 Hello, we are going to discuss this electron nuclear hyperfine interaction. I introduced this last time without explaining what it means. So today we are going to figure out what it is. Electron nuclear hyperfine interaction. I said that this interaction produces lots of hyperfine lines. Let us try to understand the origin of that. So for that, let us say the simplest possible system we can think of which is pre-hydrigal is hydrogen atom. It has one electron of course and one proton. So what we have discussed so far, if I try to look at the energy level diagram of this in a given magnetic field for the electron, I get a splitting of this kind which is the electron Ziemann splitting corresponding to the electron spin. Let us call it S of MS equal to minus half and here MS equal to plus half. And as we discussed earlier, if this is the case, then I see a transition which looks like this. Delta E is equal to B and this will be equal to H nu. So for a first frequency experiment, the EPS spectrum will look like and draw it here and this is the magnetic field. So this corresponds to the energy of transition which satisfies this relationship. But what it is found that if the experiment is done, this spectrum does not look like this, but it appears to be this type of functional magnetic field. It gives one line here, another line there. There is no line exactly at the middle of that. So this line, if your line is split into two in case of hydrogen atom. So we can call them of course the hyperfine line of hydrogen atom. Whether this gap is actually almost 506 drops or so, approximately quite a large splitting. What is happening there? Why does hydrogen atom give two line and not one as expected from this side of thing? So one simplest way to look at is that this is probably not the complete energy level diagram of hydrogen atom. Something else is happening. So if I get two lines, naturally there will be two transitions involved there. What is the origin of that? Try to look at it. There is electron there and a proton there. Proton also has a magnetic moment. So in a magnetic field, this will also align according to its nuclear spin state which is i equal to half m i equal to plus minus half. So it will be either along the magnetic field or opposite direction. Now the electron which is present there in the atom, it finds itself in the presence of two magnetic field. One is external magnetic field that is applied by the spectrometer and second one is the local magnetic field the proton produces. You see that the two are eminent of the proton spin. So that means the local magnetic field that the proton produces either adds to the external field or it subtracts the external magnetic field. So depending upon therefore the proton spin state, this energy level is going to split now. So here suppose it appears this way that this splits into two. Similarly this will also split into two depending upon what nuclear spin is associated with the electron speed of minus half. So this could be m s equal to minus half and m of i equal to plus half let us say. This could be m s equal to minus half and m i could be also remember that electron z 1 interaction is much stronger than the nuclear z 1 interaction. But here we are not looking at the electron z 1 interaction but it is the interaction of the electron magnetic movement interacting with the nuclear magnetic movement. So this splitting is given by this. Similarly here m s equal to minus half m i equal to sorry m s equal to plus half and m i equal to minus half and this will be m s equal to plus half m i equal to plus half. So why this arrangement is of this kind that we will see again later very soon in fact. But let us take it to be true. Now there are four variables now where earlier we had two. An EPS spectroscopy it is electron spin which changes, nuclear spin does not. An EPS spectroscopy is the nuclear spin that changes not the electron spin. So here if you draw the line correspondingly without flipping the nuclear spin you see that one transition will go from here to there m s equal to minus half plus half m i remains plus half and plus half. Here another one m s equal to minus half to plus half m i remains same and minus half and minus half. So you see the two transitions corresponds to two different energies and that will reflected here in two different types of magnetic field. So this splitting that you see here is therefore coming from this possible orientation of the nuclear spin in a magnetic field. In one proton I get possible orientation of two types I get two lines of this kind. So I get two lines here and these are called therefore hyperfine lines coming from the one proton to spin i equal to. Now suppose I have got more than one such nuclei which has got spin halves. What could be the simplest example pro hydrogen atom I can go to this hydrogen molecule and it has got two protons and they are identical in all aspects we call them two equivalent protons. So we go through similar argument now that in a magnetic field these two protons can align let us say all of them can align this way this one can be up other can be down or this could be down this could be up or both could be down. These are the possible nuclear spin orientation in a magnetic field and each of them give us to a local magnetic field that the electron will see. So this energy level that I have drawn here will be modified in this fashion I have got this two energy levels for this electron spin of minus half these are the possible nuclear spin. So this will split into this. Imagine here the electron's nuclear spin of one is opposite to second one. So net local magnetic field due to this combination will be 0 here it will be higher this will be lower or this will be lower or this could be higher. So let us write in this fashion that this will be total spin of nuclei could be the plus half minus half these are the possible in my component there. Similarly here the electron spin is plus half this will also have three types of nuclear spin arrangement. So I write in this way so I can draw the splitting of the energy levels due to the nuclei could be drawn in this fashion that this is a split into three. So this energy level will also split into three. Now we follow the same convention earlier that when the electron parametric resonance transition takes place nuclear spin do not change electron spin does. So transition from this to this one possibility another possibility is from this to this the third one is from this to this. We see that since the energy gaps are different as if they will appear different magnetic field but the middle one is exactly same as here because the nuclear spins are just opposite to each other. So that does not contribute to any change in the local magnetic field. So this spectrum therefore if I have to what we predict from here is that appear spectrum of this hydrogen molecule ion should look like this function of magnetic field this will look like this though. But this is not quite true the three lines of equal intensity that comes from this self argument is not quite true. The reason is that see the number of possible allowed nuclear configuration is of this kind. So even though there are three energy levels you see that number of molecules that will appear here will be I said two of them there similarly two of them there because the first proton spin could be up second could be down or first could be down second could be up both are equally likely. So in a sense there are either call it the two energy levels are there which are degenerate or the number of particles that are distributed here will be such that there will be twice of a number of particles here compared to this and this. So the intensity of the appear line will be twice now this kind. So this is therefore I expect one is to two is to one type of intensity for this system there what we had here that this four energy levels are there all levels are almost equally likely. So intensity of number of particles that this transition will be seen a number of molecules that this transition will be seen will be exactly equal this will be one is to one. So you see how therefore the relative intensity change as the number of nuclei which are producing the splitting change here both are spin half system. So I get a splitting of one is to two is to one now here I could therefore take the example that I saw last time is that this two five diracivital semi-quinone and a radical gives a spectrum which in actually intensity is one is to two is to one type of thing. And the origin of this now could be easy to visualize the one proton is here another proton is here and they are equivalent in all respects. So I have can have argument of this kind the way this nuclear spins of this two protons aligned gives rise to this sort of spectrum with intensity one is to two is to one. It also shows lot of other things is that electron though we draw here in structure electron is sitting here but that is only the crudest possible picture. The unfair electron is present throughout the pipe cloud of the benzene ring and interacting with all the ring protons. The effect of spectrum shows this. So in a sense the major of the splitting here or here is a major of the strength of the interaction of the electron and the nucleus. So we can continue further with this sort of argument that if I were more than two spin half nuclear what will be the corresponding if your spectrum is to one more and then we will generalize after that. Suppose we have system of three protons and that is also a radical something I can think of is let us say methyl radical these three protons are equivalent in all respect. So I could draw similar energy diagram as earlier for electron m s equal to minus half and here m s equal to plus half. Now we have three protons and all of them will align in a manner according to plus half or minus of each of them. So I can get this sort of possible orientation of all three could be up one possibility or this way up down only one down two ups or two down one up or all three could be down. These are the possible nuclear spin in a magnetic field now. See here that each of them now gives us to local magnetic field that the electron will see is different for all four. Maximum difference will be here maximum here and minimum there but nevertheless each of them will be appearing as a splitting of energy levels here. So what we expect now considering what we have discussed here this each of these levels will be split into four lines now. So similar here so if you write the corresponding total mi the total nuclear spin this becomes plus 3 by 2 this becomes plus half this becomes minus half this becomes minus 3 by 2. So this will split so it will be this is the possible values of the proton spin total angular momentum quantum number. Similarly here this will be minus 3 by 2 minus half plus half plus 3 by 2. Now same argument is exactly same the transition will be from the level 1 to the other level which does not involve any change of the nuclear spin. So that transition is here to here otherwise from here to there then this four transition is possible there. So the EPS spectrum will consist of four lines but will the intensity be equal certainly not see the way we had this 1 is to 2 is to 1 intensity ratios can be now extended to this number of possible arrangement for this this and this is this is one type if you say one unit this will be thrice the number of molecule will be having this sort of nuclear spin quantum equation this becomes 3 this also becomes 3 this becomes 1. So the EPS spectrum will therefore look like comes a magnetic field this will look like with the intensity ratio of 1, 3, 3, 1. So it gives four hyperfine line and the energy gap of this will be same as this will same as this because the energy gap between this, this, this same as then this, this and this and that in turn comes from the magnetic field that this sort of arrangement of nuclear spin produces at the electron. So you can see now pattern now let us summarize now how we can get this various line and then we will generalize. So bare electron it gives 1 line spectrum spin half nucleus gives a splitting of this kind. So this is the center of this one. So this is a splitting due to the 1 spin half nucleus in our case it is a proton. If the 2 of them then the splitting looks like so here it was 1 is to 1, 1, 2, 1. I have got 3 spin half nuclei then I get a splitting which looks like this 1. So this sort of splitting that comes from the equivalent nuclei of spin half has certain pattern. These numbers are supposed to be familiar to some of you is that they come from the binomial expansion of a plus b to the power n if you read the binomial coefficients have exactly same number. So these numbers can be generated in former triangle we call Pascal triangle. Let us I have got 1 here then 1 is to 1 then 1, 2, 1 see then I have got 1, 3, 3, 1. The triangle is generated in this fashion that if we starting from 1 let us go 1 step below I add the number which is directly the top left and top right take some of these 2 assuming there is 0 kept there. So here this plus this gives this number 0 and 1 gives 1 so here 1 and 0 gives 1. Go down here again I add the numbers which are directly to the left and to the right imagine there is 0 there. So this 1 comes from the sum of this and this is 1 this 2 comes from the sum of this 1 and 1, 2 this comes 1, 1, 0 and 1. So every step we just add the 2 numbers which are left side and right hand side and add the 2 to get a number below that. So this triangle is called Pascal triangle. Next one same way it goes here this is 0 there gives 1 and this and this gives 3 and this and this gives 1 and this and this which is 0 gives 1, 3, 3, 1 we can therefore generalize that. See if I have got now 4 spin off nuclei I can easily predict the spectrum which will look like from this Pascal triangle will be 1 add the 2, 4 this will be 6 add the 2 this will become 4, 1. So this is the Pascal triangle. So you can keep on generalizing this. This example is given here we saw earlier this one this radical has 4 equivalent protons they are equivalent in all respects 4 equivalent proton gives rise to 5 lines as predicted by this generalization for intensity ratio is 1, 4, 6, 4, 1. So here if you measure carefully this is almost equal to 1 is to 4 is to 6 is to 4 is to 1. See how beautifully this agrees with our prediction. So this way we can predict the spectrum of more complicated radicals and that sort of example and their interpretation will take up later and right now we just summarize that from the EPR spectrum when you get lots of such hyperfine line we can very easily characterize or even guess very intelligently the nature of the radical. What sort of nuclei which are present there which can give rise to such hyperfine lines and from there we can characterize it almost unambiguously what the radical that could be. So before we close we just take one more example which is somewhat different from spin half this generalization only for spin half nucleus. Suppose the nuclear spin is not half it could be 1 carbon 13 spin half per nitrogen 14 is spin 1. So if what happens if the nuclear spin is 1 then I equal to 1 and its component M i will be minus 1, 0 plus 1. So in a magnetic field this nuclear will also orient in three possible directions and it will produce local magnetic field which will be having three different values. This will not have no magnetic field. So if I draw the hypothetically anizable diagram of one electron interacting with the nucleus of this kind what do I see. First the electron C 1 splitting will look like this. This is the M s equal to minus half M s equal to plus half. So when this electron sees the presence of the nuclear spin it will split into three energy levels M i equal to plus 1. This will split into three M i equal to minus 1, 0 M i equal to 1 and then based on this condition that when the IPR transition takes place electron spin changes from minus half to plus half but the nuclear spin do not change. So that allows me to have these three transitions from one to here to here to here and third one is from here to here. So these three transitions will be reflected here in the spectrum that earlier the spectrum was here when there was no electron nuclear interaction now this is split into three because of these three energy levels here so I get transition which looks like this one is exactly below the other because this energy level is same as that is here. What about the intensity ratio you see that each of these three levels have exactly equal population one is to one is to one here also one is to one is to one. So that will be reflected in the intensity of this will also be one is to one is to one. So with the total nuclear spin is one here I get three levels compare that with sorry H 2 type of system where was the total nuclear spin was one but they were coming from two protons of M i equal to half and half or minus half or plus half or minus half or minus half. So these are given as to one is 0 minus 1 which is very similar to this but here the spectrum look like this one is to two is to one but on the other here we have got one is to one is to one reason is again obvious that here this transition involved energy levels where twice the number of particles that could stay here but here all the levels have just same number of particles. So I get one is to one is to one here one is to two is to one this also shows the power of your spectroscopy that by knowing the not only the number of lines and what is the pattern of relative indices I can guess very accurately the type of nuclei which are causing the splitting. If I say one is to two is to one I will most certainly be able to say that two spin half nuclei are causing the splitting I may not be able to know say that it is coming from the two protons if I do not know the chemical nature of that but the fact that I see this in the ratio that they must come from the two spin half nuclei. Similarly if I see a pattern of this kind one is to one is to one here most certainly say that this comes from the nucleus whose spin is one it may be in origin 15 I am sorry in origin 14 but unless I know the chemical nature I would not be able to say that the example of that is given here this one here if you see that this free radical center O dot is near a nitrogen this nitrogen nucleus spin is 14. So, that is exactly what we have discussed here and we predict that the intensity will be one is to one is to one and that is what is seen here. So, with this we stop our discussion of the hyperfine interaction and later we will continue to find the origin of this interaction.