 Hello everyone. So, I start that next part of the lecture on Nuclear Reaction Mechanism by discussing much more details about the compound nuclear reaction. Let us give you examples of the compound nuclear reactions like neutron capture reactions are a very fine example of compound nuclear reactions. When the neutron is captured by the nucleus, it is amalgamated with the nucleus and the compound nucleus is excited to an excited state that X-ethicin energy will be equivalent to the binding energy for the thermal neutrons. So, essentially X-ethicin energy is ECM plus Q. Q value is for the neutron capture reaction is the binding energy of the neutron in the compound nucleus. And if you take thermal neutrons, then the set of R energy is close to 0. So, that is why in the case of thermal neutrons, you can say X-ethicin energy equal to the binding energy of neutron. And the angular momentum that is given to the compound nucleus will be target nucleus plus the neutron spin. So, neutron spin is half and target nucleus per covalent 59 is 7 by 2. So, 7 by 2 plus minus half j of the compound nucleus can be 7 by 2 plus half or 7 by 2 minus half. Whatever is the state of the exited nucleus is available, that will be populated because angular momentum has to be conserved. So, we will discuss the decay of a compound nucleus in terms of the available ex-ethicin energy and the angular movement. These are the two important properties which will govern the ultimate fate of the compound nucleus. In the case of charge particle induced reaction as I discussed earlier also, for the charge particle will bring in some energy because you have to cross the coulomb barrier of the target nucleus. So, the same example I discussed earlier, we calculate the coulomb barrier for this 1 by 44 z1 z2 by r0 a raise to 1 third plus a to 1 third. So, let us say for alpha or niobium-93 coulomb barriers we can calculate 13.8 MeV and the Q value for this reaction will be mass of alpha plus mass of niobium-93 minus mass of tecnesium-97. You can again put the value delta m value and you get 2.437 MeV. So, you have to have the energy the center of mass energy of the alpha more than or equal to 13.8 MeV. Normally, you will be much above the coulomb barrier. Let us say we take a 30 MeV alpha beam and for a 30 MeV alpha beam again calculate what will be the external energy and coulomb. So, the external energy of the compound nucleus will be ECM plus Q for 30 MeV ECM will be 30 into 93 by 97. So, here you take the masses actual masses because it is not the difference in the masses that the A will get cancelled out. So, when you are calculating center mass energy or required energy you take a mass number. Mass numbers are very close to the masses. So, you take here actual mass numbers plus Q value that is 31.2 MeV and the angular momentum now the projectile can bring in a large range of angular momentum from 0 to l max and so, there is a distribution of angular momentum in the compound nucleus unlike in the case of Newton induced reaction where we have only one spin populated. You have a distribution of angular momentum and that distribution is given by 2j plus 1 exponential minus jj plus 1 upon 2 sigma square. So, that is like a complete distribution of j values from 0 to some max value. These are the two properties we will be discussing more how the compound nucleus will decay. So, I will now discuss the compound nucleus behavior in two regions. As you know, this is the compound nucleus. I have now again shown here the formation of the compound nucleus by two different reaction groups alpha on boron 10 and deuterium on carbon 12 forming nitrogen 14 at the compound nucleus. Now, for the low mass projectiles and target you will find that the excited states you see this you can equate to that cell model states. So, there is a large gap in the excited state of this nuclei. So, if you populate a compound nucleus when the excited states are discrete they are well separated from each other then that particular suppose you populate this state then the subsequent decay of this excited compound nucleus will be governed by the width of this level and the available states in the product nucleus. So, it can emit a proton to form carbon 13. So, this when you populate a compound nucleus in discrete states we call it as a resonance region. That means, you can achieve resonances in the cross sections whenever there is a population of a particular energy state of the compound nucleus. So, this is what I have tried to explain here that for two different reactions you can populate the same state of a compound nucleus which can then decay by another channel to the every residue and the particle that is emitted and this. So, this there can be now even gamma emission from the excited states of carbon 13. So, that will give you resonances in the gamma ray as a function of projectile energy. So, what you do you measure the gamma ray yield vary the alpha energy you measure the gamma ray yield vary the deuteron energy and you will see these gamma rays in the spectrum. What these people found out is what in fact it has been done quite early in the 1950s or 60s it was found that the characteristic gamma ray spectrum and their yield are independent of the with which way that compound nucleus was found. Both these reactions alpha plus boron and deuterium plus carbon 12 give rise to the same energy states. So, again in the resonance region where you populate a nucleus in a discrete state there are resonances in the cross section also of the compound nucleus and there are gamma spectrum also shows the. So, this is the kind of experiment have been done to prove the independence of the compound nucleus reaction formation and decay and in the subsequently I will see even in the continuum region. So, here actually I try to explain here the width of this level is much less than the spacing between the levels. You go to still higher and higher energy the width of this level become more than the spacing between the levels. So, the spacing between the levels are low then they will start overlapping and you see what is called as the overlapping levels will be looking like a continuum. So, the continuum now we cannot apply the discrete states formalism where we apply there the statistical model formalism. So, this was for the resonance region and now we come to the continuum region where the energy states is the gap between the excited states is decreasing as you go up and up. So, finally, what will happen this the spacing between the states becomes so low that they will overlap and then you will see a continuum region it is like you know is like a Fermi gas. A Fermi gas where you have a you know all the nucleons are populating states which are overlapping. So, they gives you the like a feeling of a Fermi gas moving in a particular chamber. So, the projectile plus target nucleus like you know charge particle induced reactions when you populate the nucleus typically you know the binding energy of the nucleon will be around 7, 8 MeV. If you populate a nucleus more than that energy invariably you will end up with the overlapping states and that will be the continuum region. So, for the continuum region compound nucleus you apply statistical model you cannot count the individual levels. So, what you talk about the density of the nuclear states you do not have how many levels are there like here you can count how many levels are there in one electron board or one MeV, but in the case of continuum region. So, you apply a level density formalism this is called number of levels in a particular unit MeV energy dN by dE equal to rho E that is the level density and it is given by the Fermi gas model a parameter constant exponential to A e star where e star is the excitation energy excitation energy and A is called the level density parameter. So, basically the level density parameter is a measure of what is the number of levels you know in one MeV. So, that is a measure of that. So, it is always associated with the level density. So, this level density parameter is related to the excitation energy to compound nucleus by a term called temperature as a nuclear temperature. So, when we have a excited nucleus we say it has got temperature and this temperature part becomes very important we will see later on you can hear this I try to give a value of temperature. So, nucleus is hot it is excited it is hot, but the hot is so hot that the temperature get us get a feeling of the temperature. So, the temperature corresponding to one electron volt excitation we can relate the temperature to kT in terms of Kelvin k is the Boltzmann constant and this is given by 8.617 10 power minus 5 electron volt per Kelvin. So, if you calculate the temperature corresponding to one electron volt one electron volt upon this much electron volt per Kelvin 1.16 bar 4 Kelvin. So, if you have a nucleus and excitation energy of one electron volt then it has got a temperature of 10 power 4 Kelvin. So, one electron volt will be close to nearly zero you populate at one MeV and straight away 10 power 6 into 10 power 4 10 power 10 Kelvin. So, that if you recollect the nuclear fusion we are talking about 10 power 8 Kelvin that is about 10 power 4 electron volt. So, 10s of KeVs. So, that this is where you try to correlate temperature of the nucleus which is behaving like a hot Fermi gas. So, this is like a pressure cooker where you have a gas molecules at very high temperature. So, we now we deal with the big suggestion of a excited nucleus as if it is a Fermi gas at high temperature. So, we will try to see what are the signature of the decay of a compound nucleus which is having high temperature which is having high angular momentum and this formalism of Fermi gas gains the cross section for a reaction a go into a a a comma that a a comma b is b. So, a and a form a compound nucleus and the compound nucleus is imitating by a particle b and they are left with residue capital B. So, we have the formation of the compound nucleus and we have the decay of the compound nucleus two steps which are independent of each other. So, let us try to see the probability of decay of the particle to the residue. Now, depending upon the energy of the particle it can populate different energy states of the residue. So, what we essentially measure is the spectrum of this particle b and if it is a compound nucleus formation what type of decay the spectrum the particle will have that we are trying to see here. So, the maximum energy of the ejectile that is small b is given by the exertion energy and let us say it is bound here e star minus the binding energy of the particle in this nucleus. So, that is called the separation energy or the binding energy because binding energy is the energy required to remove a nucleon from the nucleus. So, this much energy we have to supply. So, exertion energy minus binding energy is the maximum kinetic energy of the particle. So, here Sb is the suppression energy of particle b in compound nucleus. So, the exertion energy of this residual nucleus will be now compound nucleus exertion energy minus the binding energy of particle in the compound nucleus minus the kinetic energy of the particle that is coming out. So, these are the four reasons for when the particle is emitted from compound nucleus energy equivalent to the binding energy is required and so, this residual nucleus may be populated in the certain energy states and that exertion energy given by this bond. So, the probability of decay to the particular energy state of the dotted residue depends upon this exertion energy higher the exertion energy higher will be the level density. So, basically why depends upon e star because level density depends upon e star. So, the probability of decay of b to a particular energy state depends upon the rho eb rho level density of the residual nucleus at a particular energy that is given by this e to the power 2 root Ae and you can now explain this eb. So, you can substitute this A in terms of A t square equal to e star. So, A equal to e by t square e by t square and so, you can substitute here e star by t square into e star all the way there. So, eb square upon t square square root. So, it becomes eb by t. So, we have taken care of the level density parameter which we do not know. So, the probability of decay of b to the residual nucleus do not exponential of 2 eb star upon t where t is the temperature and we do not know the e star. What we know is the binding energy of the particle in the nucleus a kinetic energy that we measure and the initial energy. You can substitute e star b in terms of compound nucleus e star minus phantic energy minus binding energy. So, the essential energy of the compound nucleus is known the binding energy of the particle in the nucleus can be depleted. So, they are constant terms. So, it becomes variable term is the particle energy, particle energy bind upon temperature. So, the probability of decay of the compound nucleus into the residual nucleus in certain states is equal to proportional to e raise to minus eb upon t where is eb epsilon b is the binding energy of the particle. So, you will find that as a function of eb the probability of decay will be this is called as the max there is some exponential pre exponential term root of eb which we have not explained, but it is the pre exponential term followed by the e raise to minus. So, this term is e raise to minus epsilon b upon t and from the slope of this actually you can find out the temperature of the nucleus when it is hot. So, this is called the Maxvillian nature of the particle spectra. So, if a nuclear reaction happens and the particles that are being evaporated by the compound nucleus they follow this pattern we say this is a signature of compound nucleus. So, one of the signatures of the compound nucleus is the Maxvillian nature of the particle spectra. So, how does the compound nucleus decay? You have an excited nucleus having certain excess energy and angular momentum it has got different pathways for the decay like neutron, proton, alpha even gamma ray emission can take place and if it is a heavy nucleus can even undergo fissure. So, there is a you know all there are different probabilities for deacertation of the compound nucleus and that as I mentioned previously it depends upon the available channels. So, you have a excess energy, you have an angular momentum that nucleus to decay you have different options particle evaporation, gamma emission and fissure. So, whichever process can carry away the large amount of angular momentum and excertation will be more favored. I will try to explain here by this diagram here. So, we if you recall the compound nucleus has got two part one is the excertic energy second is the J here. So, here is the angular momentum population I have given here 2L plus one TL that is what I have done here. So, it is actually sigma L here and this is J and here it is E star. So, it is a mixed diagram of excertic energy and angular momentum. Now, this triangular distribution of the compound nucleus in excertic energy and angular momentum space will be now emitting it says dissipates energy and what it will dissipate by emission of particles. So, neutrons it could be neutrons for heavy nucleus proton alpha emission is not favored much because there is a excertic emission barrier and so, the particles the residual nucleus that are formed you know if you see A plus A going to C going to B plus B. So, these are the states of the B that is the residual nucleus will have certain angular momentum and when you have a nucleus with some angular momentum certain energy is tied up in its rotational energy. So, that rotational energy is given by x cross upon 2I, I is the moment of inertia into J J plus 1 where J are the angular momentum. So, this states of different E J values as a first of J certain energy is tied up in rotation this is called the iras plan. So, you cannot have a level nucleus with this energy in the below the iras plan. So, energy has to be more than the iras plan. So, if when it is populated when emission emitting neutrons let us say then it will come to this energy state it will populate a in E J plane, this is E J plane. So, this population will lead to certain population in the residual nucleus and there is a called the drip line. That means, this gap is called the binding energy that means, if you populate the nucleus here then this cannot emit neutron. Whereas, this if you populate here it can emit neutron because its energy is more than binding energy. And so, any population which will come in this iras band will be now due to the particular product. So, 1n if you recall the acceleration function 1n, 2n, 3n are founded this population now is frozen to 1n product. Similarly, and this is emitting this population is frozen to 2n product and so on. So, in the angular momentum space different product nuclei the evaporation residues are formed as compound nucleus is emitting 1n, this product is formed 2n, this is formed 3n, this is formed. And similarly for protons and alpha can also take place. So, once it is in this domain below the binding energy now this nucleus are still excited they will be emitting now gamma rays. And by emission of gamma ray they come to their ground state and that that is what we say in the when we measure the cross section for evaporation residue this is formed by 1n means this nucleus could not further deexcite because this has come below the binding energy of neutron in that particular nucleus. So, this is the schematic of explaining how the evaporation of particles lead to different residues and the competition in fact what is actually known is that if a nucleus can emit particle it will favor particle emission then gamma emission why because the gamma ray cannot carry much angular momentum gamma ray will carry 1 or 2 spins whereas the particles can carry much higher amount of angular momentum. So, particle emission is favored for higher angular momentum states provided excelsis energy is more than the binding energy of the particular particle in the components. In fact, if the excelsis energy is very high fission is the first predominant reaction, but fission has got a variant. So, if fission is possible first is the fission then particle emission and then the gamma emission that is the sequence. So, if E star is more than binding energy particle is favored in E star less than binding energy gamma it is only more along the erased band now in this band this is called the erased band along the erased band only gamma emission is possible for particle emission not possible only above the erased band states can emit particles. So, what are the signature of the compound nucleus mechanism at the time to explain? First one I already explained the maxillian nature of the particle spectra and that is coming from. So, the number of particles having energy E upon E. So, this is the kinetic energy of the particle and number of particles will be having a maxillian nature. This is a typical signature of compound nucleus this is the term pre exponential term and e raised to minus epsilon upon t. So, from this slope of this in fact, we can find out what is the temperature of this compound nucleus that is emitting these particles. And the second important signature of the compound nucleus formation is forward backwards emitting angular distribution of particles. So, as if you recall the compound nucleus emission does not know what way it was formed it looses the memory of its formation. Let me call the experiment by Gushal that the two reaction channels a same product formed by two reaction channels cross-section cells. And so, if the cross-sections are if they does not know the memory of the previous internal channel then we can say that the emission of particle is isotropic because there is no preferred reaction. So, the angular distribution we call d sigma by d theta at a particular theta the constant there is a same number of particles at all angles. So, d sigma by d theta is equal to constant and I try to explain what is the what does it translate into angular distribution. So, I try to give here the see like it is a sphere at any theta the cross-section is safe d sigma by d theta. So, let us take a particular theta here and at this theta you have a disc. So, that is the a disc of same theta. So, you try to find out the solid angle because what you essentially measure is the d sigma by d omega. You are plotting here d sigma by d omega where d omega theta solid angle subtended by this strip. So, it is a strip of the sphere which is having the same angle theta. So, how do you know the solid angle? Solid angle subtended by a sphere at this point is 4 pi surface area of the sphere 4 pi r square upon r square that is the solid angle 4 pi. So, this solid angle by this strip of the sphere will be area of this. So, it is like a circle into the strip. So, it is like this actually and so, you have this this is a circular strip to see the cross-section this is this one pi r square, but r is now r sin theta. So, this is r this is r sin theta. So, 2 pi r sin theta, which is the circumference of this strip into the thickness of the strip r d theta. So, this is the area of this strip if you see from this side and so, this strip is a part of the sphere 2 pi r sin theta into r d theta upon r square the solid angle subtended by this strip at this point. And so, it becomes if you say d sigma by d omega will be d sigma by d omega into d omega by d theta will be 1 upon sin theta. And so, the angular distribution as a function of per unit solid angle follows this is what is 1 by theta graph. So, more than the sin theta graph because this it is it is not always 1 by sin theta it is more of forward backwards symmetry because the angular momentum can change this 1 by sin theta difference. Most important aspect is that forward backwards symmetry whatever number of particles are limited in forward angle same number of particles are limited in the backward angle. And so, this forward backward symmetry is a signature of the components. So, Maxwellian nature of the particle spectra and the forward backward angular distributions these are the two signatures of components. So, I will try to give you some examples of the compound nucleus evaporation particle spectra is our own paper of 1997. So, Florian plus niobium giving rise to 110 alpha particle spectra you see here Maxwellian nature this is for protons this is for alpha. So, this is a clear cut signature of a compound nucleus decay this is how the nuclear physics community nuclear chemistry community will verify this particular reaction follows the compound nucleus mechanism if the particles spectra are Maxwellian in nature. Similarly, the angular distribution of evaporation particles forward backward symmetry as a function of theta or even if you can write cos theta then whatever particles are then forward is a symmetric in the two both the n-spheres. So, again they want forward backward symmetric angular distributions and Maxwellian nature of particle spectra are signature of compound nucleus reaction. Okay. So, lastly I have I do not have time to discuss the compound nuclear reaction mechanism, but just I wanted to tell you that this is not the complete story the compound other than component compound nucleus formation there are many other mechanisms particularly when you have projectiles like carbon 12, oxygen 16, neon 20 and so on. So, when you have a heavy nuclear heavy ion as a projectile the heavy ion induced reactions the nuclear potential plays an important role the angular momentum plays an important role apart from the coulomb barrier and so you will see that the the nuclear potential the coulomb potential will know the coulomb potential is anyway there but there will be one more called centrifugal BL centrifugal barrier also will contribute and so the the fusion of the projectile target gets limited. So, in relation to the complete fusion there are non-compound nuclear reactions like incomplete fusion, deep inelastic collision, harsh fission and so many other reactions open up. So, heavy ion reactions is a domain of research pro study what are the reactions they do not follow a complete fusion. So, that 2L plus 1 distribution if you recall sigma L versus L complete fusion is this compound nucleus part in addition to that there are competing modes of incomplete fusion deep inelastic collision and quasi elastic transfer. So, elastic reactions so they all encompass non-compound nuclear reactions and there is so much of research actually opened for the nuclear industry or nuclear physics community to study the limitations of the compound nucleus fusion reactions. Yes, if I have time let me see we have we have been done some experiments on incomplete fusion reactions that means a same compound same product can be formed by a complete fusion as well as incomplete fusion reactions. So, these incomplete fusion reactions are similar to complete fusion reactions, but the momentum transfer is different. Linear momentum transfer in the formation of the same product is different that we have tried to find out by measuring the range of the requires. So, this is the required institution of developed cell-based use formed in the formation of these products. So, the complete fusion products are stopped at much later part of the required range feed than the incomplete fusion means the kinetic energy of the requires products that are formed it is much higher for the complete fusion than for incomplete fusion that is reflected in there. So, this is the momentum this is the range of the particle in the aluminum these are the compound nucleus products, but the non-compound nucleus products have got the lower range. So, the momentum transfer is not complete for incomplete fusion reactions that is revealed in the required end distribution of the evaporation procedure. So, I wanted to just give you a glimpse of what are the different reaction mechanisms other than compound nucleus you see them all opening up in avian induced reactions. So, I will stop here and then discuss the later part subsequently in the next lecture about the nuclear fusion and nuclear fusion reactions. Thank you very much.