 Now, here is a problem. The problem is that the gamma ray is a huge amount of energy and due to this huge amount of energy it has something called recoil. What do I mean by recoil? Because this nucleus one, once it is leaving the gamma ray because it is huge energy this will go back. It is very similar what we learned in our I think class 6 or 7 physics book that if I fired a gun I will fill a recoil back. Because the velocity at which the gamma ray energy is emitting out with a similar momentum it will felt on the backward by the nucleus and because it is heavier it will move but at a lower velocity but it will move. And over here it will lose some energy because this recoil energy that it is losing on the backward side it is a momentum it has to create it has to lose some energy. So where the energy is actually going there then. So what is actually going to happen is the following. So over there i equal to 3 by 2 i equal to half this is leaving E gamma but this gamma ray energy will not be same as this difference which I can say E transition energy. Because E transition energy will not be equal to E gamma because some of the energy will be lost for this recoil energy. So I can write in the following way it will be E gamma minus E r. So some of the energy will be lost as recoil energy. On the other hand when this gamma ray is going further and interacting with my sample nipley eye this is a source nipley eye it is not going to absorb only but it will also move backwards. And this will be also E r I am writing it E r because they are the same nipley eye all the rest of the parameters are more or less same. So the recoil less this recoil energy will also be very similar. So it is kind of like you are absorbing a very heavy punch on yourself so you will be you have to move a little bit on the backwards. So this huge energy when it comes over here it is not only going to do the transition from E equal to half to E equal to 3 half but it also going to move it backwards. Previously it was going to the other side. So over there this E transition E transition will also be not equal to E gamma. Because this transition will happen such a way that this E gamma plus E r so some of the energy will be absorbed over there and that is what is going to happen. So over there sorry my bad I actually made a mistake over here. So this E gamma energy will be E transition plus E rotation and the same thing over here my god this E gamma will be E transition minus E r my god. So it is the gamma energy that it is coming out from the source or it is absorbed by the sample it will be equal to the transition plus E r the recoil energy have to give to the sample or minus E r that is lost during the energy transition in the source. So that is what is actually happening. So this recoil is going to happen you cannot stop it it is going to happen if you leave the sample like that and what is the effect of this on the overall energy transition. So what happens in the overall energy transition is the following. Now if I want to plot energy versus absorbance. So say this is the energy of transition which should be same for both the source and sample but what we are going to see is the following. This is for the source this will be the energy and this is the recoil energy that it is going to loss on the other hand this is the sample and this is the energy of recoil that is going to loss. So the energy of transition although remains same because of the recoil the sample moves this way the source moves this way and overall over here this is the overlap you are going to see and this overlap you can see it is actually very poor very poor resonance unless it is going to show the resonance you cannot have a spectroscopic signal and over here due to this you are going to see a very very poor resonance and that is why in the beginning if you want to do this experiment and try to find if I use the same source and try to produce a gamma ray and absorb it by a sample you most of the time you do not see any signal at all because this overlap is very poor because for a perfect absorbance you want to have something like that where you have a very huge area of overlap and that is not happening over here and that is why it was termed as a very impractical spectroscopic technique that it is not going to work because it is not at all going to work because of the two problems first all together this interaction between the nucleus and electron the hyperfine interaction is pretty weak secondly when we try to do that because of this recoil of energy you are losing much energy in creating the momentum and that is why it is ending up with a very poor overlap so that is not going to work and this is the situation in the 1950s and at that time one particular person was working on this particular project his name was Rudolf Mosbauer and he was a PhD student at that time and he was trying to find out how to improve this experiment and he was trying different things and the one thing he tried he actually during this experiment lowered the temperature of the system the whole experiment he did at a very low temperature and interestingly he found that the signal started improving previously it was almost no signal but at low temperature he started to see some signals and at that time he go back to the blackboard and try to find out exactly what possibly is happening and then he understand this recoil was actually happening and as you decrease the temperature your overall movement of the molecule go it is slowed down because of the temperature and that is getting affected so if I can remove the recoil energy I can probably get a very good spectrum with that thing in his mind what he did is the following he took a nucleus the source and here he took the sample but instead of leaving them as it is he actually embedded it in a solid matrix like a crystal so he put that in a crystal solid matrix so what is going to happen if I put that in a solid matrix now if I want to move this system when it is leaving the gamma ray and it is absorbing over here you have to move the recoil energy over here now you don't need to move only the nucleus but this full system all together because they're all connected because it is a part of a crystal lattice you have to move this whole system again and now because it is a not only one nuclear but it is a part of a huge system its effective mass is huge and if you have a huge effective mass because you're actually losing the energy as a momentum because of this huge mass your momentum you are creating the actual velocity you are creating is not going to be lost a lot because it is very tough to move a heavy thing completely a smaller thing like a simple nuclear no string attached and at that time when they put that around that with this particular matrices what they found that this recoil energy actually decreases and then he did this particular mathematics that I'm showing you over there he found the recoil energy has two particular component one is the translation component that we generally think that it would move backward or forward either this way or that way this is the translation motion the whole thing is going to move in one particular axis and then there's another function called vibration because the movement doesn't only walker occur in a directional movement in one particular translation motion it can also vibrate and that is also you can create a particular loss in the momentum and this vibrational motion over here because it's a part of the lattice it is nothing but a lattice vibration and this lattice vibration can be written in the following term i n h cross omega i where n is equal to zero one two three different numbers and i define how many different spots you have on the lattice and omega is the frequency of the vibration so that means you if you create a vibration in a molecule in a lattice it is also quantized and this particular system is also known as phonon analogous to photon it is known as phonon and this is the phonon which actually believed to be the main reasons of seeing temperature in a particular system when you say this system is hard that means their internal vibration in a lattice is very high if i say something is cold that means it is not vibrating all together at all and when we say there is a heat change happening for an example you put your hand on a ice you feel oh it's very cold that means it is actually start vibrating inside the ice by taking some vibration from your hand so this heat transfer can be thought in quantum mechanically with respect to this phonon transfer and if you can do this experiment at a very low temperature obviously this frequency of vibration will be happening at lower n values and e translation will be close to zero if you are doing this system in a lattice and if a n equal to zero you're all together e r will be equal to zero and you will go to a condition where you are doing this full phenomena which is can be turned as a recoil less and this recoil less event if it can work you would see a very good absorbance because now once you put this systems in the lattice both this source and this sample because of this recoil less event what will happen now the energy gap will be closed down so now this is the energy this is the energy and this is the transition energy now you can see the recoil less energy is actually pretty low a little bit of will obviously happen because some of the transition will happen through the phonon so it's very tricky to do a totally recoil less system unless you go to zero Kelvin temperature you will have some vibrational motion and it will get some energy but still with compared to the recoil added phenomena now you have a very good chance of overlap and with respect to that you have a much better chance to see a signal when you are doing this transition with respect to the nuclear state and that is exactly is the work as we said of this person Rudolf Mosfauer he did this work in 1957 he submitted his thesis and in 1961 he was awarded the Nobel Prize and along with that he got a spectroscopy named after him so if you do your study in a totally new field where there is not much success by others don't take it as a bad thing or a disadvantageous thing because that means you have more opportunity to do something groundbreaking and if you can do it perfectly you can also have submitted your thesis and within a decade you can get your Nobel Prize and a spectroscopy named after you so that is what was happening in the Mosberg spectroscopy now so if this is the scenario what is happening in Mosberg spectroscopy that you can create an energy transition between two different nuclei and you can see a transition from the source and sample then I should be able to do that for any nuclei possible but in reality it is not only a few handful of the elements can show very good signal in the Mosberg spectroscopy why it is so so that is what we are going to learn in this particular part so over here you see that when I am talking about the line of an absorbance I am drawing it like this why I am drawing it like this because it is a transition between i equal to half to i equal to three half to take an example so if this is the transition either way when I am drawing it it should be line but instead of that I am drawing a structure like that which is actually a you can say is a Lorentzian mode and over here at the half maxima of the peak so if this is the maximum this is the half maxima what is the width of this system which is known as the line width which is given by this term gamma this is actually a capital gamma and this is actually a quite a high number why so so why cannot I see only a single line why I also have to see a this kind of a band and that answer that in this transition whether it is happening between i have to three half very well defined state still I am seeing a broad Lorentzian band the question is why now that is happening due to this called the line width happening due to one particular phenomena that all of us probably heard and learn known as Heisenberg's uncertainty principle now most of us have learned this Heisenberg uncertainty principle so Richard, can you tell me what is Heisenberg uncertainty principle yes sir yes sir it is impossible to determine the position and momentum of a two electrons simultaneously yes so that is very fine so now if I want to go to Heisenberg uncertainty principle it is what you said that if a molecule has a position at a particular three dimensional axis say x and if it is momentum is p the uncertainty that means the difference in the absolute value if it is given as delta x and the absolute value in the momentum if it is delta p they can be written in this formula so over here what I am saying the uncertainty in position delta x and uncertainty in momentum delta p we cannot define each of them specifically because if I can say that I am measuring something perfectly there is no error at all so one of them has to be zero now put one of them zero you can see the other will become undefined because you have to subtract this finite term with zero so that is why perfect measurement or a measurement without any uncertainty is impossible for each of them there will be some uncertainty present but is this the only two parameters that can be defined with Heisenberg uncertainty principle or some other terms are also connected in the similar equation anyone want to comment on that sir angle of frequency and time anything else so there energy yes yes so there are multiple of them actually possible and there are known as complementary variable or which is known as canonically conjugate variables don't worry with this particular terms I am going to explain it what does it mean so what does it mean that if you have two particular parameters that can define all the different dimensions all the different dimensions possible to explain that particular system those two parameters will be known as complemented or variable so now what is the different dimensions we can see in our real world I mean how many dimensions we can see there are four dimensions right three dimensions x y z and one is the time dimension so if I have two different parameters by which I can define all this four dimension that will be complementary or conjugate variables for an example position it gives you the three different position dimensions x y z and the momentum what is the unit of momentum over there you can see there is mass into velocity in the velocity you have the time dimension so with respect to that you can say those are actually conjugate variables a similar conjugate variables in the same way is energy and lifetime and they also follow this particular equation so one particular system goes to an excited state and it tried to come back to the ground state so it also depends how long it wants to stay in the excited state and depending on that the actual energy gap between them and their uncertainty will be defined so if I say I am seeing a perfect line in an absorbent spectroscopy that means if I'm seeing a perfect line what is the difference in the delta e it is going to be zero because it's only one perfect line that means if delta e becomes zero what will be delta t infinity that means that electron should or the system should always remain in the excited state forever it shouldn't never come back so that is also impossible so that is why you never see a line graph you always see a broad graph like that because that actually defines that you have some finite delta t that means the lifetime whether it's a fast or slow it has a finite lifetime if it has a finite lifetime then there will be some uncertainty in its lifetime the delta t and due to this finite lifetime due to a particular finite uncertainty in the delta t your delta e cannot be zero if your delta e cannot be zero you should always have some non zero delta e and if you have a non zero delta e that means you're going to see a broad picture a broad graph now this line width which is written as the line width at the half maxima which is given as the capital gamma it is connected in the following way with respect to the uncertainty in the energy delta e is equal to gamma by two if that is one equation I have say it is equation number two this is equation number one put this all together what I am going to see is delta e into delta t is greater than equal to h cross by two in presence of delta e I am going to put gamma by two into delta t with an h cross by two which can be written as tau into delta t is greater than equal to h cross so that means I am going to see some line width all the time no matter what because this delta t is a non zero there will be a finite lifetime now let's do it a quick mathematics taking an example of a 57 iron isotope which exchange between i equal to 3 half to i equal to half and for this transition in excited state it has a lifetime of 141 nanoseconds so that is the 141 nanosecond it can stay there in the excited state so if it is there then you can easily find out what will be the gamma because we found gamma into delta t is greater than equal to h cross so I want to find what will be the line width so by putting taking it equal to just removing the greater than equal term h cross by delta t and put all the values over there you'll find this value comes out around 5 into 10 to the power minus 9 electron volt so that will be the line width present all the time now a line resolution depends on this line width divided by the energy gap the actual energy difference between the states I am looking into and that if you look into you'll find it will be 5 into 10 to the power minus 9 electron volt into 14.4 into 10 to the power 3 electron volt that is I am writing because the energy gap for this i equal to 3 by 2 to i equal to half transition is actually 14.4 kilo electron volt so if you put all these things together you find the value will be coming 3 into 3 into 10 to the power minus 13 electron volt so this will be the resolution in that kilo electron volt energy gap we are trying to find 10 to the power minus 13 electron volt energy so that is why it is very tricky to find out and that is why not all the isotopes can show very good signal so what are the things makes a nuclei mosbaric that we will cover in the next class