 I need a system which will be emitting gamma ray. That means I have to have the system start with an excited state. Now the question is unless I excited how it is going to happen. So that is how it is happening because now we are talking about nuclei. So that is why this iron 57 we are going to get from I sorry a cobalt 57. So that is called apparent nuclear and during this iron 57 this cobalt 57 is generally pretty unstable. So slowly it will produce the iron 57 through a nuclear reaction. So something is going to change in the nuclear. Now the question is before we go into the details how the changes are happening from iron 57 to cobalt 57 to iron 57. The question is where do we get cobalt 57? What is the source of that? Because it is actually a radioisotope. It is radioactive. That is why it is going to slowly degrade to iron 57. Now where can I found a radioactive 57 cobalt? Now that is actually prepared from easily available source of iron 56. Which is easily available and with that pupil bombard that iron 52nd sample 56 sample they bombard that with a deuterium. This is the deuterium I am writing one is the how many proton it has two is the overall mass number. So it has one proton one unit. So if we add them together then what it is going to produce is 57 cobalt. So iron is 26 cobalt is 27 and if you can balance you can see it is not totally balanced because all together 56 plus 2 should be 58 and you have only 57 and 26 plus 1 27 that is balanced but this 57 versus 50 it is not balanced. To balance it I need someone one mass unit but no charge and that is nothing but neutron. So that is the reaction is actually done and where how I can make a deuterium to go and bind inside the nucleus of the iron 56 that has been done by a system called cyclotron. So cyclotron is a system which can develop a very fast moving rapid projectile of this kind of small atoms or small subatomic particles like electrons propons neutrons it can produce how it is produced. So cyclotron let us look into Wikipedia or anything or in the internet what is cyclotron. So that is actually run by a huge radii of an electrical and magnetical feet. So it is kind of like a circular way which is spread around at least in kilometer wide and it is actually over there with producing enough high energy field by electrical or magnetic field we actually accelerate the electrons protons neutrons deuteriums even oxygen atoms small atoms at a very fast pace and once it is moving very fast then it can penetrate and at that time once we achieve a very fast rate for the deuterium by creating this electrical field we expose that deuterium to a 56 iron sample and at that point of time it actually clashes and this deuterium get inside the nucleus of iron 56 and over there this extra deuterium with their neutron and proton they actually go to their very basic form of quartz they again rearrange and over there they pop out this iron this cobalt 57 and an extra neutron would come out. So that is how the system has been developed. Now anyways nuclear chemistry is not our goal we are mostly interested what happens to this 57 cobalt. Now as we just said it is radioactive and any radioactive system we can write what is the half life and the half life for this is 270 days that means if you get one kilo of radioactive cobalt 57 after 270 days or almost nine months you are going to get only 500 drops. So it is radioactive but it is still a little bit slower on the side of the radioactivity decay so that you can plan your experiment and run your experiment with this iron 57 sample. So that is how this cobalt 57 is actually prepared with the respect of cyclotron and deuterium and this full reaction in shorthand it is written as following that is how it is written. Now coming back to this cobalt what happens to this cobalt and how it goes to iron so we start with 57 cobalt which has a t half of 270 days and the i value that means the nuclear state value for this particular nuclear spin value for this cobalt 57 is i equal to 7 by 2. So cobalt has a ground state nuclear spin of i equal to 7 by 2 and that is why if you ever actually see any EPR signal of cobalt you can see there are eight small lines because the hyper fence splitting with the nucleus with the spin of i equal to 7 by 2. But what happens once this cobalt because it is radioactive it wants to decay it wants to relax and it relaxes by something called electron capture it captures an electron from the environment. So another question is who is giving the electron that so someone who is giving electron which is actually getting ionized for an example any gas molecule that can get ionized by leaving one electron to this cobalt system and once it get the electron the electron is not going to the outer sphere so don't think it is a redox reaction because the electron is not going to the electron shell the electron is actually going inside the nuclear because the nuclear it is actually unstable not the outside electrons for cobalt 57. So cobalt 57 absorb one electron so it is already iron 57 cobalt 57 so if it captures one electron what is going to happen it is going to create 57 iron because the mass is actually pretty low for electron so you can ignore that but the charge is minus one so you can imagine it is actually coming over there such a way that the electron is balancing the overall proton and neutron balance and now it has one less proton compared to cobalt so that is why it becomes iron 56 iron 57 and 26 is the atomic number so now once it forms that by electron capture the i value it forms is 5 by 2 because it is coming from 7 by 2 so obviously it cannot go to the ground state at first once it captures the electron it goes to a very highly excited i equal to 5 by 2 state and during that it actually changed from cobalt to iron so now if i draw the energy graph over there so this is say the state of i equal to 5 by 2 now i equal to 5 by 2 very unstable so it wants to come back to some lower energy state the first it relaxes to i equal to 3 by 2 state this is also iron 57 26 now there is only change happening inside the nucleus but it is not interacting with the environment anymore it is not capturing or not releasing any subatomic particles it is not now only handles with the energy and over here it comes to i equal to 3 by 2 now what is the energy gap between them so you can measure the absolute energy the energy is 136.32 kilo electron volt for i equal to 5 by 2 and the same for i equal to 3 by 2 is 14.4 kilo electron volt so now you can imagine when it relaxes from i equal to 5 by 2 to i equal to 3 by 2 that is releasing a gamma ray equivalent to the energy gap between this and that energy gap is close to 122 kilo electron volt a very high gamma ray comes out first then this i equal to 3 by 2 is still not stable so it comes down further to i equal to half state which still iron 57 system and this energy is the ground state so it is the zero that we define so the difference of this gap is 14.4 kilo electron volt so there is another gamma ray will come out so that is why there are two gamma rays comes out almost 10 times more energy for the first one and the 14.4 kilo electron volt at the bottom and we generally look into this particular system so when we talk about cobalt 57 has a half day half life of 270 days in the 270 days if you keep it perfectly it first captures an electron and then slowly move to i equal to 5 by 2 iron then 3 by 2 and then iron half so this full process take 270 days and slowly and slowly every time it is flickering some gamma ray coming out so you can actually detect two different gamma rays one is 14.4 kilo electron volt other is 122 kilo electron so that is how we can get a system where we are having an iron already staying in this excited state and slowly coming to ground state so this particular picture we actually generally draw when we talk about that I have a i equal to 3 by 2 state for iron which is coming to i equal to half state leaving the gamma ray and this is actually coming out and this gamma ray is going to excite another i equal to half ground state system and going to the excited state so this is going to happen in the sample and this is going to happen in the source so when you see the source we generally show that it is coming from i equal to 3 by 2 question optimistically comes how it is already in the excited state because we are showing only this part of this full energy picture of this system because it is originally coming from iron scoil cobalt 57 and slowly it comes to this 3 by 2 iron 57 and coming back to the ground state of i equal to half so only this particular picture over here as I showed is actually shown so this is the picture over here so when you talk about this source and sample this is what is actually happened okay now the question is okay so now I know how the source is emitting the gamma ray and now what is actually happening during the interaction between the source and sample and how I can detect even a minute difference in the electron density between the sample and the source and here comes the second part of our discussion how the experiment is actually done so again we are going to come over here so first over there what you are actually having we are having a source which you already discussed earlier we want to put it in a matrix otherwise if you leave them in a free state it is going to show a huge recoil energy and you have very less chance of a very nice absorption so we already put that in a solid matrix so that we have a better chance of the system to absorb and the similar system we also take for the sample and over here we have something called detector so what is the idea over here the gamma ray will come out which is coming from i equal to 3 by 2 equal to half over here it is going to the opposite way i equal to half ground state is going to i 3 by 2 and it is going to absorb if it absorbs my detector will see nothing if it doesn't absorb my this system will go there the gamma ray and I will detect some gamma ray in my detector that is the very simple system we have now what has been found because the sample and the source although they are talking about the same nuclei iron 57 the electronic environment around the nuclei is not same and we haven't still discussed yet like why the electron environment is affecting the nuclei but for now just assume that it is true it is actually going to different and that is the effect we are going to see now what is going to happen over there give to this now say over here I am drawing a graph this is intensity and this is energy so over here this is the band of the source that is how the gamma ray distribution look like for the source and so let me just draw that one more time in a different color so say this is the source and say this is the sample so I have drawn it such a way that they are now not really on top of each other now at this moment if I look into the detector and the detector generally give in the same of transmittance very similar to FTI data that means it starts from hundred it can go below to zero hundred percent transmitters means whatever the energy gamma ray I am giving I am going to detect hundred percent zero means everything is getting absorbed so now what happens if this is the scenario that my source is over there it is not at all matching to my sample what I am going to see in the detector in the detector I am going to see a line like this that it is absorbed this is respect to time so the transmitters will be hundred percent nothing is detected so it may happen because they are originally this is actually not that broad as we just discussed because actually they are pretty much sharp lines so over there they have a very least amount of chance that you actually going to be that lucky that the source and sample they are exactly top on each other and you expect to see a band a band means some transmittance is going to fall out that means something getting absorbed it is very less chance so over there how I can ensure that no matter what whatever the source energy and whatever the sample energy I can still see some data so remember earlier we discussed that during the gamma ray energy emission or gamma ray energy absorption in the free system the system has a recoil energy so the system the source actually move on the backwards the sample actually moves on the forward side and that is actually absorbing some energy and due to that of the loss of energy or absorbance of the energy you might not have a total fit in the absorbance over here people thought about that can I use this phenomena for my advantage so say my systems are not actually on top of each other over here however I can introduce some movement so that they can actually match and what I am talking about over here it is known as something called Doppler effect all of you know what is a Doppler effect that if you move the source with respect to a stationary system where I actually measuring the system or say sample if the source moves towards you you will feel the energy is high if you move the source from the sample you should experience a lower energy system and that is exactly what is used in this original MOSBIRS spectroscopy to avoid this factor that the source and sample may not match and how it is done it is the following way so what people have done they take that source they take their sample and they have their detector so this is the detector this is the source and this is the sample what they did they actually put all of them on a platform and over there they fixed the detector they fixed the sample however they keep the source on a wheel that you can move forward or backward and by that you can move it forward to match the energy or move backward to match the energy and by that they actually created a Doppler effect to ensure that there is a good matching between the source and sample image now how it looks like so I am going to take a few examples how it is happening and over there for this I am going to move it right so to introduce that Doppler effect and that is why I am going to show this one intensity but this one the velocity the velocity is how much it is moving towards or against the sample if it is towards the sample I would say it is plus d if it is against the sample it is minus now first case let us say this is my absorbance energy diagram or from the sample how the sample absorbance and over there now say I am moving the sample such a way this is the source I am moving it at minus v1 that means I am moving backwards with a v1 velocity and then if I want to draw how that will look like with respect to transmittance and over here I am again putting the velocity so I am moving at minus d1 this is minus d1 point this is the energy how it looks like so do you expect any absorbance the answer is no it should be 100 percent transmittance so over there there is the first point come v1 so at v1 velocity if you are moving minus d1 you are not going to see any absorbance only transmittance then say I start moving it over here at minus v2 and over there you can see I have some common area that means I expect to see some absorbance that means the transmittance will lower down from whatever the value it has earlier from v1 so say it is having a value over here this is minus v2 then say I start moving it in a velocity of such a way that they actually perfectly match on top each other so say this is v3 I am not writing minus or plus because it might be minus it might be positive so I am just writing simple v3 and over there you expect that it is going to match perfectly and because we are taking an assumption that my source and sample has similar amount of system present I expect that all the gamma rate is coming it is going to absorb over there so I am not going to see any transmittance all the gamma rate will be absorbed the detector will not detect anything then say I start moving on the other hand side to v4 now I am moving towards it as I am moving towards it you can see now I have again losing the overall match now I have actually go down a little bit so somewhere around here this is the v4 and then I go further down to v5 and over there there is no matching between the red color sample and blue color source so that means I am not going to see anything so that is how the data will look like and over there I am showing you just the glimpses in reality I we actually measured at all different velocity points and at the end the data point actually looks like like exactly opposite of an absorbance band okay so this actual figure over here that is going to be reflected over there and very interestingly what is the y axis is transmittance and intensity and what is the x unit it is velocity but generally we discussed about that in the previous class that is spectroscopy we see it with respect to intensity versus energy so why this velocity is coming over here because this velocity over here is actually due to this Doppler effect giving an idea of the energy so that is why although it is velocity it is actually a function of the energy and what is the velocity we actually have to put it over there that depends on what is the particular isotope I am talking about what is the energy gap what is the line width and what is the energy resolution and for iron 57 specifically this velocity we are talking about it in the range of millimeter per second that is the energy I am talking about now you can think about what is millimeter per second means now gamma ray is actually moves very fast in the region of 10 to the power 10 to the power 10 centimeter per second right and over here I am talking about millimeter per second so what is the difference between them you can find it is in the region of 10 to the power 9 unit which is trying to give you an idea that we are actually trying to get as close as possible to the line resolution so that means this is if you have a very good system and if we can say measure point one millimeter per second or something we are going to be very close to the original line resolution but before going to into the details of it again going back and reiterate how the actual experiment is done first of all we actually start with cobalt 57 which is actually generated from a cyclotron that we have discussed iron 56 and cyclotron and this actually slowly capture electron produce iron 57 with i equal to 5 by 2 which comes down to iron i equal to 3 by 2 then comes to the ground state i equal to half and this is the region what we are more interested in which actually releases energy and the sample is going to absorb now the sample we actually put it in a solid matrix and fix it the source also could be solid matrix but with a system that we can make it mobile either towards the sample or against the sample how it is happening it is happening with respect to this particular system where we have some wheels over there on a platform the rest of those things are generally fixed and why it is why we are doing that is because we want to introduce the top level because we have found that the movement actually hampers the energy matching but if you are smart enough you can think about that means the effect is also on the other side with controlling the velocity perfectly you can control if the energy is matched on and that is how people have developed the setup for the MOSBUS spectroscopy they put the sample and the source such a way the source is mobile and by controlling the movement negative means against going far from the sample plus means moving towards the sample you can move it back and forth and try to see what it actually is matching and how to notice that it is leaving the gamma rays coming out over here and then on the other side we have a detector to catch it if it actually not matching at all you will see 100 percent of the gamma ray hitting 100 percent transmitters but if it is start matching you will see some drop in the transmitters and then if you do the experiment properly you will see all the points over here and see a nice transmitter graph to give you an idea what is the energy profile of the nuclear transition for the sample any questions up to here sir yes please go the sample is iron based sample right yes so the iron should be iron 56 iron 57 because if you have iron 56 this i equal to 3 by 2 to i equal to half transition we are talking about it is for iron 57 nuclear if you take a iron 56 nuclei the same transition doesn't even exist because this i equal to half value generally comes out if you have odd value of the overall nuclear mass that means if you have odd value of neutron plus proton if you say iron 56 it is i equal to 0 is the ground state so you don't going to see the same transition so it is kind of are you going to see nmr with the same system which can show proton nmr and if you try to measure deuterium over there you're not going to see that because they're two different nuclear okay sir if iron come like if i if we want to measure the MOSFET of an iron based compound so that the complex the metal that should be iron 57 then we can measure that yes so that is why we have to do this iron 57 and that is why I was talking about we have to do this artificial enrichment otherwise if you do that naturally only two atoms present in every hundred nuclei is going to absorb the rest of them are not even interacting with them so it is very similar you can think about you are trying to measure deuterium in the presence of proton only if the deuterium energy is matched only then you are going to see that and how much only one percent deuterium is there so that much of the deuterium you are going to see okay so that is how it is going to be any more question anyone yes so it is actually the electron actually goes inside the nucleus such a way that you can imagine in a very simple system that a new electron and proton absorbance and a matching it gives us a neutron very roughly if I say so one electron plus one proton gives one neutron so that means if you include one electron inside the nucleus it is interacting with a proton and that proton is becoming a neutron so what will be the overall effect the mass number is not going to change only the atomic number is going to be decreased by one and that is what is happening okay any more question sir for the 57 iron there is three excited means there is three state i is equal to 5 by 2 3 by 2 and half so from the transition 5 by 2 to 3 by 2 that is the energy is 122 kiloelectron volt and 4 by 2 to 1 by 2 it is 14.4 so how will means regulate that which gamma energy we have to take for this yeah so that you don't have to really pour it because for an example you are doing optical spectrophotometry and say you are showing light from 200 to 1500 nanometer but you know your sample actually absorb in the visible region 400 to 700 so the rest of them will it even absorb they will not absorb right so what will happen when we are doing this experiment this is happening in the region of 14.4 kiloelectron volt and now imagine in the same region you are passing to 122 kiloelectron volt gamma you are not even going to detect that because it is so high it is not going to affect the ground state of that because it is quantized right so unless and until you go to icon to 3 by 2 and then absorb 122 only then you are going to absorb so that is first thing and second thing they will generally use some filters over here so they actually allow only that particular region of gamma ray to heat on the sample so that is the other point does it answer yes