 You can follow along with this presentation using printed slides from the Nano Hub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Let's get started then. So today is lecture 8. We'll be talking about density of electronic states. And this is a way of summarizing the band structure information that we had been discussing for last few classes. We have solved the Schrodinger equation and have seen where the electrons sit and they do not all contribute to conduction equally. Now we are trying to not really remember the whole band structure. That's too much. And rather encapsulate the information in the information about the band gap and the effective masses, sort of how they vary. And this is sort of the third piece of information that again comes from the effective masses. So that is how we are going to summarize the information and you will see how it works. So we'll talk about calculation of density of electronic states and then give examples. I'll again talk a little bit about characterization of effective masses, putting a magnetic field in and letting the wing electrons swing around a little bit. But this time for a little bit more complicated material for silicon or germanium and then I will end. Now you must remember or you may remember that when you solve the chronic penny model for one dimensional solid, we got a solution which stayed between pi over A and minus pi over A, the Brillouin zone, and then there were a series of bands. Now in each band I pointed out at that time that if you have n number of atoms in that chain, then the total number of states in every band is n without spin. If you have spin included then you will have 2n because each space, two electrons, one with up and another down spin can sit. But in general n atoms, n states, 35 atoms, 35 states for each band. So for example here if you have 4 bands then you will have 35 multiplied by 4, 140 states sitting right here. Now what I want to know in most cases is not the total number because the total number is not very interesting. Really what I really need to know is the electrons that move, how many do I have in a given region? So whole band of course, but remember when the whole band is all occupied or all empty it doesn't carry any current anyway. So the one that we are interested in are partially full and partially empty. So in that case we are interested in a very little sliver of energy band where those electrons sit. And so in that case we will have to know how many per unit energy near the bottom of the bands or near the top of the bands. And that is where this density of state will come in very handy. So for example what we want to calculate many times is how many states are occupied up to a certain energy E? Or how many states per unit energy do I have? I will give you an example, things will become clearer. So this is the density of state, this is an information that you will use often and for any device analysis that you do, remember the MOSFET, the bipolar transistor that we will subsequently discuss, all of them this is a key quantity to remember. Again I have a chain of N atoms, A is the spacing, the total length will be L is N multiplied by A. The solution is given here on the right hand side and I want to know how many states do I have between K1 and K1 plus delta K. Now here I have just shown you for schematic purposes just three states. You realize that these individual states are separated by 2 pi over N A. Now N is 100 million and so therefore these states are actually very tiny, very closely spaced. I have just shown you two or three just for representation. This will actually be almost continuous number of state dots on that energy band diagram. Very easy to calculate how many states I have between K1 and K1 plus delta K or equivalently between E1 and E1 plus delta E. Because K and E are related to each other therefore the solution is the same. The same number will come out. So how many states do I have between those two boundaries E1 and E1 plus delta E? So first of all I will have to know what range I am talking about. So I have a range of delta K, this capital delta K that is the range of K vectors that I am interested in. And the points are separated by well that is the little delta K that is how much it is separated by. Now what is that factor of 2 sitting doing there? It is not spin. What it is you can see I just considered from 0 to pi over A the right branch. But of course I have 0 to minus pi over A the left branch I also have to consider because when I am talking about a given energy range both the plus and the minus both contribute. So therefore I will have to multiply by that factor of 2. Now this delta K, little delta K I know already 2 pi over NA. So that is how many states I have between for a energy range of delta E right? That is how many I calculated. So if I wanted to know how many per unit energy all I have to do is divide by delta E, the range of energy. And you can see that is exactly what I have done. The NA has flipped over on the top and the 2 have cancelled. I have a pi and delta E over delta K over delta E. So I am almost done. Somehow if I can evaluate this delta K over delta E how the slope changes? Is it going to be different for every band right? You realize because some bands are flatter some bands have more curvature. So this number is going to be different for every band. And as a result density of state for every band is going to be a different number. We will see. So let's continue I have just copied that same information state per unit energy. Now remember NA is L. L is the length of the chain, total length of the chain. And delta E over delta K how should I evaluate that? Well you remember that many times near the bottom of the band I can easily express this curvature information in terms of the effective mass H square K square divided by 2M star. And that relates E to K and taking a derivative is no problem that you can immediately see here. Now one question for you here. If I couldn't express the E K relationship with an effective mass how would I have defined the density of state? In this case this state how would I have done it? That's the question for you it should be able to answer that easily. Let's say it's a constant E K is a constant relationship. How should I have defined it? Anyway for this particular case when it's a parabolic density of state a parabolic band structure then I have this particular relationship. I insert this D E D K in the first equation. So that gives me an expression that has solely energy in it. Now if I wanted to know that instead of this sample of L what is per unit length? How much per unit length how many states I have? I'll just divide it by L. The bottom one is just a previous one divided by L per unit length at E that is density of state. So density of state is number of states I have per unit energy and per unit length. Now if I have 2 meter material I'll just take this number multiply by 2. So I'll know the total number of states. If I have 3 EV band energy I'll just multiply with 3 and that way I will get the density of state information. So this is a sort of par it's like a density per unit energy per unit length. Then I can manipulate it in any way I want. Now this is a material property once I calculate it this number is not going to change. It's silicon or germanium for a particular band whatever this number this number remains as is. And it will not change whether you apply a voltage here or operate the device in one way or other you see. So therefore this will number will appear as a constant in many of the textbooks and also tables let's say or you can calculate it. 1D is fine right so you can see how 1D works. Now how does it behave we can get a graphical look. So for example if I wanted to look how does the density of state behave as a function of energy. You can see that it goes as 1 over square root of E. And so starting from the bottom it will gradually decay away from the bottom you realize that right so 1 over E. And one thing is you can ask this question that what happens at E equals E naught right. E equals E naught does it blow up not really because remember these states are discrete points. So when you are actually close to E equals 0 you cannot just use this one and assign exactly set it equal to E 0 E equals E 0. But rather you will have to calculate the number of states just around that point. Now what will happen for the density of state for the red curves now there will be a corresponding similarly 1 over E. But this time E going below and then same for the blue one. And one thing you realize that this shape must be there because you realize because the area under the curve. The density of state multiplied by the width of the width of this region must be equal to how much N right. Remember N is the total number of states I have. So therefore if the band is very skinny my height the density of state must be larger than before right. And you can easily see that where that comes about from this from this expression. Now this is something this 1 over square root of E dependence you have seen it before. You may not remember do you remember that when you have a well potential well then the first bound level was at 1. Second bound level at 4. Third bound level at 9 N squared. So the spacing in one dimensional solid were gradually getting spaced apart. So when you wanted to do per unit energy density you realize that as you go up in energy your density of state will gradually come down. That is the information that is being reflected here. Now what about two dimensional density of states when you have a material like graphene or some other material that is two dimensional solid. In that case the spacing let's say is A and the other side is B let's say this is one of the five Bravais lattices do you remember. A not equal to B and the angle being 90 degrees for example. Now this I will not do this is your homework you should be able to do it yourself but convince yourself that amazingly density of state in a two dimensional solid does not depend on energy at all. That's a very interesting thing so you should be able to show that as a homework problem. Now let me move on to 3D because that's what is there in the textbook and many practical devices your Pentium and various computers of course are based on three dimensional solid. These are sort of material of the future one dimension and two dimension let's say three dimension is what we often need. Now there is a series of pictures here. Let's go through them one at a time. On the left hand top side I have taken a macroscopic cube meaning the side maybe let's say one millimeter the length is let's say one millimeter the width maybe you know two millimeter and height is three millimeter something big. I have taken that and once I have taken that I could have constructed the Brillouin zone. How do I kind of construct that if the lattice spacing is a then 2 pi pi over a and pi over minus pi over a remember the Brillouin zone. And within each one of them each solution point will be given by 2 pi over L that is the little spacing between the points the dots. This is 2 pi over NA and remember NA is L right the length and so one side will be 2 pi over L the other side will be 2 pi over W and the third side 2 pi over H and within this little box little tiny box there will be one solution sitting in the in the middle right. So if you want to calculate how many states you have for a given energy range of delta E what do you have to do well that is what is being done on the right most corner on the bottom. What I have drawn here is two concentric spheres two concentric spheres one with energy E1 and another with energy E1 plus delta E. This is just a shell of a cylinder right and we will try to see how many states do I have that I can fit in of those little boxes within that shell of that sphere that will give me how many states I have per unit energy. Okay now let's see this you are definitely know let's see whether you can understand this on the numerator what I have done is 4 by 3 pi k plus delta k cube what is that that is the volume of the red sphere that you can see on the bottom right. And 4 by 3 minus 4 by 3 pi k cube well that's the slightly smaller sphere that you have we have to take it out because we are just interested in the shell of the material. And what is there in the bottom I have this 8 pi cube divided by L, W and H what is that that's the volume of where individual one solution set individual point set. Now if you compare it with the one dimensional one there I had 2 pi over L because it was just one D so that's what I have. Then you can see L, W and H together we will call that a volume V this is the big volume volume V and this is a simple algebra well per unit energy what do you do divide both sides with delta E because that was the range right that was the range. And again you know how to relate K to E looks complicated but this is not rocket science here and once we have the DEDK for a parabolic band we are done. It says density of state per unit energy per unit volume now because it is 3D 1D per unit length 2D per unit area 3D per unit volume. And you can see the V has disappeared right look at the numerator compare and compare with the ones that are in the top then you will see the V has disappeared because this is per unit volume so therefore it's not there anymore. Now again there is something very interesting here do you see in 1D the density of state went as 1 over square root of E right. But in 3D and 2D it was constant I didn't show it you will do it in homework in 3D it goes as square root of E increases. And you can plot it out very nicely like this you can see that how the density of state changes for various bands. Every band has a different density of state because you can see that for band 3 I have effective mass M3. For band 2 I will have an effective mass M2 and therefore the density of states will also be different for different bands. Now have you seen this before this square root of E dependence have you seen this before. Now do you remember that I pointed out that when you look at in hydrogen atom and where the energy levels can sit. I mentioned that the difference between 1 and 3D there that in 1D they get spaced out as 1 over N squared. But in 3D they get closer together as 1 over N squared. Do you remember 1 over N squared minus 1 over M squared that is how we got the energy spacing. So further out they go out in energy the energy levels gets closer together. And therefore density per unit energy this keeps going up and that's what is being reflected. Of course not the same thing but it's a 3 dimensional solid the idea is very similar that's why it goes up as square root of E. You remember this right don't have to you know always don't have to derive things. You should be able to remember the formulas almost physically so that even when you don't have any notes or anything. If you had to write down a bunch of basic calculation you don't have to always hunt for textbooks or you know your notes. You should be able to say okay physically what is the content of this material and I should you should be able to at least in graduate level be able to remember that. What about density of states in specific material these are the general idea right we didn't talk about whether the material we just talked about whether silicon germanium we didn't say that. What about actual material well actual material sometimes are a little bit more complicated we'll see the easiest one to think about is gallium arsenide. Do you remember that gallium arsenide on the right hand side I have shown the E.K. diagram and then there are these two panels which has been pushed together the panel being gamma to X which was one panel and gamma to L being another panel. Now if you look at the conduction band around gamma 6 point the blue circle then you can see that going along the 100 direction along the X direction and going along the L direction the energy about comes out about the same right. And so at a given energy the amount of delta K you have to travel for that band is about the same either in the X direction or in the L direction that says that the constant energy surface for that band is approximately a spherical one. And that is what I have shown on the blue sphere on the slightly left to the figure and that's the constant energy surface for the conduction band of gallium arsenide. Well if it is a sphere then we have just calculated in three dimensional a big sphere inside the small sphere and that's the density of state for the conduction band you see I have the same formula. Now one thing I want to point out that you have to be a little careful about that I have a two in the denominator in the textbook sometimes you may or may not see the factor two sometimes they include the spin in there already if they include the spin the total number state will be two times larger so that two will not be there. That's maybe in many textbooks but in order to be consistent I am doing all the calculations without spin you see okay. So conduction band taken care of you have you know the conduction band the effective mass for the electrons you can put it in there and calculate this number. What about the valence band? Now in the valence band let's see it's a little bit carefully that's near the gamma eight point. In the gamma eight point you will see that there is actually two bands sitting on top of each other one is called a light hole band. Now why it is a hole you know right these are almost full and a few empty spaces I can take care of those to calculate transport. Now why is it called light hole and why is another called a heavy hole band? The reason is that if you have smaller curvature second derivative of energy with respect to k effective masses inversely proportional to the curvature. Shallow at the curvature electrons don't want to go from one place to another right. So in that case you can see that the light hole and heavy hole hand they have slightly different curvature and they have therefore slightly different effective masses. You can again see that this is approximately spherical and actually one can prove that as well and correspondingly the density of state will have that two components one from the heavy hole another from the light hole. Now do you see why I have to add them because if I am sitting at a constant energy I could either sit in the heavy hole band or I could sit in the light hole band. I could those states are all occupied it's like chairs here if you have two sets of chair on this side both have the same energy. Then some students could sit on the left side some student could sit on the right side and when I want to know where you could sit I will have to add both of them. But assume one set of chairs is significantly higher than the other one then of course the probability of those states being occupied are not the same. So therefore here I have to add it not always. Now in some cases you will have to also include the split of band but that's not necessary here if you go up in energy or go down in energy you can see at some point at a constant energy you not only have light hole heavy hole but you also have the split off. So then you have to add three pieces together for the density of state. You see this now this was good but these materials that are very useful like germanium and silicon they are turned out to be a little bit more complicated. Do you remember that the constant energy surface for germanium occur in the along the L direction and then there are eight L directions. Do you remember that's on the corner side that was from the top of the valence band that was the easiest point to access and therefore I have at that point a energy minimum and how many do I have I have eight. Not really because that minimum point is cut by the Brillouin zone in half and therefore instead of eight I actually just have four eight halves that's what I have. How do I calculate here that's a little bit more tricky that we are going to discuss next it's not sphere that's the problem you see had it been a sphere I could take a big sphere put a little small sphere small sphere inside. And I'll be done I could use that formula but it's not a sphere you say ellipsoid or spheroid what is the call ellipsoid let's say and then I'll have to I'll have to do the same calculation for them let's do that. So if I have a ellipsoids like that then I can always relate the energy e to the wave vector k1 k2 and k3. Now why am I writing it as k1 k2 k3 rather than kx ky kz this is because that's where ellipsoid could be oriented in any random direction you see some of them are in L direction some of them are in X direction so this is a general form I write it this way. Now I also have in the longitudinal side you see this has three accesses in the longitudinal side I have ml and on the transverse side I have the effective mass mt. Now in general this doesn't have to be the case you could have three different masses in three different direction for this particular material germanium silicon to just happens to be the same. Now you see I could divide e minus ec throughout take it to the right hand side and flip the h bar square from the top to the bottom couldn't I do that that's simple. And I could call this one this whole thing inside the bracket alpha square is just a name and the whole thing on the second and third one as beta square. So now I have k1 divided by alpha square k2 k2 square divided by beta square and the third term that's the equation of course for ellipsoid that we know right. So from here we can now we'll do now do a trick we'll say yes we could take two such ellipsoids and find the skin of it. And then try to see how many states I can put in there or I could take this football and squish it and make it a sphere. But you see the same than the points that I had in the original one they are sitting there I'm just squishing it so essentially the number of states will not change even when I squish it and make it a sphere right. Therefore I can do that trick and very easily calculate this so the top one in the green and the bottom one in the blue has the same volume I am defining it so so they have the same volume and if I want them to have the same volume what should I do. You see the for the green one on the top the volume is 4 by 3 alpha pi alpha beta square you know that right if the three axis is if the two axis is alpha beta and beta. Then in general I will have alpha beta square if three were different alpha beta gamma then I will put the three alpha beta gamma there this two happens to be the same. Now the N equivalent is that three equivalent the six equivalent values in the Brillouin zone because we are interested in six of them I will show you in a second. Now if I want to transform them to have the same volume I will define it as 4 by 3 pi k cubed right definition so from here I can easily calculate what the k effective is. Now I can insert the values of alpha and beta in this long expression and define and so called effective mass on the right hand side in blue which sort of represents how the electrons in the states equivalently move right I am just making it an equivalent description. Now this looks complicated but you can see this E minus EC business in the numerator lots of h bar squared they will all disappear. And in the end you will have that little expression over there that ML longitudinal effective mass that you will measure in a few seconds I will show how it's done. MT those are measured as soon as they are measured I will know the effective mass and when I know the effective mass I also know the density of state is that right I also know the density of state right there because as soon as I know the effective mass I can put that effective mass in the density of expression. You remember because now I can use the sphere equation right because I have just transformed between the sphere I can convert it and I can calculate it. You remember the 4 right half of 8 because the Brillouin zone cut it. Now what about silicon silicon is 6 apart from that everything else is the same why is it 6 why is it not 3. Why doesn't Brillouin zone just cut it in the middle for this case also and I shouldn't I have 3 right 6 in the direction half cut. Not really and that is because you have to be a little careful look at the yellow region that I have marked. The minimum point is really not at the zone boundary it is a little bit inside about 20% inside from the Brillouin zone so actually the constant energy surface starting from the bottom point is wholly within the Brillouin zone for low energy. As the energy gets gradually bigger what will happen the part of this football will be cut by the Brillouin zone. As it gets bigger and bigger and bigger you will see more and more sort of half is getting cut you see sort of a balloon type thing gradually getting bigger until it hits the Brillouin zone and when the Brillouin zone touches it that part it cuts. And bigger it gets more part it cuts and eventually it will become almost like half but for low energy where we are interested in it is wholly within the Brillouin zone and therefore the sort of the football stays in I have a factor of 6. Now you will have to do a complicated homework if you don't pay attention to how these things are being done you will not be able to do that but that will clarify all these concepts in one shot. Now finally I want to talk a little bit about the effective mass I have already told you about how to measure effective mass by experiment right in the last class but I want to tell you a little bit about how to calculate effective mass or major effective mass when you have this ellipsoidal type density of state right that's something we haven't discussed the previous class. Now very quickly if you remember then the measurement of effective mass involves that we apply a microwave frequency about 24 gigahertz let's say and then apply a magnetic field as the electron swings around then we correspondingly find that at particular magnetic field. There is a sharp absorption when the rotation of the electrons exactly matches that of the microwave signal then I have that and I could calculate from that information. Mu naught is known the magnetic field the resonance is known I could calculate the m star and that's something I have done before done in the last class you can see the electrons also swinging around in the case space. Now this part some of you have asked questions about that why is it when something goes around in real space in circles why is it in the case space also. You see in the right hand side is the case space that should it go around the circle also right that's something we should be able to understand see how it works out. Assume in the real space of x and y I have an electron the magnetic field is perpendicular to it and the electron is going around just like that. Now when electron is going around you can see when it is on the top it has a velocity along the x direction but it has no velocity in the y direction right tangent on that one. So if you look at the corresponding phase space in kx and ky you will represent that point in blue by a point on kx a finite value and ky equals zero you see that. Now when the electron comes here at the green point then it has only the negative ky velocity and therefore when you want to plot it in the phase space you will plot it here because that's negative ky and kx equals zero. And I don't have to explain you can see as the electrons goes around then on the phase space also this one will go around that should have been a little bit up. But this one goes around and as the electron is going in the real space the other electron is also going in the k space in the same way. Do you also realize that this must be a constant energy surface because h square k square divided by 2m plus h square k y square divided by 2m that's this. So that's why they go around in constant energy surface right. Now another thing very quickly that when this frequency of going around equals the microwave frequency at that point you have resonance right. And another thing if you wanted to plot at any point what was the ky component and if it's making an angle of theta you will say it's ky cosine theta right that is the ky how it the projection moves up and down on this particular y axis. Okay remember these two things because it is going to get a little complicated before it gets simplified. So that's why I just mentioned in the real space this when it executes a circle in the phase space on a constant energy it also an executions circle. By the way these experiments have to be done at very low temperature so that it doesn't get scattered the electrons when it wants to finish a orbit it doesn't get scattered. So these are generally done at liquid helium temperature these are not easy measurements to make. Now so these we saw in the last class that how to make the frequency and how to derive that formula. I just want to go and generalize this on a little bit I described it in the last class please see if you're not clear please see the notes from the last class. Now I have this germanium not little spheres on which electrons can go around. Now let's say I take the sample and put a magnetic field B along a particular direction. How many angles will it make? I have four axis right four body diagonal axis for germanium only for silicon how many? I have three right this lateral perpendicular and front and back three here I have four. So any magnetic field I bring in it is going to make four angles with respect to the four axis and what are those four angles? By the way so as soon as you put the magnetic field on the electrons now is going to go around this constant energy surface that's what I just explained right. But they are going to go around with different curvature the red ones has a different angle they are going to measure the transverse mass and the blue ones are going to measure the longitudinal mass and it will be all a mess. And how we'll have to deconvolve it and I'll show you how. But first see that there are this sort of eight equivalent direction right one one one direction you know you have done this homework before. You have done it when you have this one point ten you know you apply a force how many preces the whole thing can break they were trying to prepare you for this example so that you know that there are equivalent of eight one one and directions and essentially four unique angles with respect to the magnetic field. Now I want to prove this formula in next two slides and then we'll be done. This formula is it looks a little complicated it's not really what we'll measure is theta and mc these will measure and I'll show you how to measure it. And what we'll get out is the information about mt the longitudinal and the transverse mass this is my output my input is theta and mc so this is how it works. Now let's think just generalize the calculation we already did in the last class that for an electron in the magnetic field in the arbitrary direction I can always write it in this form. v cross b m dv dt the right hand side is mass multiplied by the acceleration right you know that and the left hand side is v cross b. Now in last time it was particularly simple it was spherical perpendicular so I could just make v cross b just a one value but in general I should write it this way. Please try try it out at home and see whether you agree just you know write out the vx as an x y and z component and y at vx by and busy component just write it out. And the m is a diagonal matrix so one element will be ml the next diagonal will be mt the third diagonal will be also mt. So this is a diagonal matrix multiply this out two minutes you'll be you'll see how this one comes about now what you want to do in the next slide is take the second equation. Equation for y and take a derivative once and replace the first derivative on the left hand side by using the other two equations so you so that you get one equation for v y again a two minute thing just check it out it's not complicated at all. So now assume that corresponding to an ellipsoid the angle it makes the magnetic field the angle it makes is theta and so the components of bx by and busy are given by that sine and cosine theta. And this is what I just mentioned take the second equation and use the other two equations to find this. And this is a equation for v y and the y w squared if you do the derivation you will see it will come out something like this that I have written over here. Now what is the solution of that equation v y equals a sine omega t is that right or a cosine omega t do you remember five minutes ago I said when the electron goes around a orbit if you take a vertical projection. And it is going to execute a sine omega t going up and down this is exactly that so from this is telling you that the electron is actually going around the orbit at that point and when it's equal to omega not at that point is going to absorb the absorb the microwave. So from here the last step is obvious you see all you need to do is in the omega squared at the resonance condition omega equals omega not use the three sets of things definition omega not omega t and omega l put it in the top equation you get the bottom equation that's it. Now it difficult to explain on a slide PowerPoint slide but I really I'm telling you it is no more than five minute exercise for you so maybe you want to check it out. Now I am done because now this is the experimental results I have seen I have let's say magnetic field as a function of absorption in a real material. In this particular case the magnetic field in 1 0 plane so I am specifying where the magnetic field is so that I can calculate the angle with respect to the three axis or four axis and you can see those are the electron absorption picks. So I will have three picks I have three masses from the magnetic field I can use the formula on the right to calculate three mcs and I have three angles what is that all about why three why not four. Because it will turn out that for this particular case when the field is in one once in direction two of the angles are exactly the same. So instead of four in general you should have four but in this particular case you have three. So how many equations and how many unknowns well you have mc1 mc2 and mc3 from the magnetic field you have that and you have theta 1 theta 2 and theta 3 from the angles. So therefore I have actually two unknowns mt and ml and three equations so I am done from here I should be able to calculate mt and ml that I should insert in the effective density of state and that is how I will get the silicon and the germanium effective masses right density of states. Now one thing that I leave it as a homework is to prove that why are those I mean if you want to know about hole bands why there should be two valence band picks why should there be two. We just discussed one is light hole another is heavy hole remember so they are going to resonate at different times. So therefore you will have for two holes and how do you know those are valence band well when you change the angle of the magnetic field because the valence band is spherical. You see when you look at a sphere from this angle or from the other angle it is not going to change it is going to absorb in the same way. So when you change the angle the three electron peak will move around because with respect to the ellipsoid looking from various angles will look different but the spherical ones will look exactly the same. As a result that will tell you which one the holes and which are the electrons. So this is how actually people calculate and that is how check out whether the theory you do whether those are correct or not. That is it so we talked about measurement of effective mass and band gaps that is the information we are going to carry through not all this solution of the Schrodinger equation and everything. Because as far as we electrical engineers are concerned for transport properties conduction band and valence band is all there that is what we have for others there may be other bands. And I have beginning to hopefully convince you that only the fraction of the bands states are important for carrier transport and that and the states and the number of available states change with energy. That you can see because effective mass changes so of course the number of states change and the density of state captures that gap information. And this is something is very important hopefully you will take a few minutes when you go home and go through this lecture step by step. It is a little complicated lecture but it is I assume it is possible to understand for each one of you it should be easily be able to understand this material. Thank you very much.