 Hello everyone. Last lecture, we discussed the liquid drop model where we derived an expression for the mass of a nucleus using the phenomenon of a liquid drop and considering the atomic energy and the asymmetry between neutrons and protons. This semi empirical mass formula we also discussed that you can predict the masses of nuclei quite accurately. Today, I will discuss the other applications of liquid drop model. One of the most important applications of liquid drop model is to predict the energetics of beta decay in an isobaric chain. We know that beta minus decay occurs in an isobaric chain when the and in this process the atomic number is increased by 1 while in beta plus decay the atomic number decreases by 1. Now, let us try to calculate the q beta for beta decay and then we will see the systematics of the q beta decay. So, for beta decay let us calculate we can in fact write the mass of a nucleus in terms of the coefficients of z and z square. So, the mass of a nucleus or an atom mza can be written as mass of z protons mass of a minus z neutrons minus the binding energy term. So, this we have already seen that derived the semi empirical mass formula and now let us try to rearrange these terms in terms of we will get try to get this formula in terms of z independent term z dependent and z square dependent terms from this above form. So, you can write this as so you can let us see you can say z mh plus a mn minus z mn minus a v a plus a s e to the power 2 third plus a c will square upon e to the power 1 third plus you can open this square. So, it will become a a and a square upon a will become a plus 4 a a and it will be so it will be 2 square 4 a a z square and minus 4 a a. So, a it will be 4 a a a 4 a a a upon e to the power 1 plus minus z. So, now let us try to rearrange the terms which are independent of a so that will be a mn minus a v a plus a s e to the power 2 third plus a a a these are the terms which are independent of z. So, that becomes the alpha. So, you can take a outside and say alpha a then we will have the z dependent term plus mh mh z minus mn z and then you will have here you will have here a term minus 4 a a z a z a it will be minus 4 a a a z. So, these are the z dependent terms. So, you can take out z and in fact, you will have here here for delta also and then lastly you will have that term which are depending upon the z square a c z square upon e to the power 1 third plus 4 a a z square by a. So, these are the z squared. So, you have actually try to rearrange this formula in terms of coefficients of a z and z square that is the purpose of rearranging this mass formula and then you can write in terms of alpha a plus beta z plus gamma z square plus minus these are the terms and you can rearrange them so that you know this equation looks familiar to you in terms of the coefficients of z. Now, this alpha beta and gamma actually are called the local constants because they are constant for that particular isobaric chain wherever they depend upon the mass gamma and if you want to find out the most stable isobar in this isobaric chain then you can differentiate this formula with respect to z and so you get so from here you get beta beta plus 2 gamma z delta m by delta 0 delta z equal to beta plus 2 gamma z z and if you equate this to 0 to get z 0 then it becomes beta plus 2 gamma z 0 equal to 0 or beta equal to minus 2 gamma z 0. So, essentially you have beta equal to minus 2 gamma z 0. So, you will have gamma and z 0 these are the two local constants if you know the gamma and z 0 then you can find out the 2 beta of the particular isobaric chain. So, that we will discuss in the next slide. So, now let us discuss the isobaric chains for the particular mass number wherein the beta plus beta minus nickel is going to take place. So, we just discussed that the mass can be written in terms of alpha a plus beta z minus plus gamma z square plus minus delta and the beta can be replaced by minus 2 gamma z 0. So, I have put here minus 2 gamma z 0 into z plus gamma z square. Now, as we discussed that for a most for the minimum mass the nucleus is most stable. So, we can write the mass of the most stable isobar by replacing z by z 0. So, it will become alpha a minus 2 gamma z 0 square plus gamma z square 0 square plus delta. So, it will becomes alpha a minus gamma z 0 square and plus minus delta. So, now let us see for odd a isobaric chain where delta equal to 0. So, for the moment we will forget about delta term because for odd a delta is equal to 0. So, let us try to find out the mass difference m z comma a minus m z 0 comma a equal to. So, you subtract from here it will be alpha a minus 2 gamma z 0 z plus gamma z 0 z z square minus alpha a plus gamma z 0 square. So, you see here alpha a will cancel out and it will become you see here gamma is out what we have is z square plus z 0 square minus 2 gamma z 0 z which is nothing but gamma z minus z 0 square. Now, you see here the delta m the mass difference between any isobar and the most stable isobar follows parabolic relations. So, we can write this in terms of if you plot this then we get delta m versus z as a parabolic where minimum corresponds to z 0. Now, this delta m is nothing but 2 beta or isobaric chains. But you can see here that this is what we mean by the mass parabola the masses of the isobars of a beta decay chain all on a isobaric on a parabola. Now, you see here as the z is increasing from left to right you will see that the up to z 0 the lower lighter z values will undergo beta minus the heavier z values will undergo beta plus. So, irrespective of where you produce for example, if you produce efficient fragment it will be having highly neutron rich. So, it will have it will undergo beta minus decay and to go at the to ultimately lead to stability. If you produce a highly neutron deficient nucleus it will undergo beta plus decay or electron capture and come to the rest at z stable. So, this is what we mean by the mass parabola and using this we can find out from this formula you can find out the systematics of beta decay. So, let us see for a odd a isobaric chain how we can derive the q beta. So, let us write now this is a typical example of a isobaric chain for odd a. So, you have one stable isobar in Sonial black and the ones on the left hand side are undergoing beta minus decay and the ones on the right hand side are undergoing beta plus decay to reach the stability. And so, this can be written this can be shown as a mass parabola this is z 0 this is delta m or q beta. Now, let us try to derive expression for the q beta from this graph. So, we write m z a equal to so, what did we write alpha a minus 2 gamma z 0 z plus gamma z square. And now you have the next isobar z plus 1 comma a equal to alpha a minus 2 gamma z z plus 1 plus gamma z plus 1 square. So, this is what is the delta m value for adjacent isobars that means you are not trying to find out the q beta for a beta decay. So, it will get cancelled out. So, what we have is delta m equal to minus 2 gamma z 0 z plus gamma z square plus 2 gamma z 0 z 0 z plus it will be minus it will be plus 2 gamma z 0 plus gamma z plus gamma plus 2 gamma z. So, this is since it is a negative sign it will become minus minus minus because z plus 1 square these are the two terms. So, now, we can find out delta m equal to so, it will be canceling out here. You see here gamma z square will cancel with gamma z 3 z square here. So, it will cancel out gamma z square and so, we are left with minus 2 gamma z z 2 this also will get cancelled out. So, we will have these three terms it will be 2 gamma z 0 minus 2 gamma z minus gamma. So, it will be 2 gamma you can take out z 0 minus z minus r this is the relationship for the q beta minus. So, we showed q beta minus because we are going from z to z plus 1. Similarly, you can do an exercise. So, this is for beta minus 2 beta minus equal to this. To do the same exercise for beta plus then q beta plus will be m z a minus m z minus 1 comma a and z minus a. So, now, you can again do the same exercise and you will find this becomes equal to 2 gamma z minus z 0 minus r. So, using these expressions you can find out the q beta using the local constant gamma and z 0. So, what essentially you need is you need 2 beta values 2 q beta values along this decay in this parabola. If you know the 2 q beta values then you can find out the local constants z 0 and gamma and thereby the beta value of any other isobaric chain. So, this is the application of liquid drop model. So, let us now discuss the q beta systematics for even a isobaric chain and an example of that is given here that is this copper 64 iron 64 decaying to copper 64 cobalt 64 decaying to nickel 64 by beta minus and then we have from terminal 64 to gallium 64 beta plus to zinc 64 and as you can see here there are 2 stable isobars in this particular isobaric chain that is very common for the even a isobaric chain. So, let we have already discussed for the odd a isobaric chains that is the delta m for beta decay will be gamma z minus z 0 square that we have seen for the case of odd a isobaric chain. Now, in the case of even even even even a isobaric chain there can be 2 situations that is you can have odd odd nucleus to even even nucleus and if this happens then there will be a term 2 delta 4 odd odd 2 even even beta decay and it can have even even 2 odd odd then there the term will be minus delta. So, it will be gamma z minus z 0 square minus 2 delta 4 even even 2 odd odd. So, what essentially you have is that you have if you see the mass parabola for even a isobars then you will have 2 parabola 1 for this even even and 1 for odd odd nuclei and so the beta decay will take place. So, this is the z 0 from even even to odd odd to even even even even even to odd odd and like similarly here you will have this way this way this way and so finally, you may end up with 2 stable isobars and this can undergo probably beta plus beta minus like in the case of copper sheet. So, in case of even a isobaric chain you have 2 parabola because of the pairing the difference. So, odd odd isobars lie on the upper parabola and the even even isobars lie on the lower parabola. So, let us now calculate the q beta for the isobaric chain. So, for beta minus decay we have already seen the q beta equal to for odd a isobaric chain it was z 0 minus z minus half 2 gamma. So, if it is ordered to even even then it will be plus 2 delta and if it is even even even to order then it will be 2 gamma z 0 minus z minus half minus 2 delta. So, that is the only difference in terms of for the odd a to even even even a isobaric chains. Similarly, for beta plus decay q beta will be 2 gamma z minus z 0 minus half plus 2 delta or odd odd to even even and q beta plus equal to 2 gamma z minus z 0 minus half minus 2 delta or even even to order. So, this is how you can write the q beta in terms of the local constants gamma and z 0 and if you know the q beta values for 2 2 decays in a isobaric chain then you can calculate gamma and z 0 and thereby you can calculate the q beta for any other decay in the isobaric chain. So, this I have tried to show here in this graph you can see here this is a isobaric chain for mass number 156 and you can see here from nidinium, promethium, samarium, european gadolinium. So, you can see the two parabola and there is a change from there is a beta decay from odd odd to even even. So, promethium to samarium, samarium to european, european to gadolinium and here again you have from rbm to holmium, holmium to dysprosphium and rbm to dysprosphium. So, you see here that the even a isobaric chain can have more than one stable isobar. This is a corollary of the two parabolas. So, you can see this new this there can be more than two stable isobars for a even isobaric even isobaric chain. So, this is the application of liquid drop model just to do again you know see how we can use this problem we solve a problem here for mass number 141 the beta values qubeta decay value for the promethium 161 to nadolinium 160 1141 r is even here and for nadolinium to proceed even the beta values are even. You have to calculate the z0 the most stable isobar from the information that is even here. So, we can write the expression for qubeta plus 2 gamma z minus z0 minus half. So, qubeta is 3.72 you can write 2 gamma 61 minus z minus 0.5 equal to 3.72. So, you can write this expression for the other beta decay beta plus decay 2 gamma 60 minus z0 minus 0.5 equal to 1.82. So, these two equations are there and you can solve them for z0 and gamma and by solving you will find gamma is 0.95 and z0 is 58.55. So, that is close to 59. So, 59 is the procedenium. So, procedenium is the most stable isobar of this isobar chain. So, this is how we can use the systematics of beta decay and the liquid drop mass formulae derived to find out the qubeta value for beta decay. Now, let us see how the liquid drop model can explain the energetics of nuclear efficient process. We know from the binary energy data, binary energy curve that is binary energy as a function of mass number. The binary energy per nucleon detaches a maximum about mass number 60 and so, heavy nuclei when they split into two equal fragments there is a gain in binary energy and so, binary energy the Q value should be positive like the light nuclei when they fuse together to form heavy nuclei there is a gain in binary energy and so, these are in the exoergic reaction. So, grossly you can say that heavy nucleus can undergo fission and because of the change in binary energy from low binary to high binary energy. Let us now calculate the Q value for fission from the liquid drop model. So, assuming that the liquid drop model is considered the highest energy release for a symmetric splitting of a heavy nucleus, we will consider the splitting of the heavy nucleus Ax into two equal halves. So, we will say z going to z by 2, a going to a by 2. So, there will be two segments of mass and charge a by 2 and z by 2. So, let us see what is the energy released in fission. We had the parent isotope the Mza fissioning nucleus splitting into two fragments of mass z by 2 and a by 2 and now this same thing can be written in terms of the binary energy because the mass numbers are constant. So, the difference in the mass is being written in terms of the difference in binary energy in other way around. Difference in masses of reactant minus product is equal to difference in binary energy of product minus reactant. So, binary energy of the products minus binary energy of the reactant that is the fissioning nucleus. So, now, we can see here we can write this in terms of twice the volumetric term of the half half the nucleus. So, 2a 2av a by 2 minus 2as a by 2 to the power 2 half minus 2ac z by 2 square into a by 2 to the power 1 third and the estimated term minus 2a a a by 2 minus 2 into z by 2 square upon a by 2. So, you just replace z by z by 2 and a by a by a by 2 for the fission products. And for the fissioning nucleus we have the standard formula as av into a plus as into a by 2 third a 2 third plus ac z square by a 1 third plus a a a 1 a 2 z. Now, you can see here this volume energy term will cancel out because it is nothing but the av 2 into av into a by 2 means 2av. Similarly, you can see here the asymmetry energy term will cancel out whereas, this surface energy and that column energy terms will not cancel out because they contain exponential power to the a by a by 2 and a. So, you can write now the q, q over the fission will be as a raise to 2 third you can take out it will be left to be 1 minus 2 to the power 1 by 3. You can see here from a by 2 to the power 2 by 3 and 2 is outside it will become 1 by 2 2 to the power 1 by 3. Similarly, ac z square by a a by 1 third you can take out and you will be left to be 1 minus 2 to the power 2 by 3. Now, if you substitute the value of the surface energy this constant no 1 minus 2 to the power 1 by 3 or 1 minus 1 by 2 for 2 by 3 these values are the values and you can substitute the value of a s and ac. So, you get minus 3.48 to the power 2 third plus 0.22 z square by a to the power 1 third in terms of MeV putting the value of a s and ac. So, let us see the magnitude of this surface energy term and the column energy term and then you can discuss the which term dominates for which nuclei. For 238 uranium substituting in this value the mass number at z the q f q fission becomes this term will become minus 130 and the column energy term become 300. So, we have 170 MeV in the predicted energy released in the nuclear fission of 238 uranium. So, you can see here that the column energy term is dominating in the fission of 238 uranium. For drink 64 q value for fission surface energy term minus 54.475 column energy term 49.5. So, that is minus 4.9. So, you can see here the q value for fission of drink 64 is negative and so that explains why the light mass nuclei do not undergo fission. For the intermediate nucleus the slag molybdenum 100 q fission will be surface energy term minus 73.25 plus 83.21 that is equal to 10. So, this molybdenum 100 fission though it is you can see q value wise it is positive, but you will find that to the for the fission to take place there is a barrier. If you recall in we explained the for this protein fission it has to cross a fission barrier which is a resultant of the change in the surface energy and the column energy and the fission barrier is much higher than this q value. And therefore, the half-life for this will be very very high you do not see them. So, energetically it may be possible, but critically you do not see it that explains how liquid drop model can explain the energetics of the nuclear fission process. The same thing I have tried to explain we will discuss this more in when we discuss nuclear fission, but the point I wanted to bring home is that the liquid drop model explains the spontaneous fission of heavy nuclei where in the competition between the surface energy and the column energy of a deforming nucleus is responsible for its fate towards the nuclear fission. So, this we will discuss more in the subsequent lectures on nuclear fission, but right now we will try to summarize that the liquid drop model can explain the masses of nuclei, it can explain the beta decay energetics, it can explain the energetics of nuclear fission process. But the picture is not that good always. So, there are some limitations of liquid drop model some of them are the unusual stability of z and n neutron proton number having these configurations and what I have shown here is the difference in the experimental and liquid drop predicted masses over a range of proton number and neutron number and you will find that there are dips that means the experimental masses are lower than the liquid drop masses at certain configurations and they correspond to these magic numbers. So, that means this magic number nuclei have lower mass than the predicted based on liquid drop. In addition to that the ground state spins at parity of the nuclei also cannot be explained by the liquid drop model and the existence of nuclear isomers also cannot be explained by the there are in fact many other things which we will discuss later on they are difficult to explain in terms of liquid drop. So, we will discuss this all in shell model in next lecture how shell model can take care of many of the limitations of liquid drop and stop here and discuss the shell model in the next lecture. Thank you very much.