 Just to kind of summarize what we have done so far in the last class we talked about basic reason for magnetism that is electrons what is the reason for electrons getting magnetic moment. So, that we have seen in some sense some details were discussed. Now, we are actually in a position to find out how an atom which is consisting of z number of electrons, how this atom just one isolated atom which consists of a z number of electrons, how the atom such an atom gets the magnetic moment. If I know the number of electrons how do I find out the atomic magnetic moment that is the issue right now we have to address. As I told you then we can slowly build it up and go to molecules and solids. So, let us start with the basic question again to find out how do electrons contribute to the magnetic moment of an atom. So, that is the question that we have to answer. So, here as I told you without quantum mechanics any meaningful discussion is impossible. I know that you have done some course in quantum mechanics something was done here also. So, with the and electromagnetism with these basic things I am sure you will be able to manage. But if somebody feels that it is definitely not possible then you do not have to really worry about that quantum mechanical part. But still there will be a qualitative description of all these magnetic properties will be there. So, I want to make it little more rigorous. So, I am actually taking a quantum mechanical route to explain how an atom which is kept in a magnetic field how it actually responds to that magnetic field. So, that is what is being done now. So, what we have to do as we have seen in the quantum mechanical lectures we have to and as I told you a classical treatment will not give you the magnetism information about the magnetism magnetization of a material. So, we straight away take the quantum mechanical formalism. In the quantum mechanical formalism as you can see the Hamiltonian that is the first thing to be written the usual Hamiltonian is the kinetic energy part which is p square by 2 m plus the potential energy which is v naught. For example, hydrogen atom your v naught is z is 1 there. So, here that is a plus when there is a magnetic field now we are applying a magnetic field that is a magnetic field b the effect of the magnetic field actually is represented it actually modifies the momentum the momentum becomes p plus e a by c I am using cgs units and a as you know is the vector potential which is related to b that is b is del cross a which you have studied in the electromagnetism course. So, that mainly what is happening is when the application of the magnetic field changes the momentum to this form and this can be expanded as shown here that is what I told you some derivation is here which of course I have written all the steps you can go through it also. So, this Hamiltonian can be expanded. So, you will get the normal kinetic energy term the potential energy term which actually is a usual Hamiltonian of an isolated hydrogen atom kind of a situation, but you have three extra terms one is this term which is e square a square divided by 2 m c square then there are two terms which consist of p dot a and a dot p. Remember p we have an it is an operator we know the form of the operator which is minus i s del as we have done in the class in the morning also. So, I can substitute this and I can write the Hamiltonian the total Hamiltonian of the atom in presence of the magnetic field as h this is h naught that is a usual Hamiltonian term plus these two terms which are there. So, I write p in terms of minus i h del. So, that is what is written here which gives you the equation number one then one can actually remember these operators are always meaningful when it is acting on a wave function that is what is shown here if I take del dot a and a dot del acting on this wave function psi you can expand this in this form and you get this result not very difficult. And as you would have seen in the electromagnetism course usually we take this the divergence of a to be 0 which actually is followed in the so called Coulomb's gauge. So, this simplification gives us the modified Hamiltonian in presence of the magnetic field for this atom given by this one. So, this h naught we do not have to worry about it h naught we know the contributions the extra contributions that are coming because of the magnetic field or one is this term and this is the second term. So, we have to worry so if you want to find out the response of the atom to the magnetic field we have to only worry about this term what is the effect of this term what is effect of this term. These two terms actually act what is known as a perturbation these two are perturbations to the original Hamiltonian. So, original Hamiltonian and that gets perturbed by these two terms. So, basically these two terms will produce some changes in the energy so this is done with the help of what is known as a perturbation approach. So, we use these are all time independent cases so we have to apply a time independent perturbation to calculate what is going to happen to the energy because of this perturbation as well as this perturbation. So, this is what is to be done once we know this contribution then we can find out what is the effect of this magnetic field on this atom. So, here we have a force to calculate these contributions through the perturbation theory actually speaking the time independent perturbation theory which I am sure you have studied in your MSc time. So, that is and before we go we take the usual thing that magnetic field is assumed to be along the z axis and you can have various ways of writing the vector potential which gives you this magnetic field along z axis one of the standard ways of taking this a is in this manner you can find that if I take del cross a for this you will get b to be along the z axis. So, using this substitution I get the Hamiltonian in this form and I know the angular momentum which is given by r cross p l is the orbital angular momentum is given by r cross p which is of course the definition from mechanics. So, we can write this z component of this angular momentum here it is a orbital angular momentum this l z is given by this operator because i is coming because of this p which actually has the form of operator. So, I can write in this manner. So, this l equal to r cross p the z component is written in this manner in the Cartesian form. So, this I can substitute here then and I know already this Bohr magneton definition this h bar will come that is already taken here e h bar divided by 2 m c is the Bohr magneton which we defined in the last lecture. So, the effect of this magnetic field on the Hamiltonian the actual Hamiltonian now is the original Hamiltonian the hydrogen Hamiltonian plus you have this term plus another term that is here. So, the we are as I mentioned we have to only worry about the extra contribution to the Hamiltonian arising from the applied magnetic field. So, that is only I am calling it as delta h that is only this term and this term please remember I am not using the magnetic field symbol as h, but I am using b to avoid any confusion with the Hamiltonian symbol which is also h. So, all the magnetic fields I am taking I am actually taking as b and h always stands for the Hamiltonian. So, we have to worry about what is effect of this term what is effect of this term these two are the perturbations which are going to act on the original Hamiltonian of the hydrogen like atom that we are discussing. Now, one can actually before I proceed I will show you or I will make this statement that the first term that you are seeing here that is essentially which is x square plus y square term this is the one which gives rise to the diamagnetism of the atom we are not at the solid. So, this is the diamagnetism of the atom and this gives rise to the usual paramagnetism that we talk about that is why I told you this diamagnetic response or the paramagnetic response of an atom in presence of a magnetic field is very fundamental and it cannot be expressed purely in a classical language it has to be done in a quantum mechanical formal quantum mechanical sense that is why I have taken this minimum steps which are essential. So, that you get a feel of quantum mechanical feel of these two terms how they give rise to the two different kinds of magnetic responses namely the diamagnetism and paramagnetism. Now, we have to identify these two terms separately and go ahead this one I already showed you that this term is the usual paramagnetic contribution this is the diamagnetic contribution. I will take you the right to the quantum mechanical aspects of this before I proceed further. We have in fact today morning also we talked about the probability density similarly I can also have a current density probability current density in quantum mechanics if I take an atom I can find for example hydrogen atom we do a three dimensional problem there and we usually take a spherical polar system and you can find out the current density the probability current density actually is related to this wave function use through this expression as shown here. I can you what I should use is here that wave functions corresponding to as I mentioned earlier the wave functions are all in terms of principle quantum number orbital angular momentum quantum number L and magnetic part ml. So, psi is all represented in terms of n L and ml. So, this if I substitute here I can corresponding terms of j I can find out for various states of the electron in the atom once I substitute I can show that this j vector the components the radial part and the theta part actually goes to 0 and only what is left is this phi part phi part actually is in the x y plane if you remember the coordinate system the spherical polar system which must have been done in the electromagnetism course that is why I thought I should show here again. So, this j phi the only the phi part of the current density survives and that is given by this expression one can actually work it out if I try the expansion of this, but I am not really doing that just wanted to give you a feel of this. So, the only part on the component of the current density that the probability current density that is going to survive is this one and so, this can be simplified again in this manner again you can see that this is a probability density that we have been talking about earlier also that is coming here psi n l m l mod square is nothing, but the probability density. If I take the probability then current density if I multiply with the charge electronic charge I will get the corresponding actual current density like if I take the probability density if I multiply with the charge I will get the corresponding charge density. So, here I am getting the current density that is a phi component is simply this one multiply with this j phi the charge which is minus then I can get the corresponding this current density and the actual current density. And if I take the corresponding this area element the current density can be converted into the actual current. So, the d i I am writing that is over this cross sectional area of d s which is what is written here. So, the d s actually represents the cross sectional area that is seen by this current which is moving along the phi direction or the phi cap direction of the orbital. So, orbital magnetic moment is mu l that is what we have been talking about. So, this is I can write once I know as you have seen in the electromagnetism course once you have a current loop the magnetic dipole moment associated with that loop. So, now as you have a loop in the x y plane the phi cap actually is in the x y plane. So, this loop is giving rise to a dipole moment and the current magnitude you know that is given by this j phi times minus e times d s and this is enclosing an area which actually is having a radius r sin theta because your r vector is like this. So, in the x y plane the radius that this loop is going to have is r square r sin theta. So, the dipole moment is the area which is pi r square in this case r sin theta square times this current that is like telling that pi r square i. So, this is giving rise to the magnetic dipole elemental magnetic dipole moment. So, that is this one. So, now, I substitute all these things whatever I have obtained in earlier steps I can get this one d s you remember d s is nothing but d s is the area which actually is in the plane which actually is perpendicular to the phi cap direction. So, that is nothing but r and actually d r and r d theta. So, I write it as r d r d theta. So, d s is substituted as r d r d theta. So, this gives me the elemental contribution to the orbital magnetic moment due to this phi component of the current density that we have talked about here. It is all very fundamentally one can show from the hydrogen atom physics itself quantum mechanics itself that is what I am showing here. So, this tells you what is a magnetic moment elemental magnetic moment that you can give. So, the classical idea of a current loop that gives rise to a magnetic moment can still be worked out that is what we are seeing by this method. So, proceeding further I substitute and I know this expression again you have seen earlier the volume element can be written in this manner. So, the d s that I have right now which is an area element can be written as d v divided by 2 pi because this actually goes around a complete loop. So, that your phi actually covers this entire range of 0 to 2 pi which is a allowed range. So, that gives me d s to be written in this manner. So, d mu L that is the orbital part of the elemental angular momentum the elemental part that actually can be written in this manner. And remember I want to find out a total magnetic moment I have to integrate it. So, that this has to be integrated over the volume and this mod square wave function integrated over the volume is actually normalized it is 1 as we have seen today morning also. So, when I substitute I see that the angular momentum if I take along the z axis because now this is in the x y plane as I mentioned phi cap direction is that means it is only in the x y plane. So, the magnetic moment direction will be along the z axis. So, when I substitute this becomes 1 and you get minus h bar m L divided by 2 m. And if we group these some these two things to be if I put a c g s unit there will be c and this simply becomes minus mu m L times mu b. So, what does it mean? So, one can visualize a orbital magnetic moment in this manner which actually is coming in the form of minus m L times mu b where m L is actually justified in calling it as the magnetic orbital quantum number that is what we have been talking about earlier. So, in a very simple quantum mechanical picture also one can see how the orbital magnetic moment rises from the kind of a more picture that we have about the atom. I mean that is a picture that we have you talking about radius of the atomic orbit and other things. So, using this picture you are able to see that you can actually find that you can get a dipole moment something like what happens in the case of classical electromagnetism where you have the current loop giving rise to a magnetic moment. So, with this identification if you go back in our original Hamiltonian this is exactly what is happening there. Such a magnetic moment which we saw just now as mu b L z L z operator corresponding and quantum number is m L. So, mu b times m L is the result that we are going to get that is Eigen value. This is interacting with the magnetic field that is applied along the z axis that is why this appears as the Hamiltonian term. So, L z is correspondingly m L in the Eigen value L z becomes its operator when we are writing the Hamiltonian form. So, we can easily identify that this term is nothing but the magnetic moment orbital part of the magnetic moment responding to the magnetic field that is through the usual Zeeman kind of term minus mu dot b because already there is a minus and minus mu dot b will give you positive sign that is what is seen here. So, you are seeing a positive term which can be easily identified as the orbital contribution of the interaction between the orbital magnetic moment and the applied magnetic field in this case b z which is purely along the z axis. So, there is a justification for writing this one. So, this is the diamagnetic contribution I am going to show this further here. So, this term I have already shown you that this is coming because of the orbital magnetic moment interacting with the applied magnetic field. So, that derivation is coming from the pure hydrogen atom quantum mechanics very elementary quantum mechanics of hydrogen atom will give you all these ideas. Now, one can also do a classical thing that I told you earlier even though the quantum mechanics is essential, but in many situations a pure classical method also works with certain approximations I am going to show that here. So, if I take a classical derivation of the orbital magnetic moment that I talked about just now, we are using a very crude classical picture here again the same kind of a current loop and so on, but no quantum mechanics here. We take a charge which is revolving around the nucleus the usual bore kind of a picture. So, you have the current is given by e by t, t is a time period of this motion in the circular orbit we are assuming the circular orbit and so on. This is nothing but 1 by t is the nu e times nu of course, this can be written and t can be written as v 2 pi r by v that is becoming v is a electron speed in the orbit of radius r. So, that is what is here. So, the magnetic moment that can be written as mu is given by the radius the area that is pi r square as I shown you earlier it is a pure classical treatment where here r is a radius of the orbit itself not the spherical polar radius here it is actual radius of the that is why there is no r sin theta here. So, pi r square times the current which is written here if I substitute here I can get e v r divided by 2 and remember m times v times r is actually the angular momentum the magnitude. So, I can use I can convert this e v r in terms of this angular momentum I will write v r I will write it as l by m l is angular momentum. So, I take it as e l divided by 2 m. So, this becomes e by 2 m times l again you can see the angular momentum coming into picture not as sophisticated as you see in the case of a quantum mechanical derivation, but still it works l is orbital angular momentum quantum number v is a electron speed. Now, the allowed values of l as I showed you quantum theory or quantum mechanics clearly show the allowed values of l are square root of l into l plus 1 because the eigenvalue of l square operator is the eigenvalues are l into l plus 1. So, l eigenvalues are l into l plus 1 h bar. So, when I substitute I get this one this is what we have seen earlier. Classical electromagnetic picture also gives you this idea about this magnetic moment orbital magnetic moment of the atom that is what you are see here which is shown little more sophisticated manner in the earlier quantum mechanical picture. Similarly, one can also have a spin magnetic moment as I told you earlier which spin is generating lot of interest in you as the questions tell. So, I told you that the magnetic moment is given by g times this gamma is a gyromagnetic ratio which is e by 2 m c in the cgs units s the eigenvalues of s operator or s into s plus 1 h bar. So, I can substitute I have already shown this I will get this one. So, basically when you have an atom an electron in an atom you have magnetic contribution coming from the spin magnetic moment and orbital contribution as shown here and of course, I showed you that a pure spin will give you one more magnet on. So, these two will interact with the magnetic field. So, when I write the extra term in the Hamiltonian extra terms in the Hamiltonian. So, this is the original term that we talked about of course, we are assuming that b to be along z axis and then the two terms that are coming this s is coming suddenly the 2 s is coming suddenly because that is not coming originally. But now we are inserting this one because we know that the way the orbital angular momentum and orbital magnetic moment interact with the magnetic field there is additional contribution coming from the from the spin that is not directly coming from our derivation, but that has to be inserted physically separately that is what is done here earlier this plus 2 s was not there. Now we are physically adding it in the light of this picture that we have regarding the spin magnetic moment. So, this is separately put in added and now this becomes the extra contribution as far as an atom is concerned which is interacting with applied magnetic field. Now we are in a position to find out what is energy shift to do to this one what is energy shift to do to this one for that as I mentioned we use a first order perturbation theory. First order perturbation theory is simple in the sense that what you are actually calculating is only the expectation value of this particular extra operators. So, here if I want to find out the energy change I simply calculates as we have done in the morning today find out the expectation value of x or p x exactly same thing is there. So, I write psi star only thing is that we use this way of writing it here in this bracket form I will not go into the details. So, this basically tells you that psi star this operator and psi. So, that is they are all operated. So, that is what is shown here and similarly this term also treated under the first order correction first order correction that very simple. So, this gives you the corresponding shift in the energy that is called the diamagnetic shift and this is the paramagnetic shift which I am going to call as the curie paramagnetic shift and this one here delta e 1 superscript 1 stands for the first order correction because I can actually do a second order correction or a higher order correction to these terms, but I will not do a first order correction is enough for us right now. So, that one this one stands for the first order correction which actually simplifies to calculating the expectation value of this extra terms in the Hamiltonian. So, it is as simple as what we calculated in the morning regarding the expectation value of x. So, this is the diamagnetic contribution and this is the paramagnetic contribution to the atom when it is subjected to the magnetic field B. I will first take the case where all orbitals are completely filled. I am going to take a situation where the atom has orbitals and all the orbitals are completely filled. In such a case one can show that the contribution from this will be 0 because when I do this corresponding to this and corresponding to this the total orbital angular momentum and total spin angular momentum will be 0 for that case which means that this term will not survive. So, it is easy if I want to recognize this particular term only you have to go to a situation where I have absolutely no contribution coming from the second term. That is realized in a situation where all the shells of the atom are completely filled. So, when you have such situations like for example inert gases all of them are completely filled then you have no contribution coming from the second term and everything is coming from the first term. So, my delta E1 is simply the first term when I substitute I get this one and remember for a closed shell x, y, z there is a symmetry as far as an isolated atom is concerned. So, this can be rearranged and this has to be done for all the electrons of the atom or if a given atom there are various I mean z number of electrons. So, all of them must be taken into account. So, the summation that is shown here or this i's that are shown here they are for all the electrons of the particular atom. So, that is what is written here. So, this can be written I know that there is a symmetry and x, y, z there are there is no particular preference for x, y or z. So, I can write x square plus y square for any i value I can write it as two thirds of your r square. So, that is what is written here because always it is convenient to write in terms of r. So, that is why x square plus y square I will not proceed further, but I will write because one third is of r square is for this x and one third is for y. So, that is why you get a two thirds of contribution that is coming here. So, I can find out if I take this contribution I have to do this summation over all the electrons that is again difficult. So, what I will do is I will take an average value that is I will calculate the expectation value if I use a quantum mechanical language I use the expectation value of r square. So, that serves as an average quantity and instead of calculating for every electron which are situated at different r values I have an r value the expectation value of this r square the mean value of r square is known. So, I take this mean value and multiply with the number of electron instead of calculating for every r square for every electron I will simply calculate the total number of electrons z and I simply multiply with the average r square value average r square value is nothing but my expectation value of r square that is why this term will be replaced by the z times r square and r square expectation value something with a good quantity as far as quantum mechanics is concerned. So, you can see I substitute this for here this one and I get this expression. So, delta E diamagnetic that diamagnetic energy shift is in a case of a completely filled orbit is given by this one you can see it is proportional to the square of the magnetic field. Now, I will specifically talk about diamagnetism here I have a little I have been clever in the sense that I took a very special case of an atom where all the orbitals are completely filled or the shells are completely filled that does not mean that diamagnetism is there only in such a case you have diamagnetism in any shell when there is an electron whether it is completely filled or not diamagnetism is always every electron in an orbit exhibits diamagnetism but the problem is if the shell is not completely filled what is going to happen is you have the contribution from the diamagnetism and there is a contribution from the second term that is the paramagnetism since I wanted to completely isolate the diamagnetic contribution I took a case where there is absolutely no paramagnetic contribution that is why I took that completely filled case that you should not think that diamagnetism can happen only in such a case no the answer is every orbit when there is an electron there is diamagnetism that is what is written in the first line this contribution is always present in all unpaired orbitals but there is a competition between paramagnetic sum and diamagnetism that is why one has to be careful so you have both these things coming in general but if you have a completely filled shell you have to worry about only the diamagnetism that we are talking about magnetic moments are induced by the applied magnetic field but here the induced magnetic field is opposite to the applied magnetic field that is what is called diamagnetism that is one can actually have a classical picture using the Faraday's law which I will show you very soon and this response when happens for a solid right now we talked about for an atom if this happens for a solid it gives rise to a bulk property namely the diamagnetic susceptibility, susceptibility we have defined in the last class you have a diamagnetic susceptibility that is something which actually opposes the magnetic field that you apply however the diamagnetic susceptibilities are very very small that is a negative response to the of the solid to the applied magnetic field in general is very very small as reflected by the very small diamagnetic susceptibility values much smaller than one if it is one that tells you that whatever you are applied the magnetic field strength exactly it will be cancelled by the produced or induced magnetic field that is on the diamagnetic substance but that generally does not happen an exact cancellation or exact competition can happen only in one type of materials they are called type on superconductors which of course I mean has been introduced to you this type on superconductors they have exactly minus one value that means if I apply a five units of magnetic field the response is also exactly minus five so in the usual diamagnetic material that we talk in magnetism the susceptibility values are much much smaller compared to one so the connection to between super conductivity and magnetism will be clear as we go along with our lectures in superconductivity also so diamagnetism that way is a phenomenon which is there for every orbit whether it is completely filled or not this is the first point one has to keep in mind so diamagnetic susceptibility one can do what we have is a energy shift if I take this second derivative of this with respect to the of course there is some minus sign I will go into that not the details second derivative with respect to BZ so you can actually find you get the susceptibility which is given by this you can see it is actually determined by the number of electrons that is a positive and there is a negative term here I should put a negative here so this R square is expectation value of this R square calculated for the atom and Z is a number of electrons so this is actually the diamagnetic susceptibility part now just to give you an idea especially again you had some discussion on the classical electromagnetism I thought I should depend on that to show you that here also a purely classical idea can give you some information regarding the diamagnetic susceptibility we do not use any quantum mechanics we use a simple electromagnetism and you can get this idea how is it done so we do this one you consider an atom in a circular orbit again like the Bohr orbit of radius R let us take radius R I apply by this condition that del cross is minus dB by dt what is this this is your this is your Faraday's law because the basically flux change I take a constant area so this flux essentially becomes B so del cross E is minus dB by dt and this can be written as I can actually convert this to a line integral so it becomes this you would have done in the other course it becomes E dot dl is minus dB by dt dot dS this is a flux B dot dS is some magnetic flux so this can be written if I take the circular loop and over that area the magnetic field is constant I can write this as 2 pi R times E I am expanding this E dot dl term here and this I can write it as minus pi R square dB by dt I am putting a C to make it in the CGS units so that is what is done here so this becomes an area element this converts into area element here also an area element so I can equate the two so that gives me this expression and this gives me an electric field that is produced at the circular orbital due to the changing magnetic field that is what is written here this is nothing but your Faraday's law this one stated in a different way so when I magnetic field is changing as represented time rate of change of magnetic field that is what happens in the Faraday's law dB by dt gives rise to an induced electric field that you have studied in the other course the induced electric field in a circular orbit of radius is what is given by this expression I am not using the vector sign here looking only at the magnitudes so the induced electric field produced by the changing magnetic field is what is represented by this equation now what is going to happen is you are having an electron let us take a particle picture which is not really true in the quantum mechanical sense but in the classical picture it is fine more picture it is fine so we are having an electron this electron in the orbit of radius R now this is subjected to this induced electric field E so what is going to happen is it is going to have a torque this is going to produce a torque so that means there will be a force and a torque about the center of this orbit the torque is actually written as d tau is r cross f r cross f where f is a force which is given by the charge times the electric field so r cross E times E so this gives me the torque that is going to happen over a time interval of dt that is what is written here so the induced electric field produces a torque on the electron and as you know a torque actually is going to produce a change in the angular momentum the connection between the torque and angular momentum is known from mechanics so the change I mean the change in the torque that is happening or the application of the torque by the induced electric field is going to change in the angular momentum of the electron in the circular orbit and the change is going to happen so the angular momentum is going to change and the torque is going to happen that is torque that is going to be applied over a time t is tau that is the integration of this one that gives me this expression very simple to integrate that the torque that is produced is E r square divided by b divided by 2c in cgs units so what happened is you change the magnetic field it has produced an electric field induced electric field induced electric field produces a force and a torque on the electron this torque that is given by this expression is going to change the angular momentum and angular momentum change can be related like this the dl is a change in the angular momentum given by this because rate of change of angular momentum is nothing but the torque we know from mechanics so that is given by this one so correspondingly when the angular momentum changes now we know that it is going to change the magnetic moment because the connection between the two is a gyromagnetic ratio so corresponding change in the magnetic moment is d mu which of course will be this multiplied by the factor e by 2 mc that is the gyromagnetic ratio already there is a c that is why this becomes c square so this is the change in the magnetic moments due to the applied magnetic field felt because of the induced electric field the Faraday's that is why whenever we classically talk about diamagnetism you always bring in Faraday's law and associated with Faraday's law you all have always have the Lenz's law which actually produces the negative sign and which actually goes down and shows you the negative susceptibility so this gives me my new that the change in the magnetic moment that is this one so this is a purely a Faraday effect classically speaking that comes here and considering all the electrons one has to replace your r square with z times r square so this contribution for the entire atom becomes and as usual we have write we can write in terms of x square plus y square is in terms of r square you can see you get exactly same expression what you get in a quantum mechanical description for the diamagnetic susceptibility so what is done here is I find out the magnetic moment I take the magnetization which is nothing but the total magnetic moment divided by the volume that is why v comes here and this is differentiated with respect to the magnetic field as I told you m by h but h in this case we are using only v so that differential with respect to v gives you the susceptibility here is a diamagnetic susceptibility given by this expression if you put the proper sign it will be negative so you have e square r z r square divided by 6 mc square is exactly same expression we got earlier when we use a proper quantum mechanical treatment so that is what even if you use a classical picture with certain approximation certain assumptions you can get the identical expressions in some situations you are lucky to have some situations of this kind here is one and since you have been talking I mean we have seen the classical electromagnetism earlier I thought I should give you this one in fact this is the explanation most of the people give when we ask how do we explain diamagnetism so the diamagnetism effect can be discussed in a very simple manner if you are ready to make certain assumptions and one can use a pure classical electromagnetism argument involving Faraday's law and Lenz's law you have to need to know only the basic Maxwell's equations divergence curl of various things Faraday's law one can substitute and you can get an expression for the susceptibility of a solid susceptibility is not for an atom it is for the solid one can get this expression so this is a very important step because this treatment also gives you an idea about the diamagnetic susceptibility without using any wave function without using any quantum mechanics but this is not a correct method I should mention that again that is why I started with a very general treatment of magnetic response of any atom kept in the magnetic field that is the correct treatment but one can actually having done that one can really get some information about how much or how far one can go if you use a purely a classical treatment that is what is shown by these derivation as I mentioned diamagnetic susceptibilities are very small for usual materials you can see that these are SI units standard SI units the values are 10 to the power of minus 11 minus 10 and so on very very small compared to one as I mentioned and this of course this is ZF don't worry about defective essentially you treat it as the number of electrons so as I told you the expression contains Z times r square that is why it is plotted as a function of Z times r square when this is more the diamagnetic susceptibility tends to be larger and larger so this is an experimental verification of this relationship which is derived both quantum mechanically and using pure classical electromagnetism. Now we will go to the other term the second term that means the second term survives only if there are unpaired electrons in the orbital if it is everything is paired if it is complete then we have seen that you have the contribution only from diamagnetism so now we are making it little general you want to see what happens when you have a general shell we do not know whether it is completely filled or not let us assume that you have unpaired electrons so unpaired electrons are there in an atom in such a case you have permanent magnetic moment so the diamagnetic moment is there only when you apply the magnetic field otherwise no but here this is an intrinsic property of the atom there are unpaired electrons there is an intrinsic magnetic moment which is called a paramagnetic moment the unpaired electrons should be present especially in order the inner shells I will come to that details little later you can have unpaired electrons let us take sodium atom you have one unpaired electron in the outermost shell but as far as you are talking about a single atom it is fine but usually these things are extended to a solid when you when it becomes a solid if the unpaired electron is in the outermost shell it does not really contribute to a paramagnetism of the kind what we are talking about so if you want to have paramagnetism atomic paramagnetism you should have unpaired electron that means not completely filled the shells in inner case now this inner shell some of the inner shells must be incompletely filled fortunately in the periodic table this happens in mainly two different series one is actually the transition metal series which actually is a D series you have a 3D, 4D and 5D in the middle of the periodic table and also in the lanthanides and actinides which are the two rows below the periodic table where you see where you have the 4F and the 5F mostly we talk about 4F elements so in the D electrons the D orbitals are not completely filled whereas in the F atoms F shells are not completely filled so you have a incompletely filled 3D shell in the case of iron cobalt and nickel and similarly for the raerers you have 4F orbitals which are incompletely filled these are the two cases where you can definitely expect intrinsic permanent atomic magnetic moment arising from the unpaired electrons that are present in those orbitals as we are going to see very soon this paramagnetism I am going to call it as curie paramagnetism the atoms must have at least one unpaired electron the magnetic moment is permanent as I mentioned susceptibility is positive not like the negative diamagnetic susceptibility the susceptibility now is positive but again it is small susceptibility is small and the examples are transition metal ions as I mentioned transition metal atoms also in some sense we can use lanthanide ions that is raerath ions isolated ion you can take hydrogen atom simplest example is hydrogen atom oxygen atom and oxygen molecule both oxygen atom and oxygen molecule interestingly are paramagnets whereas hydrogen molecule is not I will come to that very soon so the examples also please note the difference it actually tells you some important information transition metal ions isolated ions lanthanides lanthanide ions for example Fe 2 plus Fe 3 plus those kind of atoms ions hydrogen atom not ion it is an the atomic form oxygen atom and oxygen molecule they are the standard examples of paramagnetism where you have permanent moments I want to find out a situation where I have two electrons I am taking a 3d shell I want to find out if I have two electrons unpaired that is I have only two electrons I want to find out what is the magnetic moment of that particular atom I know what happens when I have one electron one electron contribution we have already found out I have to find out what is orbital part what is the spin part I know that I know through the g factor now I have two unpaired electrons in the one of the inner shells namely the d shell so that is what I am taking a 3d 2 3d 2 means I have two electrons in the 3d shell 3d shell is what is shown here if you remember l l value corresponding to a d shell is actually 2 and the correspondingly ml values there are 5 as I mentioned in the last class the values are 2 1 0 minus 1 and minus 2 so they are shown as 5 boxes here now the question at what is shown here is the electron moments are shown here to find out what is the actual magnetic moment of the ground state we are always initially we are interested in the ground state that is the lowest energy situation what is the magnetic moment of this particular atom that is what is to be found out for that there is a there are there are a set of rules given by Hoons that is called Hoons rules and only that one can give you a simple idea or simple estimate of the magnetic moment in the ground state I will demonstrate with this example when I have two electrons and this can be extended when you have more number of electrons also the 3d so what you have to do is you have to put the first electron with the highest ml value that means I cannot start anywhere here the first electron must come here the and it has to be an up spin the second electron has to go to the next one where the ml value is just lower than this because I can put one more electron here actually speaking because I can put a this is an up spin electron arrow represents an up spin electron as I mentioned in the morning I cannot put another up spin electron here because that will violate the Pauli's principle exclusion principle will be violated if I put here because then you will have orbital quantum number principle quantum number spin quantum number everything will be the same and that cannot happen for two electrons so I cannot have an up spin in this two box so if I want I can only put this up spin in the next one so what is to be done is the rule tells first of all you should try to make all the spins to be in the same direction that means if first is up you put everything up so that is to be done without violating the Pauli's principle that means this cannot come here it has to go to the next one so you have to maximize this ml value and so that you put in this manner and then you have to have this condition satisfied so that gives me two new quantum numbers which is L which actually is a summation of the ml value of this and the ml value of this in this case 2 plus 1 that gives me 3 and similarly I have an s value new quantum number that is I mean again summation of this plus this this arrow represents m s always an up spin arrow is a plus m s half the other thing is minus m s half is minus half so this is plus half this is plus half when I add the two I will get one so I have j so I have an s l quantum number and s quantum number for the two electron system where this atom has a 3d2 configuration so I have l value and s value remember l n s basically they represent the angular momentum as we have been seeing the symbols are I mean changing from small letters to capital letters but otherwise they have the same physical meaning because now it is a two electron system not a single electron system so I have l and s both represent corresponding angular momentum so you have l getting as 3 and s getting as 1. Now as we have seen j, j is result of this interaction between the spin and orbital and as I showed you the vector j equal to vector l plus vector s can give you many values starting from l plus s to l minus s as mentioned earlier so the rule is your ground state will be j equal to l minus s when your number of electrons is less than 5 in the case of the electrons so that means up to 5 that means this is completely half filled case so when the number is less than half filled for example if it is 1, 2, 3, 4 and 5 your j result and the ground state j value will be given by l minus s if it is 6, 7, 8, 9 or 10 of course 10 is extreme you will see that the ground state is given by j equal to l plus s so you have j l minus s as a ground state in one case that is when the shell is less than half filled when it is more than half filled the ground state j is given by l plus s remember j, l, s are all here quantum numbers not operators not vectors so and I know j, l, s and j I can find out my g if I substitute in this case it works out to be 2 by 3 so I have j value I have my g value I can find out what is the total magnetic moment the z component I can easily find out that is given by g times this actually is mu z z component is g times j times mu b so that when I substitute I see that it is 4 by 3 Bohr magneton this is the magnetic moment which is intrinsically there without applying any magnetic field just because you have two electrons and that is in an unpaired shell I mean in a incompletely filled shell this gives rise to the magnetic moment of 4 by 3 Bohr magneton so what does it mean just because you know that how to calculate the electronic magnetic moment it is not very straightforward to find out how an atom which has a certain number of electrons what will be the atomic contribution to the magnetic moment you have to find if I want to find out the ground state magnetic moment you have to really go through the Hund's rule Hund's rules because there are three rules as you are seeing you are actually trying to maximize your m s or s you are trying to maximize your m l or l and then the rule tells what is your j value that is a third rule that gives you what is a ground state magnetic moment so this actually tells you the Hund's rule gives you an estimate of the ground state magnetic moment of an atom it is very very important this is not for the solid all my discussion right now is for a single isolated atom where the 3d shell has incompletely filled situation in this case I have two electrons only if it is 10 then it is completely filled if it is completely filled as we have seen it does not have any intrinsic paramagnetism it has only a diamagnetism as we have seen earlier when you have a number less than the completely filled case like 2 1 3 5 7 you have a magnetic moment that is coming from the atom purely an intrinsic property that and the Hund's rule gives you that value remember it is only the ground state magnetic moment of an isolated atom do not just jump and try to calculate the magnetic moment just by drawing these boxes using the Hund's rule that in general it is not expected to be correct if you are talking about solids and other things if it is only an isolated atom this rule works and it gives you the correct intrinsic magnetic moment the so-called curie paramagnetic moment that I have been talking about one can have other examples also as an exercise you can try what happens interesting to see what happens when I have 5 electrons 5 is exactly a half filled situation because 10 is maximum 5 means you will have here so 5 will allow you to put all of them to be up without violating the Pauli's principle you can put all of them here but what you are seeing is that at that time there will be no orbital contribution your L will become 0 because all the negative sites and positive sites will be 0 I am adding together you will have a pure spin moment as we have seen sometime earlier it will be a pure spin case so please try to work out 1 or 2 more examples like a 3d 4 configuration and 3d 5 configuration using the same principles something connected with this not exactly connected with the paramagnetism but it is of interest to discuss at this point so now it is very clear that any atom if it has got unpaired electrons in the especially in the inner shells definitely there is an intrinsic magnetic moment this has nothing to do with the applied magnetic field there is no applied magnetic field this is the intrinsic property whereas diamagnetism is not intrinsic you have to have an applied magnetic field then the response is there magnetic field is removed it has gone back there is no intrinsic there is nothing like a diamagnetic moment which is going to stay permanently there whereas if you have unpaired electrons in some shells especially inner shells you have a permanent moment atom has a magnetic moment so this is clear very much now there are various experiments which actually show that this is true the atom has a magnetic moment has been demonstrated with the help of various experiments I will not go into those details of those experiments but I just wanted to talk about one case which actually is interesting in more than one sense that is what actually such an atom which has got a magnetic moment I am going to put it in a magnetic field that is what is happening here this is probably you have heard about it this is a famous larmer precision larmer precision happens because the first thing is it has got a magnetic moment atom has a magnetic moment and I just one isolated atom has a magnetic moment only if it has a magnetic moment intrinsic moment it undergoes the larmer precision when does the larmer precision happen you have this atom which has got a magnetic moment as shown by this L vector this is representing the angular momentum or as actually represents a magnetic moment as I told you earlier because electronic charge is negative the angular momentum is like this magnetic moment is exactly in the opposite direction do not worry about that right now you can take them to be in the same direction for our argument such an atom is kept in a magnetic field the magnetic field as usual we have taken to be along the z axis what is going to happen remember the z axis is our axis of quantization and I have already shown that when you tell that the magnetic moment is kept in a magnetic field the L vector cannot align along the z axis this is true when the magnetic field is present also that means there is always going to be an angle between the applied magnetic field along the z axis and the orientation of the magnetic moment or the angular momentum in this case I am not distinguishing between the two so there is an angle that is what is shown the angle this is the magnetic field along the z axis and there is an angle that is a semi vertex angle of this cone that you are seeing I will come to the cone assume that this is a magnetic moment and there is an angle between the two when there is an angle between the angular momentum or the magnetic moment and the magnetic field naturally there is a torque that torque is mu cross B that is tau is mu cross B so what happens when the torque is there this magnetic moment has to so because this torque will produce an angular momentum change but this is an atom and the angular momentum this L originally and remember what is the direction of the torque this is a cross product the angle the torque direction will be along the direction which is perpendicular to the plane containing mu and B mu is in this direction B is in this direction so the plane containing mu and B I mean mu is L actually is this one and hence the torque will be in a direction which is perpendicular to this and remember this magnitude of this L cannot change but now the torque has to do some job what is the job it can do without changing the length of this vector changing the length of the vector is changing the angular momentum magnitude which is not possible so what happens this vector L is changed by this delta L or this goes on and hence this L vector essentially precess about the z axis or about the magnetic field axis like this the classical picture but is important because of the fact that there is an angle originally there is an angle between the magnetic field or the z axis and the magnetic moment atomic magnetic moment when you apply the magnetic field itself a torque is produced and this torque is going to make this L vector precess about the field direction this goes on like this and I have written all the steps here I am not going to the details here it is one can actually work it out not it is actually self-explanatory one can see that this precision frequency this is going to go like this about the magnetic field axis this precision frequency actually is calculated to be e times b divided by 2 mc in the CGS units this frequency is called a Larmor precision frequency whenever you have a magnetic moment of an atom and if you apply a magnetic field you get Larmor precision and this Larmor precision is very important this gives rise to techniques like electron spin resonance or your nuclear magnetic resonance which we talked about in the case of MRI instead of having atomic magnetic moment if I take the nuclear magnetic moment for this L vector or the mu vector that I am talking about is corresponding to the nucleus all the in all the things that I am talking about here will be true when I apply a magnetic field the nuclear magnetic moment only difference is instead of atomic magnetic moment that I am talking about now the nuclear magnetic moment at that time will precess about the magnetic field and that is the basic idea behind this nuclear magnetic resonance technique so what happens is when you when this is precessing with this kind of I mean frequencies if you apply external frequencies electromagnetic field having that correct frequency there can be energy absorption energy exchange between the system and you the source and this is giving rise to electrons spin resonance in the case of atomic magnetic moments and nuclear magnetic resonance in the case of nuclear moments that is NMR nuclei so you are using and you can see the mass in the Larmor precision frequency mass is in the denominator here mass of the atom so in here if I talk about here this is the mass of the electron if I have one electron otherwise the number of electrons if I talk about the nuclear magnetic resonance the M will be mass of the nucleons which will be much larger the Larmor precision frequency given by this expression event of the form is the same the Larmor precession frequency for an electron is much larger it is almost 2000 times larger compared to the Larmor precision frequency of the nuclear moments when the same magnetic field applied is applied in both the cases so you have precision frequencies different in the case of electron moment and the nuclear moment so one case you will have one set of frequencies in the other case use much lower frequencies in the case of nuclear you will use much lower frequencies the rf frequencies this is what actually is used for energy exchange absorption and emission between the system that is the system which actually produces the electromagnetic waves and the nuclear system or the atomic system the electronic system and this exchange is what is actually used in the case of these resonance techniques of magnetic resonance techniques both in the case of esr and nmr in fact esr electron spin resonance is also called electron paramagnetic resonance or epr that is why I am just putting it here since we are discussing paramagnetism so electron paramagnetic resonance and electron spin resonance they mean the same and the corresponding thing for the nuclear moments is called nuclear magnetic resonance which actually is the fundamental thing in the case of mri magnetic resonance imaging both are magnetic resonance what is a resonance the magnetic moments actually which are undergoing the precision in presence of a magnetic field they are able to have an resonance absorption or emission with the external agency which is nothing but the so the electric field oscillating electric field generator so that is what is used in these both the techniques which are very important today so this essentially tells now we get a somewhat a clear picture of what happens I have when I have an atom I have an atom under what condition the atom has a permanent dipole moment so called the curie paramagnetic moment under what condition I can expect only a diamagnetic thing that means not permanent not intrinsic and in general if I have unpaired electrons I can expect both because diamagnetism will always be there whether the number of electrons is 1 2 3 or whatever it is so the diamagnetic response will always be there but if I have unpaired electrons the paramagnetic effect will be much larger much stronger so you do not essentially feel the diamagnetism that is the only difference that is why if you want to look at diamagnetism you have to look at a case where there is absolutely no paramagnetic intrinsic contribution which can be seen only when all the shells are completely filled in any other case seeing diamagnetism probing diamagnetism will be difficult because you have a much stronger positive contribution so seeing this negative contribution which is very feeble will be difficult so if the other contribution the positive contribution is 0 then whatever small value you have for the negative thing you can actually see that is the main thing so this tells you how the atom has a magnetic moment so atomic part is completed now before we go to the solid it is interesting to see what happens in the case of a molecule I will just take one or two simple examples just to show you the difference especially what I mentioned earlier I talked about hydrogen atom as a simplest case of paramagnetic atom but I told you that hydrogen molecule is not then I also gave the example of oxygen atom and oxygen molecule both being paramagnetic so there is a slight apparent contradiction between the two so in the case of hydrogen then hydrogen atom becomes two hydrogen atoms in the form of a molecule the paramagnetism goes away and it gets a diamagnetic the same thing does not happen when two oxygen atoms pair up and becomes an oxygen molecule why is this difference that is what is shown in this beautiful picture here so first thing you have to remember you cannot explain magnetism of a molecule with the help of the atomic orbitals we talked about atomic orbital 3d 4d and other things are atomic orbitals when you go to a molecule that is not true you have to bring in what is known as a molecular orbital you must have studied in chemistry you have to bring in molecular orbitals and see how the filling happens again without violating the Pauli's principles and things like that so what happens in the case of hydrogen molecule hydrogen molecule is I mean hydrogen atom let us start this is one just one electron in the oneness shell electronic configuration of hydrogen atom is one as one so it has got a spin like this it has got a magnetic moment of one more magnet on as you have seen earlier another identical hydrogen atom has the same thing that is what is shown here when these two things bond together to form the hydrogen molecule H2 molecule what is happening is that I will not be able to go into the details of that in this limited time so you have two possibilities what you have to construct is a molecular orbital using the atomic orbitals that is the thing that is done in chemistry so you have two possibilities you have a possibility of getting what is known as a bonding orbital which is representing a stable molecule like what we know in the case of hydrogen molecule or you can also get an anti-bonding molecule which is something like an excited state the ground state will be obviously this one and this to avoid violation of the Pauli's principle when you have a bonding orbital and if you have to now you have two electrons to be accommodated both of them have to be accommodated here and since the Pauli's principle has to be kept in mind they cannot be both of them cannot be having the same spin up that means this is ms half this is ms half that cannot be accepted here so you can accept a ms half and ms minus half that is one is up and the other thing is down this net magnetic moment of this will be zero so this is why the atom hydrogen atom is paramagnetic whereas the hydrogen molecule is diamagnetic very simple you bring in a bonding orbital and you see that diamagnetism has to be the only solution here the question now is why not this get extended for oxygen case that is what is shown in the next one in the case of oxygen the number of electrons is more so your molecular orbitals are little more complicated here again you have bonding and the only bonding orbitals you can see that these are the situations you have 2s 1s is not shown here you have 2s 2 electrons here 2p you have 4 electrons here that corresponds to the one hydrogen atom similarly other oxygen atom here now you bring in these two oxygen atoms so these two pair up here here it is paired up here then this is 2px and 2py and 2pz so there are these two separated now if you bring in these two together to form oxygen molecule what you see here is this is the bonding orbital of course it can take two but it will be non magnetic because they will cancel each other here also it will cancel each other here also again it is sigma this all called sigma bonds and pi bonds this is a chemistry language you must have gone through that so this again it will cancel each other here again you have pi orbitals of 2py and 2pz again you can see they are canceling each other but now what is remaining so how many you accommodated 2 2 2 there are 6 of them 6 8 and 10 are accommodated here then you have 2 more because each of them have 6 electrons each 6 each so you have to accommodate 12 so you have up to this you have come accommodated 10 and these 10 are actually not giving rise to any magnetic moment because they are completely canceling each other but you are left with two more these two more have to be accommodated here only in the and so called and d bonding orbital because they are all completely when it goes here these states but here also there is a degeneracy here these two states even though it is appearing as a same energy level they correspond to two different orbitals that means there is an orbital degeneracy here so you can actually put two up spins here in fact here doing it here also see there are two up spins and two down spins here because there is an orbital degeneracy of two here similarly the same thing is happening here there is an orbital degeneracy of two here that means I can put two up spins here because orbital is anyway different so there is no question of Pauli's principle being violated here so I can put two up spins here which means that here it is completely cancelled here it is cancelled here it is cancelled this is also cancelled whereas here it is not cancelled because there is a degenerate state there double degeneracy there I can put them together so this is what makes oxygen molecule magnetic because there is a uncompensated moment coming from these two electrons which are occupying the and d bonding pi orbitals shown here in there is no question of going here because you do not have those many electrons electrons are completely over by the time you reach here so this is why oxygen molecule and oxygen atom both of them are non-diamagnetic they are paramagnetic one can actually extend this argument to other molecules like nitrogen same thing works nitrogen also one can show similar picture the difference is in the case of atoms two molecules when you go you have to really go and find out what is a molecular energy level scheme and try to accommodate the total number of electrons that are belonging to both the original atoms and try to accommodate them without violating the Pauli's principle after considering the degeneracies and see whether you are able to see some net magnetic moment if you are able to see some net magnetic moment that means it is going to be paramagnetic otherwise it is diamagnetic one case the diamagnetic case is the one where it is hydrogen and it is paramagnetic case when you have oxygen that is a difference between the two now next topic which we have to do now we from atoms you are going to molecule the next thing obviously is to see what happens in the case of a the solid state so I think I will stop here maybe some questions there are there I can you have mentioned that paramagnetism exhibit due to the unpowered electrons in the inertials yeah why it is so sir why the inertial unpowered electrons contribute to paramagnetism yeah only in unpowered electrons of the inertials can contribute to paramagnetism especially for a solid take the as I mentioned take the sodium example so sodium has 281 two is a completely filled shell eight is a completely filled shell one is the unpowered electron which is in the outermost shell suppose you are making a sodium solid what is going to happen this unpowered electron which is residing in the outermost shell this is going to become a bond this is going to bond with other electrons so that it becomes not an atom but it becomes a solid so this is going to get paired up with other electrons so effectively that will not be able to contribute to magnetism whereas if you are having this unpowered electrons in one of the inertials like a 3d or 4f it is not affected by bonding because bonding is all done by the outermost electron no d electron will be involved in any bonding in any of these solids when the atoms are becoming solids so as far as possible they should be unaffected these electrons must be unaffected due to this process of bonding and so on this can happen only if the electrons are in the inner shells that is why the in fact the higher I mean the if it is more and more to the core it is better if I if you take 3d and 4f I will tell the 4f is better choice than 3d because 4f is really unaffected by anything that is happening outside the atom 3d is that way not that good because site I mean interference will be there I am not I am using a very colloquial language there but as you go to inner and inner part of the core and if you are able to see an unpowered electron there it is much better as far as an intrinsic paramagnetic moment is concerned thank you sir sir how theoretically and practically be interpreted the magnetic susceptibility of materials okay theoretically we have done in some sense I will complete for a solid see susceptibility you can define only for a solid susceptibility you do not talk about for an atom most of the thing I have been talking about for the atom I will go to the solid very soon as I told you the next lecture is on the solids but idea is the same you theoretically speaking you find out what is the energy shift that is happening from that you can find out what is the magnetization response and from that I can actually do one more derivative with respect to the magnetic field I can actually find out what is the susceptibility so the general formalism I have shown you that is nothing but a theoretical formalism of finding out what is that what are the different susceptibilities the first term energy shift is giving rise to the diamagnetic susceptibility finally if I do a second derivative with respect to the magnetic field similarly the energy shift due to the second term in the perturbation will give if I do a second derivative with respect to the field it will give me the susceptibility the paramagnetic this is as far as theory is concerned nothing more is needed this is absolutely the fine theory as far as quantum mechanics is concerned nothing is assumed it is all fine experimentally I will come to the details when I actually deal with the solids but since you are asking I will tell so what you do is susceptibility can be done in two ways there are certain instruments where you actually directly measure the susceptibility you can measure the susceptibility straight away methods like the standard equipment is what is known as faraday balance faraday balance is something which actually can give you a the susceptibility of a paramagnetic material straight away but these days faraday balance is not very often used what one uses use a magnetometer use a magnetometer but magnetometer can give you only the magnetic moment of the magnetization if you know the volume of the material you can find you can convert the magnetic moment to magnetization then what you do is you divide with the applied magnetic field you can get a susceptibility because it is chi is m by h m by b in my terminology so this is what is done so there are two ways of finding out one is directly one is indirectly as far as experimentally susceptibility measurement is concerned I will show you some data next time while we talk about the bonding and antibonding orbitals yeah we know that the antibonding orbitals belong to the virtual state when we carry on the problems of molecules having more 50 atoms how to study the virtual state of that molecule sir can you tell me sir how to study the virtual state yeah so I have taken very simple examples of hydrogen and oxygen where I mean diatomic molecules very easy for molecules of this 50 and all it is I mean not this way you have to do a proper theoretical treatment there are I mean these days lot of packages are available one can actually find out the energy scheme I mean there are many things available even I mean chemistry people it is I do not work on that but I know that there are many packages available you do not have to really be a theoretician to do that so then you can actually find out how the filling happens you need to know lot of more things when you are molecule is complex 50 atom is not a very small molecule and you have to be very careful what kind of atoms you are going to have in that molecule so lot of things are there so 50 is definitely not a thing that I am talking about I was talking about the basic idea of finding out how the filling happens that much only I have told you in this thing for this one my answer is that you have to have first you have to identify what is the energy level scheme for such a molecule you have to identify the ground state and then you have to see how the various spin populations can be accommodated how the filling happens what are the degeneracies because the lot of symmetry will be there in those molecules there are other interactions you are talking about a realistic molecule there are other interactions like the ligand fields lot of issues have to be worried about it depends on what kind of a molecule you have I cannot give a general answer to that depending on the particular molecule particular structure what kind of a ligand that is there what kind of magnetic ions are there what kind of degeneracies are there for various levels this will determine where your filling is going to end and then you have to see whether it is going to give you a final net uncompensated moment in that case of course it is going to be paramagnetic otherwise no so it is not a very simple question but your question is very important it has to be done by essentially by a theoretical chemist in your previous lecture you say that no any material is non-magnetic my question is how any liquid like a water molecule behaves when kept in uniform magnetic field it means that what is the effect of magnetic field on the polarity of water molecule please tell me so what I meant is in my classification I include diamagnetism also as one of the categories when I do that anything that you tell any material that you tell is magnetic something will be diamagnetic most of them actually like the water that you are talking about and many many some of them are paramagnetic at room temperature some of them are ferromagnetic some of them are anti-ferromagnetic all this thing so what kind of a response this system has when the magnetic field is applied is the question if a diamagnet is there naturally there is a repulsion if in attraction definitely this is a ferromagnet where example if I take a magnet and put an ion piece definitely there is an attraction because it is experiencing a positive effect the response is positive whereas anything diamagnetic in fact the superconducting levitation that I am going to talk about our professor Gorsh is going to show is a very extreme effect of diamagnetism because there it is minus one but water the susceptibility is much much smaller it is negative but it is much smaller nowhere near minus one so that is the difference so I mean somebody else also ask this question I make it very clear if you include a diamagnetism also into our categories then everything is magnetic I am repeating my statement many people do not do that then they will tell okay this is magnetic this is non-magnetic in my language there is nothing non-magnetic everything is magnetic maybe diamagnetic that is fine I do not worry about it my question is related with complex ions whose effective atomic mass is close to that of inert gases so what will be their behavior so the magnetism is actually determined by whether you are going to have some unpaired electrons or not if you are going to have unpaired electrons definitely it must have at least some weak paramagnetic effect if all the shells are completely filled you can expect only diamagnetic effect there is no question on that so depending on what our system that you have you have to is very essential to find out what kind of electronic configuration you have for that particular atom atom is that way very good because there is there are the interactions are very minimal see when you go into solid it is at the extreme the interactions are very large so many interaction gas that way is very good I will be very happy to talk about the gas but there are other issues let us not worry about it so gas I can consider it as an atom which is completely non-interacting so then it is very easy you have to find out what kind of a the electronic configuration it has got that will simply determine what kind of magnetic response it will have you talked about rare earths that comparison is the other side because rare earths are strongly paramagnetic at least because they have unpaired electrons in 4f shell 4f is very much inside so they are not at all affected by the bonding and any kind of interactions this atom will have with the neighbors and other things so whatever is there intrinsically will be protected as in the case of an isolated rare earth atom so rare earth solid and rare earth atom as far as the atomic the magnetic properties especially the magnetic moment is concerned there is not much of a difference because this 4f shell of a solid is essentially similar to what happens in the case of an atom there is no much effect of these external things the bonding and other things are not going to affect the magnetic moment so it is able to retain the atomic property considerably sir some related question yeah please we know that the body organs are supposed to be non-magnetic if I made some material whose z effective is equivalent moment issue so in that organ if it may be artificial or biocompatible what you call it yeah can we have some treatment regarding that magnetic material yeah your question is yeah yes yeah I do not know whether I am answering exactly same thing what you are telling but these days for example it is I think I mentioned in the first lecture also magnetic hypothermia which actually is used for cancer treatment at least I know that at least in rats and other animals it is being tested I do not know whether it is done in human beings at the moment so what they do is they use magnetic nanoparticles locally they inject wherever the tumor or tissue is affected and they actually use I am going to talk about in the later lectures what you use you apply oscillating magnetic fields so there is a heat local heat produced and this is able to cure the that part so this is actually is a very important field in biomagnetism today so not just for the drug delivery this is actually the treatment itself so this you use magnetic particles as you said rightly said the it has to be biocompatible you cannot inject anything into the body it is very very important so once it is done fortunately many of these ferrites are that they biocompatible unfortunately the rarer things which are actually very good but they are not biocompatible they are very good magnetic systems but they are not biocompatible ferrites are all right so they are actually they can be injected and locally and locally they can produce a heat by having a external source outside so this is being I am sure I know that definitely it is done in animals it may happen to the human beings maybe in the near future if it is not done so far so this is called a magnetic hyperthermia it is a very important field today yeah I have a question can you just give some compare and contrast points or features between a permanent magnet and electromagnet which is having same magnetic field B okay magnetic field intensity do they exhibit the same magnetic properties or is there any difference I am going to talk about in the next coming lectures but still I will mention it electromagnet the problem is there is a limit I mean electromagnet the magnetic field is determined by the current when you apply current above certain thing you will go you are going to have the heating so you have to have cooling arrangement and it becomes extremely complicated and it becomes very very bulky today the electromagnets typically produce the highest is something like if you have put the best thing it is something like pre tesla that is 30 kilo austere is a field that can be produced by an electromagnet superconducting magnet can be used of course it has got its own problem that is a different thing low temperature and other issues typically even in our department we can actually go to something like 9 tesla that is 90 kilo austere without any problem so electromagnets can be used provided your field requirements are something like 1 tesla 1.5 tesla anything more than that you have to depend on superconducting magnets only there is no other way permanent magnets yes permanent magnets one can use but again it cannot be I mean my rough estimate is you cannot generate fields more than 3 tesla in whatever configuration you are going to use it is not possible so higher field definitely you have to depend on superconducting magnets only I will be talking about in my superconductivity lecture later on I have one more question sir yeah please as we know that oxygen is a para magnetic is it possible to attract oxygen in atmosphere with the help of strong magnet problem will be as I mentioned the susceptibility is very very small para magnetic susceptibility like the magnetic susceptibility is very very small I have not thought about it but I am sure the fields required will be very very large I do not know whether such a thing happens in the atmosphere because I am I do not have any idea about the atmospheric sensors and so on I will look at it and tell you what kind of a field can actually make some attraction of something but definitely the susceptibility are very very small so only thing is that it is positive that is only thing unlike the diamagnetic case but otherwise it is very very weak effect but it is a good point let me see whether something of that kind happens in the nature okay thank you sir