 Okay, so let us slightly change the gear, we go to the solid state fixed portion and whatever I had discussed with you, so today's lecture I will give one free electron theory and tomorrow's lecture I will give more on the basic lattice, the Bravais lattice and the reciprocal lattice and Bragg's law and the Lawyer's law, that is what I am planning to do for tomorrow. Okay, of course this particular part we also do normally in our first year course, we include this particular part but this normally we do after when we do a particle in a three dimensional box and we derive the energy states for a particle in a three dimensional box. Then we talk a little bit about the statistics, then we differentiate what is the difference between both science and statistics, fermi direct statistics and also the Maxwell Boltzmann statistics and then eventually we come to the solid state physics. So last three or four lectures are basically based on free electron theory, so I mean from whatever we have been doing in first year probably there is a little bit of gap between that and that but I have tried to make it up by adding additional transparencies but you know maybe there are certain deficiencies so let us hope you will accept it. So generally when we talk of solids, the first thing that we normally talk is about metals because metals out of all the solids had been the most interesting part historically because the ages have been named after metals and also as far as solid state physicists are concerned they are known to have very high electrical and thermal conductivities. These are the two things which signifies metal from many other different types of solids, they have very high electrical conductivity and high thermal conductivity that is the reason most of the electrical wire connections are all metallic. Now 1897 was the time when J. J. Thomson discovered electron and 1900 was the year when Brute gave his theory for the metals and it is very interesting that just after three years he could tell something which are of course not 100% correct from today's word but nevertheless he could approach the problem reasonably correctly. Let me also add at this particular point that some of the things which I am going to talk about it you know are from today's perspective of course at that time of Brute probably they were not I mean for example the electronic structure etc. they were not known at that particular time they are much later known Bohr's theory came much later so there was no clarity but that is why it makes it much more interesting that you know Brute could think you know so much into this particular material these metals so he developed what is now known as a free electron theory. So what he assumed I mean at that time as I was telling that most fashionable theory was kinetic theory of gases so he also talked about kinetic theory of gases and he assumed that in these metals they are large number of free electrons free electron means they are under the influence of no force which essentially means that he neglected the interaction between electron and electron because if they are large number of electrons electrons are electrostatically negatively charged therefore they are always going to be interacting with amongst each other also because overall metal is neutral so if electrons are free so whatever has been left behind which we can call as core must be positively charged particles and therefore these electrons when they are moving they must also be getting influenced or a force must be exerted on them because of the positive charges but all those things were neglected by Brute just assume that these electrons are totally free and applied essentially all the concepts of kinetic theory of gases that we apply on gas molecules even though the density here turns out to be much larger than the density of gas at NTP but even then many orders of magnitude I think 4 or 3 orders of magnitude still he applied the same thing. So now what was thought as a positively charged core can be thought of as a valence electrons and I mean it is not a valence as core electrons and it is the valence electrons which in today's world or today's time we can think that's the final outermost electrons are the which are the loosely bound and they become other free and that's what I said he neglected electron electron interaction he neglected electron core interaction there is only one difference which he made from the normal kinetic theory of gases which normally people do in kinetic theory of gases they are only gas molecules so the only collision which takes place takes place between the molecules themselves here he assumed that the collisions that these electrons make while they are traveling are not amongst themselves he did not include that particular aspect but he assumed that these collisions are taking place from the course because he realized that probably electrons are much lighter particle so heavier particle must be the one which are left behind which are thought free and therefore when these electrons are moving the chances that they will get scattered by these heavier particles are much more probably that was the thinking so he assumed that the scattering of these electrons are taking place from the positive course that is one difference from the standard kinetic theory of gases whereas we said as I said we assume that molecules collide amongst themselves here he assumed that electrons are colliding with the positive course which has been left behind so that is what I have written the only way electrons sees core is by scattering otherwise there is no force obviously in approximation you know because you know that like static charges have to be present and that inter atomic distances are fairly small so there has to be force in but it was all neglected now like kinetic theory of gases introduce concept of a Lexation time introduce that the probability that electron will get scattered in time dt is proportional to dt upon tau tau was supposed to be constant which is generally called a Lexation time like kinetic theory of gases and this less than time was assumed to be independent of position and the velocity of the electrons irrespective of whatever is the velocity of the electron irrespective of what is the position of the electron the Lexation time was assumed to be same it was not supposed to depend on that now using this particular idea he could derive ohm's law is actually ohm's law is one of the very very fundamental laws of electricity almost like Newton's law of motion and let us focus our attention on one electron which has just come out of a collision and let us assume that velocity is V naught consider electron whose velocity immediately after collision let us call that time is time t is equal to 0 is V naught and let us assume that we have applied a electrical field to the system when we apply electrical field to the system this particular electron remember salt classical mechanics there was no quantum mechanics at that time so this particular electron will experience a force as per Newton's law and therefore this electrons velocity would change under the influence of electric field until the time it suffers the next collision that's basically the idea so electron will collide then after collision its velocity becomes random then again accelerates because of the electric field then again collides then again it becomes random situation something like this so let us suppose it at a time t it has not suffered a collision where we at time t is equal to 0 it suffered a collision at a time t when it has not suffered a collision let us try to calculate what will be its velocity if the electric field is e the force on the electron will be e times e of course with the negative sign okay and therefore acceleration would be force divided by mass so e upon m will be the acceleration of this particular particle and during time t it is getting accelerated its initial velocity was V naught this is basically applying the kinematics equation it will be the velocity at time t will be given by V naught minus ee e is the force divided by m which is makes it acceleration multiplied by the time so the velocity at time t of this electron will be given by this thing 2 ideal of course electron does not suffer a collision now what I am interested in calculating the average of this particular velocity not only on this particular electron but over all the electrons because that is what we are going to give rise in the current so that is what he called as a drift velocity averaged over all the particles so when you average this particular expression because in principle the velocity of the particle after the collision is going to become random therefore average of V naught I can make is 0 because it is equal probability that the particle will get scattered in all the directions so average of V naught I take it to be 0 now this we take the average this is constant this is constant this is constant we take the average of t with that it will be the relaxation time so the drift velocity from that very simple expression he gets as minus ee tau upon m this is what we call as a drift velocity because the electron is drifting it is almost in the typical way as the air drifts though actual it may be the electron may still be moving but their average velocity may be 0 it is almost like air drift so there are molecules which are still moving at a great velocity in this particular all air molecules but we do not see when there is a net drift of this particular thing then we feel that there is a breeze going on exactly the same way now let us assume that the total number of n electrons per unit volume into the system then we can define the current density standard expression of current density is minus n e times drift velocity we put the expression of drift velocity if you have this if you have just now calculated it becomes n e square tau by m so this n e square tau by m is called conducting the material and you get the expression j is equal to sigma e where sigma is n e square tau by m it is called the conductivity of the material and resistivity of the material is gone upon sigma now what I insist that this expression is nothing but the ohm's law normally ohm's law is given in the form of v is equal to i r but you know from point of view of solid state phases okay generally current is not very important current density is more important okay the voltage is not important the field is more important resistance is not more important difference in the shape of the sample okay what is more important is the resistivity or conductivity this is called the resistivity of the material as I said that you know v is equal to i r in principle is exactly the same thing if you convert this electrical field into the potential and take particular shape of the material okay let us suppose later it could be wire it could be whatever it is okay then same thing can be written in the form of v is equal to i r so it is nothing but ohm's law now general these are the typical values of the resistivities which normally we observe of course resistivity depends on a lot of factors it depends on temperature depends on how pure you are able to make the material it depends on purity so it depends on the purity of the material but a typical value these are just the typical values I would say which I have taken from a textbook the resistivity of copper at 77 k which is the boiling point of liquid hydrogen is 0.3 micro ohm centimeter just take care of the units because these units are neither SI units nor Gaussian units okay ohm is supposed to be a SI unit but centimeter is actually Gaussian unit somehow in solid state physics people are used to using the units ohm centimeter probably due to semiconductor guys because you know they always find if they express an ohm centimeter one ohm centimeter two ohm centimeter that is the type of order of the resistivities whatever it is generally resistivities are expressed in terms of ohm centimeter which is not a very good thing so this is what I have used and if you go to 273 k which is room temperature the resistivity becomes higher which is 1.56 micro ohm centimeter now you use this particular values in this particular expression okay now what you can do you can approximately estimate what will be the value of N okay of course these days we can estimate easily assuming that for example let us take alkali metals okay the outermost electron which is the last s shell electron becomes free so each atom gives you one particular electron okay so you know how many number of atoms are there per unit volume if you know the density of the material if you know the atomic weight of the material you can always calculate total number of atoms per unit volume and assume that each of these atoms is contributing one free electron so you can estimate what will be your value of N and you can estimate what is your value of N now I have already got the experimental value of resistivity so I can know what is the experimental value of conductivity just take inverse of that particular thing okay e we know fundamental constant m we know what is the mass of the electron okay you can estimate what will be the value of tau which is the relaxation time okay now if you do that particular thing the tau turns out to be of the order of 10 to the power minus 9 second you know it is fairly small it means the electron is able to accelerate only about 10 to the power minus 9 second no not really okay please minus I am probably confused with numbers I will number is much smaller than that thing yeah j is equal to sigma i is equal to sigma i I don't think this is also known as microscopic forms of ohms law at least I have never heard about this particular uses of the term or nor I have myself ever used it we always call it ohms law because as far as I am concerned there is no difference between this and ohms law okay I can give you an example you take particular wire just convert no electric field into you know voltage okay just e by d okay and just take the you know current density take you know current density is current divided by area from area of cross section it will just come out to be is equal to a r at least I do not form I mean I do not know maybe there's some person who likes to call it but I do not know hello sir yeah it is called as microscopic form because the equation carries a term sigma which can be represented in terms of carrier concentration mobility of charge carriers yeah so the conventional v is equal to IR law it doesn't talk about the charge carriers involved in conduction now when you write the same law for semiconductors you talk about both electron holes contributing for the current and that's the reason it is generally preferred as a microscopy I mean I at least I have at least I say as I say this is a matter of again semantics I don't I mean personally I don't see any reason why it should be called different when I can derive one from the other where they're exactly the same identical things you know you're putting in a different form okay I personally don't I mean maybe there's something who prefer it to call it because of whatever reason because of what reason you have called it but at least as I say I have not I'm sorry I have not come across any book who calls it microscopic form of Ohm's law I mean take any standard solid state Facebook you know take it a Shroft murmur etc everyone will call it Ohm's law this is Ohm's law but maybe there's some authors who have used it I will not deny that okay now we define something what is called a mean free path approximately the distance that a particular electron travels before it suffers a collision now this particular thing is defined not as drift velocity multiplied by time but the rms velocity multiplied by time because as I said the electrons actually speed is much different okay the average velocity is different see like as I said in this particular hall their molecules air molecules which are traveling the reasonably large velocities but I don't feel them because average velocity all these molecules become zero but if you look at the particular molecule if I want to ask the question how much a molecule actually travels before it suffers a collision then I have to take its actual speed into consideration not the average speed and that actual speed is supposed to be given by the rms speed okay so therefore we should write half mv square is equal to 3 by 2 kT using again classical our kinetic theory of gases and from that you can calculate what will be the value of rms if you do that I'm sorry that's not I was telling that I thought I was getting confused the relaxation time turns out to be of the order of minus 14 seconds and not minus 9 seconds and rms speed turns out to be of the order of 10 to power 5 meters per second if you just do that calculation order of magnitude and therefore the mean free path turns out to be of the order of 10 to power minus 9 meters which is typically 10 angstrom a nanometer which is typically of the order of 10 angstrom this particular value is a very very reasonable value because generally it's known that interatomic distances are of the order of few angstrom okay most of the lattice constants are 4 5 angstrom so inter distance between the atoms are of that particular order and that's what Drude had assumed that when it suffers a when electron suffers a collision because of this positive course so having a mean free path of the order of interatomic distances is a very very reasonable assumption it appears to be whole thing appears to be fairly consistent here's the assumption that these electrons suffer a collision from positive charges positive charges being approximated distance of nanometer and mean free path also turning out to be of the order of nanometer so everything looks to be consistent how are there certain problems this is what I'm trying to tell about the problems first few problems that I'm going to tell are probably we are not known at that particular time they are much better known now because the data and understanding of the subject become more clear but one thing which for example I have already told you that resistance or resistivity for example depends quite drastically on temperature the value which I had given you it changes only almost by an order of magnitude as we go from 77 k to room temperature for copper now here we do not we are not very clear because the only factor on which the resistivity depends is relaxation time why relaxation time between 77 k and room temperature will become so large that resistivity becomes one order of magnitude larger after all those atoms I mean those course are still there at 77 k they are also present at room temperature okay probably they are vibrating a little bit more okay I can understand probably their scattering will go up slightly but I cannot understand that that will change almost by an order of magnitude not only that it also depends on the crystalline quality and purity if you make impure material if you take pure metal as pure as the material you will find conductivity to be much better of course as I said these data were not available at that particular time if you make a material pure okay conductivity will be much larger if you put certain impurities you suddenly find that conductivity goes on the resistivity increases okay it also depends on the crystalline quality if you make a very good quality single crystal of the material okay then you will find that resistivity is very low conductivity is very large if you create certain defects this locations etc you make material polycrystalline or you make material amorphous okay you will find the resistivity goes up now again it is not very clear because these are course always there whether you are talking crystalline material whether you are talking polycrystalline material or whether you are talking amorphous material okay why the resistivity depends so much on the crystalline quality why it depends on purity if instead of copper I have added some silver there okay why silver item of course silver core similar to copper core why the resistivity will change by so much of order thank you this was not clear okay I can just tell that these days of course possible to get almost mean free path of the order of a centimeter it is possible to do it you know you cannot understand how mean free paths of that larger by magnitude can all can be obtained but as I said many of these things are known now at that particular time during the time of dude the biggest objection came from the last statement which I am making overall small contribution to the specific heat of solid this was the major objection against the two theory which people could not understand or root could not explain let me try to explain this particular part little bit in more detail I have not done currently theory of gaseous for that matter the classical part in detail but I am pretty sure that all of you know about Deolong and Patith law which is a very very famous law which essentially tells that the molar specific heat at constant volume for all the solids is 3R normally when it is taught in high school it is taught in the form of product of molecular weight and the specific heat but there normally in high school we talk of specific heat per gram but in solid state physics we always talk of molar specific heat one mole of the material and then we talk of molar specific heat at constant volume and that depends on what we call as derivative internal energy with respect to temperature because at constant volume PDV term is 0 so it is just du by dt that is the specific heat at constant volume and the total energy of the system can always be calculated by equipartition law which actually have flexure transparency rather hurriedly last time which tells you that because a particular atom inside is solid if you are talking about a conventional solid it has 3 degrees of freedom per atom because it can vibrate in 3 different directions independently and cause all these are vibration so there is a half kT of kinetic energy half kT of potential energy so there is a total of kT energy per degree of freedom so there is a 3 times kT and because they are NA in a mole they are NA that is avocado number of atoms so 3 NA into kT will be the total energy you differentiate with respect to temperature you get 3 NA times k which is 3R now problem is that in the case of metals now we had just now said that there is another possibility there is another channel for heat to be absorbed because these particular atoms are also now leaving let us assume one electron per atom so each atom is contributing one electron which becomes free so when I am heating the material not only the heat is being consumed for these particular atoms to vibrate but will also be used to supply kinetic energy to the pre electrons okay in fact when I have used half mv rms square is equal to 3 by 2 kT that is not why I was assuming that when I have heated the system then these particular electrons have also gained energy because of the heat so if I am looking at the total internal energy of the system not only I should use what we call as a lattice contribution or contribution due to vibration of the atom but I must also use assuming that each atom is contributing one free electron the kinetic energy gained by these electrons when I heated them to a temperature T now again an electron will have 3 degrees of freedom because it can move in 3 different directions but it cannot vibrate because it is free okay so there is only half kT of kinetic energy contribution therefore for each electron I must have a half kT additional amount of energy so therefore if everything whatever I have said is true then what I should see that for the case of metals I must get a specific heat of 4.5 R knock 3 R so in principle if everything is consistent if everything is consistent because of the kinetic theory of gases then what I should observe that if I am having an insulating solid then the specific heat of that particular solid should be 3 R and specific heat for a metal should be 4.5 R but that was never observed it was observed of course let us be very very clear that this value of 3 R is only the high temperature limit of the specific heat because specific heat goes tends to zero as we lower the temperature okay once we have much more modern theories we know that 3 R is only the high temperature limit of specific heat for the solids but nevertheless whether you are talking about metals or whether you are talking about insulators you still get the value 3 R you do not get 4.5 R as if these electrons are not gaining energy from the heat okay if the free electrons are not gaining energy from the heat then whatever I have said earlier about half mv square is going to 3 by 2 get into the meaningless this was one of the biggest objection against the Drude model. After that when the quantum mechanics came and the concept of quantum statistics came there was a huge period of time the second model which is also a free electron model which got developed which is called Somerfeld model again free electron all the assumptions of Somerfeld were exactly identical as the original Drude's model no changes okay only difference is that he applied quantum mechanics to the problem and quantum statistics to the problem. He realized that when we are talking of these electrons electrons cannot be treated as a classical particle even though they may be free I must treat these electrons as if they are bound inside a box and therefore whatever we have found out the energy states of a particle in a three-dimensional box those are the energy states I must use because I am still assuming that electron electron interaction is zero I am still assuming that electron core interaction is zero so I can treat whatever was the energy states of a particle in a three-dimensional box same energy states can be used for these particular electrons of course the number of electrons will be very large if you are taking one mole of the substance will be typically 10 to 423 electrons that is the order of Avigato number okay but nevertheless whatever they are okay they are going to occupy only those states which are the quantum energy states. The second thing that he mentioned he proposed that electrons are the particles which are non-integral spins so they must obey Fermi direct statistics it means one particular state cannot be occupied by more than one electron or if you include spin two electron all right so the states that we have to remember in the normal Schrodinger equation we do not talk about spins so whatever state that we have derived from quantum mechanics in principle can be occupied by two electron one by spin up and one by spin down so if we take the ground state of let us say particle in three-dimensional box the ground state will be given will be occupied by two electrons. Now that makes things quite different because though we have 10 to power 23 electrons okay if you go to the ground state because these are fermions which obey Fermi direct statistics even at 0 degree Kelvin all the 10 to power 23 electrons cannot go and occupy the ground state because they are fermions so only two electrons will be occupying the ground state all right now next electron that we are supplying to the system okay have necessary to go to the higher energy states and remember we have to accommodate something like 10 to power 23 electrons you will find that even we are at 0 degree Kelvin you will find a states I mean a large number of states even at 0 degree Kelvin will get filled so that is what makes things different as far as Somerfeld model is concerned so let us just look into slightly more detail about the Somerfeld model let me just read uses all the assumption of good model but applied quantum mechanics and quantum statistics to the problem assume that the energy states of a free electron in a metal are similar to that of a particle in the three-dimensional box electrons obey Fermi direct statistics so these are the only differences which he made otherwise everything is same as free electron I mean this is also called free electron theory Somerfeld is also free electron theory now there is only one small difference many times we use the same old boundary condition which we normally use for a particle in a three-dimensional box but then we have some difficulty if you start talking about dynamics so let me just mention a few minutes about this particular boundary condition when we solve particle in one-dimensional box the boundary condition that we apply is that wave function is 0 at x is equal to 0 and x is equal to L probably all of you are aware about it that when we solve the quantum mechanics problem of a particle in a box which I could not do I did it because you expect that wave function will disappear at the two ends where potential energy becomes infinity so you apply psi is equal to 0 at x is equal to 0 and psi is equal to 0 at x is equal to L similarly when you solve the problem for a particle in three-dimensional box in a quantum mechanics course okay we apply exactly the same boundary condition that wave function disappears at x is equal to 0 and x is equal to L then y is equal to 0 and y is equal to L z is equal to 0 and z is equal to L okay that is the way we apply and then we get certain energy states however when we want to apply the same idea to solve the state physics we land up into a small problem so let me just illustrate this particular problem so that you know there is certain amount of clarity the problem is that the solutions are that those equations for example if you take particle even let us talk about particle in one-dimensional box because that is much more we are much more familiar with that okay the solutions of that we get are stationary wave solutions now what happens when we are trying to describe the motion of electrons inside a solid we have to assign a particular direction we cannot accept solutions which are stationary wave because in a stationary wave there is no net transfer of energy so if we are going to associate a wave function to the electron okay we never like to associate a wave function to the electron which is in the form of a standing wave see like when we are talking of a scattering from a particular you know barrier we wanted to put them in the form of periods for ikx because we want to assign a direction therefore we also like to obtain as a result of solution of Schrodinger equation solutions which are not really standing there so Born and von Karman had proposed a different type of boundary conditions these are also called periodic boundary conditions okay strictly speaking there is no clear logic of these particular boundary conditions though people give a lot of logic if you look at any book on solid state physics for that matter for any book even on modern physics they will give that solid may be treated as circle or whatever it is and you know the ends may not be free but if it I like one of the books which are very very boldly writes the sticking spring these are only arguments for argument sake let us accept that this works that is why we are going to do so the boundary condition that we apply that instead of saying that wave function at this end and this end is 0 we say wave function at the two ends are equal that is all we do okay so rather than making it 0 we say that whatever is the wave function here same should be the wave function here now you can say what is the reason the wave function should be here it is just to somehow address that there is a boundary there okay it could be something different but I mean as I said one of the arguments which is given if the normally the solid is so large in comparison to inter dormitiveness so these ends are immaterial so if the ends are immaterial so I take them to be the same but as I said these if you go deeper into that the arguments are not really excellent arguments but these are the arguments which people have been using almost for almost 100 years these are the boundary conditions again one of the dark area where you cannot really 100% explain but nevertheless it works so these are the boundary conditions that we use they the net advantage of this particular using boundary condition is that we do not get stationary waves even negative values of k become allowed and therefore you can assign a direction to the motion of the electron so if you use this particular boundary condition this is what you get in fact the density of state turns out to be exactly identical so even if you use that stationary wave boundary condition density of state there is never a problem so if we apply these boundary conditions which we I mean remember in the case of particle in three-dimensional box you apply this particular thing that at x this is equal to 0 then put l and then put it equal to 0 all I am putting here that this plus this is equal to this plus this I think that bz term I have probably missed it there should have been a plus you know kz term also does not matter no so basically what you are getting kx is equal to 2n pi by l in that case you are getting kx is equal to n pi by l here you get 2n pi by l but here the negative values of k also become allowed so overall the total number of states allowed remain same except that in the normal particle in a box you have only positive values of k's are allowed but they are half spaced here both positive and negative values are allowed but the spacing becomes double now using that you can always find out density of state so you take all the allowed values of k and plot into it because eventually we have to deal with a very large number of states okay because they are going to almost end up out on the field particles that we have to accommodate so we start with quantum states but then we realize because l is going to be very very large here okay so therefore the spacing between them is going to be extremely fine it is going to be extremely fine mesh okay so for the purpose of density of state we calculate them and in the form of a continuous distribution so that is what we do so we plot of course this is highly exaggerated and it is only in 2 dimension we have to do in 3 dimension we take the values of kx ky kz and plot into a 3 dimensional graph paper okay so all values are allowed so you have one point allowed here one point allowed here one point allowed here like that you plot all of them once you plot all of them then you take a particular value of k and draw a sphere in this particular thing you take another value of k which is slightly larger than k which is k plus dk you draw another sphere then find out how many number of these points are going to lie in this small shell that is going to be created by these two spheres because of the intersection of these two spheres one of radius k another of radius k plus dk now whatever are the number of points which are falling in this particular shell okay that will be the number of points or number of states allowed between in between wave vector k and k plus dk one thing which I would like to always insist on this particular thing and make I mean we basically point it out that boundary conditions are always applied in the real space at x is equal to this something happens the wave function becomes 0 or wave function becomes equal so whenever we are applying quantization condition in solid state physics we always limit the values of ks okay so from that you will always get that these are the allowed values of wave vectors okay on the other hand statistics like fermi state both sides and status depends on the energy okay it always gives you what is the occupancy of a level of a particular energy okay so I have to always convert from k space to e space energy space for that we use what we call as a dispersionization a relationship between k and e which is one of the most important part of any solid state physics problem relationship between k and e of course this is a free electron k so the e versus k relationship becomes very very simple because e is just equal to p square by 2 m h square k square by 2 m k represents the momentum so e is just equal to h square k square by 2 m now we find out what will be the volume of this particular shell volume of this particular shell this surface area is 4 pi k square and thickness is dk so the volume of this particular shell will be 4 pi k square dk now how many points are expected to be following in this particular thing we realize that this particular thing is essentially like a simple cubic lattice with a lattice constant of 2 pi by l because the distance between these two atoms is these two points is 2 pi by l this is in the inverse of length units this distance is also 2 pi by l okay now so it is essentially like a simple cubic lattice where the lattice constant is 2 pi by l of course this is in the inverse of length of space now in simple cubic lattice we know that on the average there is one particular point per unit cube because there are eight corners of a cube if we assume this particular room to be a cube and each is being shared by eight cubes so on the average there is always one point per unit cube okay so whatever are the number of what is the volume whatever is the volume okay if we divide by l cube upon 2 pi by 2 pi cube that is the total number of points which are expected to follow this so this is what I have said 4 pi k square dk is the volume of this particular shell and on the average you expect one point in the volume 2 pi by l cube therefore total number of points that are expected to fall is this particular factor l cube this particular thing this thing this particular factor we multiplied by 2 because each of the state in principle can be occupied by 2 electrons one by spin up and another by spin down so spin generation is taken into consideration so you multiplied by t the relationship between e and energy and momentum is very simple e is equal to h square k square by 2 m so you convert this particular thing from k space to energy space you just change the variable this is what you will get and this is the well known density of state function this is generally called density of state which goes as e to the power half pre-electron density of state goes as e to the power half very well known now if I have to calculate how many number of points are lying between energy e and e plus d see these are the number of states these are the total number of chairs in the room okay if they are occupied or they are not occupied depends on statistics so if I have to find out what is the probability of what are the numbers of electrons which are likely to lie between energy e and e plus d I must multiply by the Fermi Dirac function so let us be very clear g e d is the total number of states available total number of chairs available okay how many of them are occupied this is given by any d and this any d has to be obtained by multiplying this by the Fermi Dirac function because Fermi Dirac tells you what is the probability of occupancy of this particular state which is as all of you know Fermi Dirac status is given by 1 upon e to the power e minus e f by k t plus 1 this is the Fermi Dirac function of course there is unknown which is called Fermi energy which has to be evaluated by knowing the total number of particles that is what we are going to discuss just now all right so this is the total number of occupancy now this particular thing is rather easy to evaluate at t is equal to 0 because at t is equal to 0 what you will find that if I put in this particular expression t is equal to 0 denominator is 0 and if e is less than e f then this becomes infinity minus infinity it becomes minus infinity this particular factor becomes 0 so overall factor becomes 1 if e is greater than e f then this becomes plus infinity therefore this one can be elected this 1 upon infinity becomes 0 which makes sense because at 0 degree Kelvin you expect states up till a particular energy to be completely occupied because you are not giving them any additional energy to the system and after you have accommodated all your electrons okay after that all the states are going to be empty so remember all people would have liked the particle would have liked to go to the lowest available energy state if the energy state is not available at 0 degree Kelvin then particle is forced to go to higher energy state but it will never happen at 0 degree Kelvin that it will leave a vacant state at a lower energy and go to higher energy state when you give them additional energy in the form of temperature then it is possible but otherwise at 0 degree Kelvin it is not possible it will always take the lowest unoccupied state okay so if I do that particular thing I can calculate that t is equal to 0 and calculate what is the e f what is for mean energy yes we can tell whether e may or may not the electron will lie at the level e f in that case see that that is precisely what I am trying to say if at all you have a particular state which is exactly at that particular energy is equal to e f okay depends on how many number of electrons you have been able to occupy okay you are talking you are you have to complete 10 power 23 electrons now that particular state at e is equal to e f may let us have a degeneracy I am just giving some number or arbitrary number it is 100 okay at the most it is going to be 200 electrons now it depends when your electrons get exhausted if you have only 5 electrons left out of 200 there will be only 5 electron states okay other will be empty but you are talking of electrons of 4 or 5 electrons out of 10 power 23 okay I mean this is again as I say taking skin out of the you know hair so we never talk about it so we know that below e f all are occupied above that all are unoccupied all are filled what happens at e f we do not bother about it in the formula also if you put e equal to e f then 0 by 0 indeterminate will come see at this time it is a discontinuity okay so if you take the Fermi energy at this is what you calculate if you have total number of n electrons okay so I calculate g e d because the Fermi factor is either 1 or 0 so normally I should have integrated up to infinite energy but we know this particular factor will be 1 for all energies up to Fermi energy and will be 0 for energies later so this integral reduces to 0 to e f into g e so this any factor is either 1 or 0 we put the expression there of g e d which we have evaluated then solve it you get this is the expression of Fermi energy it means h square you know 2 m you know you are supposed to know total number of atoms per unit volume total number of electrons per unit volume okay which for example at least for alkali metals that there is 1 electron per atom so no total number of atoms per unit volume okay from that you can calculate what will be the Fermi energy so this is just for typically for sodium for sodium the density turns out to be 0.971 gram per cc atomic weight is 22.9 so you can calculate n by v so this is the factor put in the expression you get for Fermi the Fermi energy to be 3.16 electron volt this number is a very very interesting number because it tells you that even at 0 degree Kelvin the electrons have energy of 3.16 electron volt in root model okay at room temperature the electron energy was of the order that you can calculate the average energy turn out to be 3 by 5 but let us not bother about it see at the time in the roots model an electron at room temperature will have a energy of 3 by 2 kT and as I have just now said kT is 1 upon 40 electron volt so the value of energy of an electron at room temperature will be equal to 3 by 2 divided by 40 okay which is typically of the order 0.04 electron volt at room temperature while in the case of Sumerfeld model even at 0 degree Kelvin you have electrons which have energies which are two higher orders two orders higher in magnitude so electrons are having such large I remember there is no potential energy here they are free so all this energy is kinetic energy so even at 0 degree Kelvin you find the electrons are moving with very very large velocities even at 0 degree Kelvin so this is a big contrast from roots model now you can also calculate Fermi energy at finite temperature this is the expression normally we do in MSc class we actually evaluate this particular expression that Fermi energy is also actually shows a small temperature dependence but this temperature depends is fairly small because at room temperature the kT is 1 upon 40 electron volt and EF0 is typically 3 to 4 electron volt and there is a square term here so this particular term will be of the order of 1 10 to power minus 4 so to a good approximation we can assume for metals remember for metals that Fermi energy is temperature independent okay though it has a very small dependence very very small dependent but it can always be ignored so if we assume that whatever is the value of Fermi energy that we have obtained at room temperature for all practical purposes okay I can use the same value of Fermi energy now Fermi energy is also expressed into various units Fermi velocity Fermi wave vector Fermi temperature let us not go into the detail this is the table which gives the Fermi energy in electron volts for most of the metals so alkali metals these are the values if you go to noble metals like copper with 7 electron volts silvers 5.51 electron volts now these are some of the observations Fermi energy is around two orders of magnitude higher than the value of kT at room temperature even though the energy of the system at T is equal to 0 is much higher than the average energy of the classical gas at room temperature increase with T is very very small now let us understand this particular factor let us go back to our Fermi function this was my Fermi function here you have E minus EF by kT in fact what I normally do ask students that why don't you calculate this particular expression for different values of E minus EF by kT remember there is an exponential factor here okay you realize that if E minus EF is let us say plus 10 kT it means Fermi energy was approximately 3 to 4 electron volt okay 10 kT will be 0.25 electron volt then this factor will be e raise power 10 e raise power 10 is a huge number okay one can be neglected probability will be essentially 0 all right if I put E minus EF is equal to minus 10 kT okay probability will be essentially 1 because this factor will be essentially minus infinity almost like that in principle because of this exponential factor you will find that E minus EF by kT will be different from 0 or 1 only around approximately kT of per minute because that is where I mean you can just do a small calculate a small calculate and calculate the values this is what we give as a problem to the student just calculate this number you will find that this will be more close to 1 or 0 only time when it will be different from 0 to 1 is only around kT of this particular Fermi energy which also makes sense physically because most of the electrons are so deep rooted that by giving them such a small energy tiny energy 0.025 only those electrons which are very very close to Fermi energy they may be able to get those energies but a electron which is now deep rooted inside and you are giving them 0.025 and you expect that this particular electron will jump to a state which is available only of 3 electron volt above chances are very very low. So, essentially only the electrons which are very close to the Fermi energy they are the one which are going to get occupied they are going to leave their states and going to occupy high energy states. So, most of the excitation takes place only close to the Fermi energy only the electrons which are close to the Fermi energy are the ones which get affected by heat they go and occupy higher energy states. Most of the electrons which are deep rooted they are not affected by temperature at all. So, what is happening that when I am at T is equal to 0 they have lot of energy but when I heat the heat is never absorbed by the system because only few electrons which are close to the Fermi energy they are the one which can actually go and go to higher energy states. Therefore, we can understand why free electron contribution to specificity small because specific heat depends on rate of increase of energy D e dt it does not depend on e it depends on D e dt e is very large but D e dt is very small. Therefore, free electrons do not contribute drastically to the specificity of the system. So, this is what I have written here. Even though the energy of the system at T is equal to 0 is much higher than the average energy of the classical gas at temperature it is increased with T is quite small. The free electron contribution to specificity is therefore quite small. It can be shown in fact again if you do calculation it can be shown that the free electron contribution to specific heat is proportional to temperature and hence can be seen at very low temperature. In fact, this lot of people do this experiments if you go to very low temperature in fact you can put the numbers only below 10 Kelvin you will start finding that free electron contribution can be seen separately. See if you are knowing about the device law the lattice contribution goes as T cube that is called device T cube law very well known the specific heat at low temperature because of the vibration goes as T cube while free electron contribution goes as T is proportional to T. So, if you are going to very low temperature it is the T contribution which will start dominating and that will happen typically around 4 to 5 degree Kelvin. So, if you are able to make measurements below that temperature you go to very very low temperature and make the measurements of course these days it is possible to do those measurements then you will find a clear cut free electron contribution. So, free electrons do contribute but they contribute very very small and at room temperature the contribution absolutely small. Now, only electrons close to Fermi energy take part. So, there is one particular part which we have been able to explain but we go off into another part because conduction also will be taken care by only a very small number of electron which are close to the Fermi energy. So, when I calculate the mean free path I cannot use RMS speed that I had calculated but I must use Fermi speed ok. So, in Sommerfeld model the mean free path is actually calculated by Fermi energy multiplied by tau. Tau we had already calculated by using our old pre-electron model ok. If you do that particular thing and because Fermi velocity is much larger ok then you will find that mean free paths have become very large. So, typically in Sommerfeld model the mean free path will turn out to be of the order of 200 to 300 angstrom 20 to 30 nanometers ok. Now, this contradicts from our earlier happiness when we said that this mean free path actually came out to be equal to almost inter atomic distances. Now, it is very difficult to believe that if you have atom every 1 nanometer ok the electron is able to travel 20 to 30 nanometers without suffering any collision very very difficult to assume. So, though we could understand on the business of Sommerfeld model why free electron contribution should be small to this specific heat but as far as mean free path is concerned we became off. There is one more experiment which I would like to mention which is Hall Effect experiment it is a very very interesting experiment is still even performed today for many of the model materials. Let us just discuss Hall Effect and its rule as far as the free electron theory is concerned. This was performed for the first time in 1879 by Hall much much older time much before the discovery of electron. It is very important research tool in fact if some of you have been working in the area of magnetism which is happens to be my area of research ok. One of the most common way of measuring magnetic field is using Hall Probe ok you put a small probe there it measures Hall Effect and from that you calculate what is the magnetic field. I mean that is not the most accurate method of calculating field but most of the common method to find out the magnetic field between any magnet ok. As far as this particular experiment is concerned why I am describing it here is that this particular Hall Effect can be used to determine the sign of charge carriers. See remember at the time of Hall we did not know that there are electrons ok. It can only tell you which were the sign charge people knew that there is a positive charge people knew that there is a negative charge. Charges are of two type it was known ok concept of electron was not normal alright. So, this particular experiment can be designed or is designed to show what is the sign of charge carrier whether it is positive charge carriers which carry the current or whether it is the negative charge carriers which carry the current. So, let us first describe this particular Hall Effect experiment. Here we use all the three dimension let us take sample in the form of a block a rectangular block and we use all the three dimensions. We allow I mean these x, y, z directions are also fixed you know the way we talk about them ok. So, conventionally we always say that current flows the conventional current flows along the positive x direction. So, there is x direction and that is the direction in which the conventional current flows alright. Then in the positive z direction we apply a magnetic field alright. Now if you have x direction you have a z direction plus y direction is fixed because you have to have a right handed system. So, you know what is plus y direction now question is what will happen in the y direction. So, let us try to understand it see there is a current flowing now there is a magnetic field applied. So, current is because of the motion of charge carriers because of Lorentz force these charge carriers will experience a force and you can very easily show that this force will be along the y direction plus y or minus y alright. Because of that there will be accumulation of charges on the y face and an electric field will be generated in the y direction until this field exactly cancels out the Lorentz field and at that time a steady state will reach. So, an outcome will be that along the y direction you will find a voltage developed which is generally called as Hall voltage ok. Let us understand this particular thing little more. Let us first assume that charge carriers are positive just for the sake of it sake of the arguments. So, if the charge carriers are positive it means they must all be flowing in the direction of the conventional current because the conventional current is plus x direction. Therefore, these charge carriers must be drifting along the plus x direction ok. Therefore, this v which is there in the Lorentz force I am writing as vx times i because this is along the i direction. Question yeah. Is it ok to write a physical to q times v cross b as Lorentz force because we have the other term. You see in this case of that electric field e times e that I am sort of ignoring in this particular case of course you know you have that. Sorry. See let me put it like that you are right you know you are 100 percent right because actually we should write e times e plus v cross b that is what is the term. This electric field has been used to cause this particular current ok. So, I am only looking at that particular force which is along the y direction you 100 percent right. If I do a more detailed analysis I must use that particular current also ok, but that does not matter the result outcome result turns out to be exactly identical. I have applied a field in x direction because of that I am getting current in the x direction ok. Now if I am getting the current in the x direction because of that there is a v otherwise drift velocity would have been 0. Yeah actually if you write e plus that thing yeah then it will automatically come. See I agree. E y will be equal to that thing I mean that force will be equal to. No you are 100 percent right in fact many times we do in that particular fashion also we also take the electric field separately we write e e and then try about x direction ok then we are talking only in terms of electric field ok. Here what I have done the electric field effect I have taken already in the form of current because that is anyway not going to give me a force along the y direction. So, I am looking only along the y direction other than the hall voltage there is a hall field also eventually in equilibrium state there will be two electrical fields. One will be along the x direction which has been responsible for me giving me the current another will be in the y direction ok which cancels this Lorentz force. So, that is my point if you keep it keep y there yeah e there then e y will be equal to the hall field. Exactly that will come automatically. I fully agree with you I mean I am that is another way of doing it I am doing it slightly different way I am I am I am not denying that you are incorrect I am saying that that is a probably more general way of doing it ok. So, my b is let us say bz times k and i cross k will be minus j. So, we will see feel that this particular forces the force on these particular charge carriers is in the minus y direction. Therefore, you will find that all the positive charges will accumulate at the bottom surface because this is the direction this has to be y direction. So, you will find that because of this a electric field will generate in the plus y direction alright. So, if you apply take a volt meter you will find the direction of the voltage. Now, if we assume negative charge carriers then this q becomes minus q and if there is a current in the plus x direction then these charge must be moving in the minus x direction ok. Therefore, this v also becomes minus i and there is a minus sign because of the charge carriers being negative. You find that the force direction is still minus e y and therefore, in this case what will happen that the negative charges will get accumulated at the bottom face and therefore, the field will be generated in the minus y direction. So, knowing the direction of the field along the along the y direction you will be able to find it out how much whether charge carriers are positive or negative. But in addition to this we can also get one more information from the Hall effect that is what I am is quite interesting. In equilibrium I expect that a field will be generated along the y direction which will be in this particular direction you must have net field equal to 0 as he has mentioned. We define a quantity which is called Hall coefficient. Hall coefficient is defined in terms of all the experimentally measured quantities. If you have allowed a current to flow in the x direction you know what will be the current density. If you applied a magnetic field you know what is the magnetic field value. Then eventually you will calculate what will be the voltage along the y direction. So, you will find out what will be e y. So, you take e y divided by g x by b z this is what is called Hall coefficient. This can be positive or negative it will be positive if the charge carriers are positive it will be negative if the charge carriers are negative. If we substitute all those numbers which I have done it earlier the magnitude of the Hall coefficient turns out to be equal to 1 upon n q where n is the total number of charge carriers per unit volume as we have discussed earlier. This comes because the current density is given by n q times v x just put these values here you get 1 upon n q ok. So, Hall effect can be used for two purposes it not only gives you the sign of charge carriers it also gives you their number the total number of charge carriers per unit volume. So, this particular experiment can be used to test free electron theory. You can find out whether this n that you have evaluated matches with our idea that there is one electron per let us say alkali metal which becomes free. Now, this is a table which I have taken again from a textbook which I will give you the difference. This table gives you the value of 1 upon r h divided by n e. This r h is experimentally measured quantity and n has been calculated just by assuming total number of atoms per unit volume alright. Now, if everything whatever I have said is correct then for all alkali metals I must get this particular expression to be equal to 1 because sodium is monovalent. So, you expect that this should be equal to 1, but when you actually do this measurement and evaluate n the way I have just now said you get for sodium the value 1.2. For potassium it is 1.1, rubidium it matches pretty well. For cesium it is 0.9. These are all alkali metals. If you go to noble metals they become slightly more off 1.3, 1.5. Alkali metals you can probably I mean say that is okay 10 percent 20 percent you know after all we have assumed electrons should be free which way they are not okay. So, we have made so many approximations which anybody can say that these are not correct approximations. So, if you are getting agreement within 10 percent you are happy. So, you would say that things seem to be matching pretty well as far as sodium, potassium or all alkali metals are concerned reasonably okay not very well though for noble metals silver, gold, copper seems to be matching okay I can tolerate it but not very happy. The biggest problem comes when we talk of metal which are in the second group of the periodic table okay or third column because that is where you would expect that you may have 2 electrons per atom or 3 electrons per atom for example if you talk of aluminum okay it should have 3 electrons per atom let us look at the table. If you take beryllium for which the valence is 2 or magnesium for which the valence is 2 you actually get hall coefficient to be positive you do not even get negative it appears that in this metals as if the charge carriers are positive and the number that we get has no bearing with the valence 0.2, 0.4, indium and aluminum they are also hall coefficient is positive okay while these are having 3 valence while what if everything of free electron theory would have been correct then I should have got here plus 3 or close to plus 3 okay here close to plus 2 close to plus 2. It clearly shows that free electron theory probably matches reasonably alright with alkali metals to somewhat tolerable limit for noble metals but definitely not for group 2 group 3 metals. This table the source is that famous Bupayesh Kraft and Berman solid state physics on that they have taken this particular time. So clearly free electron theory does not work completely definitely not for group 2 and group 3 metals. This is the failures of free electron theory can still not explain the variation of resistivity with impurities crystalline purity and temperature this problem existed even the roots model case cannot explain positive hall coefficient and number of charge carriers in some metals. Obviously things are not alright for a free electron theory we must go to a little more important theory a different theory. So another 10 minutes I will talk about that particular theory and then we will close this particular chapter. Next step is to assume that electron is not really free but it is actually in a periodic potential we will be talking about the crystalline solids in the next lecture. See in a regular crystal we arrange the atoms to be arranged in a periodic way. So each of these positive charges once these electrons have left them should be positive charges and there has to be a periodicity maintained in them because whatever happens close to this particular atom exactly same thing should happen close to this particular atom because this atom and this atom is perfectly I mean plays in exactly identical fashion that is what is the starting point of a lattice as we call it that each atom or each point of the lattice must be having identical environment. So the next thing is to talk about to discard the assumptions that electron in the solids are free. Next step is to consider that electron experiences a periodic potential inside a solid. Let us consider one dimensional case for simplicity. So let us assume that my potential energy is not 0 but it obeys this particular condition that v of x plus na must be equal to vx where a is what we call as a lattice constant. So at each distance of a things are exactly identical. So if you have one atom here another atom distance between them is a so whatever you expect as a potential here must be at same here whatever is the potential here must be same here because this atom and this atom are perfectly identically placed. So to solve this particular equation you have a potential energy function which you assume to be of the form vx plus na is equal to vx. I will just give the outcome of what happens I will not go into the details of this particular thing. In fact block for the first time tried to solve this particular problem for mathematics there are some independent I think flow cave other person who solved similar type of differential equation in the mathematics but in physics it is generally called as a Bloch's theorem and he found that the wave function of this particular electron in such a type of potential will be given in the form of yx into ux of e raised power i kx. Remember for free electron it was just e raised power i kx. Now all that has happened that instead of constant times e raised power i kx it becomes u of x into e raised power i kx it becomes a function of x and not any other function of x but this particular function must also obey the periodicity of the lattice. It must be ux plus na is equal to ux. Now actual shape of u would depend on what is the actual shape of v. We have not talked what is the shape of v all that I have assumed is a periodicity condition. Okay from that I say get that u will also be periodic what will be the actual function of u of x will depend on what is v of x. Let me come to this particular thing and then probably I will stop. I will summarize this particular thing in one or two statements just to tell what happened in earlier transmissions but let me come to this particular thing. See it can be shown that in the case of block electron the speed of the electron will be given by 1 upon h cross d e d energy as a function of wave vector. Actually what I have just run through the earlier transmissions which I have not displayed one can show that the wave vector of this particular electrons or what we call as a block electron does not have a unique value. And in the case of free electron h cross k actually represents a momentum of the particle here h cross k will not represent the momentum because momentum has to have a unique value. So it is generally called crystal momentum but what is the most important aspect which I want to emphasize and where I would like to end this particular thing is that this particular group velocity would be present even in the presence of completely periodic potential. That is the most important conclusion of block theorem which essentially tells that if potential is perfectly periodic the group velocity would not change. It means electron will not get scattered. Just mere presence of the potential would not cause electron to scatter. Electron scattering takes place when there is a departure from periodicity. That is why we always found out that when we add impurity the resistivity went up because if for example instead of silver I put in a gold atom there the charges in the nucleus of the gold atom is different. So the potential which will experience the electron will experience when it comes close to the gold atom will not be periodic. It is a departure from periodicity therefore this will cause scattering. If there is a defect in the crystal, if the material has become amorphous there is no periodicity therefore electrons will get scattered. If there is a surface it will cause scattering. When we heat the material the electrons are no longer there the ions or atoms are no longer at their regular position they start vibrating. This is the departure from periodicity. All these will cause actually the scattering of the electrons and that is why you will find that resistivity depends on these factors which we have already always found out but never are they able to understand. So essentially it is the change from periodicity that cause the scattering of the electron and not just mere presence of periodic potential. A perfectly periodic potential will not scatter electrons. I think I will stop here. Actually last is that when you were considering harmonic vibration of the lattice, is it like that harmonic phonons are not scattering or kind of thing? Any phonon whether harmonic or non-harmonic does not matter. If there is a presence of phonon there is always electron phonon scattering where you can still assume phonons to be because of harmonic interaction. See anharmonicity comes because of some other reasons. When there are two lattice waves and when they have to mix up that is caused by anharmonicity. But as far as electron phonon scattering is concerned I do not think anharmonicity is needed. When two lattice waves sort of get scattered with each other. When we talk of thermal conductivity we have to talk about anharmonicity. But when we talk of electrical conductivity as far as I know I do not think anharmonicity is needed. See crystal momentum is just h cross k. See let me just tell this point of transparency. See what you can show that any value of k if you add what we call as the reciprocal lattice vector which we will talk later can equally well represent the wave vector. So k that see remember what is k here? This is the k. Now this whole wave function can be rewritten in different fashion that instead of k here there is another k appearing and instead of that ux there is another ux which also has the same periodicity. It means the k that you are assigning the wave vector that you are assigning to the electron is not a unique value. It can have different values. Therefore k is not related to its momentum. All right. So we call h cross k as crystal momentum of the particle. But what happens when we talk of the semi-classical equation of motion which we talk? Essentially this crystal momentum almost acts as momentum. All right. So generally I mean in solid state physics whenever you talk about momentum people always mean crystal momentum. Always. See what is the band structure if we say what is the band structure of silicon or band structure of germanium. Band structure means relationship between e and k. What we call as a dispersion relation. When we say relationship between e and k it is always the relationship between energy and crystal momentum. In solid state physics we only talk about crystal momentum which are basically the wave vectors. Any other question? They are talking about the degenerate and non-degenerate semi-semi-semi-semi-semi-semi vector doped in that case the density of state is high and low. No, no. Well when we talk of the degenerate semi-semi-semi-semi-semi if I remember right we talk of highly doped semi-conductor. See what happens when you talk of Fermi energy in the case of semi-conductors things have to look slightly differently because let us say a talk of t is equal to 0. t is equal to 0 all the states in the valence band are filled and all the states in conduction band are empty. So in principle your Fermi energy can be anywhere in the band gap. Only thing it has to be in the band gap. All right because above this all the states are empty at t is equal to 0 below that every state are occupied. Now in order to calculate Fermi energy in the case of semi-conductors you use a different boundary condition altogether. You take the charge neutrality condition and then you determine Fermi energy. Now what happens when you heavily doped it you find that this particular Fermi energy for example if it is n type of doping the Fermi energy comes closer to the conduction band. If it is p type it comes closer to the valence band. If you doped it very very heavily then it is possible that it goes into one of the bands. Okay that is what is called generally degeneracy if I remember right. Okay thank you.