 Hello everyone. So, in the previous lecture, we discussed about the liquid drop model, which explained the masses of the nuclei, the energetics of the beta decay and fission. But then we found that there are certain observations it could not explain and they are essentially the fluctuations in the gross properties of the nucleus. So, to explain those other phenomena which could not be explained by liquid drop model, another model has been proposed and that is the nuclear shell model. So, today we will discuss the nuclear shell model. Thus to recapitulate what are the limitations that the liquid drop model had, which it could not explain certain observations I have tried to list here. So, the most important aspect that the liquid drop model could not explain was the extra stability associated with the magic number of protons and protons. In fact, what we discussed in the previous one that if you draw, to draw the difference in the mass number and the mass liquid drop mass and the experimental masses, then we found that the experimental masses are lower than the liquid masses for certain number of neutrons and protons and these corresponds to let us say 2A, 20, 50, 82 minutes. So, that means these nuclei have lower mass that means they have higher binding energy and therefore, there is extra stability associated with this magic number. So, this part could not be explained using the liquid drop. And associated with this magic numbers there are other properties, other observations like the very high suppression energy. Suppression energy means the energy required to remove a neutron or a proton for nuclei having these number of protons and neutrons. Again the magic numbers 50, 82, 126. So, this is 126. Similarly, now if you have a nucleus having one neutron more or one proton more than the magic numbers then that nucleus, that neutron or nucleon is very easy to remove. That means the suppression energy is very low for this nuclei which have magic number plus one nucleon axis. Again associated with the magic number. And also the nuclei having the magic number of neutrons and protons have particularly magic number of neutrons have very low neutron absorption cross-axis. So, that the next level is very high. So, they are unable to, their probability of capturing a neutron become very small. Then the again if the in the case of alpha decay which we will discuss later on in the probably subsequent lectures. If the radioisotope has got 126 neutrons that is again a magic number then the alpha energy for such a nucleus is very low. Also in fact, if the after the alpha decay the nucleus is having the daughter product is having 126 neutrons then the alpha energy is very high. So, again associated with the masses. Again the magic number of neutrons and protons they have very large number of isotopes particularly if you have the 50 proton, 13 isotopes. Thin has got a large number of stable isotopes. Again extra stability. Other than that there are there are other observations like beta delayed neutron emission. Certain isotopes like present-fragged present products 137 iodine 87 bromine these isotopes emit delayed neutrons means they undergo beta minus decay and after beta minus decay the daughter product is left with an excited state that is more than the binding energy of neutron. So, instead of gamma emission that certain state emits a neutron they are called delayed neutrons because they follow the half life of these precursors. Also the isomers the nuclear isomers are those certain states of nuclei which have a long half life and they decay by gamma ray. Sometimes they are they decayed by gamma ray hindered they can also emit beta minus or beta plus. So, the nuclear isomers also could not be explained by the liquid model and that is why there is urgent need to have another model that we call as the shell. So, this is the shell model which was proposed by Mare and Jensen in 1949 and the shell model has certain assumptions like you know here now it is more like atomic structure where the nucleon the electrons are evolving around the nucleus in a certain fixed orbits stationary orbits. Here the nucleons move now and like in the atom where we have a central potential offered by the nucleus the nucleus is positively charged and the electrons are negatively charged and electrons move around the nucleus because of the central potential effective potential of the nucleus and so you have the assumptions about these stationary states of nuclear needs more. Here in the case of nucleus there is nothing like the central potentials because the nucleons themselves constitute other potential. So, what is assumed that every nucleon moves inside the nucleus under the attractive potential generated by the remaining nucleons. So, any nucleon you take it is it is in a potential well that is formed by the other nucleons. Now in the case of protons and neutrons now they have they move in they are in separate potential well. So, you start seeing two potential wells one for neutron one for protons and like you know for the atomic structure you have you solve the Schrodinger equation for an electron you find out the energy levels of electron k, l, m and shells. Similarly, in this case of nucleus you will have the which you solve the Schrodinger equation for the neutron and the proton in a particular potential well you will get the energy states and that energy states are analogous to that for the electrons. Now as you know this nucleon neutrons and protons are fermions. So, they follow the Pauli exclusion principle and so then you will find that in a particular state you know only two nucleons will be occupied if they have to be paired up. So, no two nucleons will have the same quantum numbers. So, this been followed and in that way when the neutrons and protons are evolving in orbiting in different orbitals you know. So, essentially you assume this nuclear force is a weak force in that sense it is not a weak force associated with the beta decay, but it is compared to the strong force that we envisaged in the liquid drop model in the in the shell model that the neutrons and protons are in separate potential wells and they do not interact with the other nucleons in that position. So, that way in that sense it is a relatively weak force. So, essentially though the shell model constitutes the solving a Schrodinger equation for the nucleon in a particular potential well. So, what are the different types of potential wells that you can envisage? The simplest potential well is the square well potential. We have the attractive potential well having the depth of minus v0 this minus v0. So, the potential well of the nucleus you know is of the order of 30 40 MVD potential well and so the the nuclear when we are solving the equation Schrodinger equation for a neutron or a proton we assume that is you can assume at the square well. So, essentially you have vr is equal to minus v0 up for r less than the nuclear radius and then for more than r it is 0. So, it is let it that is the typical attractive square well potential. So, you can in fact solve the Schrodinger equation assuming the square well potential or more realistically you can use as a harmonic oscillator. In the harmonic oscillator again you have the attractive the minus v0 the depth of the potential and now it is a parabolic expression with r. So, you have the potential minus v0 plus half and omega square r square that is the energy of the oscillator. And so, the frequency of this oscillator is nothing but k the force constant upon the mass of the nucleon. So, the essentially the cell model will in the cell model we solve the Schrodinger equation for this harmonic oscillator potential and then we get the energy eigenstates and the eigenvalue. So, let us try to know generate the cell model scheme by solving the Schrodinger equation for a three-dimensional harmonic oscillator. So, the potential for the harmonic oscillator can be given in terms of vr equal to minus v0 plus the half m omega square r square where omega is the oscillatory frequency that is given by the force constant upon the mass of the nucleon. And assuming that this oscillator is isotropic that means the oscillation frequency in x, y and z direction is same wx equal wy equal to w. So, the Schrodinger equation for this will be xi equal to e xi will not go into details of the solving Schrodinger equation. Our interest is to get the eigenvalues for this equation and the eigenvalue of this Hamiltonian will be given in terms of the oscillator quantum number n plus 3 by 2 x cross omega 3 by 2 is coming for is the three-dimensional harmonic oscillator. And this is the unit of the oscillator energy. So, the frequency of oscillation can be given by 2 v0 upon m r square where r is the radius of the nucleus and m is the oscillator quantum number. And so, the energy the astronomer can start from 0 to higher values 0, 1, 2, 3, 4. And so, as the n value increases the energy of the cell increases. Now, let us try to set up the eigenstates the different cell model states. For that the like I told that the electrons are occupying the KLM cells in the atoms. Similarly, the nucleons, neutrons and protons are separately occupying the orbitals like different orbitals like S, P, D, F and so on. And so, there is a formula for the oscillator quantum number in terms of the different other quantum numbers. They are the n small n principal quantum number and small l is the orbital quantum number. The principal quantum number n can take values 1, 2, 3, 4 and so on. Where is the orbital quantum number can take the value 0, 1, 2, 3, 4. And so, corresponding to the value of l we have S, P, D, F, G and so on. So, the different combinations of small n and small l can give you the same value of oscillatory quantum number. And so, you will find for the same oscillatory quantum number there will be multiple orbitals contributing to that particular oscillator quantum number that we can see in the next one. So, if you see the let us try to generate these cell model states for a harmonic oscillator, we will try to fill the values of small n and small l which will satisfy this condition starting from n equal to 0. So, for 0 ground state of the oscillator we have you can have, you can see here n small n equal to 0 and small l equal to 0, you can see here n minus n minus n equal to 1, 1 minus 1, 0 into 2, 0 and plus l equal to 0 also. S way, so l equal to 0 means S orbital and S orbital can occupy two nucleons. So, the occupancy of the occupancy of this hospital is 2 and the cumulative is also 2. For the next oscillator quantum number 1, again we can have how can you get n equal to 1. So, you start from n small n equal to 1, 1 minus 1 equal to 0. So, this term is 0. So, we have l equal to 1 that is the P orbital. So, N capital n equal to 1 will have again 1 orbital P and the occupancy of P is P 6, 6, 6 nucleons and the total occupancy becomes 8. Now, let us go to the next higher oscillator quantum number n equal to 2. Here you can have now two possible combinations of small n and small l. You can see here when you take n equal to 1, then 1 minus 1 is 0 and so, you will have when we have n equal to 1. So, you can see here, ok. So, here we are going to n equal to 2. So, you will have n equal to 1 and 1 minus 1, 0. So, into 2, 0. So, l equal to 2, l equal to 2 is the D state and it can occupy 10 nucleons and in combination of that. So, n equal to now you can have n equal to 2, 2 minus 1, 1 into 2, 2 plus 0. So, you can have a 2 S state. So, basically you will have 2 S, 1 D, 1 P, 1 S. So, you will have a 2 S orbital close to the 1 D orbital which can take 2 nucleons and the total occupancy becomes, you can see here 6 plus 2 plus 10 plus 2 will have 20 nucleons. Let us come to the higher oscillator quantum number. So, you can again you can have small n equal to 1, 1 minus 1. So, small n will take 1, 2, 3, 4 values. So, 1 equal n equal to 1, 1 minus 1, 0. The system is 0, l equal to 3. So, you will have 3 f orbital, 1 f orbital and f orbital can take 14 nucleons and in combination with that you can have now n equal to 2. So, you will have n 2 minus 1 is 1, 1 into 2 2, 2 plus 1 is 3. So, you have 2 P orbital and so, these two actually they are corresponding to same capital N. So, they are very close and same P will can take 6 nucleons. So, total is 43. Let us come to the N capital N equal to 4 and here you can have now 3 combinations n equal to small n equal to 1. So, l equal to 4 we have 1 G, n equal to 2 and this l equal to 2 you have 2 D and n equal to 3 it will have 3 S and so, the occupancy of G orbital 18, D orbital 10 and you have 70. So, you can see here that by filling the orbitals using this formula and their occupancy you can see that we are not able to reproduce their magic numbers of 2, 8, up to 20 you can reproduce, but beyond that the important magic number 50, 82, 126 we are not able to reproduce. So, that means there is something missing in the cell model which we still we have not taken into and so, that is where the spin orbit coupling becomes important. You know in the atoms in chemistry we have heard about the LS coupling that means the small l orbital angular momentum of all electrons combined to give the capital L that is the total angular orbital angular momentum and the spins of all electrons combined to form the capital S that the spin angular momentum and the capital S and capital L combined that is called the LS coupling or Russell Santos coupling and that nicely explains the spins of the atoms. In the case of nuclei because the spin orbit coupling is very, very, very strong the L and S of the individual nucleons combined to form what is called the J and so, this is called the JJ coupling. That means the J angular momentum of each nucleon they combine to form the total spin of the nucleus. So, it is not the individual L and individual S, L and S coupled together to form a J and so, if you in the potential for this harmonic oscillator you take a LS angular momentum spin orbit coupling term that is minus U R L dot S then you can let us see how the label scheme will change. So, you can see the vector product of vector sum of the L and S will be J and so, you can write J square equal to L square plus S square plus 2 L dot S and so, L dot S we want to get the magnitude of L dot S J square minus L square minus S square by 2 and you can now write the magnitude of J, J square L square and S square means of JJ plus 1 L L plus 1 and SS plus 1 that is half into 3 by 2 and so, let us just see what happened to the spin orbit coupling what does it do to the particular state. So, we have now instead of LS states we have a J state and so, for J, J can be now L plus half or J can be L minus half. So, you have a this spin this L it can be L plus half or it can be L minus half and so, we will have different energy states. So, each L state splits into two states L plus half L minus half. So, when J equal to L plus half then L dot S you can calculate from here substitute J equal to L plus half you get L plus half L by 2 and so, this L plus half state is that means, if you put L by 2 here then this is a negative term minus U L dot S. So, this is the lower energy state and for J equal to L minus half L dot S you put it in this formula you get minus L plus half by 2. So, essentially meaning that L minus half state is raised because if you add to put it here the minus minus half plus and so, L plus half L minus half state is lowered. So, the net result of the spin orbit coupling is a particular L state you have splitting of the L state L plus half being lowered and L minus half being raised. The splitting between these two states is you can calculate from here L plus half by 2 minus of minus half L minus L by 2. So, that is L plus 1 by 2. So, this is the gap between the two states as we can see here that as you go to higher and higher angular momentum values orbital angular momentum values the gap between the L plus half and L minus half is increasing and this is what actually explains later on you can see that the higher cell higher magic numbers or the this the spin orbit because of spin orbit coupling the higher cells are much more split and that leads to rebunching of the orbitals in different sense and that we will see in the next slide. So, higher L value higher the splitting and that the percussion of that we will see in this particular and now we will fill the cell model states in the view of spin orbit coupling. So, let us see how do we start filling the spin orbit opposite states. First thing is that we follow the polyics which are principle that means the the nucleons. So, no no nucleons will have the same quantum numbers that is the polyics principle, but the the main point is that when the filling of the electrons in the atoms and filling of nucleons and nucleus is quite different. In the case of atoms the maximum multiplicity is prevails the state having maximum unpaired electrons having lower energy, but in the case of nucleus the moment we have two unpaired nucleons they get paired up because their pairing energy is very high. Okay, so the L orbital as you can we have seen previously can accommodate 2 into L2 and plus 1 nucleons like S2, P6, D10, F14 and so on. Accordingly the occupancy of J state because L has been L plus half and L minus half. So, each L has split into 2 and the total occupancy is equal to 2A plus 1 into 2. So, the J states will have occupancy 2J plus 1. So, 2J plus 1 every J state you will see 2J plus 1 occupancy. Then the pairing energy of nucleons is much more than the pairing energy of electrons, the electron pairing energy of the order of few electron volt whereas in the case of nucleons we have seen that delta value for pairing energy is 1 to 2 and maybe so it is a very high pairing energy and the two nucleons of same type having same J are always paired up because of this high pairing energy which is unlike the electrons in the atoms. So, we will consider these facts by filling the orbitals by nucleons in the next slide. Now, you can see here. So, we have a J state which is arriving from L plus half or L minus half and occupancy of this is equal to 1. So, this spin states whatever I am showing they are the J states of that particular orbit. So, up to this we have already seen. So, the S orbital is not split because L equal to 0 and so it occupies the two nucleons. The P orbital is split into 1P 3 by 2 and 1P half and occupancy again 2J plus 1. So, 3 by 2 is 4, 1 by 2 is 2. The D orbital will split into 1D 5 by 2, 1D 3 by 2 as a orbital it is not splitting and 5 by 2 will have 6, 3 by 2 will have 4, 1 by 2 will have 2. F orbital again will split into 2, 1S half and 5 by 2 and P orbital 2P half and 2P 3 by 2. So, this is 1, 1, 1, 2, 1, 2, 1, 2, 3, so on. So, these are the value of small n. So, now you can see here J, the G orbital will split into G 9 by 2 and G 7 by 2, D orbital will split into D 5 by 2 and 3 by 2 and S will not split and so on, H 11 by 2, H 9 by 2. Now, you can see the occupancy. So, what happens whenever the F 7 by 2 is lowered, this S, so up to 20 we have explained in the previous scheme, but now in fact, this S will give you 28 and 28 also there is a gap here. So, this is called a semi magic number, meaning 28 neutron, 28 protons also have extra stability, but this is not a complete magic number, but 28 nickel, 28 proton, around 28 proton there is a extra stability in the nucleus. So, this is called a semi magic number, but the major magic numbers at 50 again because of the lowering of G 9 by 2 level. So, this G 9 by 2 is a part of the lower level, lower tail and so, this bunching of these levels leads to what we call as the shell. So, G 9 by 2 lowered gives you magic number 50, then give this stability. Similarly, H 11 by 2 when it is lowered compared to H 9 by 2, it explains the magic number H 2. So, this bunching of this H 11 by 2 in the lower shell explains the magic number. So, that is how similarly you will find H 13 by 2 will give you 1.6. So, the major factor responsible for reproducing the magic number is the spin orbit coupling when the splitting of the J state, L state into L plus half and L minus half with the L plus half being lowered significantly for high L value that it is overlapping, it is in close to the lower shell and so, the magic numbers can be produced. So, this is now the up to scale, you can see here the cell model states without spin orbit coupling 1S, 1P, 1D, 2S, 1F, 2P, 1G and so on. And then you can now apply spin orbit coupling here. So, as half 1P 3 by 2, 1P 3 by half. Now, as you go to, so here until 20 there was no problem, you have now F 7 by 2 being lowered. In fact, there is something called a semi magic number of 28. So, there is extra slight higher stability for 28 protons or neutrons. And now you see here because of the lowering of 1G 9 by 2, this there are no additional 10 nucleons here. So, there is a 50 nucleon shell magic number and again lowering of the H 11 by 2 you get 82, lowering of the I 13 by 2 you get 126 and so on. So, you can see at the gap this because of the gap between the L plus half and L minus half states for higher L values the the differences are rising. And because of that we are going to explain the magic numbers of the nucleon. So, the orbitals that are responsible for this magic numbers are the L plus half state of D F G H and I. So, this is how you can see that L plus half state gets lowered and then it bunches with the previous shell and so you get the magic number of 20. So, this is what the cell model explains and the other aspects how we can apply the cell model in predicting the different properties that we could not use for the in the liquid drop model I will discuss in the next next part of the my lecture. So, I will stop here and take up the application of cell model in technical. Thank you.