 Okay, hello everyone. Previously, we discussed the cross sections for nuclear reactions, particularly for the Newton induced reactions and the charge particle induced reactions. And now, we will discuss how to determine the cross sections experimentally. In fact, when you irradiate a target in an accelerator or a charge or a reactor, how the active atoms are formed or you can even determine the activity. So, if you determine the activity, you can determine the cross sections. So, first let us discuss how to determine how much will be the activity of a reaction product formed when you bombard a target with the neutrons or charge particles. So, let us discuss in detail the irradiations in the facilities like nuclear reactor and an accelerator. So, let us first discuss the nuclear reactor. So, in nuclear reactor, you will have the neutrons all along the moderator system. So, you will have a sea, it is like a sea of neutrons, large amount of neutrons just diffusing through the medium. And so, when you have a sample, the sample dimension may be 1 centimeter cube or so on. Where is the nucleus, neutrons are there all along in a, you may just say about the meter cube. So, it is like a sea of neutrons and you put a target, the entire target, all the atoms target are exposed to the neutrons. So, the nuclear reactor, so, how do you define, how many neutrons are there in the reactor? You define in terms of plots. So, across when you are having a neutron in the reactor, we say this many neutrons are passing through a 1 centimeter area, 1 centimeter, centimeter square per second. So, you have the neutrons going and 1 centimeter area, how many neutrons are passing in 1 second, that is called the plots. You can actually, we also call this NV sometimes in reactor physics, we call NV, N is the number of neutrons per cc into its velocity. So, neutrons per centimeter cube into centimeter per second. So, it will be neutrons per centimeter square per second. Sometimes the neutrons are not normalized, they are having half component. So, velocities are different. So, you define neutron flux as NV, number of neutrons per cc into velocity of neutrons. So, this is also defined sometimes as neutron flux. So, you irradiate the sample. So, this whole thing is a like a nuclear reactor. Entire target is exposed to neutrons. All the atoms in the target are exposed to neutrons. And so, when you have the target atoms, this is the rate of reaction, target atoms, total terminal target atoms cross section into flux N sigma phi rate of reaction. And when the target is getting activated, then the active atoms start decaying with the rate N lambda. So, the net rate of reaction is N sigma phi minus N lambda. So, this is the, if you are producing a radioactive isotope as a product of a nuclear reaction, then the rate of formation of the, the net rate of formation is Tn by gt, this is the growth minus degree part, like the radioactive decaying. So, Nt is the number of target atoms in the sample. If you have the weight in gram, then you can multiply by the number divided by the mass number, if it is a monoisotropic target. If it is a multi isotropic target, then you have to find the abundance of that isotope. So, in that particular isotope, how many atoms will be there, you have to calculate. So, weight is the weight of target, W is the weight of target in gram, A is the mass number of the target. For example, if it is a monoisotropic like cobalt, A will be 59. But if it is a magnesium target, magnesium, 5, 24, 25, 26, you have to take the average atomic weight, 24. Something. And a particular isotope of magnesium 25, 24, which one is of that you have to accordingly calculate the number of atoms of that target. Sigma is the cross section in terms of centimeter square or 1 and phi is the flux which we define in terms of neutrons per centimeter square per second. So, this if you solve this equation, you get activity at the end of radiation, rate of reaction into 1 minus e raise to minus lambda, Ti, this is called the saturation factor. It will also call this S. So, the saturation factor essentially tells you for how much time you should irradiate the target. There is no point irradiating for a long time. So, you can optimize the irradiation time from this graph. So, actually if you see here, this if you plot this S here, then this factor will become 1. And when Ti equal to Ti equal to 0.5, this is Ti. So, if you irradiate for one half life of the radioactivity that you will produce, you will get 50 percent activity. And after about 3, 4 half lives, you will find that you will saturate the activity. So, you cannot, you do not gain anything by radiating more than 3, 4 half lives. So, usually you irradiate for 1 to 2 half lives of the radioactivity that is sufficient to produce optimum activity. So, this factor is deciding how much is the irradiation time and this decides the rate is formed. So, activity at the end of irradiation will be following again the same graph as the irradiation factor. So, let us do an exercise to know. Now, it is better to get a feel of how many buckles or how many curie we produce when we irradiate certain amount of target. So, that will also tell you how much we can irradiate in this in the reactor. So, let us calculate the activity of cobalt 60, cobalt 60 and gamma 60 cobalt, this is the reactions. 459 is a monobicectopic target, we capture a neutron and by emission of prompt gamma, we come to the ground street of cobalt 60. This ground state will emit, this is the half life of 5.27 year to 60 nickel by emission of 2 gamma is 1172 and 30th. So, I have given a problem here, 1 gram of cobalt is irradiated in Drubar reactor the flux is 510 power 13 neutron per centimeter square per second and time of irradiation is 1 year, half life of cobalt 60 5.27 years. So, this is the equation we use for neutron neutral reactions n sigma 5 the rate of reaction into irradiation factor. So, Nt number of target atoms 1 gram will contain this much upon 59 atoms. So, this many atoms are irradiated sigma 37.2 bonds. So, 37 3.7 minus 23 centimeter square flux is 510 power 13 neutron per centimeter square per second and the saturation factor now 1 minus e raised to minus lambda ti. So, you can write 1 minus e raised to minus 0.693 into time of irradiation 1 hour 1 year on 5.27 years. So, the quantity in exponential has to be dimensionless. So, the time of irradiation and half life should have the same units. If it is in years, then it will be also 3 years. So, we have to take care of the units. So, this factor becomes 0.1236. So, if you multiply these factors, you get activity and the end of irradiation 7.03 10 power 11 factors or 19.01 Q e divide this by 3.7 into 10 power 10. This is the 1 back 1 Q. So, you can see here you irradiate 1 gram of cobalt in a reactor for 1 year and you get about 19 Q e of cobalt 60. So, that gives you an idea the how much activity of cobalt 60 you will get if you irradiate certain amount of cobalt in a reactor for that some amount of time. Now, let us see the irradiation of targets in the accelerator. Unlike in the reactors where we have a pond like a pool like a swimming pool, we have a lot of neutrons moving around and the sample is having very small dimensions. So, all the target atoms are exposed. In the accelerator the target the projectile is a beam of dimensions 1 or 2 millimeters. So, you have a very thin beam of projectile bombarding the target and this is the so this is the vertically you. So, if you have a target like this in a cross sectional area beam will be like this if you see in the perpendicular direction. So, the entire target is not exposed to all the charged particles very small area is exposed to actually the particles and so you normally know the charged particles cannot travel much in the target. So, you use very thin targets of the few microns and if you recall the your lectures on thickness then thickness you write in terms of if it is centimeter then you write rho gram per centimeter cube. So, you write gram per centimeter square, gram per centimeter square is very thick. So, normally you will write milligram per centimeter square. So, in terms of actual thickness it will be in microns. So, because you do not want to stop the beam if the beam will stop it will generate the heat and the energy will be reduced in that. So, you use very thin targets and again how do you produce the activity? So, the rate of production of radioisotopes dn by dt reaction rate n sigma i we do not have the flux now you have a intensity of the targets a projectile minus n lambda a rate at which the radioisotope is decaying and again the solution of it is the similar to that in neutron intrusion reaction rate into saturation fact. Now, that empty the number of target atoms per centimeter square which you can get from the thickness. i is in all that intensity we do not call it a flux, but number of particles per second. So, if you have a target here you put a faraday cup. So, how many particles are heating the faraday cup you can count them. So, you like normally you know when you irradiate you would call it a current beam current will be in nano ampere, micro ampere, milli ampere and so you dump all the beam. So, the particles charge particle will not be stopped in the target they will induce reaction and go ahead with lower energy and will be collected in the faraday cup here. So, you can have the integrated charge and find out how many particles were heating the target and the divide by time you get the particle intensity per second how many particles are heating the target in one second that is what is called the beam intensity. Many times you may have thick target suppose the projectiles energy is not very high then the targetile may stop in the target and then all the beam will be stopped. So, you have to determine the current from the faraday cup this itself can be taken as a faraday cup. So, the saturation factor again 1 minus raised to minus lambda ti will again vary the same way as in the case of neutrons. So, you can decide at what time you should store the irradiation. So, ultimately the formula for the activity of the target the projectile radiostope is again 1 sigma i 1 minus raised to minus lambda ti. Now, let us again do an exercise to see how the activity can be generated in the accelerator irradiation and prior to that it is important to know how to choose the energy of projectile and what energy and what projectile we use to how to produce a particular radiative isotope. So, this is the important exercise to know if you have to produce a particular isotope by charge particle induced reaction what target what projectile and what energy we should use. So, let us give and do an exercise we want to produce thalium 201 which is used in the oncology for the stress test. If someone undergoes some heart problem if the doctor want to know what part of the heart will become intractuous that means the blood is not flowing and you can inject thalium 201 activity and see that we can monitor the activity of this isotope in the blood stream and see whether the heart is receiving the blood all part of the heart is receiving blood or not. So, this isotope we want to produce the reaction that can be used to produce this isotope is thalium 203 P3N proton induced reaction on thalium 203 followed by emission of 3 neutrons giving rise to 201 lead and which is emitting beta minus 2 thalium 201. So, we have proton beam accelerators from aculotrons and now let us see how do we what is the energy of projectile proton that we should be using. So, target is fixed 203 thalium projectile is proton how to fix the energy what energy will give you 201 lead because there are the there are actually the different channels like 3 neutron, 2 neutron, 1 neutron depends upon what the energy brought. So, let us try to see first is the coulomb barrier the proton should cross the coulomb barrier of thalium 203. So, for that we use the reaction the equation R this one this is the factor 1.44 Z1 get to R0 A113 plus A2 one third that is 13.2 is the coulomb barrier. Now, when you bombard thalium with proton 204 lead will be formed and it should be excited such an extent that it should emit 3 neutrons. So, the exercising energy of the compound nucleus that is 204 lead would be sufficiently high emit through typically you know to emit 1 neutron about 10 MeV energy is utilized because the binding energy of neutron in heavy nuclei will be of the row 7, 8 MeV and the neutron will carry some kinetic energy 1, 2, 2 MeV is a total roughly about 10 MeV energy is required to emit any through. So, the exercising energy should be of the row 30 MeV in the compound nucleus so that 3 neutrons are limited. So, the exercising energy of the compound nucleus will be given by ECM plus Q. ECM is the energy available in the center of mass system and Q is the Q value. So, let us calculate the Q value. Q value for this reaction mass of the proton plus mass of thalium 203 minus the mass of 204 lead. So, that is in terms of delta M value the excess mass mass defect values proton thalium 203 and lead 204 these are the delta M into C square values not the actual masses in atomic mass units. So, that becomes plus 6.743 MeV this is the Q value which is positive and as I mentioned for emission of 3 neutrons roughly 10 MeV per neutron that means about 30 MeV should be the exercising energy. So, if 30 MeV is the exercising energy then central mass energy should be ECM plus Q is 30 MeV. So, ECM will be 30 minus 6.73 about 23.2 MeV should be the central mass energy and accordingly the laboratory energy of proton will be ECM into the mass factor mass factors compound nucleus upon target. So, it is slightly different slightly higher than ECM. So, it is 23.37. So, if you take a 23.7 MeV proton beam bombard the thalium 203 target then excess energy of the compound it will be about 30 MeV and by emission of 3 neutrons you will get 201 lead then thereby beta minus decay will lead to 201. So, that is the kind of exercise you can do a priority. In fact, there are no control codes by which you can generate even the cross sections and you can simulate the activity how much activity you will get to irradiate this much intensity. Let us do an exercise for the activity that we can get in the particle induced reaction. So, I have given you an active problem. A 0.01 centimeter thick magnesium foil is irradiated with for 1 hour with deuterium beam of current 100 micro ampere. So, currents will be either nano ampere, macro ampere, milli ampere that can be measured by a factor cup. Calculate the activity of sodium 24 that is produced at the end of bombardment. So, this is the reaction 26 magnesium D alpha 24 sodium and magnesium has got 24, 25, 26 isotopes. Magnesium 26 is about 11 percent hardness and atomic weight of magnesium is 24.3. Now, you have taken 0.01 centimeter thick in terms of milligram per centimeter square. You can multiply by the density 1.74 gram per cc find out the gram. So, you can say 0.01 centimeter into this much gram per cc will give you gram per centimeter square or milligram per centimeter square you can convert. So, the number of target atoms per centimeter square will be thickness into abundance in the Avogadro number upon the atomic weight 4.74 by 19 atoms per centimeter square number of atoms. The current is 100 micro ampere. So, you can see here one coulomb per second is one ampere and so, one electron will give you 1.602 kilo minus 19 coulomb. So, one coulomb per second will be 1 upon 1.602 kilo 990. So, it will be this this will be this many 1 1 ampere it will be 6.24 into 10 power 18 particles per second. So, one ampere gives you 6.24 10 power 18 and if it is micro ampere then 10 power minus 6. So, it will be 6.24 10 power 14 particles per second. So, now, you can see here that the current you can convert into particles per second. Then the saturation factor 1 minus e to the power lambda t i the half life is 14.95 hours for sodium 24 and you are eroding for 1 hour. So, half life would be in the same units half life and eroding time the same it will become 0.045297. So, now, you can calculate the cross section is 25 milliband 1 milliband the power minus 27 centimeter square. So, all of them you can put and you can calculate the activity of sodium 24 or n sigma i that and the saturation factor will be 3.33 and power 7. So, you can see here that in the case of accelerators where the wave intensities are low targets are very thin you get about military levels of activity compared to the curie levels of activity in reactors. So, accelerators we will discuss more on the accelerators this comparing the new reactor radiation and accelerators radiation in general in accelerators we get lower activity because of the the limitations of the target thickness as well as the particle intensity. Now, we have now got the idea what is that level of activity that you can produce. Let us now what we will think discuss the determination of cross section for the charge particle induced reaction. So, let we will take an example of alpha induced reaction on niobium-93 giving rise to compound nucleus-97 and this excited state of nucleus can emit now one neutron, two neutron, three neutron depending upon the excitation energy. So, the variation of the cross section with the energy of the projectile which call as the excitation function. So, we discussed the reaction cross section which is varying as this one sigma r equal to pi r square 1 minus e c upon e c m. So, that is this this is the total reaction cross section, but the compound nucleus that is formed it can emit one neutron at low energy, still high energy can emit second neutron and still high energy it can emit third neutron. So, as you it is the energy of the projectile the compound nucleus is formed with the higher and higher excitation energy and therefore the different channels are opening up the different products are formed depending upon the energy of the projectile. So, therefore, when you are measuring the cross section one needs to measure the total cross section. So, at a particular energy suppose you want to measure the total cross section then you will find at this energy you have one product and two product the sum total of these two will give you the total reaction cross section. So, the excitation functions normally what you do if you do an offline experiment means you radiate the sample you produce reductive isotopes and count in the level three you generate this excitation functions by varying the energy of the projectile. So, for this this measurement what you do you measure the excitation function. Excitation functions means cross section as a function of energy of the projectile sigma e and for that if the products are reductive you can do offline experiment. By offline I mean you radiate the sample in the accelerator from the radiation take out the sample from the accelerator and count the activity in the laboratory on a gamma spectrometry setup and we can form that. So, the same equation is used to find out the cross section. The activity at the end at any time after the radiation is n sigma i sigma is the one we want to determine saturation factor and the after radiation it is maybe decaying with time gamma ray intensity of the isotope and the detection efficiency for counting. So, if you know the target what target you are radiating you know the intensity from the paradigm of radiation you know how much time elapsed after radiation gamma ray intensity and the efficiency if you know you can find out the sigma and the energy at what energy you are going to experiment determined by this. So, what we have here is projectile is bombarding the stack of target and catch. So, this is the target the red ones are the target and the blue ones are the catch. Why this catch is required that when the target is bombarded the products may come out of the target by the coil energy and get stopped in the catch. So, if you do not put this catcher files the products of this target will follow on this target and so on. So, it is important that you stop the required products in the next file which can be a catcher file. This catcher file which also used to reduce the energy of the projectile. So, subsequent you will find 1, 2, 3, 4, 5. The six the five targets will be facing the beam of different energy. So, you have got the cross section measurements at high projectile energies in one of your products. So, this couple you can count them together in the gamma spectrometry setup and measure the activity and from the activity you can find out the process. So, if you are doing the offline experiment for the measurement of radioactive evaporation residue. Evaporation residue means the one after evaporation of neutrons whatever regimes are formed like Technician 97 giving rise to Technician 96, 95, 94. But these products are radioactive and you can measure their activity find out the cross section for alpha n, alpha 2n, alpha 3n channels. So, this is how the offline experiments are done. And lastly many a times you know the physics community particularly they would not like to do offline measurement they want to do online measurements. They want to find out the cross section by online experiment. So, here I have tried to explain the online experiment. The online experiment you irradiate a target. So, I have shown a photograph of an experimental setup in electron at TIFR. We have this target ladder and a particular target you to suppose this is the target you are exposing to the beam and in the forward direction. So, beam will go out on this way. See there are these stainless steel cups you know there are two telescopes. Telescopes means delta EV telescopes you have a thin sapphire detector and you have a thick sapphire detector. The thin one will stop will reduce the energy of the charged particles like proton, alpha and so on. And the thick one will stop permanently. So, a two-dimensional plot of delta E and E the signal of this delta E and E will if you see if you recall the formula d by dx mz square by E the stopic power formula. So, D versus E is a hyperbola. D versus E will give you this hyperbola for different masses like proton, neutron, elem3, elem4 and so on. And you can put a gate on this bananas to get the energy spectra. You can sum the delta E plus E to get energy spectra. You can take the total area to get the total counts. So, whatever counts you get from these experiments these counts are related to n again n sigma phi instead of n sigma phi number of target atoms per centimeter square. Now, you are doing it at a particular angle. So, you do not have sigma, you have the d sigma by d omega that is called the differential cross section and this is the total charge total number of particles which you can measure on your faraday cup. So, you identify the different particles by delta E telescope this assembly as you have shown here and you generate the two-dimensional plots for different particles that are produced. So, it is the particle that are imitated in a nuclear reaction and then you can find out the energy spectra of these particles. You can integrate over the energy to get the total cross section. So, cross section for particle induced reaction by online mode will give you differential cross section at a particular angle and that you can then try to integrate to get the total cross section over the angle. So, that is the kind of experiments one does if you are doing online experiment you do in delta E telescope if you are doing offline you can do by measurement of activity of the radioactive procedures. So, this is how you can measure the cross section for charge particle induced reaction. So, I will stop here and the next lecture I will give the different type of reactions that take place that type of reaction mechanism in the next lecture. Thank you very much.