The Kronig - Penney Model | Band Theory of Solids | By Dr. Kavita Segwal | @DrKavitaSegwal

[Music] foreign [Music] so today in this video lecture we'll see what is chronic Penny model the KP model and how it explains the original bands in the solids so let us see what is the use of this model so it explains origin of bands in the solids and it is classifies the solid on the basis of Band Theory as conductors semiconductors and insulator we have seen that the solids are categorized on the basis of the band Gap the bands in the solids as conductor semiconductor and insulators so this model basically provides a theoretical explanation for the origin of the bands in the solids so let us consider a 1D lattice for the sake of simplicity so we have seen that in a lattice in a crystal the atoms are arranged regularly so there is a periodicity in the crystal so let's see these are the atoms which are arranged periodically so if we talk about the atoms in the atom in KP model it is considered that an atom consists of two things one is a positive nucleus other another is the electronic shell surrounding it okay so the electrons which are in the electronic shells they experience a nuclear potential so nuclear potential it varies as distance from the nucleus so we have seen that the nuclear potential when Z is the charge of the nucleus the electron will experience electron will have this potential energy which varies as R so let's see this when R is small that is the electron is close to the nucleus it is having the r is small then potential energy VX this will be large but because this is negative so this value will be negative large so you can see from this figure this value is negative and a large value so v x is negative negative potential shows the attraction so it is in the attractive potential of the nucleus but as this R increases it means we are talking about the electrons far away from the nucleus so as the r increases the value of this VX this will decrease it means this will become less negative less negative it means it is moving towards the positive it is moving towards the positive so it is less negative so we can see that the potential is varying from negative more negative to less negative it is going from negative to positive side we can see so this is the case of one atom so as you know that in a crystal there is a series of atoms so let's see these are a few atoms in a 1D crystal lattice so let's say this is nucleus and electron are electrons are in its so if we talk about the potential behavior of the potential so close to the nucleus the potential is negative as it moves far from the nucleus the potential become positive this is the case for one atom similar case will be for the another atom if we talk about this atom the electrons which are close to this they are having a negative potential as the electrons move far away from the nucleus they are having a positive potential so the shape of potential for two atom system takes this form similarly for the next group of atoms the potential will take this form and same will be for the whole chain of the atoms moreover you can see that when the electron is close to its nucleus it is having an attractive potential but when it is far from the nucleus the attraction decreases but the repulsive forces due to the nearby electronic shells of the another atom you can see that this atom will have its electronic shell so as we if we talk about this electron the repulsive forces due to the another atom this will also be experienced by this electron so repulsive forces because you know that the potential repulsive potential is always positive so this potential is higher than the attractive potential so potential is moving from lower to higher value similarly when close to the nucleus it is lower then it is moving to the higher value this one so our potential takes the shape of this now if we consider this close to the nucleus the electron is in a well and away from the nucleus it experiences a barrier so this is what the barrier is well and a barrier similar case will be for another atom this is a barrier this is a bit the electrons are in the will close to the nucleus it experiences a barrier because why this is a barrier because if the electron has to move from its atom to the another atom it has to move freely in the sea of another atoms it has to cover this barrier and move to The Well of another atom then again it has to cover this barrier potential barrier and move to The Well of another atom okay so we can see that a 1D Crystal itself can be considered as an array of wells separated by the barriers this is one will this is another will separated by a barrier this is again another Well separated by a barrier so particles are in its Wells they have to cross this barrier if they have to move from one well to another Well it means if the electron has to move freely in the sea of another atoms then it has to cross the barrier then only it can move from its well to another so we have seen that in a one day lattice we have the potential function which is an infinite array of potential well separated by the barriers like this so this is the nucleus surrounding this electrons they are in wind and they are experiencing a barrier if they have to move in the will of another ad okay so let us consider this is our potential function so for the sake of Simplicity we are considering the rectangular Wells and barriers so this is the rectangular will of width a and this is the rectangular barrier of width B this B is the width of variant and this is the width of the well weaker so our potential function is repeating itself after the period a plus b so a plus b is the period of our potential after that again we have a barrier then we'll barrier will similarly here also well in the barrier so we can say that the electrons are moving in the periodic potential of a 1D lattice so our potential is periodic in nature so we can say that potential function is repeating itself after the period so this kind of periodic potential in this kind of periodic potential the behavior of the electron is described by a function called block function which was given by the block F block so he gave that the behavior of the function is described by a wave function which consider a function which is also periodic in nature so the block function is described by this the wave function for that kind of problems is given by this PSI X is u x e raised to power plus minus Iota KX where this u x this is the block function which is periodic in nature which repeats after the period so this ukx will have the same value after the period a plus b which is the period of our potential so to describe this problem for 1D lattice time independent Schrodinger wave equation will be used so let us see this is Schrodinger wave equation in Monday we have already studied this in quantum mechanics so this is d square PSI by DX square plus 2 m by H cross square e minus VX this v x is our potential function this is PSI wave function of our system so now in this periodic potential we have two regions which are which keeps on repeating after a fixed interval of time let's say this is first region this is well region and this is second region this is barrier region again this is Barry region this is this behavior of the electron in this region will be same as the second region similarly behavior of the electron in this region will region this will be same as the first region so we will describe the two regions our system is the infinite array or repetition of these two systems as a whole so v x if we take the potential for the two regions in the well region the potential is zero here the potential is considered as 0 in the well region from 0 to a and the potential in the barrier region in the barrier region the potential is V naught it means our barrier is of height V naught and width B this is the height of the barrier and this is the width of the barrier so this is second region is barrier region from minus B to 0 and here the potential is V naught so we have to write down the Schrodinger equation for these two regions and we have to find the solutions for these two regions to describe the behavior of our electron in the periodic potential so we have seen that the Schrodinger wave equation will be used to describe the two regions well in barrier region so let us write the Schrodinger equation for well region in well region which is from 0 to a that is the first region we will have the Schrodinger equation d square PSI 1 by DX square plus 2 m by H cross square e minus v x v x is 0 so this will be e sine X is equal to 0 so let us consider this 2 m e by H cross Square as Alpha Square so our equation will become d square PSI 1 by d x square plus Alpha Square PSI 1 x is equal to 0 the general solution of this equation in terms of the block function as we have discussed earlier will be given by PSI when X is equal to U1 x e raised to power Iota KX similarly we will write down the Schrodinger equation for the barrier region the barrier region is from minus B to 0 so we'll have Schrodinger equation as this T Square PSI 2 per second region the wave function is PSI 2 these d square PSI 2 by DX square plus 2 m by H Pro square e minus here VX is V naught for the barrier region VX is V naught so this is e minus V naught PSI 2x is equal to 0. so this 2 m by H cross square e minus V naught let us consider this 2 m by H cross Square V naught minus E this is equal to Beta Square so putting here you will get this equation b square PSI 2 by DX Square minus beta Square PSI 2 is equal to 0 again the general solution for this second wave function PSI 2 in terms of the periodic wave function that is the block wave function that this can be given by this this can be given by PSI 2x is equal to U2 X this is periodic function that is block function e raised to power Iota K so we have the two wave functions one for the well region another for the barrier region as PSI 1 given by equation 3 and PSI 2 as given by equation number 7 so now we have to see what is the value of this U1 and U2 which are the periodic wave functions so to do so to find the values of the u n and U2 we will put the values of PSI 1 and PSI 2 in equations so using three in equation number two we will get an equation in terms of U1 and using 7 in equation number 6 we will get an equation in terms of U2 to find the values of the solutions for U1 and U2 Now using the values of PSI 1 and PSI 2 in Schrodinger equations we get the equation in terms of u n and U2 as b square U1 by DX square plus 2 outer k d u 1 by DX plus Alpha Square minus K Square U1 X is equal to 0 in the well region that is the first region where the block function is U1 in the barrier region that is the second region where the block function is U2 we get an equation in terms of U2 by substituting the value of PSI 2 in terms of U2 as d square U2 by DX square plus 2 Iota k d u 2 by DX minus beta square plus K Square u 2 x is equal to 0 let's say this is equation number nine the general solutions for U1 and U2 from these two equations 8 and 9 they are given by U1 is equal to a e raised to power Iota Alpha minus K so from where this Alpha minus K Comes This Alpha Square minus K Square we will take the root of this this is the root of the equation so from here we will get one value Alpha minus K another value Alpha plus K so that appears in the solution of U1 so that is a e raised to power right Alpha minus k x plus p e raised to power minus Iota Alpha plus K into X similarly from this equation the roots from this part this comes out to be Iota beta plus attack and beta minus Iota K from there you will get U2 is equal to c e raised to power minus Iota into Iota this will give you minus 1 so this is minus beta minus Iota K into X plus d e raised to power Iota part this one when you will apply again Iota part from this minus 1 I can write this minus 1 as plus Iota Square so from here I will have the roots I'll put it here I will get first part c e raised to power minus Iota beta minus Iota K into X and D plus d e raised to power beta plus Iota K into X so equation 10 and 11 are the general solution for the block functions given and U2 U1 is the function for the well region Q2 is the wave function for the second region that is the battery region so in the equation 10 and 11 the coefficients a b c and d they can be obtained by applying the boundary conditions so what are the boundary condition let's see the wave function at this region that is the first boundary say x is equal to 0 this is the common boundary for the battery region as well as the well region Central one so at this the wave functions are same so U1 at X is equal to 0 is same as U2 at X is equal to 0 that is the central boundary similarly the wave function for the second boundary second boundary is this right side of the Bell region is same as the left side of the bearing region so here we are considering the well region right side of the well region at X is equal to a and left side of the wail region at X is equal to minus B so boundary condition will be U1 at X is equal to a this one this will be same as the wave function for the barrier region at the left boundary left boundary is at X is equal to minus E similarly for the derivatives also the supplies third is first derivative of U1 at X is equal to 0 that is the central boundary this will be same as the derivative of seeking function as at X is equal to 0 and the DU 1 by D X at X is equal to a that is the second boundary this will be same as the U2 by DX at the second boundary of the battery region so we have two regions one is well region other is Barry region the two have two boundaries one is this boundary that is the left side of this well region other is the right side of this well region the right side of the well region is same as the left side of the barrier region and left side of the battery region is same as the right side of the barrel region so these are the common boundaries so at the boundaries we have all these four equations or the all these four conditions when we use the values of U1 and U2 in these equation for equation we will get four equation in terms of a b c and b to find the non-trivial solutions we will put the determinant equal to 0 to obtain the non-trivial solutions for the a b c and b so let's see so I'm putting the values of U1 and U2 in the boundary condition we will get the four equations in terms of a b c d from first boundary condition U1 and U2 at X is equal to 0 are same we will get a plus b is equal to C plus d similarly from the second equation when u 1 at X is equal to a is same as U2 at X is equal to minus B we will put it in the equations so this will be this will give us a e raised to power Alpha minus K into a plus b e raised to power minus Iota Alpha plus K into a is equal to c e raised to power beta minus after K into B plus d e raised to power minus beta plus Iota K into B similarly from third equation we will get this part Iota Alpha minus K into a minus Iota Alpha plus K into B is equal to minus beta minus Iota K into C plus beta plus Iota K into D and from the fourth boundary condition we will get this outer Alpha minus K into e raised to power out Alpha minus K into a at X is equal to a U1 into a minus Iota Alpha mine Alpha plus k e raised to power minus I to Alpha plus K into a into B is equal to minus beta minus to power beta minus Iota K into b c plus beta plus Iota k e raised to power minus beta plus Iota KB into D on solving these four equation in form of the determinant to obtain the known trivial solutions for a b c d we will get this equation beta square plus Alpha Square upon 2 Alpha Beta into sine of alpha a sine of hyperbolic beta B plus cos Alpha a COS hyperbolic beta B is equal to cosine of K A plus b this is the final result obtained for The Chronic Penny model now we have to discuss this result to see how it explains the origin of bands in the solids so let us see the saline features of the KP model on the basis of the last result obtained from the application of periodic potential using Schrodinger wave equation okay the end result of chronic Penny model was this beta square plus Alpha Square upon 2 Alpha Beta sine of alpha a sine of hyperbolic beta B plus cos Alpha a or hyperbolic beta B cos a plus b now let the barrier is thin so B tends to 0 beta B will be will tend to zero if beta B is 0 sine of hyperbolic beta B this will be simply beta B and cos hyperbolic beta B this will be Unity putting it here we'll get this equation beta square plus Alpha Square upon 2 Alpha Beta sine of alpha a instead of this sine hyperbolic beta B will have beta B plus COS of alpha a COS hyperbolic beta B is one here cos k a because b is z so now putting the values of Alpha and beta from here you can see that beta will be canceled out also put the values of beta Square beta square is 2 m by H cross Square V naught minus E and Alpha square is 2 m e by H cross Square so here we have to put the value of beta square plus Alpha Square so beta square plus Alpha Square from these two when we add we will get this 2 m by H cross Square V naught put it here we will get 2 m v naught by H cross Square into 1 upon 2 Alpha from here sine of alpha a into B B I am keeping here B by 2 Alpha Plus cos Alpha a is equal to cos k from here what I am doing is I'm multiplying it with a numerator and denominator both multiplying with a 2 will cancel out so in this we have now this equation has been modified to this equation now has been modified to now let's see this part is m v naught b a m v naught b a upon H cross Square H cross Square from here this Alpha a I'm keeping it with sine Alpha a so this sine Alpha a by Alpha a sine hyperbolic beta B there is no term like that now and then cos Alpha e Plus cos Alpha a only and this term is equal to cos okay so we have this equation now this m v naught B A by H cross Square this is represented by P which is extent of binding energy of the electron to the lattice or it's it also represents the strength of the barrier so this m v naught B A this is represented by a term P so our equation becomes P sine Alpha a by Alpha a plus cos Alpha a is equal to cos a this is the final expression of KP body where where P this is m v naught B a by H cross Square and V naught B this is extent of binding of electron this V naught B V naught B this is area of barrier and also represents strength of the barrier so we have this equation this is final result of KP model now we will use this equation to explain the features of KP model so let us discuss the features of the chronic pen model so we are discussing the features now these are important so this is the final result of KP model P sine Alpha by Alpha a plus cos Alpha a is equal to cos a a so this equation is used to describe the behavior of the electron whether they are moving in the as a free particle in the sea of lattice or they are perfectly bound or they have some energy or something in C to move in the lattice so let us discuss first case when p is equal to 0 t is represent basically The Binding energy of the electron to the lattice or the strength of it represent also the extent of strength of the barrier so when barrier is thin is negligible when p is equal to 0 so we have when p is equal to 0 from this equation you can see that cos Alpha a will be equal to cos k a so Alpha a will be equal to k a or we can say that Alpha will be equal to K so as we know the alpha Square we have assumed 2 m e by H cross Square this will be equal to K Square from here you will get the value of energy e is equal to H cross Square K Square by 2 m which is the energy of the three particle so this energy represents the continuous range of energy it means when p is equal to 0 that is the strength of the barrier is negligible the particle behave as a free particle and it has a continuous range of energy in the lattice second cases when strength is infinity the particle is perfectly bound to the lattice that is where it is very high strength barrier strength is very high when p is infinity this term will become Infinity if this term is infinity the solution it is not valid so what would be the only solution for this equation only solution is sine Alpha is equal to 0 this term should be 0 if this is infinity so when sine Alpha is equal to 0 it means Alpha a will be n pi so if we put the value of Alpha from here Alpha will be n Pi by a so if we use the value of alpha in the relation for energy using this equation 2 m e by H cross square is equal to Alpha Square from here e will be H cross Square by 2 m into Alpha Square so using the value of alpha Square n Square Pi Square by a square we will get this relation n Square Pi Square H cross Square by 2 m a square we have seen this result in the energy for particle in a box in an infinite way so this is the energy for the particle which is which is in an infinite will that is a perfectly bound electron so this is the particle which is perfectly bound to the lattice so these are the two cases when barrier is negligible or when barrier is very strong when barrier is negligible the particle is free to move in the lattice when barrier is very strong the particle is perfectly Bound in the lattice it is not free to move in the lattice it means when it is free the conduction will be large when it is not free the conduction will not be large now we have to discuss the cases which are in between these two when the barrier strength is significant not 0 and not infinite so let us see what will be the behavior of the particle when it is moving in a barrier of a finite strength the next case is when we consider the strength to any random value so the let's say p is equal to 3 Pi by 2 it is any general value considered for p is equal to 3 Pi by 2 if we consider p is equal to 3 Pi by 2 that is a finite value if we plot the left hand side of this equation as a function of alpha a because you can see that this left hand side of the equation that varies as Alpha so if we plot the equation this equation the left hand side is a cosine curve because this is cosine curve whereas between plus minus 1 and this is the limit when the value of alpha a will increase this limit sine Alpha a by Alpha a it will decrease so this term will modify the values of alpha a the values of alpha a because of this term it can go beyond plus one and it can go beyond minus one the left hand side can take the values beyond plus minus 1 because of this term but you can see that the right hand side it can only take the values plus minus one so if this is a valid equation the values of left hand side should be in the range of right hand side and the right hand side this can take only the value if I consider this this cos k a this can only take the values between plus minus 1 or I can say that plus minus 1 it means this k a this can be plus minus n Pi or k can take the values between plus minus n Pi by a this is this right hand side basically is doing what it is applying the limit on to the left hand side okay so these limits provided by the right hand side they are called Zone boundaries they apply the boundaries to the zones with Zone boundaries allowed Zone boundaries or allowed values they are applying a limit to the allowed values so only the allowed values for the equation or for the left hand side will be the those which have the values between plus minus 1 because only then left hand side will be equal to the right hand side if the values goes beyond plus minus 1 this left-hand side will not give you the allowed values because according to this equation if this is this model is a valid solution but this model provides a valid solution this left-hand side if it has the values beyond plus minus 1 those will not represent the allowed values according to the KP model so let us plot this as a function of alpha a if we see the behavior of this this is a cosine function the values can go beyond plus minus 1 this plus minus 1 is actually the values that the cosine function can have but due to this term it can have the values beyond plus and minus 1 it means it is going beyond the allowed values as provided by the right hand side of this equation so if we see let's say first of all let's see for Alpha is equal to zero for Alpha a is equal to 0 this sine Alpha by Alpha a this term will be Unity so this term is 3 Pi by 2 it has it has some finite value cos Alpha a this is equal to 0 it means it has value one so one plus something it means at Alpha is equal to 0 this left hand side is going beyond plus one so it has an upper value than plus one this one so the function will start from this beyond plus 1 it has some value it is a cosine function the behavior will be like this again when Alpha a is equal to minus 1 due to this term this value will be Beyond minus 1. so it will have a value Beyond minus 1 2 similar Behavior will be for the other values of alpha a the Curve will go beyond plus 1 and Beyond minus 1 due to this term as the value of alpha a increases the value of this limit will decreases the value of the limit will decreases so the values they will be they will go beyond plus minus 1 but not to the larger extent so the behavior will be like this now see but according to Chronic Penny model the valid Solutions are the valid Solutions are when left hand side is equal to right hand side it means if the values of this left hand side are going beyond the right hand side values of the right hand side as you can see that the values of right hand side can only be between plus minus one so if this term don't be your go beyond plus minus 1 these values represent not allowed values that is the forbidden regions so if we see from this curve we can see that from this point to this point it means I can see this from this point to this point the values of my function the left hand side it lies between plus one and minus 1 but similarly from here the values between lies between plus and minus 1 again but there are certain regions where the values are beyond plus minus 1 where are the regions let's see this is the region this one is the region where the values are beyond plus minus 1 this is plus one beyond plus one similarly from here I can see that this is the region where the value is beyond minus 1 again this is the region which represent and invalid solution of the left hand side so we can see that there are loud regions this is the loud region where left hand side lies between plus minus 1 but this this region represent the region which is not allowed because the values are Beyond minus one or beyond plus one so we can see that this equation represents again the series of allowed regions separated by some not loud regions those are called forbidden regions as these regions represent the value of alpha a and Alpha is a measure of energy so it means these represent the band of energy which is allowed for the particle and this represents the band of energy which is not allowed for the particle it means we can see that from this Behavior we can see that these represent the allowed regions and this regions represented by the Red Band they are the Forbidden reasons so we can see that solids can be considered as having the allowed bands a number of Allowed by not a single allowed band a number of allowed bands separated by The Forbidden bands so not all the energies are allowed for the particle there are forbidden range of energy also which are not allowed for the particle those ranges are called forbidden energies and form the orb form the Forbidden bands or what we can say forbidden gaps between the allowed bands so as the value of alpha increases as we move on to the higher energies the width of a loud band it goes on increasing and width of forbidden bands these goes on decreasing a loud bands give it or forbidden band keep it as we move on to the higher values of alpha a so this is basically the KP model where we started if the strength of barrier is very weak the particle act as a free particle if the strength of barrier is very high or the barrier is very strong the particle will behave as a bound particle it is perfectly bound to the lattice but if the strength is finite it means if we have certain value of this P then the particle it will experience a series of bands so in the crystal the particle will have allowed bands separated by The Forbidden bands as the values of alpha a increases the width of allowed bands goes on increasing and width of forbidden bands it goes on decreasing why this is so because this right hand side can only take the values between plus minus 1 and we have the allowed Solutions only when our left hand side will take the values between plus minus 1. if it will go beyond plus minus 1 these will represent the not allowed regions these are not allowed regions these red regions and the regions which are represented by the black they are the allowed regions okay let's summarize the Salient features of chronic Penny model so we use this equation the final result of KP model to summarize the saline features let us first discuss when p is 0 when p is 0 it means various strength is negligible the energy of the particle is same as the energy for a free particle in that case the particle is perfectly free to move inside the lattice and its energy is a continuous energy second cases when p is infinity when p is in finite it means that barrier is very strong in that case the energy of the particle will be same as the energy for a bound particle in the infinite will so it represents a discrete set of energies for the bound particle now third case is when barrier is finite it's not 0 and not Infinity it is any value so in that case the right hand side because the right hand side can only take the values between plus minus 1 so it applies to a limit to the left hand side so this limit is represented by K is equal to plus minus n Pi by a it represents the allowed Zone boundaries so it represents the boundaries of the allowed zones these allowed zones are called brilliant zones the allowed bands of energy will be when the left hand side will be same as the right hand side as right hand side can only take the values between plus minus one so if our left hand side will take the values between plus minus 1 it represents the allowed Banks and if the left hand side will take the values beyond plus minus 1 it will represent the Forbidden band of energies as we have seen in the graph previously now width of the allowed bands it goes on increasing with increase in the values of alpha a that is the energy of the system and that of forbidden bands it decreases with increasing the values of alpha I so as we move on to the higher energy states the width of allowed bands keeps on increasing and width of forbidden bands keeps on decreasing so gap between the allowed band that is the forbidden bands this keeps on decreasing so let's have a final look at the graph as plotted for this p is equal to 3 Pi by 2 so this is the final graph obtained from the equation when we consider p is equal to 3 Pi by 2 so this equation this graph represent all the regions this is these plus 1 and minus 1 they represent the boundaries provided by the right hand side to the left hand side we have plotted this left hand side with respect to Alpha a so from this graph it is seen that the values of this left inside are going beyond plus minus 1 upward and downward you can see that these are going beyond plus minus 1 so when these values are beyond plus minus 1 they represent the not allowed regions that is the forbidden regions represented by these red boxes red regions these are forbidden bands and these forbidden bands between these two forbidden bands there is an allowed region or we can say that between two allowed regions there is a forbidden band so we have a series of allowed regions first loud region second loud region third loud region and so on similarly on this side also first loud second loud third loud and so on separated by the series of forbidden bands so a solid can be considered as a series of allowed bands separated by The Forbidden bands if it goes on to the increasing energies of the system the width of loud bands this keeps on increasing and that of forbidden bands keeps on decreasing so this is all about the saline features of chronic Penny model we have seen how this model explains whether the particle will behave as a perfectly free particle that is the case of metals or the particle will behave as a perfectly bound particle that is the case of insulators or the particle will behave as if it is moving in the sea of bands allowed and forbidden bands that is the case of semiconductors so I hope you understood will the concept please like this video And subscribe to my channel to get further updates thank you [Music]