find at which energy I create this nucleus at the most, which means I have to look for counts coming from barium 130, OK? Now, at 60 MeV I measure 2 hours, because I have a good student and he's really working hard. At 70 MeV is night and my student is sleeping and it measures 6 hours, OK? How can I compare the two spectra? Here is my spectrum, blah, blah, blah, blah. One peak that comes from 130 barium and here is a spectrum that is very high, because this is 6 hours. I don't believe the current. I don't believe the current, I forgot to know the time. No, you cannot do this because the background is different. What do you do? If your target here is backed, just a second, not to lose more time, OK? This is 120 and behind you have, let's say lead, OK? Blah. How is lead? When your beam comes here, it continues and makes current excitation, OK? And you know the energy. So you also count the current excitation events and you know how these events go with the time, OK? So here you have a correction factor, let's say C2 and this is C1. So what do you do? You take the integral here, OK? Times C1, C2 and you have the correlation. Is this measurement relatively absolute? This is totally relative, OK? I insist in that point, please guys, if you want to be a credible scientist in your work, you have to understand what it means to measure absolute and what it means to measure relative. In most of the applications, in materials analysis, for example, you are not necessarily need an absolute measurement, OK? But that's OK, because there, you are interested in relative phenomena. However, when you want to, for example, determine absolute number of atoms of an element, OK? There you really need to have an absolute measurement or a good reference point value that is very, very well measured with a very small error, OK, so that you feel comfortable. OK, that's why I'm saying here this is the general expression for cross-section, OK? For example, when you do RBS, what do you really need, don't need here? Do you need everything here? You measure relative always, OK? And because you know the Coulomb scattering term, which is, you know, it's a number, bam, kinematics, it's on so angle, I get so much counts, OK? Fine, let's proceed. What gives me the reaction cross-section? It gives me the, it is a measure of the probability, OK? That a certain reaction takes place at a certain energy, which means the cross-section depends on energy, fine? Now I will show you another very important quantity apart from the reaction, apart from the cross-section, there is the so-called reaction rate, OK? The cross-section gives you the reactions per time, projectiles per area per time, target nuclei, OK? What gives you the reaction rate? The reaction rate gives you the number of reactions per unit time, per unit volume. And because of that, it is the most important reaction when you do, let's say, nuclear synthesis, so nuclear astrophysics, why? Because if you take that cross-section of the reaction, OK, and insert it in this expression, you get the reaction rate. Now all these symbols here are to explain as follows. Assuming you have a stellar site, OK? This stellar site has a temperature, which means kinetic energy, OK? A density, a mass and a distribution of the elements. Now, KT is this temperature, but T is your lab energy. Thermal energy is kinetic energy, which means if I know that, let's say, in that star a specific reaction takes place and I want to see how this reaction contributes to evolution of the star, what quantity do I need? I need to know how fast this reaction occurs, OK? What does it mean how fast? How many reactions per time per second? And at the same time, I need to confine this in a specific volume. That's why I divide per volume. And look here, nuclear physics, reaction cross-section, provides data to nuclear astrophysics. Because if you insert the cross-section of the reaction and it takes place in the star, you can calculate the reaction rate. And with the reaction rate, you can make modeling and calculations how fast or how slow a star or a stellar environment is proceeding. OK? Now, apart from the cross-section or the total cross-section or whatever, of a reaction, we have the differential cross-section and no differential cross-section. What are these cross-sections? A total cross-section means that I create here a reaction that emits particles or gamas or whatever all over the place, all over the volume. So total cross-section, absolute cross-section means that I know what is the yield, the absolute yield, the total yield of the reaction. What does it mean yield? It means the number of produced particles, which can be also gamas, in the whole volume in 4 pi. Differential equation means what? I want to know at a specific angle from the beam direction, where is the so-called z-axis. How many particles are produced here? To do this, I need the so-called angular distributions. No matter if they are gamas or particles or electrons or whatever, if you want to measure differential cross-section, you have to do angular distribution measurements, which means you have to measure at specific angles, your counting rate, so to say. Now, there is also the double differential equation, which is what? At a specific angle, specific energy. Your products here are emitted all over the place with most probably different energies. So then your differential equation gives you the angular distribution of the products. And the double differential equation gives you also not only the angular distribution of the products, but also the energies at the angular distribution. Fine? Is it clear? Question. I haven't put here the so-called partial cross-section. Can you imagine what a partial cross-section can be? Partial cross-section. The fact that I am saying partial, what does it mean? I fragment my total in many others. But what others? For example, if you have a reaction where you have, let's say, three channels, then you have sigma total equals sigma 1 first channel. Plus sigma 2, plus sigma 3. In some sense, these are the partial cross-sections of your total cross-section. However, when we talk about reactions and exit channels, we don't call them partial cross-section. We call them the cross-section of the channel X. Now, guess what? You create aluminum 27 here, here, somewhere, and this guy cannot emit particles, emits only gamas. One case. Then you can have this gamma, you can have this gamma, you can have this gamma. How would you measure the total cross-section of the reaction that gives you aluminum? No particle emission, just gamas. Don't go under spectra. Tell me the principle. Suppose we can measure all over the volume. What do you need? What would you measure? What give these gamas here? Only? The energies, they are clear. They have epsilon, whatever, gamma 0, whatever. The number of these gamas, what gives this number? No. What does it mean to measure the transition? Decay. The number of nuclei of aluminum that have produced here, and decay. So, if you know the absolute number of the photons of the gamas here that have been emitted by the reaction, what can you calculate? Come on, guys, this is the total cross-section. If you are able to find, I don't know, a medium that will tell you, you know, I'm falling down and there are coming another 300 guys that are falling down. So, how many guys have been produced? 301. So, you take this number. You connect it in your telecontroller, whatever, if you see them all. And you say, these 301 guys that are falling down, there was a criminal, or how many criminals were somewhere and they throw them away. This is your beam. So, then you say 301 divided by 20 lunatics. This is a number, and I have to correct because I don't see them all. This is my efficiency. You take this formula, roughly speaking. And you get your cross-section. But let me ask you something. You take all of those, all these gamas, these are one, two, three, four, five, six gamas. How many you take? All the six? Which one? Which one? You say three. Correct. Which one? You have two choices. These you have to take anyway. Because it leaves the balcony and it arrives at the earth. So, you do either this guy, that guy, and that guy. OK, here is not serious, cannot be two. OK? Or you take this guy, that guy, and that guy. This gamma transition doesn't count because it's a part of that. OK? So, if you take this and this gamma ray, you have it all. However, if your device is not, let's say, is obscured by background and you cannot see this one, and you see the other ones, you go there. OK? That way, what do you do? You determine the cross-section of the reaction that produces aluminum in its ground state. Question. Assuming that this is a metastable state. Metastable means long-lived. OK? And you don't have any gamas here, you have an E0. An E0 means a transition that does not emit gamma. OK? Can you determine the cross-section for the production of aluminum in this level? Of course you do. You do the same thing. You don't take everything that goes away. You just take everything that comes up. OK? Now, this is sigma metastable. OK? And this is sigma ground. What will be the total? Sigma ground plus sigma metastable. This is the total, this is the partial, and this is the partial. In general, you can determine for every gamma a cross-section. And if you make the correct addition, you end up with a total, with a correct total cross-section. And these are the partial cross-sections. I have an example for later. A real one. OK, where do we are now? OK, we said, we understood what is a Q value, and what is a threshold, OK, of a reaction? And what is a cross-section? Now, let's look in that picture here. This is a picture of a proton that is impaging on a nucleus, indium 113. OK? The black curve of what you see is the proton capture, so to say, reaction where only a gamma is emitted. OK? So, the blue one is where an alpha is emitted. OK? Neutron is emitted, and pp prime, OK, this is an inelastic scattering. Does this has a Q value? It has a threshold, but not a Q value. It's a matter of interpretation. You can say it's a Q value, but in fact it's a threshold. OK? The P gamma reaction has a Q value of 8.4 MeV. The P alpha reaction has 5.8 MeV. And the Pn reaction has minus 1.8 MeV. What does it mean, this? That in order to have this reaction, your proton has to have at least 1.819 MeV. Now, if you look at 1.8, you see the Pn goes down to Hades, to zero. That's why you see this one. Whereas all the other reactions, they have a positive Q value, which means they are occurring no matter what. OK? Oh, come on, here is 10 to the zero, and this is 10 to the minus 18, which means we are already zero. OK? Now, let's take the example where I have the emission of only one type of a product, but one time, two times, three times, and so on. You see here. This is a reaction Pn on something. Then P2n. What does it mean, 2n? I, you know, I suited out two neutrons, three neutrons, four neutrons, and so on. You see that they are moving to the right, as expected, because in order to take two neutrons out of your target, you need more energy, so you move that way, and they all have a threshold. OK? Here is the threshold for the Pn. It goes to zero. Here is the other threshold, and so on, and so on. OK? Try, please try to understand the meaning of the threshold of the q-value and the cross-section. They are totally correlated, and if you understand what these quantities mean, you can understand a lot of things in nuclear reactions. OK? Now, let's classify the nuclear reactions. OK? How many types of reactions we have? This classification is in line with my age, I would say, so it's before 90. OK? So how many reactions we have? We have elastic scattering, inelastic scattering. Rearrangement collisions, this doesn't exist anymore as a term. It's simply a nuclear reaction. We have many body reactions. We have photonuclear reactions, and we have radiative capture. OK? Let's see what are these. Elastic scattering means A plus A capital gives A capital plus A. q-value is zero. OK? What is this? This reaction is always possible. It can be because of Coulomb repulsion, Coulomb effect. OK? And when the Coulomb forces are dominant, then we have radiative scattering. And this reaction, the elastic scattering, plays a key role in surface analysis. Inelastic scattering. Here, inelastic scattering means that both products, A and A prime, OK, can be in excited state. What does it mean? This projectile, after the event, will be in excited state. OK? And the product, the heavy product, could also be in excited state. OK? Here we have a threshold. That's why I told you before. The q-value is negative, and when the A is in excited state, usually it decays with a gamma. OK? And that's why because of the reasons that we said before, that's why these reactions inelastic scattering are often used for analysis, but using the gamma. OK? Rearrangement. This is simply nuclear reaction, OK? Which means a small A and a big A give me a B in different excited states. And, of course, the corresponding q-value is different, OK? What does it mean in practice? If you have a deuteron on a nitrogen 14, you produce carbon 12 plus an alpha plus 13.5 mV energy. But you can have the same reaction, producing carbon in an excited state, OK? Plus the alpha. In that case, if the carbon is in the first excited state, which is at 4.4 mV, then you produce less energy, OK? Many body reactions. This is also overdoing because this means that at the end you have not only one light product, but many, OK? This is something like this. Gold plus alpha gives you thulium. 198 plus plus 3 neutrons, OK? And this reaction is like this. Photonuclear reactions. High energy gamma, OK? Impeginzon and nucleus and Brexit, OK? It gives you a new product plus particle. Gamma plus carbon 12 gives you carbon 11 plus neutron. This was supposed to be one of the most sensitive methods to detect carbon, OK? But if you see here, 18 mV, which means your gamma should be 18-something mV, so an enormous amount of energy. So photonuclear reactions are possible when your gamma energy is really high. And radiative capture is simple. NA and the capital A produces a system, the so-called compound nucleus, that is excited, and then it de-excites by emission of gammas. The most famous reaction is 27-aluminum, p-gamma, and this reaction is used for calibrating your accelerator, OK? It has a resonance, it is a very well-known resonance, sharp, so you can really run your beam, your proton beam on a relatively thick target, and then you see, when you see the step of the cross-section, of the counting grade, you can say, here is my resonance, here is your counts, here is your energy, and it goes like this, OK? So it's a step, and in the middle is your resonance energy, and from here to there, OK? This is your energy uncertainty. In fact, here is not the energy of the proton, is the magnetic field of your magnet. But, OK, you can translate it in energy. As I said, this classification is old, and in fact, all these are thought as reactions, OK? So we have elastic in elastic and reactions. Now, the point is that since recent years, these both are also reactions, because we say that in any reaction you can have this exit channel and this exit channel. So we talk only about reactions. Fine, now. When do we have to go for lunch? Sorry? 12.30. OK, I still have my 1.10. OK, good. Reaction mechanisms. We have two major mechanisms, the direct reactions and the compound reactions, OK? In the first case, we have the case where a projectile interacts with one or a few nucleons of the target, OK? Within a very short time of the order of 10 to the 21 seconds, minus 21 seconds, OK? And either it transfers energy to the target or, as I wrote there, it picks up or loses, OK? And then it turns, leaving them to the nucleus, to the target nucleus. Here you see, a juteron is imaging on berylium 10, OK? If you have an elastic scattering, at the end you will have berylium 10 and juteron. This is elastic. What is a transfer reaction? A transfer reaction means that the juteron goes to berylium, catches one of the nucleons, and leaves the other one moving. In that case, the neutron. There is, of course, the other possibility that the juteron approaches berylium 10, interacts and is breaking up. These are direct reactions. They are very fast, OK? Now, in the category of direct reactions you have elastic scattering, in elastic scattering, the so-called transfer reactions, stripping or pick-up, OK? Two types, knock-out, break-up, and direct capture reactions. I tried here to give you some explanation what is in elastic scattering. You have individual collisions between the particle and the single target nucleon, OK? And the incident particle emerges with reduced energy. This is in elastic scattering. What is not conserved in elastic scattering? The kinetic energy. That's it. Pick-up reactions. The projectile collects additional nucleons from the target. For example, djuteron plus oxygen 16, OK? Get what? Another proton, sorry, another neutron and leaves oxygen in oxygen 15 state, isotope. The same here. Stripping reaction is the inverse. What does it mean? The projectile leaves one or more nucleons behind to the target. Djuteron plus 90 zirkonium gives you 91 zirkonium plus a proton. These are all typical direct reactions. Now, what are the compound reactions? It's the second category of reactions. Here we have following situation. The projectile with enough energy, or let's say, not high, but enough to do what? To go inside the target nucleus, start interacting with the nucleons, OK? Having collisions, sharing energy. But, as I wrote here, during this process one or more particles can be emitted and they form with residual nucleus, the products of reaction known as pre-equilibrium, OK? So it's an intermediate phase where you don't reach equilibrium. Equilibrium means all the energy that you can give is shared by all nucleons. You are somehow selective, OK? This is the first phase of compound reactions. The compound nucleus, which is the next phase when this equilibrium has been reached, OK? The largest possibility is the continuation of the aforementioned process so that the initial energy is distributed among all nucleons with no emitted particle, OK? In this case you create a system with one more extra nucleon, OK? And you have the compound nucleus. In this picture you see the direct. In the case of the direct reactions, you have a single interaction before emitting a particle. Pre-equilibrium, you have a small number of interactions before emission, OK? In compound nucleus you have a large number of interactions before emission, and in this case, in most of the cases, it's gamma, OK? Now here you have again in the pre-equilibrium phase in elastic scattering, transfer reactions, knockout or heavy ion reactions. I saw you before, OK? And this phase is favored above 10 MeV, OK? And the compound nucleus contains resonance scattering evaporation, including radiative capture and fission. We don't have time to go all this through, OK? I just want you to make clear in your mind that the reactions, OK, they differ because of their time and because of their mechanism. Don't ask me, I don't know. There is some prompt. You tag your product, OK, in coincidence with your projectile. Yeah, but not in that scale. Yeah, but you make delays and whatever, or indirectly. You have a way to find this. You cannot measure the direct lifetime. That's true. No, no, no, no. No, no, no, 10 to the minus 20. Nanosecond is 10 to the minus 9. Picosecond is 10 to the minus 12. Femtosecond is 10 to the minus. So it's, you get this number indirectly, OK? What is my technique? OK. Yes. No, the pre-equilibrium, let's put it that way. At low energies, when your particle enters your target, you don't have anyway enough energy to kick out some particle. Which means what you have, you are the good summary. You just give them all, you know. You share your energy with all the nucleons. And because you share the energy with the old nucleons, you have to make a lot of collisions with them. So it takes time. Whereas if you have a little bit more energy and you can send some nucleons out, OK? But it remains some energy in your pocket. You want to share this. So it's an intermediate step. So now let's see what do we expect theoretically at least. OK? These are my reactants, OK? And they give me enough energy to go that high. Now, if I emit neutrons, OK? If I have, let's say, the direct mechanism, what does it mean? I emit particles, and these particles are populating my product, OK? In direct reactions, 99,9% is particle emission. And these particles usually end up in an excited state of a product nucleus. And because they are in excited state, then you see transitions between these excited states. And you see in your spectrum, just please look at like this, OK? You see discrete lines. If you have a compound nucleus with emission of neutron, you end up in a highly excited state, OK? Here, don't see the lines. This is really continuum. So any transition between these states is almost continuum. And that's why you don't see discrete lines. You see a bump, OK? And this part is the compound nucleus mechanism, and this is the direct mechanism, OK? So if you can create, if you have, let's say, the advantage or the privilege to have an accelerator and start shooting a target with the same projectile, up to 50 MeV, and you make the so-called excitation function, and you measure for every energy, the cross-section, at the end you will end up with something like this, OK? Just try to imagine it like this. Fine. Now, we said something about angular distributions, OK? And we want now to understand what happens if I have the three different cases, the compound, the pre-equilibrium, and the direct. How my angular distribution should look, is expected to look like. And here it is. In the direct case, OK, the differential cross-section, the cross-section versus angle, OK, will be around zero the maximum. It will drop. It will make something like this, a subtle point, close to 90, and then will drop completely, which means at the backward angles, I won't see anything, at the very backward angles. I will not measure any counting rate. Something similar is the pre-equilibrium. Why? Because the pre-equilibrium is something like in between the compound and the direct. And what is the key feature in the compound reactions that the angular distribution, the differential equation is symmetric around 90 degrees. OK? You see it here. Now, look here. This is the case. We measured the nothings. We measured e to the power of 89 p gamma. So we had the compound reaction with the emission of gammas. And what you see there are the angular distributions of some transitions. OK? You see that when I have an angular dependency, which is not always the case, OK, you see that I'm, this is 90. I'm symmetric around 90. It could be also like this, but it will be symmetric around 90. Now, if I take the double differential cross-section and look what happened, what happens in a reaction with protons where I can start from very low energies up to let's say 60 MeV. As I told you in the previous picture that was like this, OK, you have the compound, you have the pre-equilibrium and then you have the direct. And the double cross-section here, it's sharp. You see all these sharp lines. These correspond to discrete states of the product, OK? Now, the point is that the energy scalar goes like this to the ascending, OK? Whereas the time goes like this, OK? As we said, very fast and direct, very slow, the compound and somewhere in between, OK? Now, this is a real excitation spectrum measured by neutron on uranium 238, OK? And here you have Xn, which means you can have various n's, 1n, 2n, 3n and so on, OK? As you see here, this is the total cross-section. The direct is this one, OK? The pre-equilibrium is this and this is fission because here we are at very massive nucleus, OK? Now, what can I learn from elastic and inelastic reactions? I measure elastic or inelastic reactions. Here you see the inelastic reaction of neutron on iron 54, OK? This is a real spectrum. You see a peak here, a second one. Two doubles here, a double here and a double here, a doublet, sorry, not double, doublet, OK? And here you see 0, 2, 4, 0, 6, 2. These are the spins of the states where the neutron, inelastically scattered neutron, ends up. Indeed, if you look the excitation spectrum of iron 54, you will find this is the zero here, the two here. The four is here. The zero is here. Another two is here and another six. This means I go there and I make inelastic scattering on something and I see these peaks. Some of those can identify. Can I identify? Some not. So if I discover this kind of peaks, what can I do? What can I say? This belongs most probably to an unobserved excited state. Fine. Now let's go and measure inelastic in the same nucleus, OK? In the same energy, OK? So this is not tungsten. It is, I would say Wolfram is in German, but tungsten, OK. So, here is the angular distribution in the elastic case, OK? You see all the differences here. Remember the previous figure. And here you see the inelastic, you see this distribution corresponds to two plus, the four plus and the six plus. And indeed, the two plus, the four plus and the six plus. Sorry. No, these are particle spectra. OK. I think this is the last one. If you want to continue tomorrow, we can continue here, OK? From here. Final slide for today. Reaction mechanisms for capture reactions. What are the capture reactions? The capture reactions is a special case of compound reactions. What happens here is that your projectile goes to your target. You create a compound system in equilibrium within 10 to the minus 18 seconds and then you emit particles or gamas, OK? So this is the famous way to writing the nuclear reaction and it was said before in the first slides. And this is the capture reaction, OK? The star is here and instead of B plus B you have the C, de-excited plus the gamas that de-exciting the nucleus. And this is your capture reaction, OK? It's 12.30. So if you want me to continue tomorrow I have in the following various experimental slides to show you, OK? It's up to you. If you don't want this, I will make a recap. It's not a recap. It's actually my way of seeing the topic of the workshop, OK? Accelerators, detectives, and so on, OK? So we will close this tomorrow and I think the second hour I can show you something maybe funny, OK? So enjoy the lunch.