 Welcome back to this course on nanostructured materials, synthesis, properties, self assembly and applications. This is the fifth lecture of module two. We have earlier done four lectures of module two in which we have looked at two synthetic methodologies using the sol-gel method and using the micro emulsion method. This lecture we will be starting with CVD technique which is chemical vapor deposition where we will be using this method to make nanostructured films. So, chemical vapor deposition is a process whereby a thin solid film is deposited onto a substrate through chemical reactions of the gaseous species. So, you have a substrate which is a solid and you will have you want to deposit a thin solid film starting from a chemical precursor from which you generate gaseous species and they will condense onto a substrate and give you a thin solid film which is nanostructured. So, why do we want to study thin films or deposition processes for thin films? Because these single and multi layer films and coatings have lot of applications. We want to synthesize nano layered materials. We want to synthesize optical films for transmission and reflection studies. You can make decorative films for various purposes. You can also make functional coatings which are having say wear resistance or some other function. For all these type of films we can use such deposition techniques like the CVD method. You can also make permeation barriers for moisture and gases. So, that they do not enter once you have this coating moisture and gases cannot enter. You can have corrosion resistant films. For example, you can have some metals which normally get corroded in oxygen or in the presence of water vapor. If you make a film which is resistant to corrosion, then you can protect those metals. You can make electrically insulating layers for microelectronics. So, if you have electrical wires which you want to insulate from adjoining wires, you can make insulating layers. You can make coatings of engine turbine blades to enhance their longevity and life. You can coat high strength steels to avoid them becoming brittle using hydrogen. Normally, hydrogen leads to brittleness in the steels. So, you can avoid that by coating some material using the CVD technique. You can make diffusion barrier layers for semiconductor metallization. You can make magnetic films for various kinds of recording media like DVDs or many other kinds of recording media. You need magnetic films and these can be processed using CVD technique. Transparent electrical conductors are required in several places and anti-static coatings are required. You can make as we discussed wear and erosion resistant hard coatings on tools. So, the tools which are used for heavy duty work, they do not get worn off. You can make coatings of certain materials on top of those tools. Then you can make films which can act as lubricants. So, these are called dry film lubricants. Then many other composite films, nano composite films can be made and thin walled freestanding structures and foils which are to be used elsewhere which are very thin can be made by the CVD technique. So, what is this CVD technique? That is the chemical vapor deposition technique. What is the principle? Typically, a CVD technique depends on the availability of a volatile chemical which normally we call the precursor, the chemical precursor. It is normally a liquid solution which can be converted by some reaction into the desired solid film. So, ideally you have a source gas and the source gas undergoes a chemical reaction and forms some monomers. These monomers nucleate to form some oligomers here and this is one process where the oligomers can coagulate and condense become large particles and they undergo Brownian diffusion etcetera and thermophoresis and then fall on the substrate or they can be these monomers which form from the chemical reaction of the gaseous phase. This can be targeted using ions and that is called ion induced nucleation. So, you target these monomers which are formed from the gas phase of which has been generated from your initial volatile chemical that if you bombard with ions, you get ion induced nucleation and this ion induced nuclei are charged particles and then you can apply an electrostatic force on the charged particle and direct them on to this substrate. So, by that method also you can make a coating. So, you have this substrate or substrate can be a disc which is polycrystalline or which is single crystalline and on top of this substrate these molecules will start adhering and condense and will result in a thin film. Now the various steps that occur in a CVD process both physical and chemical steps are as follows. You have mass transport of the reactant gaseous species. So, that volatile precursor gives out gaseous species and these gaseous species are transported. So, there is mass transport near close to the vicinity of the substrate. So, that mass transport is important. So, you have to move these gaseous molecules which are coming from the precursor which may be a liquid which is normally liquid solution and then those gaseous molecules are transported close to the substrate. The next thing is diffusion of these reactant species through the boundary layer to the substrate surface or homogeneous chemical reactions to form intermediates that is another part of the CVD process and then adsorption of the reactant species or intermediates on the substrate surface. So, these mainly are some of the key steps which occur during the CVD process. After that you may have surface migration that means the clusters or monomers or oligomers which are coming on to the substrate migrate on the surface of the substrate. They may undergo heterogeneous reaction, they may have inclusion of coating atoms into the growing surface and it may also form by product species. These by product species are then desorbed from the surface and then diffusion of the by product species to the bulk gas and transport of the by product gaseous species away from the substrate through an exhaust. So, these are subsequent steps after the charged nuclei or neutral monomers or oligomers come on to the substrate surface then these surface migration and subsequent reactions and the desorption of the by product take place. The by product has to be eliminated or transported away from the substrate. Now, if you look at the instrumentation required for chemical vapor deposition you have your gas or the source for the vapor. So, it may be a liquid or some source and there is a panel where you use for purging and then you come it goes through some meters where you can control the flow of the gas in or out. So, there is this in and then it is out and then again the gas is sent to the final deposition chamber where the substrate or the vapor if it is a silicon substrate or a silicon vapor we call or it can be a quartz substrate there can be many substrates they are in this deposition chamber. So, the molecules have to come from this specialized gas liquid which if there is a liquid then you have to the gaseous molecules have to be generated and then go through this various valves where you can control the flow of the gas inside the deposition chamber and typically there are two types of transport analysis. One is when the reactant which is coming out and the reactant which is going in the ratio is nearly 1 is to 1. So, that is called a differential reactor. However, if the reactant coming out is much less than what is going in then it is called a starved reactor. So, this gas in and gas out the ratio of this leads to these two terms the differential reactor and the starved reactor and the exhaust gases are taken out from the exhaust pump and if they are toxic they have to be treated in the waste treatment before releasing less toxic gases outside. So, this is a schematic diagram of a typical chemical vapor deposition process where you have the gaseous source the valves and the deposition chamber the exhaust and the release of the waste gases. Now, to understand the physics and chemistry of the chemical vapor deposition we have to undergo or understand some of the basic principles of physical chemistry or gas laws which are involved during this process of the gas molecules going on top of the substrate and what kind of nucleation or whether it is homogeneous or heterogeneous nucleation and how the growth of the film takes place etcetera. The basic concepts in CVD start from approximating the gases and the vapors that you use to be ideal and so they obey the ideal gas law. So, this is an approximation and the ideal gas law as all of you know is p v is equal to n r t where p is pressure v is volume n is the number of moles of the gas r is the universal gas constant and t is the absolute temperature in Kelvin. So, the units are important. So, when you are taking volume in meters so the volume unit is meter cube and then you have to take the r the gas constant in s i units which is 8.3 joules per mole Kelvin. So, these are typical quantities one needs to know to apply the ideal gas law which is important for understanding CVD process. The gas flow as you know you will have to flow gases through these walls etcetera. So, the gas flow is very important and that gas flow are usually measured and reported as either standard liters per minute which is SLPM or standard cubic centimeters per minute which is SCCM. So, these commonly you will find numbers so many SLPM so many SCCM. So, these are units to tell you about the rate of flow of the gas going through the reactor. So, SLPM and SCCM both are related to the gas flows and they measure gas volume at 0 degree Celsius and one atmospheric pressure. So, these are measures of the molar flow of gases through the reactor. An important consequence of the ideal gas law is for CVD reactors is that volume flow increases tremendously for the same molar flow at low pressures. So, volume flow will increase significantly at low pressures at the same SLPM. So, if you have the same SLPM that is the same flow a standard liters per minute same SLPM, but you change the pressure from high pressure to low pressure the amount of volume flow will increase tremendously. There are some other useful conversions one needs to know which are normally available in textbooks or in all kinds of manuals where CVD is discussed and these are present in many other textbooks of physics and chemistry or material science and one needs to know these conversions because many times you have to convert pressure say from atmospheres to tor or to Pascal's and you need to know these conversion factors. So, this is for pressure volume and then you have one CC is equal to how many moles at STP. Similarly, this rate of impinging suppose you have how many moles per centimeter square is falling in 1 minute. So, that is this number 1 a per minute how many molecules or moles per unit area is falling per minute on a particular substrate that is given by this quantity and what we just described the gas flow as 1 SLPM and 1 SCCM they are given here and you see 1 SLPM and 1 SCCM are different and they are 1000 times different this is 1 SLPM is 1000 times 1 SCCM. Now, some other useful conversions are between the Boltzmann constant K and the universal gas constant R which can be written in several different units and one has to be careful when you are doing mathematical problems or converting numbers from one unit to another unit you need to be careful what to take the value of R the value of R can be 8.3 or 62 depending on what units you are taking and so these conversions are useful in doing some calculations with respect to pressure volume and basically using the ideal gas law to understand the gas flows etcetera. Now, the next thing that we try to understand because gas molecules will be flowing so we have to understand the velocity of these molecules assuming that this is an ideal gas and hence it will have a Maxwellian velocity distribution and so that is given by the rate of change of the number of molecules where n is the number of molecules as with respect to velocity that is given by this quantity and as you see there is a term where temperature to the power minus 3 by 2 and also there is on the there is an exponential with minus 1 by T. So, if you this is the Maxwellian velocity distribution and this will be the distribution of the gas molecules assuming that it is an ideal gas and from there we can calculate the mean velocity of these molecules which is given by C mean to be the square root of 8 RT by pi m and for this R must be in joules per kilogram per mole. Now, from this then we can calculate the number of molecules which are following on a plane which is say 1 square centimeter and in 1 second. So, that can be calculated as J which is the number of molecules impinging on a plane per square centimeter per second and that is given by this number n C mean by 4 and where the small n is the molar volume and C mean is the velocity here the mean velocity divide by 4 will give you the number of molecules which are following on a plane per square centimeter per second. Now, from that we can calculate something which is very important or which is the maximum possible deposition flux for a given partial pressure. So, given a partial pressure P what is the maximum deposition flux that we can have and that is J is given by this equation and if you know at some pressure P and temperature T for a gas where with a molecular weight m which is given in grams per mole you will have the J which is the maximum possible deposition flux by this equation. Now, the another quantity which is important is the mean free path of each molecule that is the distance the molecule travels before it is hit by another molecule. The mean free path given by lambda is given by this equation where the mean free path is now inversely proportional to the square of the molecular diameter and also to the inversely proportional to the molar volume. So, n is the molar volume and alpha here is the molecular diameter and the mean free path is related to it. That means, the smaller the molecular diameter. So, smaller the molecule the larger will be the mean free path. That means, the molecule will be moving without collisions for much longer distances if the molecule is smaller. Similarly, the molecule will be moving much longer distances if the molar volume of the molecules is smaller. So, lambda the mean free path is inversely proportional to both the molecular diameter and the molar volume. Now, from these we can now compare some situations like if you compare three different pressures for the same gas you have a pressure of 0.1 torr, 10 torr and 760 torr. So, if you assume these three pressures we can calculate what will be the density is of this order and the mean free path for at this pressure is around 500 microns. Whereas, at a higher pressure of 760 which is normal atmospheric pressure you will have a mean free path of around 0.07 microns which is equal to 70 nanometers. So, as the pressure is increasing the mean free path is decreasing. Similarly, the mean velocity it is kept constant at 47000 at these pressures and you can calculate how many molecules will be impinging on a surface under these different pressures. So, at low pressures you will have less number of molecules impinging per second which is say of the order of 10 to the power 19. Whereas, at atmospheric pressure at 760 millimeter pressure of the same gas keeping the mean velocity same you will have 10,000 times more molecules impinging on the surface. So, depending on the number of molecules which are following on the surface per unit time if that is large then the growth rate will be larger. And that is seen here the number of molecules falling on a surface per unit time here is much less than here. And you see the growth rate how much thickness the film grows per minute that is the thickness is in microns. So, how many microns per minute does the film grow can be compared under these three conditions of 0.1 torr of pressure you have 10 micron thick film growing per minute. If you increase the pressure to 10 torr this thickness goes up to 1000 micron per minute. And at atmospheric pressure you will have atmospheric pressure it is an enormous number of 77,000 microns per minute. So, you have kept the same mean velocity, but you have changed that pressure of the gas and the other parameters change. And finally, what you require is the film and the thickness of the film is you can control by controlling the partial pressure of the gas. So, this slide shows you how some of these numbers change by changing the partial pressure of the gas. So, as you are going from low pressure to high pressure you can change the thickness of the films which is the right control that you want on the growth of the films using the CVD technique. Now the growth rate that we just now discussed shows this is the key point it shows with those numbers is that high pressure leads to high growth rate. So, at one thing what we have kept here is a mean velocity of around 47,000 centimeters per second. Now what it really means 47,000 centimeters per second means that a molecule will cross a 1 meter chamber. So, if you have a 1 meter chamber in which you are doing this experiment the molecule will be moving if you say at 2 meters per second. So, in 2 meters 2 millisecond 1 meter long chamber a molecule will be moving in 2 milliseconds at 1000 miles per hour. So, that would be the speed kind of speed if you trans if you change the units here from 47,000 centimeter per second to miles per hour you will get a speed of around 1000 miles per hour. So, this is the kind of speed of molecules that you are talking when you are studying the impingement of gaseous molecules on the substrate or on the target. Now the impingement limited growth greatly exceeds actual growth rates that means whatever we are calculating is much larger. So, what we are showing here is say 10 microns per minute this is the calculated growth rate in reality this will be lesser in actual experiment and the reason is that the actual deposition will not be the same because the partial pressure at the surface will be less than the pressure inside total pressure inside the gas. The other probability is that there can be the incorporation of the gaseous molecules on to the surface is small. So, there are two things one that the partial pressure what we are assuming for the total gas is not same and the actual partial pressure at the surface is much lower or the probability of incorporation on to the surface of these gaseous molecules is small. The other thing that we have to discuss so we discuss some fundamental concepts using Maxwellian distribution of the velocities of these gaseous molecules and assuming ideal gas behavior then we also have to consider the thermodynamic aspects. So, if some chemical reactions are going on we have to incorporate the Gibbs free energy function which is G and G is a function of enthalpy and entropy which is S and the relation between free energy enthalpy and entropy is given here where T is the temperature and by the second law of thermodynamics a spontaneous reaction will occur at a pressure P and a temperature T kept constant if the free energy is less than 0. So, this is a standard second law of thermodynamics. So, if you have any chemical reaction which is going to happen during the formation of the film it has to obey this relation of free energy going to be less than 0 and the free energy calculation is done typically like this. For example, consider a reaction where a gaseous molecule of A n moles of A react with m moles of another gaseous system B to give P moles of a solid say C which is the thin film which will form and r moles of another gas say D. Now, the delta G or the change in the free energy or the change of the free energy of the reaction is the free energy of the product minus the free energy of the reactants. This is a standard free energies delta G is not delta G not and that can be equated to minus r T l n the product of the activities of all the products divided by the product of activity of all the reactants. So, if there are two products it will be A 1 and A 2 multiplied and if there are two reactants A 3 and A 4 so this is A 3 multiplied by A 4 that means the activity of 3 and activity of 4 and that gives rise to an equation like this. This from here you come here because you have been given this reaction. So, if A is say water and B is ammonia then accordingly you have to choose the concentration or activity of water and ammonia raised to the stoichiometric or the mole ratios in which they are reacting. So, here n moles are reacting. So, the activity of A will be raised to the power n where n is the number of moles of A reacting with the number of m moles of B and so here in the product of the activities you will have B to the power m. The other important thermodynamic aspect is well known chemical equilibrium principle given by Lee Chattelier that if the conditions of a system initially at equilibrium are changed the equilibrium will shift in such a direction as to restore the original conditions. So, what happens typically in reactions where gaseous molecules are involved you can assume this reaction which is well known reaction for the ammonia formation where all the three the reactants and products are gaseous species. So, you have nitrogen gas, hydrogen gas and ammonia gas this reaction at under high pressure is moves the equilibrium towards the right side and at high temperature it moves the equilibrium towards this side. So, that is basically Lee Chattelier's principle because it says that the equilibrium will shift in such a way if you change something. So, you change pressure you go from one pressure to higher pressure the equilibrium will shift and more of ammonia will be formed from nitrogen and hydrogen. If you lower the pressure or you increase the temperature then some of the ammonia will dissociate to give nitrogen and hydrogen that means the equilibrium is shifting to the left side. So, this is a very important principle given by Lee Chattelier and is very important wherever gaseous species are involved to understand how the chemical equilibrium will shift in the when you change a pressure or temperature in a particular manner. Now, you can write down these reactions which are the law of mass action for the forward reaction and the reverse reaction and you can write down the at equilibrium the forward reaction will be equal to the backward reaction. This is an example of that that was one example we showed was nitrogen and hydrogen reacting to give ammonia. This is another reaction where hydrogen and iodine are giving you hydrogen iodide and the reaction is basically due to the collision of hydrogen iodine molecules and the rate of reaction how fast the reaction will happen is proportional to the number of such collisions and the number of collisions is proportional to the density of the hydrogen gas and the iodine gas. So, the density is proportional to pressure and hence finally, it boils down to that the reaction rate is proportional to the partial pressures of hydrogen and iodine and this is the reaction rate where the reaction rate is proportional to that means you can write it is equal to a constant k 1 and multiplied by the partial pressure of hydrogen multiplied by partial pressure of iodine that will give you the rate of the reaction. So, the partial pressure of the gaseous species is important. Now, similarly the reverse reaction rate will depend on the reverse constant the k minus 1 and will depend on the partial pressure of H i which is in equilibrium. Now, we can define this just like we defined earlier the equilibrium constant which is a function of temperature taking a ratio of k minus 1 by k 1 and being shown by this the partial pressures of the product and in the reverse reaction the product is hydrogen and iodine gas divided by the partial pressure of the reactant gases in this case it will be hydrogen iodide. So, you can write down this kind of reactions and find out the equilibrium constant of course, which is a function of temperature. So, at a particular temperature you will have one particular equilibrium constant. Now, this law of mass action can be rewritten in terms of partial pressures and the partial pressures of this particular reaction where this is equal to a 1 small a 1 multiplied by capital A 1 plus small a 2 multiplied by capital A 2. This can be rewritten as a product of the pressures raised to the powers of these coefficients a 1 a 2 a 3 etcetera and that product will give rise to the equilibrium constant at a particular temperature. So, if you continue on this product which is equal to the equilibrium constant for ammonia we can rewrite this like this and from instead of activities then we choose concentrations for certain cases where we can take the activity to be equal to the concentration. We can rewrite this equation in terms of concentrations and you can find out that the k is related to the concentration of the nitrogen and hydrogen and ammonia by p square. So, an increase of p here means increase of ammonia. So, if you increase p the concentration of ammonia will increase and this is in accordance with Lee-Shetley's principle. Ultimately delta G naught of course is given by minus RT ln k or minus delta G naught is given by RT ln k. This is a very useful equation. Now, to understand the mechanism at the equilibrium we used we understood this reaction of n moles of a plus m moles of b reacting to form p moles of c and r moles of d. We used the fact that the equilibrium vapor pressure of a gas over its condensed phase is determined by the minimization of the free enthalpy or the Gibbs free energy. Now, so this can be shown in this way. This is the mechanistic pathway where you have this a molecule interacting with b molecule and it undergoes a bimolecular association reaction and it forms a transition state which is in quasi equilibrium with the reactants. So, you have a and b forming this transition state where a and b are collided. Now, the next step from the transition step where a and b have collided there can be two ways that the reaction can proceed. One is it may dissociate back to give you a and b or it can because this has some extra energy than a plus b. This can transfer its energy to some other molecule and the entire unit can remain as such which is c. So, the third atom which is m here can collide with the transition state and take away this excess energy and give back this new product which is c which is a combination of a and b or the other simple ways it falls back to the original state of energy with a and b dissociated. So, if you fall from the transition state you can go back to the original reactant state or from the transition state you can go to the product which is a new molecule by the interaction through a third molecule which takes away part of its energy. Now, there are various methods of doing C V D. One is the pyrolysis method and then you have reduction then oxidation and compound formation you can have disproportionation reaction or reversible transfer reaction. So, there are variety of methods by which you can do this chemical vapour deposition techniques. So, if you take the first case which we said the pyrolysis reaction. So, this is a typical pyrolysis reaction where you want to make polysilicon. So, you take a gaseous species this is called silane silicon there are SiH4 and when you heat it at 600 degree Celsius then it will give rise to silicon and hydrogen. So, basically you are heating this gas molecules over a substrate and this decomposes to give you silicon which forms the film on the substrate and this hydrogen gas has to be taken out as an exhaust through the exhaust. So, this is a typical reaction where a pyrolysis is taking place that means breaking down of a compound under some giving some heat that means at high temperatures you are breaking this up to give you these molecules these molecules will sit on the will be transported to the substrate and form the film. So, this is a CVD process which uses pyrolysis reaction. Then coming to an example for a CVD where a reduction reaction is taking place can be seen in the case of tungsten metal preparation. So, W is the symbol for tungsten and you start from a gaseous species hexafluorotungsten hexafluoride and you because this is a reduction reaction the name through you heat in the presence of hydrogen and you get tungsten which deposits on the substrate and HF forms during the reaction and this hydrogen fluoride gas is then taken out through the exhaust. So, this is a reaction which is a reduction of the reactant to give rise to the product which then is deposited or as a thin film on the substrate. This is the third example where you are doing an oxidation reaction to prepare a film using the CVD technique and the example here is of aluminum trichloride you are using hydrogen and carbon dioxide to oxidize aluminum chloride to aluminum oxide and carbon monoxide and HCL gas are formed. So, these two are gases and this is the only solid material. So, this will deposit on the substrate and give you the film. So, you are using an oxidation condition to make aluminum oxide or alumina films from aluminum chloride and this is again a typical chemical vapor deposition process is aluminum chloride molecules will be deposited on the substrate. The next methodology is the compound formation methodology where the example that we show here is of titanium tetrachloride reacting with methane. So, there are two gases titanium tetrachloride and methane which are brought together near a substrate and at a temperature of 1000 degree Celsius and at that temperature titanium carbide which is a solid will deposit on the substrate and form a film. So, this is an example of a CVD procedure where you have a compound formation starting from two gaseous precursors and they interact and form a compound which deposits on the substrate and forms a thin film of titanium carbide. The next example is a disproportionation reaction as you understand from the word disproportionation means it is breaking up. So, it is a chemical reaction in which the same chemical from a more scientific point of view a disproportionation reaction is basically the substance itself acting as both oxidizing and reducing agents. So, you start with two moles of germanium iodide one of them oxidizes the other and the other one reduces the other. So, G E I where germanium is in oxidation state one when two molecules of G E I come together in the gaseous state they react and these G E I molecules then disproportionate that means one of them becomes higher or becomes oxidized to G E 2 plus and the other germanium gets reduced to G E 0. So, here disproportionation the word is used because you are having a metal ion in a particular oxidation state. In this case it is in plus one oxidation state in germanium iodide molecule which disproportionates that means two molecules of germanium iodide both having plus one ions will give rise to one ion of 0 oxidation state which is the metal and which deposits on the on the substrate and forms the film of germanium and the other germanium from here becomes germanium plus 2 and reacts with iodine to form germanium iodide or diiodide I should say and this germanium diiodide is a gaseous species and is taken out of the exhaust. So, that is why this reaction is called a disproportionation reaction because from two plus one species we are getting one zero valence species and one di valence species and the zero valence species deposits on the substrate and gives you a film of germanium on the substrate. So, this is a CVD technique to make to quote germanium films on several substrates. Similarly, you can quote aluminum boron gallium large number of other metals and non metal solids on substrates as films using this disproportionation CVD method. Then this is an example of a reversible transfer. So, here what we see is that you have a gas molecule it is a tetramer 4 atoms of arsenic are there in this cluster 2 atoms of arsenic are there in this cluster. So, it is a tetramer and a dimer and they react with gallium chloride where gallium chloride is here having plus one oxidation state and this gallium chloride in the presence of hydrogen at 750 degree centigrade gives rise to gallium arsenide and gallium arsenide as you know is one of the most well known materials for semiconducting devices and has a lot of applications and this technique has been used in the CVD technique where it is a reversible transfer because if you use 750 degree centigrade you can go from the left side to the right side where you produce gallium arsenide on the substrate as a film it is a solid and you get HCL gas. Now, if you start from this side you start with gallium arsenide and HCL and heat at 850 degrees you will get back the arsenic and the gallium chloride. So, that is why it is called reversible transfer reaction but it is a CVD technique because you are starting for molecular precursors and ending up with a solid film on a substrate. Now, so we saw that there are lot of precursors which are used which have which are either gaseous molecules or they are gaseous which are pressurized into liquids and then gas molecules come out of them and can be controlled before they are sent to the reaction chamber. Now, the one of the important requirements for this precursor which is either the pressurized gas as a liquid or just the gas is there should be sufficient volatility to achieve acceptable growth rates at moderate evaporation temperatures. That means, the volatility of that liquid or to form gas should be very high otherwise you will need high temperatures to volatilize gaseous molecules from those liquids. So, this is a very important requirement for the precursor that it should be very highly volatil or sufficiently volatility should be there to allow for gaseous molecules at moderate temperatures. You do not want to heat them at very high temperatures of course, you do not even want them to be very volatile that comes to the next point these should be stable. So, that decomposition does not occur during evaporation the molecules which are going should not decompose while moving close to the substrate and they should be able to be transported without any decomposition from the point of entry to the reaction chamber to this vicinity of the substrate they should be stable and should not decompose. So, this is the same thing we are mentioning that the stability should be high and it should not decompose while evaporation. Some other requirements are like there should be high chemical purity, there should be clean decomposition without the incorporation of residual impurities and there should be good compatibility with co precursors during the growth of complex materials. So, this means that suppose you are trying to grow a bimetallic film. So, you have a precursor for one metal and you have a precursor for another metal and that is called the co precursor and during the growth of this complex metal film there should be some compatibility within these precursor molecules. So, good compatibility with co precursors during the growth of complex materials. So, materials which I have two metals or three metals or two metals and oxygen or carbon or carbon dioxide has to be injected to make a carbide or nitrogen for a nitride. So, these co precursors should be compatible among themselves this is an important point of course, I mentioned about high chemical purity and also the clean decomposition there should be no other impurities which solidify on the surface of the substrate. Now, so they should these precursors again talking of these precursors because this is the most important thing in the CVD process should have long shelf life with lot of stability at ambient conditions and should not be affected by air or moisture. They should be able people should be able to manufacture them in large yield at a low cost and they should be non-hazardous or with very low hazard risk. So, if you have to make an industrial process you have to consider all these points for making CVD films using molecular precursors. Now, the type of materials normally used for deposited by CVD techniques have applications in several areas of technology. Microelectronics is one of the most important areas where CVD techniques are used. Optoelectronics protective and decorative coatings and optical coatings all of them sometime or the other people use CVD processes to make films which will be acting as components or devices in many of these applications. And with this I come to an end to the lecture 5 and we will be continuing our discussion about various methods of deposition of films nanostructured films by different methods in the coming lectures. Thank you very much.