 After talking about temperature dependence of diffusivity, let us move to the important aspect of this set of lectures on diffusion, which is the atomic models of diffusion. These models have to be understood because these will play an important role in understanding how species diffuses, what are the critical issues, what is the activation energy for diffusion and certain what we may call the short cut or short paths for diffusion which have a high diffusivity. So, if you look at atomic models of diffusion, two kinds of species which are diffusing are of importance, one are interstitial atoms and one are atoms sitting in the substitutional site. Now, of course, this diffusion we are talking about focusing in these set of lectures is diffusion in metals and diffusion in ionic solids and other kind of solids may have other aspects which are not covered in this basic set of lectures. Now, when you are talking about interstitial diffusion, usually we have to note that the solubility of the interstitial atoms and these interstitial atoms for instance could be carbon and steel, it could be hydrogen in palladium or such kind of elements which is sit in interstitial sites that is they do not occupy the lattice position. And because the solubility is small, it implies that most of the interstitial sites are vacant, only a fraction of the interstitial sites are occupied actually by atoms for instance carbon in B.C. Sian if you look at, we know that the carbon actually occupies as we have discussed before the smaller octahedral void, but the fraction of these voids occupied is small and therefore, most of the interstitial sites are vacant. Now, if therefore, if an interstitial species wants to jump, then most likely the neighboring site would be vacant and jump of the atomic species can take place. So, this jump will require some thing known as an activation barrier and we will try to find out what this activation barrier depends on in the coming slides, but also there are cases wherein the jump or the motion of the interstitial atom takes place by tunneling mechanisms. This happens specially for hydrogen which can diffuse very fast and for a correct description of the diffusivity of hydrogen, certain quantum under barrier effects may become very important. Though especially this is true at low temperatures for diffusion of hydrogen wherein over the barrier activated jump is restricted, but this kind of under barrier tunneling jumps we will not consider or tunneling diffusion will not consider in these set of lectures and we will only talk about jumps which involve above the barrier transport. So, we note that there is an atom sitting in the interstitial site and this is of course a schematic. This interstitial atom has many options to jump when it wants to jump and why does it want to jump because at any finite temperatures this atom is vibrating in all directions. So, let me show this schematically in the board that there is an atom sitting in an interstitial site and this atom because of its temperature is actually vibrating in all three directions, three dimensions and when this atom is vibrating about these positions there is a chance that it will actually jump. Now if you talk about a jump of an atom then once suppose these are it jumps to its neighboring interstitial position from the first position say 1 to 2 then further it may jump from 2 to an interstitial site 3, 3 to it may go to some other site 4 and therefore this may undergo what is called a random walk process and this species keeps on jumping. Now in the absence of a concentration gradient there is no net flow of this atom this atom essentially remains in its vicinity and keeps on jumping, but if you track the average position over a period of time it is as good as the atom remaining in same place. But suppose there is a concentration gradient which has been applied of course this gradient we have noted can be a concentration gradient can be an electric field or can be a driven by stress, but all these random jumps in the presence of a field and for now we are considering concentration gradient then these random jumps in the presence of a concentration gradient leads to net flow of matter. That means though there are jumps against a concentration gradient the overall motion of the atom is down the concentration gradient or as the relevant quantity may be the chemical potential gradient. So, we see that an atom sitting in its lattice position makes its vibrating because of temperature and because of these vibration once in a while it will actually many of these vibrations it does not actually jump, but sometimes it jumps over the barrier and gets into its neighboring position for instance it could be this site from site A it could go to site B, but then further it could again jump to site C and then to A and in the absence of a concentration gradient it will not move anywhere and there will be no net flow of this matter, but if there is a concentration gradient then the atoms will tend to move in spite because these random jumps in a particular direction which we called diffusion the long range diffusion. So, this is true for interstitial atom which is sitting in the interstitial void and we the most important thing we noted from this slide is that most of the interstitial voids are vacant because the solubility of most of these interstitial atoms in the given system is actually small for instance carbon has very low solubility as we have noted before in BCC ion. On the other hand when you are talking about diffusion of an substitutional atom and the concentration of substitutional atoms could be pretty large for instance suppose I am talking about an A B system it could so happen that there is 50 percent of A and 50 percent of B and suppose I am talking about A B C system there could be one third of each one of these atoms forming a solid solution and therefore, there could be I am not saying in every cases is possible, but there could be and that implies that the species which is diffusing is then present in a very high concentration. Now, an lattice atom that even an atom present in a lattice site can only jump to the neighboring site if the neighboring position is vacant suppose I am talking about an atom is marked in blue here and I am talking about the jump of this atom to a neighboring site here this jump can only take place if the neighboring site is vacant and this implies that the basic mechanism giving rise to diffusion of an atom in a lattice site which we call a substitutional atom is the vacancy mechanism in which case if the atom jumps from site A to site B for instance for this is site A and it jumps to site B this can equivalently be thought of as if the vacancy is jumping from site B to site A that means after the jump has taken place the site A will become vacant here. This implies that the vacancy concentration plays a very important role in the diffusion of species at lattice sites via the vacancy mechanism. Though we are focusing here on the vacancy mechanism and the interstitial mechanisms there are other mechanisms which are also been proposed for diffusion, but typically these other mechanisms involve higher activation energies and we will not discuss them in this basic elementary course. And therefore, two things come into picture when you are talking about a vacancy mechanism for diffusion number one is a fact that the substitutional atom has to find a vacancy neighboring it only then can it jump that means in this schematic picture the atoms which satisfy that criteria would be the atoms which is here, here, here or here for instance an atom sitting here cannot diffuse this atom cannot diffuse because there is no vacant site next to it. So, these atoms which I have marked by a tick can jump into this vacant site which is labeled B and for instance these labels could be like for instance C D E. So, this A C D E site atoms can jump into site B and diffusion can be caused and that automatically implies that what is we have to add on the estimate what is the probability of finding a neighbor a site which is vacant in a crystal. So, that means, I have to find the concentration of vacancies in a crystal at any given temperature. Of course, there could be a non equilibrium concentration of vacancies in a crystal this could happen for instance if we are talking about as we have seen before there are two kinds of vacancies one is structural vacancies which comes from of stoichiometry. One is thermal vacancies which come from the entropic stabilization of vacancy in a crystal and suppose I am talking about this thermal vacancy then normally there will be an equilibrium concentration of vacancies in a crystal, but additionally it is also possible that I take I take a crystal to higher temperature and then quench it to a lower temperature in which case I can obtain an excess concentration of vacancies at room temperature as compared to what would be present in terms of the equilibrium concentration. Point to be noted here is that vacancy clusters and vacancy defect complexes can alter the simple picture of diffusion involving vacancies. So, we have to note that there are complications which can arise, but for this point of time we just focus upon the simple basics which we shall elucidate further. So, therefore, when I am talking about the two mechanisms of diffusion one involves no vacancies which is for the interstitial diffusion and one which involves vacancies which is the substitutional diffusion. So, let us dwell a little further into this concept of interstitial diffusion in which case an atom sitting in site 1 which is shown here a red atom in the interstitial site jumps to a neighboring site say from 1 to a neighboring site which is labeled 2. The energy of the system in these two positions is identical and if you are talking about an energy coordinate diagram and this is now coordinates or what you might call the reaction coordinates then you would notice that the energy in these two positions is 1 and 2 is exactly identical, but between the two positions when actually this two atoms if you want to talk it in a more physical picture have to be pried apart. So, that this jump can take place that is an activated state that is a state in which the system goes to an higher energy and this higher energy state which has to be the barrier which has to be crossed is called the enthalpy of migration or enthalpy of motion for this interstitial atom. And we said that at any temperature which is a positive Kelvin temperature atoms are vibrating and this atomic vibration provides the energy which can overcome the barrier which is the delta H m the enthalpy of motion. Now, suppose an atom is vibrating with the frequency nu sitting in its interstitial position and this is the frequency of vibrations it is not necessary that every one of those atoms or the oscillation will lead to actually a successful jump only a small fraction of that which is labeled as nu prime can actually jump. And this nu prime is related to nu by the activation energy by Arrhenius kind of an equation that means nu prime is nu exponential minus delta H m by k t. That means, at higher temperature more of these jumps will be successful and if you have a lower barrier delta H m then more of these jumps jumps will be successful. So, these two are obvious. Now, on the other hand suppose I am talking about an substitutional atom not only I have to worry about the number of these successful jumps cost by this thermal vibration, but additionally I have to worry about the fact that is there an neighboring site which is vacant. So, therefore, the probability of the jump is not only proportional to the probability that the atom has sufficient energy to jump to the neighboring site the vibrations, but additionally if the there is a chance that the neighboring site is actually vacant. Therefore, when I am talking about the number of successful frequency jumps as compared to the total frequency then I would note that there is a term which is based on delta H m by k t. And this delta H m by k t term is very very similar to that what we encountered for the case of the interstitial diffusion. But in this case additional to this term which is coming from the fact that the atoms are vibrating and they want to jump. We have to note that the neighboring site is only vacant is the probability is given by exponential minus delta H m by k t which we have encountered before. In other words if there are total number of atomic sites is n capital N the number of sites which are vacant is small n then this is given by exponential minus delta H f by k t where delta H f is the enthalpy of formation of an vacancy. So, higher the enthalpy of formation of a vacancy the lower with the vacancy concentration lower will be the chance of a successful jump and that implies that lower will be the diffusion rate. Therefore, now when I am talking about interstitial diffusion I have two terms which are going to play an important role the one which is based on the vacancy concentration and the one which is based on the usual enthalpy of migration. Therefore, nu prime can be written as nu exponential minus delta H f minus delta H m that is minus of delta H f plus delta H m by k t. And this quantity using a small derivation which I am not showing here can be related to the diffusivity. In other the diffusivity of a species is a substitutional species is directly proportional to the exponential of the delta H f plus delta H m by k t. Therefore, now since if you look at the case of the interstitial diffusion the term opposing the diffusion is delta H m, but in this case there is a additional term opposing your diffusion which is delta H f. In this case the delta is the jump distance. Now, this additional term in the exponential plays an important role in determining the diffusivity and finally, of course, the diffusion rate at a given temperature. Now, suppose I am comparing for instance the diffusion diffusivity of carbon in FCC ion at say 1000 degree Celsius vis-a-vis nickel in FCC ion at 1000 degree Celsius. Carbon sits in the interstitial side, nickel is a substitutional alloying element and sits in the lattice side. Now, I am comparing the diffusivity. In the first case the diffusivity d is proportional to exponential minus delta H m by k t. In the second case the diffusivity is proportional to minus delta H f minus delta H m by k t. And now therefore, if I compute using known values from literature I find that the diffusivity for carbon in ion at 1000 degree Celsius comes out to be of the order of about 10 power minus 11 meter square per second. On the other hand if I compare it with the nickel diffusivity using data from literature then I notice that the diffusivity comes to the order of 10 power minus 16 meter square per second. In other word carbon would diffuse a few orders of magnitude faster since the diffusivity is higher it will diffuse much faster compared to nickel at a given same temperature given the medium is same that means its ion at 1000 degree Celsius and both ions are at FCC form. Therefore, the medium is same only the diffusing species either in the first case happens to be carbon sitting in the interstitial side. In the second case it is nickel sitting in the substitutional side I can see that the diffusivity changes orders of magnitude. This implies that the diffusion rate also for ion nickel in ion will be much smaller. This is why when you are trying to homogenize an alloy for instance suppose you made a casting and the alloy turns out to be inhomogeneous you want to homogenize it the homogenizing hold has to be much longer time as compared to for instance suppose you are doing a carburization experiment in which case carbon is diffusing from the surface. So, we will actually do a computation to find out how much typically it takes time it takes for instance to carbon to be carburized on the surface in one of the coming slides. So, to summarize this slide the diffusivity of an interstitial species is typically much higher and therefore, if you want to diffuse an interstitial species then you have to carry out the diffusion anneal for a smaller time. So, far what we talked about was what is called lattice diffusion that means the species is diffusing through the lattice it is non diffusing through any of the defects in the crystal. And typically if you know that if there are defects present in the crystal that means we are not talking about a perfect single crystal, but we are talking and even in the perfect single crystal we do tolerate vacancies we are talking about other presence of other defects and these defects of course, one of them is inevitable which is a surface the second could be a grain boundary like in a poly crystalline material or there could be dislocations which leads to something known as the pipe diffusion. So, if you experimentally observed the activation energies then you notice that the surface as the lowest activation energy for diffusion the grain boundary as slightly higher energy the pipe as even slightly higher energy, but the lattice diffusion cost the maximum activation energy. That means suppose I have some short circuit path for diffusion like the grain boundary the dislocation core and typically dislocation core of edge dislocations or the surface there is a free surface available then these short circuit parts will play a very important role in diffusion. For instance suppose I had a precipitate two precipitate let me consider a scenario and this actually has been experiment has been done I had a larger precipitate and a smaller precipitate. Now, we know that this if you give hold the system at a higher temperature then because of the Gibbs Thompson effect followed by diffusion this smaller precipitate has to shrink and this larger precipitate has to grow which is called the precipitate coarsening effect. For this the mass transport has to take place across from here to here. In the absence of any defects this mass transport would take place for instance by the lattice, but suppose I am able to consider that for some reason this is a plasticly worked specimen and there is actually a dislocation which connects the two precipitates. Then and this is an actual experiment which has been carried out then what you would find that this is short circuit path connecting the two precipitates and mass transport can actually take place from this precipitate to the other leading to shrinking of this precipitates and may be growing of this precipitate. So, if you are talking about a coarsening experiment then you can notice that this kind of a short circuit path actually gives you an enhanced diffusivity, but a point to be noted that it is not merely the diffusivity or the activation energy giving rise to a lower diffusivity which matters, but diffusivity for a given path along with the available cross section will determine the diffusion rate of that path. That means even though I may have a higher diffusivity for for instance dislocation core or a green boundary as compared to the lattice diffusion, but the problem is that the amount of the dislocation density the cross section area available through the dislocation is much smaller compared to the lattice. And therefore, the overall diffusivity may be dominated by the lattice at high temperatures, but suppose I go down to low temperatures wherein the activation energy for diffusion through a lattice becomes prohibitive then in these cases the other other what might call the short circuit path for diffusion become more prominent and we will take up an example in this slide. Suppose I am comparing the diffusivity of cell diffusion of silver in a single crystal versus poly crystal. So, now this is silver diffusing into silver and I have two specimens one is a single crystal another is a poly crystal. And now I plot log of diffusivity versus 1 by temperature and the reason for doing so is because we know that the diffusivity depends exponentially on 1 by T. So, a log of 1 by T versus diffusivity log d plot should give a straight line. Now, suppose I see do the experiment on a straight line I find that this green curve is the curve corresponding to the single crystal diffusivity log of the single crystal diffusivity. And the axis is such that since 1 by T is increasing on the right hand side the axis of increasing temperature is towards the left. So, these are high temperatures here and these are low temperatures on the right hand side. So, you notice that there is a straight line especially at high temperatures, but if you come to low temperatures you notice that you will actually see that there is a branch of diffusivity comes out like this. That means the slope of the curve changes and this slope here in the high temperature regime corresponds to diffusion through the lattice while this slope on the right hand side corresponds to green boundary diffusion. The reason that green boundary diffusion plays a prominent role at low temperatures is because the lattice diffusivity Q for lattice is say for about say 200 kilo joule per mole and that of the green boundary is approximately half of that. That means at low temperatures there is not sufficient activation energy available for silver atoms to jump from one side to the other. And therefore, it will try to look out for a short circuit path or a low activation energy path which is now the green boundary path and that starts to dominate at low temperatures and therefore, the slope of the log diffusivity versus one by temperature changes. So, this is an important experiment which shows that at high temperatures even though the for instance the green boundary assuming that the green boundary area available cross sectional area available is the same for diffusivity, but that does not dominate the diffusion it is a lattice diffusion which dominates in spite of the higher activation energy. But when at low temperatures the system does not have sufficient activation energy to drive diffusion to the lattice then these other mechanisms take over and they can play an important role in for instance as an example we considered coarsening of a precipitate or as we can see in a poly crystal the green boundary diffusion driving the motion of atoms. So, we have seen that not only that we have to consider normal diffusion in a crystal, but we also have to worry about additional paths for diffusion which can play a very important role in the diffusion of a species. Next we consider an important application of the fixed second law which is the carburizing of steel. Now, in carburizing of steel we start with a specimen which has an initial carbon concentration for instance C 0 typically that means we talk about a small carbon concentration and as we shall solve in the next example this carbon concentration could about be about 0.2 percent. But then we want to introduce more carbon on the surface and the reason for doing so is that we would like an internal toughness in the material along with the surface hardness and the whole process takes place like this. I take a material which is a 0.2 percent carbon I expose it to a carburizing atmosphere which could be gaseous or solid and after carburizing for a certain time I would note that the surface carbon concentration would be increased and the next example considered for instance the surface carbon concentration could go up to for instance 1 percent. So, we increase the surface carbon concentration then we take the material out of that bath and slowly cool it and later on we take this whole material heat it to high temperature and quench it. In the process of quenching the surface the material becomes martensite since the hardness of the martensite depends on the amount of carbon present and now there is a surface carbon concentration which is higher compared to the bulk therefore, I would notice that the surface hardness would increase. Now, there is a subtle point involved in this understanding for instance I would do this kind of a process with the component like a gear wheel. So, I want a gear wheel which is tough on the interior and on the surface it is hard. Now, suppose I had taken a high carbon steel for instance like I already took a 1.1 percent carbon steel and did this entire process of hardening then though I would get a very hard material uniformly hard such a material would tend to be brittle. But a subtle point which is often not emphasized in this context is the fact that suppose I have a gear component gear wheel here and it has got the gear tooth. So, I am drawing a small section of that and so forth. So, what I do is that I build up a high surface concentration of carbon here and this material after quenching becomes hard on the surface. The interiors hardness is also increased, but not as much as that on the surface because the hardness of the martensite depends on the carbon concentration. Now, as I pointed out I want a tough interior I want the material inside to be tough. Now, if you take even usually so called a brittle material and we know the toughness is a measure of the energy absorbed either in a tensile test or in impact test before fracture. So, it is an energy absorbed which means it is an area below the stress strain diagram. Now, even if you are talking about a brittle material a brittle material typically will at least give you some small elongation for instance in tension of say 1 percent. And if you notice that the kind of dimensional tolerances a gear wheel needs to have is extremely precise because this would be meshing with another gear component somewhere else. And therefore, even this 1 percent activity is way beyond what you would normally tolerate in service. Therefore, why is that I am talking about toughness in the interior while this component is never going to face that kind of a deformation. The reason is that this toughness helps in what you might call damage tolerance or fracture toughness suppose in the interior somewhere here because of manufacturing processes there is a crack. So, there is a crack and as you know a crack is nothing but a stress amplifier. So, suppose I am having a far field stress which is like this. So, this is my crack here in the material and I am having a far field stress which is been employed somewhere of this value close to the crack tip the stress say for instance I am talking about a crack opening mode stress sigma x x and this crack opening mode stress will have a higher value. And therefore, even though the far field stresses are small the stress presence say for instance far away from the flaw close to the flaw which is I am located here this is my crack here these stresses get amplified here at the crack tip. This implies that if the material is brittle and the crack tip is sharp then the material tends to fail. But, suppose this material is ductile then I will blow up my crack tip here. So, originally the crack tip was for instance sharp then in the presence of this ductile material the crack tip will get blunted. And if you plot the stress for a sharp crack it goes like this that means it tends to blow up close to the crack tip. But, suppose I am talking about a ductile material whose crack tip has been blunted in that case you would notice that the stress field goes through maxima and then comes down. That means that you do not have that kind of a singularity at the crack tip. And for this kind of a blunting you may require large percentage ductility in the material. Though I am talking about a very low gross plastic deformation at the large scale that means large lot of ductility is not required at the overall specimen level because the tolerance with which at gear component would mesh with the other gear component would be very high. And therefore, that kind of a plastic deformation is not tolerable and is not required. But, at the crack tip I need this kind of a what you might call a plasticity. So, that the crack tip can be blunted and the material can be tough. And for this high toughness and high plastic deformation and high ductility I need a material in the interior which has a low carbon concentration and which can have this kind of a ductility. On the other hand on the surface which is meshing with another component where there is a chance of high aberration there is a chance of you know touching with other components you need certain hardness. And that is why I would like to maintain a surface at a high hardness and this is achieved by introducing high carbon on the concentration on the surface. And finally, quenching it to produce as microstructure which has martensite. Now, this component which has having a high surface hardness and as a decreasing hardness into the interior can be thought of as an what you might call a functionally graded material. And what you might call a primitive or an original example of a functionally graded material. So, to summarize the things we have talked so far the surface hardness is a surface the most important part of the component which is prone to degradation. Surface hardening of steel component like gears is done by carburizing or nitriding. And in the current example we take up carburizing and this carburizing can be achieved using solid sources carbon sources or even can be achieved in the gas medium where in for instance you can take the methane gas and produce carbon which diffuses into the steel. So, essentially you have a this during the carburization process during which you are actually introducing carbon into the material. So, the blue part on the right hand side in the diagram is the material the gray part is the carburizing medium which is imposing a surface concentration on this material as C s. It is important to note that this C s is not the carbon concentration in the carburizing medium it is the carbon concentration which is in equilibrium with the surface of the steel. So, therefore, this line for instance cannot be extrapolated to the left. So, instance I cannot draw this line to the left this is only at the surface this C s is a concentration of carbon in equilibrium with a metal which is imposed on the surface and for now we will assume that that carbon concentration C s remains a constant. So, that is a constant as imposed by the carburizing medium. Now, since we start with the material the which is shown in blue with the initial carbon concentration C 0 like for instance in the example to be considered in the next slide this carbon concentration could be about 0.2 percent carbon. Now, as we initially the carbon concentration profile would be that the surface would have a concentration C s the interior would have a concentration C 0 and it will be a some kind of a profile which is like this it falls off the C s on throughout the surface and then suddenly it falls to C 0. But, as you hold this material for longer and longer time in this carburizing atmosphere the steel then you would notice that carbon would diffuse into the material and a carbon concentration profile would develop as shown in this curve. So, this curve shows you how the carbon concentration profile would develop after a certain time. Therefore, if I want to write down my initial and boundary conditions and I want to compute what you might call the constants from the fix second law solution to the error function solution to the fix second law which we have encountered before here. So, we encountered this fix equation to the fix second law here right here which says that C x t is a minus b error function of x by 2 root d t. So, in this if I want to determine the a's and b's I need to know my initial and boundary conditions which I am going to substitute for now and get the constants. Now, C at any positive x at time 0 is C 0 of course, this is wrongly written as C 1, but you can also call I will just now define the C 0 also equal to C 1 for simplicity. So, therefore, that is the concentration which is present in the material before I started my carburetion experiment and that is true for any positive x in the material because this is my origin 0. Now, on the surface at time t equal to 0 we notice that C x equal to 0 at any time because at any if you hover long you wait the surface carbon concentration remains constant which is C s. Therefore, using these conditions substituting in the error function solution I notice that a is equal to C s and b equal to C s minus C 0 or C s minus C 1 I can get these constants. Now, let us consider an actual example a salt example to understand this carburetion process. In other words I ask myself this question how long do I have to wait or how long do I have to do a carburetion experiment to build up a certain carburetion carbon concentration to a certain depth in the material. So, for instance in this example suppose I am starting with a 0.2 percent carbon steel and I want to surface carburize this such that the concentration at a depth of about 0.2 m m which is a small depth is about 1 percent. So, I build up a high concentration about 1 percent to a depth of 0.2 m m if you look at a depth below 0.2 m m closer to the surface then the concentration will be higher at whatever the time may be and beyond that the concentration will be lower and as we already seen when we talked about the nature of the error function solution there will be a certain depth at every time beyond which the carbon has not penetrated and you will find that the carbon concentration is the original carbon concentration of 0.2 percent. So, the carburizing medium imposes and surface concentration of about 1.4 percent which is the C s as we saw in the previous slide this is my C s and this is 0.2 percent is my C 0 or C 1 as I called it and the desired final product I want is at a depth of 0.2 m m I want a carbon concentration of 1 percent. So, the process we carry out at high temperatures and the temperature we carry out is at 900 degree Celsius and the reason for carrying out this process at 900 degree Celsius is primarily because of the fact that the solubility of carbon in gamma ion in much larger compared to the solubility of carbon in alpha ion and at 900 degree Celsius ion is in the gamma form the FCC form that implies I can actually put in this larger amount of carbon I am talking. If I try to put this high percentage of carbon say 1 percent of carbon or 1.4 percent of carbon at the surface starts to diffuse inside in a BCC ion it will tend to prostate out as F e 3 c and therefore, it will not be present in the form of a solid solution and I need to retain it in the solid solution form because only then can when I quench it from high temperatures I would get a what you might call a martensitic structure. Therefore, I need to carry out the primary this process at above 900 degree Celsius so that I can diffuse this carbon into gamma ion which has a high solubility for carbon additionally of course, doing it at high temperature gives you a faster kinetics because now we know that the diffusivity is exponentially dependent on the temperature. So, the data we have to need is that what is the diffusivity d 0 the diffusion constant d 0 in the equation d equal to d 0 exponential minus q by r t and d 0 for carbon in gamma ion is about 0.7 into 10 power minus 4 meter square per second the activation energy for diffusion of carbon in FCC ion is 157 kilo joule per mole and we are working at 900 degree Celsius the C 0 is C x at time 0 or equivalently it is at any time at infinite distance of course, infinite really does not mean infinite means that distance beyond with there is no penetration is 0.2 percent. So, this is a concentration of material carbon in the original material. Now, what we want is a C final which is about at 0.2 mm we want a carbon concentration of 1 percent and we want to know how long do I have to wait for this how long do I need to carry out my carbon carburizing experiment so that I can get a 1 percent carbon concentration on the surface we are already also given that C s which is C 0 t is also 1.4 percent. So, the solution to fix second law we know C x t is a minus b error function of x by 2 root d t the constants I determine from initial conditions and boundary conditions are C 0 t that means that is C s and that is equal to a and that happens to be 0.014 because this 1.4 percent has to be written in fraction terms and C infinity t is C 0 as we saw here before and which happens to be 0.2 percent which is equal to a minus b because now suppose in the error function I substitute x equal to infinity then I get the error function value of 1. So, a minus b C 0 is 0.02. So, giving the boundary conditions and initial conditions I determine that C 0 is a minus b and C s is a and I can substitute it back into the fix equation that C x t is equal to C s which is a now minus C s minus C 0 which is b error function of x by 2 root d t which I can further write down as the equation below which is C x t which is the concentration profile I want to know minus C s divided by C 0 minus C s is error function of x by 2 root d t. So, since we have the values of C s and C 0 I can write down the equation as C x t equal to 0.014 minus 0.012 error function of x by 2 root d t and I want to now determine we I know the x at which I want to know the carbon concentration and that x happens to be 0.2 mm which is 2 into 10 power 4 minus meters is equal to the right hand side as before 0.01 which is equal to 0.014 minus 0.012 error function of this x which is now the same as this x substituted here by 2 root d t 1. I am now labeling this t as t 1 because that is the time I want to know I have to wait I have to do this heat treatment at 900 degree Celsius. So, that I build a carbon concentration of 1 percent at a distance of 0.2 millimeter. So, simplifying this I can find that one third is error function of 2 into 10 power minus 4 divided by this error function of 2 into 10 power minus 4 divided by 2 root d t 1. So, now I need to know the diffusivity at 900 degree Celsius for carbon in gamma n given the fact that my d 0 value has been given here to be 0.7 to 10 power minus 4 meter square per second. So, to summarize some important points which we have used so far. The mechanism is interstitial mechanism which is diffusion of carbon in gamma n and the diffusivity d has to be evaluated at 900 degrees using the expression d equal to d 0 exponential minus q by r t. So, of course, we are using r t because we are talking about a mole of material instead of a single atom d equal to d 0 exponential minus q by r t that is equal to d 0 value is given to us and activation energy as we saw in the data before is 157 kilo joule per mole. I substitute that minus 157 here and distance power 3 is because it is kilo joule per mole I substitute the value of the gas constant and the temperature now in Kelvin I negate the diffusivity to be 7.14 into 10 power minus 12 meter square per second. Now, I have to substitute this value of diffusivity at 900 degree Celsius into equation which I call the equation 2 here which is a simplified form of the solution written here and I take this and substitute this into for d into this equation which now reads that one third is equal error function of this diffusivity 7.14 10 power minus 12. Now, if I take the error function on the other side the error function error in function inverse of 0.3333 is about 0.309. So, I get 0.309 is equal to 2 into 10 power minus 4 divided by 2 root d t and therefore, I can solve for t 1 to be about 14580 seconds. That means, I have to wait approximately for 4 hours suppose I divide this by. So, this 14580 I have to divide by 60 minutes into 60 seconds to get it in hours and this is approximately about 4 hours. So, I have to do this experiment for 4 hours such that I build up a carbon concentration of about 1 percent at a depth of about 0.2 millimeter. So, this depth to which we are carburized is called the case depth in the case of carburizing experiment that means, this is the now the case which is higher carbon concentration and if you do a cross sectional optical metallography you will see this region to have a distinct color as compared to the interior of the material. So, suppose now I plot my the way the diffusion profile C x t or the carbon concentration profile C x t varies with time I would notice initially of course, the surface concentration is 1.4 percent the y axis being carbon concentration and the x axis is the distance from the surface and I would obtain a square function which is at t equal to 0 which is shown by this green curve. Now, as time progresses now suppose that I am working at 1000 seconds then I can substitute for 1000 in this error function solution given this diffusivity at 900 degree Celsius and I would get a blue curve as shown here. So, that means, that carbon is penetrated into the material, but after a certain depth I would note that say this depth approximately about 0.4 mm there is practically no carbon and the material has not been affected by this influx of carbon from the surface. Now, suppose I wait for even longer and I wait about 7000 seconds then I would notice that the carbon concentration profile is increasing along this orange arrow and now I am working on this purple curve that means that actually the carbon is penetrating more into the material and not only the at if you take a given x for instance suppose I am working at a given x which is say 0.1 mm then I would notice that initially the carbon concentration of 0.2 percent at time t equal to 0 then a certain time later time. So, I can draw the distance at point I would notice that the carbon concentration already increased to about more than 0.6 percent at 7000 at about 1000 seconds then at 7000 seconds the carbon concentration already increased to about more than about 1.1 percent and if I wait still longer then I would get a curve which is shown the red curve and what I have specifically chosen this red curve to be drawn exactly at the time t equal to t 1 wherein my solution lies this solution at 1400 and 14580 seconds. So, the solution lies such that now I get a red curve for waiting for this amount of time which is about 4 hours then I would notice that if I take a depth of about 0.2 mm from the surface then I plot the carbon concentration profile it would be exactly about 1 percent of carbon further into the material beyond this point to mm the carbon concentration dies off and about and, but still carbon has penetrated up to about 1.4 mm into the material, but suppose I use a definition of a case depth as being 1 percent carbon then I would call this that point to mm is the case depth. Suppose I want to use a different definition of the case depth for instance suppose I use the fact that I want half the surface concentration which is now 0.7 then I would use a case depth which comes out to be about say 0.4 mm. So, I can use the definition of case depth based on the kind of criteria for concentration which I want, but you see that the below about 0.2 mm into the surface the carbon concentration is even higher than 1 percent and reaches about 1.5 percent when you reach the surface, but beyond about 1.4 mm the material is the original material we start off with about 0.2 percent carbon which means that it is going to be tough beyond the material will have a tough property or higher toughness beyond about this 1.4 mm. So, these are this is a nice illustrative example wherein we are taking interstitial species which is diffusing into the substrate and this is we are doing using in an engineering application to harden the surface and this is involves the concept of phase transformation which we will come to in one of the coming lectures wherein we make the surface instead of keeping it in the gamma ion form we make it martensite by quenching it and this and we will study more about how this martensite is obtained why is that the martensite is tough etcetera in the coming lectures, but essentially we note that this martensite is an hard phase or a hard produces a microstructure which is very hard and that is what you want as a surface property. So, this is a nice example of diffusion used in an engineering example equivalently you could also talk about another nice example which is actually taking for instance diffusion of for instance a dopant into a material in this case of course, the boundary conditions would be slightly different for instance suppose I am talking about a diffusion experiment wherein I am putting in certain quantity of dopant on the semiconductor once in the substrate could be semiconductor in this example and I am putting a fixed quantity of material on the surface which is for instance a p type dopant which is going to diffuse into the substrate for instance which could be silicon and after putting this fixed quantity of material I do a diffusion anneal that means, I hold this material at a higher temperature and this material is slowly diffuse into the material, but for this the boundary conditions are different and we can actually obtain a different solution. So, for instance suppose I am talking about this kind of a system wherein the fixed quantity of material which does not change in the system throughout the diffusion process once in the previous example when carburizing of steel we note that as the carburizing media is is held at longer time in contact with the metal or the steel more and more of carbon is getting into the material and therefore, the you are actually push pumping in more and more of carbon into the material, but here if you notice suppose the p type material which is put on the surface which is shown in gray color here this p type of material is a fixed quantity of material and the overall quantity does not change with time therefore, I can actually write down one of the conditions as integral 0 to infinity which means of course, 0 is here to infinity of course, I can actually place my 0 here because initially I start with the material which is a thin film of concentrate c d x that is integration with respect to x which is the coordinate is this fixed quantity of material m which have fixed quantity of mass m material which have imposed on the surface. Additionally of course, I notice that at time t equal to 0 at any positive x the concentration of this dopant is 0. So, that is another condition I can use and for such kind of circumstances instead of using the error function solution. So, this is a different boundary condition problem I use a solution which is an exponential solution where I can write down c x t is equal to m by root of pi d t exponential minus x square by 4 d that implies that based on the boundary conditions I have multiple solutions of fixed second law and there are nice examples wherein I can go and use the solution to find out that how say for instance how much time do I hold what is the concentration profile into the material etcetera and I can design my process parameters taking knock on the time of diffusion. Towards the end of these lectures let us consider one another important thing which is called the approximate formula for depth of penetration. So, we will define appropriately what is called depth of penetration and this approximate formula is useful because often we want to quickly calculate how long do I have to carry on an experiment and for instance a diffusion anneal experiment or a carburization experiment. So, that I can quickly calculate how long do I await for a given concentration I need to obtain a certain distance. So, let us again work with the example of a carburizing medium and steel to be carburized and you already noted this solution to this the error function solution given the boundary conditions and the initial condition is c x t minus c s by c 0 minus c s is error function of x by 2 root d t. Now, let us call the distance at which this ratio the concentration ratio c x c minus c s by c 0 minus c s and if of course, I assume that that c s c 0 is actually 0 then I can think of this as about half as a ratio more physically visualizable, but then I assume that this ratio is half and I call the distance at which this is obtained is x half which I call which I can visualize as an effective depth of penetration. So, I need to carry on an diffusion experiment such that this concentration ratio turns out to be half and I call that distance to be x half at which it is obtained after certain time t. So, I substitute for the left hand side as half and I write down the error function solution. So, I call since I am calling this x half I am calling this error function of x half by 2 root d t is equal to half and we noted that for small values of x error function of x is approximately x therefore, I can write down error function of x by x half by 2 root d t is half from this I can see that I can cancel these two halves and x half which I call now the depth of penetration. So, this x depth of penetration is x half which is based on this concentration ratio is can be approximately written as root of d t. That means, if I quickly want to calculate the time say for instance I have to do an annealing such that say for instance to a depth like in the previous experiment to a depth of about 0.2 mm I want to get a concentration which is given as half for this ratio then I need to do an annealing time which will be given the time will be given t 2 will be given by x depth of penetration which is x half square by d. So, using this I can quickly calculate my time to which I need to do my diffusion experiment. Of course, using the same kind of formula I can also know as I pointed out the depth at which c x is nearly c 0 and this again we noted is because of the nature of the error function solution the way the error function solution works. We noted that previously that as x tends to 2 the error function of solution almost tends to 1. So, the area under the curve beyond about this x equal to 2 is negligible in the error function solution. Therefore, there is a region beyond which practically no material would have penetrated by diffusion. So, therefore, I can now write down I want to know where the beyond the distance beyond which there is no carbon concentration is what you might call penetrated and that distance I called x infinity. So, therefore, I put 0 is equal to 1 minus error function of x infinity by 2 root d t and error function of u is approximately u 1 when u is equal to 2 you have noted this before. Therefore, x infinity by 2 root d t is 2 and that implies x infinity is about 4 d t. So, if you look at this x infinity versus x penetration you note that this is of the order of d t while this is of the order of 4 d t. So, at d t you can think of it approximately that it is the case depth the depth to which you want your penetration and beyond 4 d t which is about 4 times this length scale you notice that there is no penetration. So, all the scene of diffusion is between the surface and 4 d t and between that the important number which is root of d t. So, with this we cannot come to a sort of a brief introduction to diffusion, but we shall note that this diffusion plays a very important role in the coming two chapters which one of which is happens to be on phase equilibria. That means, if I want a system to be in equilibrium then diffusion plays a very important role to achieve that equilibrium and later on we will talk about phase transformations wherein again many of the mechanisms of phase transformation which are called diffusional transformations involves this process of diffusion and therefore, I need to understand the basics of this definition keeping in mind especially the mechanisms involved in diffusion.