 Hello everyone. In the previous module, we discussed the conservation laws that govern a nuclear reaction and also the Q value, how to calculate the Q value from the masses of the nuclei. Q value could be positive or negative and accordingly they are called a exoergic and indoergic. Now, we will discuss the energetics of nuclear reactions particularly what is that energy that is actually available for inducing the nuclear reactions that is called as the energy available in the central path system and also the kinematics of the nuclear reactions whereby you can in fact calculate the energy and the of a particular reaction product at a particular angle. So, that will be the main focus of the particular lecture. So, let us discuss the kinematics of a nuclear reaction. We will set up the equation which will solve to find out the energy of the ejectile as a function of angle and energy of the projectile. So, we will see here we have the projectile of mass M1 and energy E1, the kinetic energy E1 bombarding a target which is at rest P2 equal to 0. So, the target is stationary in the here and then it can form a compound nucleus, composite nucleus or a compound nucleus. We will discuss the compound nucleus in more details in subsequent lectures. So, MC is the composite M1 plus M2 will form a compound nucleus. It may not form a compound nucleus, but for the time being we will say it is a composite nucleus and after these reactions, then you have an ejectile found of mass M3 kinetic energy E3 emitted at angle theta and the heavy residue of mass M4 kinetic energy E4 emitted at angle 5. So, now we use the conservation laws to determine a relationship between E3 a function of E1 and theta. So, that is the problem that if you know the projectile and its energy hitting a target, then at a particular angle theta, what is the energy of the ejectile for a particular energy of the projectile. Once we set up this equation, then for any reaction. In fact, it is independent of the reaction mechanism. We can calculate at a, suppose you are putting a project detector at a particular angle, you can know what the energy of the ejectile. So, that is the purpose of this reaction and there are many corollaries of this derivation which may become apparent subsequently. So, let us set up the equation for the conservation of mass and energy which we derived in the previous lecture. So, M1 plus E1 is the mass and energy of the projectile plus M2, M2 targeting the stationery. So, E2 equal to 0. So, M1 plus E1 plus M2 equal to M3 plus E3 plus M4 plus E4. So, this is the conservation of mass and energy before and after the reaction. So, now we can rearrange these equations in terms of the Q value. The Q value is nothing but M1 plus M2 minus M3 plus M4, mass of the reactant minus mass of product or the in kinetic energy of products minus kinetic energy of reactants. You can arrange it this way also. Second equation is the conservation of linear momentum parallel to the beam because linear momentum is a vector quantity. So, it has got components along the beam and perpendicular to the beam. So, for parallel to the beam along this direction, so you will have the cos theta component along the beam direction and the sin theta component perpendicular to the beam direction. So, the momentum P equal to MV or you can write equal to 2M with P square upon 2M is equal to E. So, P square equal to ME or P equal to 2M E. So, you can write in terms of this. So, for the incoming projectile it is coming at 0 degree. So, you can write root 2M1E1 equal to initial momentum is root 2M1E1 equal to the momentum of the ejectile along the beam direction root 2M3E3 cos theta term and root 2M4E4 cos phi. So, this is for the ejectile that is for the heavy residue. Linear momentum is conserved along the beam direction. Similarly, the linear momentum is conserved perpendicular to the beam direction perpendicular to the beam direction there is no momentum for the projectile. So, it is 0 equal to root M3E3 sin theta. So, this component is sin theta component for the ejectile and this component for the heavy residue that is negative sign because it will be in the negative direction. So, root M4E4 sin phi. So, now these are the three equations and you can eliminate E4 and phi from these equations to get a relationship between E3, M1, M2, M3, M4 and E1 and theta of course. So, how do you do you eliminate it? So, what you can write this in terms of so, this equation you can write as here you can write 2M4E4 sin phi equal to 2M3E3 sin theta and you can write as 2M4E4 cos phi equal to 2M1E1 minus 2M3E3 cos theta. So, essentially what you do you bring the sin theta sin phi and cos phi term E4 phi term on the left hand side and then the remaining in the other right hand side now you square and add. So, this M4 sin 4 this square this even you add them you will be sin square phi plus cos square phi that will become 1. So, you will have 2M4E4 and then the rest will be on the right hand side. So, the result of that will be M3 plus M4E3 you can rearrange the terms minus 2M1M3E1E3 cos theta plus M4Q plus M4 minus M1E1 equal to 0. So, this is the result of adding. So, you can substitute E4 in terms of Q and this one. So, you see here you can replace Q equal to E4 plus E3 minus E1 equal to Q. So, you can write E4 equal to Q plus E1 minus E3. So, you can substitute for E4 in terms of Q E1 E3. So, all you will be left with terms corresponding to E1 M E3 M1 M2 M3 M4 and theta. So, this equation if you see here it is a quadratic equation in the square root of E energy of the ejectile and so, A square A x square plus B x plus c equal to 0 where x equal to root of E3 and the solution of this will be the familiar equation. You can write the equation a solution root of a quadratic equation in terms of ABC. So, that is what is I have tried to explain here. The equation which is quadratic in root E3. So, what you write minus B plus minus B square minus 4 AC upon 2A is the root of that quadratic equation. But here I am slightly deviating from you write V plus minus B square plus W. So, V square B is nothing but B and 4 AC is W or rather V upon 2A is B and B square minus 4 AC upon 2A is W. So, that is how you can write the and so, V equal to root M1 M3 E1 cos theta and W equal to M4 Q plus E1 M4 minus M1 upon M3 plus M4. So, now, this equation which depends upon the Q value, the Q value is inbuilt in this equation and the E1 and cos theta. So, if you want to find out the energy of ejectile E3 at a particular angle and for a particular energy of the projectile, you can use this equation irrespective of the mechanism. It can be it is valid for elastic scattering, elastic scattering or any type of nuclear reaction, direct reaction or compound nuclear reaction. So, for energetically possible reaction, root of E3, root E3 has to be real and positive. So, there are sometimes there will be situations where some roots are in imaginary. So, that and negative. So, that those roots are ruled out. So, based on this equation, you can calculate the energy of reaction products for any angle and projectile energy. In fact, you can also rearrange this equation in terms of the Q value. If you put Q equal to, then this is called the Q value equation. Q value is nothing but E1 M1 by M4 minus 1 plus E3 M3 by M4 plus 1 minus 2 root M1 M3 E1 E3 cos theta upon M4. So, this Q value equation essentially it is the same equation in a different fashion it has been represented here. And the solution of the Q value equation again they are the root E3 equal to V plus minus V is 1 plus W where V and W are given by these expressions. So, this is a generalized solution of the Q value equation and for any type of equation mechanism reaction mechanism, you can use the same formulations. Now, when a projectile is providing a target, we will as we will go along, we will see that all the energy of projectile is not available for the nuclear reaction to take place. So, certain energy is lost in moving the whole central mass system and therefore, the discussion of the collisions in the laboratory and central mass system becomes important. So, we will discuss the solution in the laboratory system and in the central mass system, how does this collisions take place. So, in the laboratory system let us see here, in the laboratory system the projectile with mass M1 energy E1 and velocity V is moving towards the target and target is stationary. So, if the projectile is moving the central mass which is in the center here is also moving in this direction, the central mass will have the mass M1 plus M2 and velocity Vcm. So, this is the econometrics of the nuclear reaction. In the laboratory, we will see the projectile is moving with velocity V towards the target which is stationary. And after the reaction, the, the, the, it is supposed to be considered a collusion, simple elastic scattering, M1 will go at theta and M2 can go at phi. So, first let us focus on the incoming reaction channel. So, let us set up the equation for the momentum and energy of the system. Before the collision, the momentum is M1 V because the target is stationary and that is equal to the momentum of the central mass system that is M1 plus M2 Vcm because the momentum has to be conserved. So, this momentum is same as the momentum of the central mass system. So, this you can now calculate the velocity of the central mass in terms of M1 V upon M1 plus M2. So, this is an important relationship we will use this subsequently. So, in the, in the laboratory with the central mass system is moving with velocity M1 V upon M1 plus M2 and the central mass system is moving with the kinetic energy of M1 plus M2 Vcm square. Velocity of central mass M1 V by M1 plus M2 energy of the central mass half MV square. So, half M1 plus M2 this is the mass into Vcm square. Now, you see you can substitute for the Vcm from this formula. So, half M1 plus M2 and Vcm square is written by M1 V square upon M1 plus M2 square. So, this M1 plus M2 will cancel with 1 M1 plus M2 we are left with half M1 V square upon M1 plus M2. So, if you if you write here you can take M1 common from outside it will be half M1 V square upon upon M1 plus M2 and this is nothing but E1. So, what we have here is that kinetic energy of central mass kinetic energy of central mass equal to M1 upon M1 plus M2 into E1. That means, out of the kinetic energy of the projectile a fraction M1 upon M1 plus M2 is involved with the motion of the central mass that is called the kinetic energy of central mass. Now, let us discuss in the central mass system we will calculate the energy in central mass system. So, this kinetic energy of central mass system actually is wasted in the motion of central mass and what we are going to discuss is what is the energy available in the central mass system that is useful to induce the reaction. So, let us see now in the central mass system in the central mass system the central mass is stationary here with the mass M1 plus M2 the energy of central mass is 0. So, Vcm is 0 here because the velocity of central mass in central mass system is 0. The mass of projectile M1 velocity will be V minus Vcm where Vcm is the velocity of the target. Now, the target is moving towards the central mass with the velocity minus Vcm because we are considering the central mass to be stationary. The target is moving towards this side where in the frame of reference of central mass target is moving towards the central mass the velocity minus Vcm and after the collision you will find the central mass remains the same place the projectile and the projectile and target or the initial projectile target move in the opposite direction because the momentum has to be 0. So, the bottom line is that in the central mass system the total internal momentum is 0 because the central mass is stationary. So, let us write the equation momentum of the projectile M1 V minus Vcm plus momentum of the target minus M2 Vcm equal to 0. So, now you can solve this M1 V equal to M1 plus V2 Vcm. So, if you see the total momentum is 0 then you can write in this way and so Vcm will be equal to M1 V upon M1 plus M2. So, this is the same formula which we got from the laboratory frame of reference the velocity of central mass is given by M1 V upon M1 plus M2. Now, in the kinetic energy or kinetic energy in the center of mass system mind you earlier we discussed the kinetic energy involved in the motion of central mass what we are discussing is now the kinetic energy available in central mass system. So, that is the kinetic energy of projectile in central mass system plus kinetic energy of target nucleus in central mass. So, kinetic energy of projectile half M1 V minus Vcm square plus half M2 Vcm square kinetic energy of projectile plus kinetic energy of target in the center of mass system. So, you can now calculate it half M1 V square it will be now you can write half M1 V square plus Vcm square minus 2 Vcm. So, you can write half M1 plus M2 Vc square. In fact, if you open it up and you substitute for the Vcm from this equation Vcm equal to M1 V upon M1 plus M2. So, you can you can write out. So, this will be equal to half M1 plus M2 Vcm square. So, it is equal to now half M1 V square into. So, you can see here if you calculate this if you solve this two you will find because you can put the Vcm equal to in terms of M1 V upon M1 plus M2 then it becomes half M1 V square upon into M2 upon M1 plus M2 see the mass of the target upon mass of projectile plus target this can also be written as in terms of reduced mass of the system M1 M2 upon M1 plus M2 is mu reduced mass. But let us not bother about it we can write it in terms of M2 upon M1 plus M2 into E1 where E1 is half M1 V square. So, here if you recall the kinetic energy of center of mass as discussed in previous one was M1 upon M1 plus M2 into E1 and kinetic energy in center of mass system equal to M2 upon M1 plus M2 E1. In the previous slide we derived the expression for kinetic energy of center of mass that means this much energy tied up in the motion of center of mass whereas here in the center of mass system the conjugate energy available in center of mass system is M2 upon M1 plus M2 into E1. So, this much fraction of the projectile kinetic energy is available for the reaction to take place this much fraction of projectile energy is not available for reaction to take place and that is why it is also called as the required energy. Required energy means the projectile hits the target in the holes whole energy is tied up that some fraction of energy tied up in moving the whole system in the forward direction or it is the target gets give the required to the from the projectile and that much energy is not available for the reaction to take place. So, out of the projectile energy E1 only this much fraction is available in the center of mass system which will be useful to induce the reaction. So, let us see now the energy available in the center of mass system just now we discussed mass of projectile and having kinetic energy E1 plus M2 going to a composite nucleus and 3 plus M4 where E1 in the projectile energy in the lower P system and the energy available in the center of mass system is now ECM equal to initial energy of projectile into M2 by M1 plus M2 this is the fraction of which is available in the center of mass system and the remaining fraction M1 by M1 plus M2 even that is called as the required energy. So, this much energy is not available for the reaction to take place. So, in fact like you know you call free energy in chemical reactions the free energy only is available useful energy to induce the reaction. Similarly, here in the nuclear reaction is the energy available in center of mass system ECM that is the useful energy to induce the reaction. Let us try to explain this more details. So, again for this reaction Ilm4 plus aluminum nitrogen neutron plus as per us 30 the Q value we have calculated earlier 2.425 is the mass of alpha particle delta M value aluminum 27 minus 17.197 neutron mass and the pass porous 30. So, this is equal to minus 2.642 MeV which is a endoergic reaction. And now the projectile energy 4.87 MeV that is the energy of alpha particle available from 4.210 reaction type of decay. So, ECM will be for this reaction. So, when the alpha is bombarding the aluminum 27 ECM will be projectile energy into M2 upon M1 plus M2 the target mass upon projectile plus target mass and they could be 4.240. So, out of the initial energy of alpha particle of 4.87 MeV only 4.24 is available to induce the nuclear reaction that is the meaning of ECM. The remaining 4.87 minus 4.24 is not useful for the reaction that will go as a required energy moving the whole system forward. We also discuss in the case of compound nucleus suppose projectile and target form a compound nucleus. For example, here they can combine to form pass porous 31 and then pass porous 31 emits a neutron. So, this path porous 31 will be called as the compound nucleus. So, in such cases what is the excitation energy of the compound nucleus which also calculated using the center of mass energy. So, A plus A, projectile plus target going to form a compound nucleus, the Q value for formation of compound nucleus mass of projectile plus mass of target minus compound mass of compound nucleus. And so, the energy available in center of mass is the energy of projectile into the ratio of projectile target pass upon projectile plus target mass. So, this is the energy available in the center of mass system and this is the Q value. So, the excitation energy of the compound nucleus, the compound nucleus let us say pass porous 31 in this particular reaction will be formed with an excitation energy of ECM plus Q. Whether Q value is positive or negative that is immaterial the compound nucleus will have excitation energy E star equal to ECM plus Q. So, that is the significance of the compound nucleus excitation energy which is governed by the energy available in the center of mass system and the Q value of the reaction. Now, depending upon the type of Q value which was Q value positive or negative if there is a positive Q value reaction then for neutron induced reaction there is no threshold even the thermal neutrons can undergo reaction. But if the Q value of a reaction is negative then for neutron induced reaction also we require a threshold energy neutron should have a minimum energy that is called as the threshold energy of a reaction. So, let us consider a nuclear reaction induced by neutrons which is having a negative Q value. So, here nitrogen 14 N tritium carbon 12. So, let us calculate the Q value of this reaction Q value will be mass of the nitrogen 2.863 mass of neutron 8.071 mass of tritium minus 14.950 and mass of carbon 12 is 0. So, that becomes minus 4.016 MeV. So, this is a negative Q value reaction means it is an endoergy. So, if a neutron has to induce this reaction then the energy in the center of mass system should be at least equal to Q value. You have to supply energy otherwise it is a negative Q value reaction. So, ECM should be at least equal to Q value. So, when we say ECM should be at least equal to Q value that means minus 2. But this is nothing but ECM you know equal to E1 M2 upon M1 plus M2. And so, you can write EC threshold energy equal to Q into we can write here MA plus MA upon A. So, threshold energy will be instead of minus Q you write ECM or instead of ECM you write minus Q into MA plus MA upon capital. So, this much additional energy you need to have in the threshold energy. So, we say threshold energy is the projectile energy in the laboratory because the accelerators giving you or the neutron that is coming out has to have that much energy in the laboratory system. So, when you say threshold energy that is the energy of projectile in the laboratory. So, threshold energy of neutron will be minus Q value. So, minus of minus 14 plus 1 upon 14 that is 4.303 MeV. So, actually you require 4.016 which is the Q value. But since the neutron will be giving a slight recoil to the nitrogen 14, neutron energy should be actually more than the Q value by this much factor. And so, this is the threshold energy in the laboratory of neutron which will be sufficient to cause the nuclear reaction. So, that is how you can calculate the Q value of nuclear reaction or those which have got negative Q value. Now, we will come to charge particle induced reaction. So, the charge particle induced reaction with the reaction is a negative Q value. Not only that you have to cross the threshold of the energetics, but you have to also cross the pulmonary barrier. And so, that is why for a projectile hitting a target charge particle projectile, then you have to see what is the pulmonary barrier. And the pulmonary barrier for charge particle induced reaction can be calculated z1, z2 e square by r1 plus r2, z1, z2 are the atomic numbers and the r1 plus r2 is the radii of the projectile and target. So, if you recall the previous lectures, we have dumped this sum of certain units into 1.44 factor so, that the energy becomes in MeV and r0 is radius constant 1.4 and this mass number of projectile target to the power one-third. So, now, the projectile has to have this much energy in centroid mass system. Then only the pulmonary barrier will be crossed. So, for the charge particle induced reaction, projectiles should have energy equivalent to, at least equivalent to Vc. And so, accordingly the pulmonary threshold will be Vc into a factor for centroid mass. So, Me plus Me upon capital N capital A, that is the target mass. So, by this much factor the centroid mass energy should be, the laboratory energy should be more than the pulmonary barrier. So, for this reaction, again 27 aluminum, alpha N, as per as 30, Vc will be equal to 1.4382 into 13 upon r0, a raise to 1.3, a1, 1.3, a2, 1.3. So, that is 2.584 MeV. And so, the energy centroid mass should be equal to 2.584 and accordingly the projectile energy in the laboratory will be higher by this much factor, 31 by 27, that means 2.96 MeV. So, the pulmonary, the pulmonary barrier is 2.584, but the projectile should have 2.96 MeV in the laboratory, so that we have Ecm, energy available in centroid mass equal to 2.454 MeV. Now, the Q value for this reaction is minus 2.4642. So, energetic point of view threshold should be 2.642 into 31 by 27, that is 3.03 MeV. So, now you have, you have a threshold based on the negative Q value, you have a threshold based on the pulmonary barrier, whichever is the higher threshold value, that will be the actual threshold. So, when it comes to a charge particle induced reaction, if the Q value is negative, then you have to calculate the threshold not only for the pulmonary barrier, but also for the energetic point of view and whichever is the higher value, that will be the threshold. So, threshold energy is the higher of the pulmonary and the energetic threshold. So, you have to compare the threshold for the pulmonary barrier as well as the negative Q value and whichever is the higher one will be the threshold for the nuclear reaction. So, today we have discussed the energetics of nuclear reaction, what are the Q values? It could be endoergic, exoergic and if it is an endoergic, then it has to, there has to be a threshold because it will have some certain minimum energy to cross that pulmonary, the threshold barrier. And if it is a charge particle, not only the negative Q value threshold has to be crossed, but the pulmonary barrier also has to be crossed. And accordingly, the energy available in the center of mass has to be calculated, which then the laboratory threshold value is higher than the energy in the center of mass by the factor that is m1 plus m2 by m2, the factor takes care of the energy conversion from laboratory to center of mass. So, I will stop here. In the next lecture, we will discuss the cross sections of the nuclear reaction. Thank you.