 Sir, please tell us about the canonical conjugate. Okay, so canonical conjugate is actually connected to Heisenberg-Eisenberg principle. So the way we have studied Heisenberg-Eisenberg principle yet, so we generally try to cover like if there is a uncertainty in the position mapping of a quantum mechanical particle and also its momentum. So if we try to find them, we'll find they actually follow this particular equation where h cross is h by 2 pi. So now what this is actually saying that if we can measure one of these parameter precisely, that means one of them will become 0, either delta x or delta p. If one of them becomes 0, the other one you can see will become undefined because you have to divide a finite number with 0, which will become an undefined number, that means it is not possible. So that is why mathematically it is saying that you cannot have either delta x or delta p close to 0. That means you cannot have a system where you can measure the momentum or the position precisely. However, these are not the only two parameters that are connected with this particular scenario. So what it is found that there can be multiple different combinations of parameters, different combinations of parameters and parameters are such that they are actually nothing but a variable that you can measure. So you actually can find that there are different combinations of parameters or variables that is possible to that they are following this particular equation. So one variable is a, one variable is b and they are following this particular equation and they are multiple vectors. Such combination of variables which actually follow this particular law or this particular rule that is actually known as the complementary canonical conjugate variables. And what is the other important term? So generally this a or b will be such parameters that they actually can define all the four different dimension for that particular system you are actually working on. So whatever the quantum mechanical system you are looking into, these two parameters will be such a way that they will be defining all the four dimensions that is required to define a system, x, y, z, three dimension and the time dimension, this four dimension. So one example over here the delta is defined x, y, z. The momentum has the velocity where there is a time hidden there. Similarly, there could be energy and lifetime, one of such parameter. So don't always look into Heisenberg-Answery principle such way that it is only valid for delta x and delta p. It is one of the examples of that, but it can be valid for any particular combinations of complementary or canonical conjugate variables. So does it clarify your doubt or no? Yes, sir. Thank you. Any more questions from anyone? Okay, so if not, we will start in a little bit where we actually left in last class about the line resolution. So now as we have just discussed the Heisenberg-Answery principle and we know delta e, that means the uncertainty in the energy value. The energy value, that means the energy difference between the two states excited and ground state and delta t is the lifetime at which the lifetime of the electron or particular nucleus in this case remaining in the excited state. So the excited state lifetime. So if this is the two parameters that we are talking about that is also following this Heisenberg-Answery principle. Now over here, what we also found that whenever we are seeing a signal because an excited state system is coming back to ground state and it is giving an energy and if I want to plot this energy with respect to intensity versus energy, that is how the graph looks like. And we can find at the half maxima, that means there is a 50% of the maxima. What is the line width? So this is known as the line width at half maxima and this is defined as the term as we said the capital gamma. And we found this capital gamma is connected with the uncertainty in energy with this following equation. So higher will be the uncertainty in the energy. Obviously this energy uncertainty increase, you will see a much more broader peak. And with respect to that, obviously this line width will also increase and because we are measuring at 50% so that is why it follows this gamma by 2 equation. So now what we are going to do, we have two equations equation number one, equation number two. We are just including this in the place of delta e. So we are going to find delta 2 into delta sorry gamma by 2 into delta t is delta nickel h plus by 2. And further that gives me this equation. So now what is happening? We are trying to find that there will be certain amount of line width always present. This line width is important because over here at the end we are actually emitting a gamma ray from a source and trying to getting it absorbed with a sample. And over there the minute difference we have on the electronic environment which is affecting my nucleus that is I am trying to measure. So that is why what I want to have that this line width should be very sharp so that if there is a certain particular change over here I can actually see that. However, just imagine if your line is already very broad and if there is similar amount of changes happening over here. So now this will be much more tough to follow because it is already very wide and the wide scale you are trying to find a very small difference. So it will be always better if your line is pretty sharp and that is why this term energy resolution coming to the picture. That means can you find the difference you are getting due to the interaction between your sample and source and can you figure it out on the background of the original absorbance. That is the point we are having. And this energy resolution is given by half with a line width at half maxima and the energy the original energy. This particular ratio gives you the energy resolution. Now what we are going to do with this thing in our hand we are trying to find a rough number where it surely comes one. So again energy resolution is gamma by energy. Now over here we already know how to find gamma because we know gamma into delta t greater than h cross. So delta t defines what is my lifetime t is my lifetime. So whatever the lifetime I can find because lifetime is a very small number delta t will be very similar to that. So now if I take iron 57 as an example that has an excited state lifetime around 141 nanosecond. So that means that delta t would be very similar to 141 nanosecond at least. And that is the largest value we can consider not the smallest. The largest value it can have is the same as the lifetime itself. And if I consider that I will find my gamma will be h cross by delta t which will come around 5 into 10 to the power minus 9. By putting the value of h cross the Planck's constant value. So this is in the region of this gamma. So don't worry about the numbers I am just trying to give you an idea like at which particular range of numbers will be given. And over here now if I look into delta by e that also should say that energy gap that you are talking about for iron 57 for a transition between i equal to 3 by 2 to i equal to half that is generally coming around 14.4 kilo electron volt. There is a kilo over here so that means it is 14.4 into 10 to the power 3 electron volt. So with respect to that now I have gamma value now I have e value I can find what is the resolution. So I am going to put this numbers over there and you can find the number is going to come around 3 into 3 into 10 to the power minus 13. It is unitless because it is canceling each other out. So it is actually you can see a very small number the resolution 3.3 into 10 to the power minus 13. So you can imagine how tough it will be to gain an idea how much difference you are going to see. Because that is the resolution value pretty low. So how this number matters. So to understand that let me compare between optical spectroscopy that we are doing all the time versus Mosbyr spectroscopy. And let compare how the energy differences and resolutions are going to come. So now if I look into an optical spectroscopy in this side an optical spectroscopy the gamma value is generally comes around the region of 10 to the power minus 8 unit. Whereas you can find that it comes around 10 to the power 9 unit for Mosbyr spectroscopy for i. I am just taking the numbers without any unit side. And then the energy that we are talking about it is around 0.121 electron volt that is optical spectroscopy energy. And this is having an energy around 10 to the power 3 or 10 to the power 4 electron volt. So all together you can see now if I take gamma this is going to be in the region of 10 to the power minus 7 to 10 to the power minus 9. Whereas this one is going to come around 10 to the power minus 13. So you can see whatever the optical spectra I am looking into this one is 10 to the power 6 to 10 to the power 4 times much weaker. That means that will be that much tough to get the resolution out. And now imagine you can measure say 1 nanometer difference in an optical spectra if you are seeing the difference. And over there this is saying the gamma ray energy resolution is at least 10 to the power 4 to 10 to the power 6 times lower. So you have to understand then you have to have a system that can even record 10 to the power minus 4 nanometer difference in a wavelength. Only then you can go to see the energy difference if it is happening or not. Or just imagine you have a ruler in your home with you where you can easily measure 1 millimeter. And then say that is the optical spectroscopy in the region of 10 to the power minus 7 to the minus 9. And in most spectroscopy it is 10 to the power minus 4 times weaker. So that means in your naked eye you have to see 100 micrometer or 1 micrometer system which will be very tricky to see. So that is the way we are talking about the energy resolution and that is why it is very much important to understand where do we stand with respect to energy resolution. So this is the reason that most spectroscopy we cannot do for every other element. Why? Because over here what it is saying that energy resolution you have should be such a number that this energy resolution is a very high. Higher number is better because 10 to the power minus 13 is actually lower number compared to 10 to the power minus 7 or 10 to the power minus 9. Because this is a minus sign is over there. So that is why what you want to have you want to have a higher value for this. And only then you can see a spectra very easily and some difference very easily. And to make sure that this is actually a high number what you need to have that there is two things. So this delta the gamma value the line width value should be high and this E value should be low. It is a normal denominator and numerator system. So you want to have your numerator should be high denominator should be low so that overall your number is actually high number. So how I can make sure that may gamma value is actually a high value. So you already know that we have the gamma into delta t is greater than equal to h cross. So that means gamma is equal to h cross by delta t. So in this way it is saying that if your delta t value is a very low value only then your gamma value will be high because h cross is a constant. So it is saying your excited state life time should be very low and only then your delta 2 will be very low gamma will be very high. So this is condition number one. Second is this E value should be low. That means that the energy change you are seeing between the excited state and ground state of the nuclear states. The nuclear state energy state that you are seeing and where it is leading the gamma energy. If this gamma energy has to be low that means they should be very close by. So that means you are looking for a low translation energy. So these are the two parameters you want. Your excited state life time should be very low and your energy of transition over here that should be also very low. If these two parameters are present only then you can have a practical chance to even look into a system where you have enough resolution to see a spectrum. And that is why in reality you don't have a lot of the systems a lot of elements that can actually be used for MOSBAR spectroscopy. Only a few of them can do that. So over here I am showing you an example of the elements that is there you can see the MOSBAR spectroscopy. So you can say it is actually a MOSBAR spectroscopy p-o-dictator. And over here the darker ones are actually inactive and white ones are actually active for MOSBAR. And over here as it is shown the number over here it is showing like how many transitions you see. That means it is 3 by 2 to half there is one transition or there is another 5 by 2 to 3 by 2 another transition. So that is showing how many transitions you have. And this one is showing how many nuclei can or isotope can actually show this particular MOSBAR active transitions. So because it is a nuclear phenomena so if you change the isotope you are changing the composition of the nucleus. So the properties will not be same for MOSBAR spectroscopy. So that is why in MOSBAR spectroscopy you have to define that particular isotope. So that is why I am saying it is iron 57 not iron 56 or just simple iron. And over there you can see what are the nuclei are actually practically possible to have it. And even among them not all of them are very easy to use. I am coming into that pipe. So there are the most common ones that is actually used worldwide. One of them is this iron 57, 119 tin, 121 antimony, 129 iodine, 197 gold. So these are one of the few ones which actually have quite a wide use for MOSBAR spectroscopy in the reality. And among them iron 57 is the most common one that has been used for the MOSBAR spectroscopy. So it has been used so regularly that nowadays if you are doing a particular iron complex chemistry where you are talking about oxidation state change or spin state change people will invariably going to look for that do you have the MOSBAR spectroscopy to support your hypothesis. So MOSBAR spectroscopy and iron chemistry right now became almost inseparable at this moment because they are wide usage. So from now on that is why we will mostly keep our discussion based on iron. So iron 57 MOSBAR spectroscopy. So that we are going to discuss. So now we are going to dissect the overall spectroscopy for iron 57 and find out how it is actually going to happen. So first of all you need an iron 57 isotope. So this is the isotope and as you know iron 56 is the most common one naturally abundant. So this is actually not really abundant a lot in nature. So the natural abundance for this particular isotope is pretty low actually. This is around only 2.14%. So if you find 100 of the iron atoms naturally only two of them will be iron 57. So that means if you want to do a chemistry with iron 57 you need to artificially enrich your sample. Otherwise you cannot even see the signals particularly. Because you need to have enough sample which actually can undergo MOSBAR spectroscopy. Because only the iron 57 work the iron 56 is actually a nuclei which will not participate in this MOSBAR spectroscopy. Now as we have discussed so far in the MOSBAR spectroscopy what happens that this figure will come again and again. You want to have an excited state already available which will come to ground state while it will leave the gamma ray. And this gamma ray will come and excite your sample. So this will be the sample, this will be the source. So this source and sample has to be identical. So both of them has to be iron 57 and only then it will work. So now the question is that if they are very same then what is the difference? There will be a minute change in this energy of the ground state and excited state depending on the electronic environment around the system. And that change we are going to capture over there. And that is the change we are seeing that should fall in the resolution range that we are actually talking about. Now first of all we need a source which actually going to leave a gamma ray. That means I have to have an iron 57. We start with a system where the nucleus is already in the excited state and how that is possible.