 Hello, today we are going to learn how to analyze the first order EPR spectra of free radicals. We have seen the first order EPR spectra comes from a Hamiltonian of this kind which we have seen for hydrogen atom for example, this is the electron Z 1 term, this is the hyperfine interaction term and this could be written as I z S z plus here up to first order calculation this term does not contribute to any energy. So characteristics of the first order spectra are that they are symmetric with respect to the center and center corresponds to the if you call b 0 then g value is given by H nu equal to g e beta e b 0. So I can get immediately g of electron is H nu by beta e b 0 and the line widths are same for all the hyperfine lines. These are the conditions which are satisfied by the first order spectra. Now this is true for only one hyperfine nucleus. Suppose there are more than one then this can be easily generalized to write the sum of all the possible nuclei which are giving rise to the hyperfine splitting plus. So I have this is the first order spectra comes from this hyperfine splitting constant a. So if I have various type of nuclear let us put I here sort of I here and sum over I. So I could be the nuclear which are giving rise to splitting. So for example, the splitting I have shown many times that this could be split in this fashion and this could be the let us say a 1 which is the bigger splitting and I have only one spin half nuclear. So this is splitting to 2. If we another one a 2 which is also a 2 this is another spin half nucleus which is also causing splitting that we can analyze the spectrum. So the analysis involves going the other around that we start with a given sort of pattern of various hyperfine line and try to find out the nuclei which are giving rise to the particular pattern and what are the corresponding coupling constants are. And if we know the frequency of microwave which was used to record the spectrum and a very precise estimate measurement of the magnetic field for the center of the spectrum we can measure the g value. So we will take some examples now and learn how to analyze such spectra. This is a spectrum of a free radical. This gives four line spectra to understand what sort of radical could give rise to this sort of pattern. Let us first find out the hyperfine splitting constant of this. Now if all the four lines are coming from the same equivalent nuclei then the gap must be same. So let us measure that first but for measuring that I need to find out where these signals are crossing 0. So first I draw a line through this 0 signal. This draw a line here and then make the measurement. I have already done that in another copy of the spectrum. Here it is. So this is the line that I have drawn through the 0 signal. So find out the coupling constant. Now let us measure the gap from the this outermost line to the next hyperfine line and further I use this divider. This is the gap between the two adjacent line. Let us say the same gap is present here also. Yes, this is same as that. This is also same as that. So this gap is same for all hyperfine lines. That means the same group of equivalent nuclei are giving rise to these four lines. So what is the coupling constant? There is nothing but this gap. What is the measure of that? Let us measure this in the scale. This is 9 cm. So the coupling constant is 9 cm. And here is the scale of the magnetic field. So here this length is the 10 gauss. So what is the length here? 4 cm. So 4 cm corresponds to 10 gauss. So now we can easily find out the coupling constant A. So A corresponds to 10 by 4 into 9 gauss which is 22.4 gauss. So we have got the coupling constant determined from these four lines. Next question is what sort of nuclei I can give rise to these sort of relative intensities? For that let us measure the intensities of all four lines. For the outermost line the height is 5.5 cm. I write it here. Next height is similarly from this to this 16.5 cm. Then the next intensity is maybe 16.7 cm. The last one is again 5.5 cm. So to normalize all the intensities with respect to the outermost line we divide this number by this 5.5. That gives the relative intensities to be let us say this is 1, 3. This is very nearly equal to 3. This is 1. So we have got the four lines with this sort of relative intensity 1, 3, 3, 1. Now this can come of course from 3 equivalent nuclei with spin half. So we have got 3 spin half nuclei that is present in the radical. Without knowing the chemical nature of the radical it is very difficult to say what precise the radical is but you can confidently say that radical contains this sort of nuclei which are giving rise to this sort of hyperfine pattern. What is the G value for this spectrum? For that I need to find out the center of the spectrum. Center of the spectrum will be somewhere here. How to find it out? Again we use the ruler. Total distance from middle of this to middle of this is 27.2 cm. So half of that will be 27.2 divided by 2. This gives 13.6. So let us find this point here. 13.6 corresponds to this place. So this is my B0, the center of the magnetic field. So if I knew the magnetic field axis I can measure the magnetic field value for this position of the spectrum. And if I know the frequency of the microwave which was used to record the spectrum, the G value can be written as G of electron will be h nu by beta i B0. That is it. So we get all the information about the spectrum. Of course this is a rather simple spectrum and one may not need all this detailed measurement. One can actually see that essentially this 1 is to 3 is to 3 is to 1. But life is not often as simple as this. So we will take the next example. Here there are 1, 2, 3, 4, 5, 6, 7 lines. And let us first measure if the gap is same among them. Using the divider, if the gap between the outermost line and the next line is this. And then this gap is same as this one. This is once more. Yes. This is same as that. This is same as this. This is same as this. This is same as this. This is same as this. That means again all the lines are coming from a group of nuclei which are all equivalent. So immediately we can measure the coupling constant from this. This is the measure of the coupling constant. What is value? This coupling constant is given by this gap which is 3.75 cm. So here A is equal to 3.75 cm. What is the scale of the magnetic field? It is given here. This 5 gauss corresponds to this length and that is 5 cm. So 5 gauss corresponds to 5 cm. So naturally this says that coupling constant A is equal to 3.75 gauss. So we measure the hyperbolic coupling constant. What are the nuclei which are given as to this splitting? For that we measure the height of all these lines. The same way as we did earlier. For that measure the height here. This is 0.85. Next one is nearly 4.9. We do it more carefully. And next one is 12.5. This is 16.8. This being symmetric. This is also variation 12.550 and again nearly 0.8. So these sort of small deviations are expected in an experimental spectrum. One does not expect them to be exactly following the relative values that we expect theoretically. So small deviations are possible. So divide all the numbers by the average of this and this. And if you do that this values will turn out to be 8.5. This will be 1, 5.8, 14.2, 19.7. This is 14.2, 5.8 and again nearly 1. So now these numbers we now round it. So that we can get the values that a Pascal triangle can give either spin half or spin of other nuclei. It is very obvious that the number that corresponds to which matches very nearly to Pascal triangle corresponding to spin half nuclei is 1, 6, 15, 20, 15, 6 and 1. So this is nothing but 6 spin half nuclei can give rise to such relative intensities. We have got 6 spin half nuclei present here in this radical. So that is the way we can explain this pattern. Now for this what is the G value? G value will be again the center of the spectrum because there are odd number of lines. The center of the spectrum is precisely here. So this is my B0. From this I can calculate the G value if I know the frequency of the microbe and the actual value of the magnetic field here. Let us take a little more complicated spectrum now. Here again we start from the outermost line which is here. They measure the gap between that and the adjacent line and this is the gap. Now from here you see the next line does not fall in line. There is no other line adjacent to that. That means that this line and this line belong to different type of nuclei. So the gap between these two actually happens to be same as this and same as this. So that gives the one hyperbolic coupling constant. Let us call it A1 whose value is 1.2 centimeter. So next coupling constant will be between let us say one of this line to the corresponding line of the other group. It is too big I cannot measure here. So let us use the ruler. This gives me 0. This is 17 centimeter. This is the 17 centimeter. I should expect that between these and the next line should also be 17 centimeter. So that is exactly what is seen here. So between these and that the gap is same as between these and that. And that is also the same as between this and this and this and this. So 17 centimeter corresponds to another hyperbolic coupling constant. Let us call it A of 2 to the 17 centimeter. So we have got two types of nuclei. What is the corresponding coupling constant in magnetic field unit? This is the scale given here 5 gauss. Again 5 gauss is equal to 5 centimeter. So 5 centimeter 5 gauss. So this gives therefore this is corresponding 1.2 gauss and this corresponds to 17.0 gauss. These are the two hyperbolic coupling constant. Now for this what will be the G value? Again we have to find the center of the spectrum for that. Center will be somewhere here. It does not fall on any of the hyperbolic line. So we measure the center of the spectrum by measuring the total extent from this to this and then we can measure the center and that will be here somewhere and we can find out the corresponding G value. How about the relative intensities? So then we can ascribe this split into sudden nuclei. For that this height is found to be 17.8 centimeter. This is also 17.8. This is 15.8. This is also 15.8. This looks very similar. This is 17.7. This is also 17.8 or so. If we divide by this number, the ratio I get is 1 is to 1. Here I get 2 is to 2. Here I can get 1 is to 1. So from this relative intensity it is very clear that this is a doublet. Doublet and doublet that comes from a one spin half nucleus. So this corresponds to one spin half nucleus and this comes from this relative intensity of 1 is to 2 is to 1 triplet. This comes from two spin half nuclei present there. So the radical has this sort of nuclei. Again, without knowing the chemistry of this formation of the radical, it is very difficult to say what precisely the radical is. All we can say that the radical has one spin half nucleus and two spin one nuclei and those two nuclei are equivalent. Now a little more complicated spectrum. Here now things are reasonably routine. So many lines are there. So we start again from the outermost line and see how many lines belong to one group of equivalent nuclei. So this is the first gap between the outermost line and the next one. So between this gap is same as this gap. But then after that there is no line here. This does not fall in line with this. That means these lines belong to one equivalent coupling constant. This does not belong there. That means this is the beginning of another equivalent coupling constant. So if you start from this, you see that this gap is actually same as this gap, same as this gap and then nothing else is there. So this again belong to therefore same type of end coupling constant. So after that this line is not accounted for. So again we start from here and measure the same gap as earlier. So this and this again nothing else is there. So these three again belong to, let us use some other symbol here. Similarly it is very obvious that they also belong to this pattern and this belong to let us say this pattern. So I have got this group, this, this, this, this. The splitting is same for all of them. So one hyperbolic coupling constant therefore corresponds to this gap which is 2.7 centimeter. And next coupling constant will be the distance between let us say one line of this group to the corresponding line of the next group. That is from this to this. That gap is here 7.5 centimeter. We can confirm that by checking that this should be consistently followed by all the lines. So this is the gap here. Now between this and then here yes it matches with that. This matches with this, this also matches or if you take the center of this group of line, they are between this and this, this, this and this. So this group they all belong to another coupling constant whose measure is 7.5 centimeter. So now from the scale again here 5 centimeter is called 5 gauss. So this gives this corresponds to 2.7 gauss. This corresponds to 7.5 gauss. Here is the splitting constant. How about the relative intensity? So for this we measure this height because we measure actually vertically. So this is nearly 1.2 centimeter and this is corresponding to 2.2 centimeter. This again corresponds to 1.2. So you see here these are approximately following this part 1 is 2.2 is to 1. Similarly this also is 1 is to 2 is to 1. We can quickly check that by without measuring. Let us say this is the height here. So between these and this is 2 and this 1. So 1 is to 2 is to 1. How about the next one? Again I measure this. This is the height. So again here. So 1 is to 2 is to 1 and same pattern continues. So triplet, triplet, triplet, triplet, triplet. So what is the relative intensity among these triplets now? So that what that I measure this, compare this height with this height. So that height is 10.7. Similarly center of this triplet corresponds to 16.7. Here 11. This is again. So here is 2.6. Let us check here. This is what I made a mistake here. This is indeed 2.6 here, 2.6 here. So this will be also 1.4 or so. So 1 is to 2 is to 1 is fine. Now here between this, this, this, this and this relative intensity turn out to be divided by 2.6 all over then this becomes 1, 4, 6, 4, 1. So again this pattern is familiar. This comes from Pascal's triangle of 4 equivalent spin half nuclei. So what I have here that this corresponds to 4 spin half nuclei and this corresponds to 2 spin half. So we have figured out all the hyperbolic complex constraint that is present here. Again what will be the g value? That will be the center of the spectrum. For this one there are odd number of lines. So this is the center of the spectrum. So this gives v0 which can give the g value. Now this is all very well if the spectra are very well resolved and all lines are seen with decent signal noise ratio. But practical difficulties are quite common. For example, if there are many hyperbolic lines, usually the outermost lines intensity becomes very, very small. Often difficult to detect them in their spectrometer. Then what does one do? You see the analysis always goes that this whether you start from the outermost line and then keep on finding out the pattern. If I do not see the outermost line, what does one do? So here is an example which sort of exemplifies that. So here one does not know if these are real line or not. In fact there are so many such lines are lying here. Difficult to pinpoint if there is a transition. So in fact there is a line here. I am saying because I know the answer. But if I do not know the answer, how does one go about this? Analyzing this. So if this is the outermost line that I take it to be true, so using our prescription that you first measure the gap and then see the next line which is here and then after that there is no more line. So these three line corresponds to one group of hyperbolic line. This, this and this. This one does not belong to that. So it is possible that this might be a triplet but we will check the later from the intensities. So we now start from the next one which is not part of the previous group. So this is the line here. So this is the gap or hyperbolic constant. So we start from here and measure the same. So you see this is matching with this. So this and this and this, this and this. So now we have got five lines now. Earlier for this group has three lines. So let us mark them as this, this, this, this and this. Let us check once again. This, this, this and this is same. So I have got five groups of line. So something probably is wrong here. There is something missing here. To be absolutely sure there is something missing, let us look at the next group of lines which are not part of this hyperbolic line. So that is here. So we start from this again, find out the lines which have the same coupling constant with this or this. That is starting from here. This one here, this, this and this. Again you see there are five lines. So let us put some sort of, this sort of sign here. So see this pattern is present here. It is natural that there must be line for these three groups of three lines also. This cannot be just three. There will be five lines. So I start from here, go back and see that it looks like there is a line which is hidden here. Similarly, there could be line here. Even if I do not see it, I have to assume that it is indeed there but buried in the noise level. So I have a line here and here. Similar to other side also that there could be a line which is here, here and somewhere here and not seen there. So that way one goes about analyzing the spectrum and even if we do not see the outermost line, one has to make intelligent guess and find out where those little lines which are buried inside the noise level are hidden and then from this pattern one can completely analyze the spectrum and get all the possible hyperbolic and coupling constant. We have just now seen how to analyze first order EPS spectra which are recorded on this paper and use these two important tools. One is this my divider and other one is this ruler. This may sound old fashioned because nowadays almost all EPS spectrometers are interfaced with computers and one rarely gets spectra printed on paper but they are available in electronic form and one can analyze on the computer. So we are going to quickly have a look how one goes about using computer programs to analyze the spectra but the principle that you have learned using this technique of measuring this height and the gap here that remains the same. So it is the same principle which are used in the computer which only helps us presumably do it somewhat faster but not necessarily easier. So you have to still measure the coupling constant to measure the height and then find out the number of a new player which are present there everything remains the same. Also sometimes the lines of various adjacent hyperfine components may be overlapping. So it may become difficult to measure using this ruler and divider. So then the computer program may be able to help us do a somewhat more accurate job. Let us see how that is done. We have seen this spectrum earlier it has got two coupling constant A1, 1.2 gauss and A2, 17.0 gauss. We measure these values using our ruler and divider technique. If you see very carefully the height of this is not matching with this height here this is slightly lower than this one and at the bottom this is slightly lower than this one. Same is true for all the lines. It means that there is a partial overlap of the intensity coming from this line with the line here. Same is true here also. Part of the signal correspond to this hyperfine line is overlapping with the signal from this hyperfine line that is why this sort of unequal heights are appearing here. So there could be more problems of overlap and then our ruler and divider technique may not be giving very accurate estimate of the coupling constant. For that we need to go to computer program and try to analyze the spectrum using the program. Now here is a computer program which will help us analyze the spectrum. Most commercial EPR spectrometer vendors provide their own software. One can find similar software in the internet also. This particular one is made in our own laboratory but nevertheless by enlarge all the programs look very similar in their operation. So to show how this work one could initially calculate a spectrum by giving some known hyperfine and coupling constant. Let us say one coupling constant is 1.2 gauss. I can enter those values here. Here one can enter various coupling constants and the nuclear spin half. This could be spin of one or half a number of such nuclei and the coupling constant. Let us give the value of let us say 1.2 gauss and I have a splitting of one of a nucleus of spin half and number of such nuclei is only one. Another coupling constant is 17.0 gauss and there are two of them and again spin half. Let us give some line width of 0.2 gauss and the spectrum source would look like first derivative bit here. Let us simulate the spectrum and the experimental range of the magnetic field scan. Let us keep it at 50 gauss. Now simulate it. That is how it looks like. Doublet, doublet, doublet and overall triplet is present here which is similar to what we saw earlier. Now we can see the effect of line width on the spectrum. Here I have used a line width of 0.2 gauss. Let us increase it to this 0.4 gauss and again simulate this. See now these two heights are not same. Same as here. So that means when the line width has become bigger there is partial overlap of the two high profile line. If we increase further let us say 0.6 gauss again simulate it. See how it is becoming unequal in their heights. Further broadening, simulate that is how it comes out to be. So things can be pretty complicated in the sense that a ruler and divided technique may not give the values accurately. If there are more than one nuclei giving similar splitting and such broad lines then it may become almost impossible to make some measurement. So to see that how the line width affects the number of high profile lines and the appearance, let us consider a case of only let us consider a case of 4 spin half nuclei and they have a fine coupling constant. Let us give a value of 2 gauss to start with and width is kept very small. Let us simulate this one and remove this nuclei. So simulate this one. So 5 line spectrum and they are following the usual 1, 4, 6, 4, type of intensity ratios. Now one could expand the spectrum by working at a smaller scan range. Let us make it 20. So total range becomes 20 gauss and simulate that is how it looks like. So it looks so cleaner. Now if I keep increasing the line width see how it changes 0.6 gauss. Already they start showing distortion is not it. So that means that here it will be very difficult to measure the line width and the coupling constant accurately. So we need to use the this simulation program to compare with our experimental spectrum and get an accurate estimate of the hyperbolic coupling constant and line width. So to do that we first load our experimental spectrum from which has been recorded earlier and this is here. For this the total scan range was kept at 40 gauss. Let us say using this cursor I can place this anywhere we like and I can measure the height of this peak and the value from there. This I can get the intensity and the distance from this to this in the horizontal axis will give me the separation. So that way I can find out the approximate value of the hyperbolic coupling constant. And let us say from this I have got an approximate value which are entered in the data. So let us say one value is 2.1 gauss and there is only one half nucleus of spin half and other is let us say 9.5. There are two nuclei of this kind and this is derivative signal. Let us start with a narrow line width. So with this I can calculate the spectrum and that is the simulation one. So this simulates this one. Of course it does not look anywhere near the observed spectrum but what we could do is that we can obviously change the line width and see how it approaches the experimental spectrum. 0.4 gauss, similar again 0.6 gauss here increasing, keep increasing, line width is increasing one gauss. Another 1.2 gauss. So it is sort of not quite similar but trying to become. Now let us watch here that this red graph is the red trace is the experimental spectrum and blue is the simulated one. So this peak are not coming near to this observed peak here. That means our values of the hyperbolic coupling constant are not quite right. So this has to go away from here, this has to go away from there. So this gauss values need to be modified. Let us make it this, increase this coupling constant, see what happens. Now at least this positions are right. So we may have made some progress, some improvement in the coupling constant. Now still this is not matching with this. So this is not matching with this one. So maybe we need to increase this coupling constant, this 2.2 let us say, then see what happens here, does not look quite right at all. So maybe we should increase the line width further, little bit more. So it is trying to be similar to the experimental spectrum. But now what has happened? This blue, the similar spectrum shows a line which is further away from the observed one. This is also further away from observed one, same is true here also. That means the big coupling constant has become more than what is supposed to be. So let us reduce this, maybe what was the value earlier 9.5, let us go back to 9.8, then simulate it. Now it is coming closer to that, little bit more 9, 10 maybe 10 here. So these are almost matching the line positions here. Now maybe the width is not yet sufficient, increase the width. It is trying to become similar to that further, not bad. But now this has gone away from this. The blue line is more than the red line. So I need to optimize another coupling constant. Let us see how about this one, 2.1. It is improving, maybe in the same direction let us continue, 0, not bad. So almost all features are reproduced except this, little bit differences one can see between red and blue. So maybe the line width is too much, reduce it further. It has disappeared. No, it has not disappeared. They are actually overlapping too. So that let us deliberately shift one spectrum here. So the exponential spectrum has moved out. So it means that now I have got the perfect matching between the red exponential spectrum and the blue which was calculated here. Let us go back and bring all the shift back to the previous value here. Now shift has become 0. So I have got perfect matching. Of course it was a contrived example because this red spectrum, though I am calling it an exponential spectrum, it is actually calculated. So I can get the exact matching. These are the values now I can report. For this the coupling constant for first nucleus was 2 gauss with one number of spin half nucleus. Coupling constant of 10 gauss was for the second group of 2 nuclei of spin half and line width was 2 gauss and gives me perfect matching. So this is the way to go about analyzing the EPR spectrum using a computer.