Assignment 7

Now, we will solve Assignment 7. So, first question, consider a one-dimensional chain of atoms of mass m. So, this is question number 1. The masses they were like here, this there, this one were blue. So, this we take took as mu s, this suppose we took as V s. This is displacement of each side due to and C 1 is the red line and C 2 is for the blue. It is given also in the question. So, what will be the equation of motion? So, equation of motion will be given for the mu s will be mu star C 2 V s-mu s-C 1 into mu s-V s-1. So, we can know that this is for the diatomic chain as Professor Harbolas has done in the lectures. We saw that it was just like that because for both them C 2 and C 1 were same. So, we took that both as C but here they are different. Simple as that, simply for V s, we will have C 1 just the nearest neighbor atom interaction. C 1 will be u s plus 1-V s-C 2 into V s-mu s. So, these are the 2 equations of motion. This is option C. Moving to the next problem 2. For question 1, for question 1, if C 1 is less than C 2, frequency of the acoustic and optical modes of the frequency of acoustic and optical, what will be their value? At k equals to pi, first belongs. So, now we have equations m u s dot equals to C 2 V s-mu s-C 1 u s-V s-1. This is 1 and we had m V s double dot equals to C 1 u s plus 1-V s-C 2 into V s-u s. This is u s 2. Now, we put the solutions at that just plane traveling wave. So, we put plane wave equations. Plane wave solution for u s, I can take the amplitude to be A 1. They are different amplitude but the waveform is same for both of them. This is A 2 into e to the power i k s a minus omega t. These are the 2 equations. Now, if the 2 equations we put, we put these 2 values of u s and u in 1 and so by putting from 1, we will get from 1, A 1 m-i omega whole square e to the power i k s a-omega t equals to C 2 A 2-A 1 e to the power i k s a-omega t-C 1 A 1 e to the power i k s a-omega t-A 2 e to the power i k. This will be s-1 A because it was s-1 A side-omega t. So, here we just make it much simpler C 2 A 2-A 1. We just cut this part from both sides. Here only this k s a term will go out and –k term will remain. So, this will be C 1 A 1-A 2 e to the power –i k. This is k due to that s-1 for that part it is written here. So, we take this as 3. Similarly by from 2, from 2 we will get and I am just writing the simplified form A 2 m omega square C 1 A 1 e to the power i k a-A 2-C 2 into A 2-A 1. We take this at equation number 4 okay. So, now rearranging question 3 from 3, I am just taking everything in one side. So, for A 1 into m omega square-C 2-C 1 plus A 2 into C 2 plus C 1 e to the power –i k a equals to 0. We take this as 5 and from 4 we can get this is 3 and from 3 we will get this from 3 rearranging and from 4 we will have A 1 into C 1 e to the power i k a plus C 2 plus A 2 m omega square-C 1-C 2 equals to 0. This is our equation 6. So, we have 5 and 6. This 5 and 6 we have. This A 1 and A 2, this A 1 A 2, A 1 A 2. So, we have these 2 equations with 2 variables. Suppose, we take this A 1 and A 2. So, the coefficient matrix of both these 2 equations will have to be 0, the determinant of the coefficient matrix. We are just doing that by doing m omega square-C 2-C 1 and C 2 plus C 1 e to the power –i k a and this will be C 1 e to the power i k a plus C 2 m omega square-C 1-C 2 that it will be 0. So, this is just the coefficient matrix I took and just I am taking the determinant to 0. This is just the eigenvalue equations. So, now the value we will get by this if we do that m omega square-we are just taking the determinant C 1-C 2 whole square will be C 2 square C 1 square plus C 1 C 2 e to the power i k a plus e to the power –i k a. So, step by step you will get this. So, this we know will be 2 cos k. So, m omega square-C 1-C 2 the whole square we will have C 1 square C 2 square 2 C 1 C 2 cos k. Now, at k equals to pi we are given. So, cos pi will be our –1. So, m omega square-C 1-C 2 whole square will have C 1 square plus C 2 square-2 C 1 C 2. So, this is my equation and here it is we have given that C 2 is greater than C 1. So, from the previous equation from this equation I can write it as m omega square-C 1-C 2 whole square equals to C 2-C 1 whole square because is C 2 greater than C 1 I am writing like this since C 2 greater than C 1. So, now if we just take the root m omega square-C 1-C 2 it will be plus minus C 2-C 1. The optical mode we know will be optical mode it will be the higher part so plus. So, m omega square-C 1-C 2 it will be plus C 2-C 1 so C 1 C 1 goes on. So, omega is your 2 C 2 by m this is for optical and for the acoustic mode I take the minus of this so m omega square to take the minus of this. So, m omega square-C 1-C 2 will be minus of C 2-C 1 or plus C 1-C 2 m omega square C 1 C 2. So, here C 2 will go on and we get omega will be 2 C 1 by m this is for acoustic. So, at k equals to at k equals to pi this will be the values so your correct option is p. Question number 3 question number 3 is for question 1 what will be the displacement ratio V s by mu s at Brillouin zone boundary for optical and acoustic mode. So, we took that V s equals to A 2 e to the power i k s a-omega t and U s I got A 1 e to the power i k s a-omega t. So, actually the ratio of V s by U s will be just the ratio of their amplitudes this part will go on. So, the eigenvalue equation now for the previous problem we saw that these 2 equations this 5 and this 6 this 5 and 6 if we can write this in a matrix form we will get m omega square-C 2-C 1 C 2 plus C 1 e to the power –i k a and C 1 e to the power –i k a plus C 2 and here m omega square-C 1-C 2 into A 1 by A 2 will be 0. This is just the equation we got and we now put k equals to pi here already we are putting k equals to pi because we have we have asked the question at Brillouin zone boundary at Brillouin zone boundary it is k equals to pi. If we put the value of k equals to pi here so we get m omega square-C 2-C 1 C 2-C 1 because e to the power because e to the power –i k a e to the power –i pi will be still –1. So, C 2-C 1 and here it will be –C 1 plus C 2 into m omega square-C 1-C 2 into A 1 by A 2 equals to 0. So now for acoustic part acoustic problem from the previous question we got omega square equals to 2C 1 by m from the previous problem acoustic mode 2C 1 by m. So, we just put the value here in the equation in this equation put the omega square value here. So, we will get m 2C 1 by m-C 2-C 1 and this will be C 2-C 1 this is –C 1 plus C 2 this is m 2C 1 by m-C 1-C 2 A 1 A 2 is 0. So, m m goes m m goes and this 2C 1-C 2-C 1 so it will be C 1-C 2 this is C 2-C 1 this is –C 1 I can write it as C 1-C 1 C 2-C 1 and this is 2C 1. So, I can write it as C 1-C 2 A 1 A 2 goes to 0. So, now if we just take one value one equation so C 1 C 2 into A 1 plus C 2-C 1 into A 2 goes to 0. So, C 1-C 2 into A 1 equals to this part will go a change sign. So, C 1-C 2 into A 2 both goes so A 1 equals to A 2. So, the ratio A 2-A 1 will be plus 1 this is for the acoustic for acoustic similarly for the optical simply again just that equation m omega square-C 2-C 1 C 2-C 1 we have already put k equals to pi C 1 plus C 2. And this will be m omega square-C 1-C 2 into A 1 A 2 goes to 0 now we put omega square equals to 2C 2 by m from the previous problem here we see that for the optical it was omega square 2C 2 by m. So, again I put that values put omega square 2C 2 by m and I will get C 2-C 1 just put the value of omega square here and C 2-C 1 C 2-C 1-C 1 plus C 2-C 1 plus C 2 into A 1 A 2 equals to 0. So, C 2-C 1 into A 1 plus C 2-C 1 into A 2 equals to 0. So, we get here A 2 by A 1 will be minus 1 for optical part for optical so the correct option is this A 1 by A 2 by A 1 will be minus 1 for optical and A 1 A 2 by A 1 will be plus 1 for acoustic. This means that here you can see that A 2 by A 1 is plus 1. So, that means that both mu s and V s will move with same amplitude and in the same direction they both will move in the same direction and here A 2 by A 1 is minus 1. So, that means they both will move u s and v s will move at the opposite direction, but with equal amplitude. Moving to the next problem the question number 4 we have in terms of Debye temperature 3D the internal energy of the lattice in the Debye model what is this the internal energy of the lattice. So, we have we know that from the derivation we have already got u equals to 3 V by 2 pi square mu s whole cube into k to the power 4 by h cut whole cube into t to the power 4 into 0 to x t dx e to the x cube e to the power x minus 1. This was also derived in the lectures by professor Hargadasan and their x is written as h cut omega by k t. This was the form of the internal energy. This is also done in the book of Charles Ketter you can chapter 5 you can see also here and here x is h cut omega t I can write x t as theta d by T where theta d is my Debye temperature Debye temperature ok. So, now this is for a 3D suppose this is your k x k y and k z this k x k y k z will have k x if we use periodic boundary condition periodic boundary condition this k x k y k z will have value 2 pi L plus minus 2 pi L into n x and this is n x for k x and k y will be value of plus minus 2 pi by L into n y. We just we can get this from the periodic boundary equation it is just trivial n z. So, the smallest the occupied by the this k x k y and k z this k x and k y and k z will be 2 pi by L whole cube. So, 2 pi whole cube by L means L cube this L is the length length and this is the total volume of the cube this is the smallest cube and this is the smallest area of in the k space one state can occur. So, this is by V. So, here our total number of modes will be n will be this is your smallest and if we consider a sphere with value k. So, with the radius k this along the sphere this in 3D this how many of this this small cube are there. So, number of there will be the number of states obviously. So, this will be given by total area now the volume is 4 3rd pi k cube this is this volume the sphere and for one smallest rectangle is given by as we got from the previous formula is 2 pi cube by V. So, 2 pi cube by V we get this. So, n will be your V by 2 pi cube into 4 3rd pi k cube. So, V by 8 pi cube into 4 3rd pi k cube 4 4 2 pi square. So, you have V by 6 pi square k cube. So, your k cube is 6 pi square n by V now we know that omega is V s by k it is for the acoustic point the acoustic approximation we can take this V s into k it is the Debye approximation. So, V s is your velocity of sound velocity of sound this is the dispersion relation. So, this is omega equals to V s into k. So, we can write k equals to omega by V s. So, this k cube is k cube we can put omega cube by V s cube just putting this value in this equation 6 pi square n by V. So, we get omega cube equals to 6 pi square n by V into mu cube. So, there will be a obviously there will be a highest frequency about which it can get the number of states they can occur. We denote that highest frequency as omega d the highest frequency or we can say the cutoff frequency above which there will be no states available highest frequency is the Debye frequency we write this as omega d this is if this a n number of states we can write omega as the omega d we can write omega d. So, omega d cube is your 6 pi square mu s cube I am sorry mu s cube into n by V. So, and theta d your Debye temperature is h cut omega d by KB. So, this will be h cut mu by KB 6 pi square from here we get the omega d value and just put the value omega d here 6 pi square n by V in one third. And sigma d cube will have h cut cube mu cube by KQ 6 pi square n by V suppose this we take as equation this is one equation and u we already got 3 V KB 4 this 3 is coming actually as there are 3 kinds of polarization 2 transverse 1 longitudinal and 3 V KB cube into T to the power 4 by 2 pi square mu cube this is Vs and this is h cut cube 0 to xt dx x cube by e to the power x-1 this we got already. So, now this we can replace this value here by putting we get u equals to I am trying to write this as this is KB T4 just watch it closely I am doing it 2 pi square Vs cube h cut cube into h cut cube mu cube by theta d cube into 6 pi square n by V. See that I can write this you can simply write this e to the power x-1 just take value I am just replacing it here suppose we I am taking this K cube here K cube from here and replacing this K cube by this form here suppose here then I can write this K cube because to your theta d cube and here I am taking 1 K and 1 another K cube part this is K to the power 4 here so there will be K cube we are replacing this K cube and again by eliminating everything you see that there will be lot of elimination here this mu s mu s there will always be and by that we will get mu equals to this 2 3 will be this pi square pi square will be and it will be 9 n KB this will be n into KB into T into T by theta d whole cube 0 to xt dx x cube by e to the power x-1. So this is R so it is option P so this is gone for problem 4. So now question number 5 we have KCL its Debye temperature is given as 230 Kelvin specific heat at 5 Kelvin is given as 3.8 into 10 to the power-2 joule mole inverse Kelvin inverse. So what will be at CV at 2 K so the CV is written as 234 T by theta d whole cube into n KB. These are the constants of pi square n and V and everything so we put just put that every values and we will get this constant number 230. So the main part is actually is your CV proportional to T by theta d cube. So here is the KCL is same so theta d will be same. So simply CV by 5 K by CV by 2 K this will be proportional to 5 by 2 this is your given as 3.8 into 10 to the power-2. So CV at 2 K will be given a 2 by 5 whole cube into 3.8 into 10 to the power-2. So this will be 2.4 into 10 to the power-3 joule mole inverse Kelvin inverse. This is option A. Now similarly for question 6 NaCl has same crystal structure as KCL. So in both case n will be same n will be same and Debye temperature of NaCl Debye temperature of NaCl is given as 310 Kelvin. So what will be the specific heat of NaCl specific heat of NaCl at 5 Kelvin. So simply CV 5 K of KCL by CV 5 K of NaCl we get 234 T by theta d KCL whole cube nKB by 234 T by theta d NaCl whole cube nKB. And for both of them are same so this will go on this will go on. T will be same because here for 5 Kelvin both are 5 Kelvin so T will be same. So they will also cut off. So theta d NaCl by theta d KCL both cube and here this is given from the previous problem we had. This is 3.8 into 10 to the power-2. This is CV 5 K 5 K NaCl NaCl. So simply just putting the value this theta d NaCl is given as 310 Kelvin and theta d KCL is also given for the previous problem as 230 Kelvin. So by calculating just put the value in the calculator and we will get 1.55 into 10 to the power-2 Joule mole inverse Kelvin. This is d. Moving to the next question, question number 7. We have for the germanium it is given 80. The thermal conductivity is given by this. Debye temperature is 360 Kelvin. Your atomic weight atomic weight is given as 72.6 and density is 5500 kg meter cube. So we know that one third CVL. So CVL this is the specific heat per unit volume. Specific heat per unit volume unit volume. This will be the velocity of sound here. Velocity of sound and L will be the mean free path mean free path what we have to calculate. So L is 3 K by C into C is this is a specific heat per unit volume. I can write this like this into V the velocity of sound. Now we know CV is given as V KB to the power 4 theta d cube 2 pi square mu s whole cube h cut cube something and the theta d cube there are relations h cube bs cube into 6 pi square n by V by KQ. So these are the relations we got and there is a lengthy calculation and like we can get also the we know that rho is mass per volume. So we got mass atomic weight with and with the density we can calculate the V and if we put every values here suppose from here we can calculate mu s theta d is everything is given and here we will calculate the CV by putting the value of Vs by putting we will just see that L will be 30 nanometer. Just put the values and we will see. Moving on to the next question. Question number 8. D by temperature of diamond theta d for diamond is 2000 Kelvin. 4 K and 5 K 50 K what the conductivity is given as K 4 and K 50. So you see here that this 4 K and 5 50 K they are very very less than this D by temperature. So we see that there will be no Oom-Klum process and all the scattering will be determined by the defects and the sample boundary. So there will be no T dependence of the thermal conductivity the material. So it will just be normal process and CV will have is proportional to simply TQ. So we have kappa I mean this K thermal conductivity this will be just proportional this is we got we know that this is one third CVL. So K will be proportional to CV I mean the specific heat. So K 4 by K 50 the ratio will be 4 by 50 whole cube. This is because K is proportional to T cube we have this phonon region is T cube region because at the higher temperature it would be just be constant of Dulong-Petit's law. CV will be 3R. So this will get almost 5 into 10 to the power minus 4. So the correct option is option A. Into question 9 by using Debye approximation in a now we have in a 1D monoatomic crystal lattice. Monoatomic lattice I have and the lattice constant is given by A inter atomic spacing is A. Speed of the sound is Vs and Debye temperature is theta T okay. So now here for the 3D we got that this is Kx, Ky, Kz everyone was 2 pi by L just using the periodic boundary condition we are getting. These values are allowed. Here 1D this will be only one K value and K is 2 pi by L. So density of states so density of states we can write density of states for 1D I can write as d omega d omega L by pi I can write d omega by d omega by dK. If we use the periodic boundary condition we will get this 2 pi and if we this long chain we will have pi by L. Just see detail if you have any confusion. So this pi by L d omega by d omega by dK. So the Debye approximation we have omega equals to Vs K. So d omega dK we will have Vs. So again up to the Debye frequency omega d d omega d omega this will be L by pi 1 by Vs. I am just putting the value of d omega here 0 to omega d d omega. So it means L by pi omega d by Vs which is 2. We can write it as equation 2 and again we know that this is equals to n the total number of modes here. So for 1D we know L is total length is n into a total number of modes into inter-atomic spacing or total number of atoms in a monatomic lattice. So here L by pi into omega d by Vs equals to n. We can write n as L by a okay. So L goes up. So omega d is your pi Vs. This is suppose 4. Now we know we know h cut omega d is your K theta d. So this is just the Debye temperature and the Debye frequency. So h by 2 pi I am putting the value of omega d pi Vs by a K into theta d this pi pi goes out. So your theta d is given as h Vs by 2 K a. So this is the Debye temperature for the 1D monatomic lattice. This is option B. And the last question, question 10. For the previous problem just for 1D calculate Cv. So total energy will have total energy u is up to this omega d. This is d omega density of states. This is number of states per along that d omega and h cut omega. This is the Debye model. So we get d omega. This d omega I can write this as for the previous like L by pi 1 by Vs. This is just 1 by this is famous Planck equation. So h cut omega by KT-1 h cut omega. So u will be L by pi 1 by Vs 0 to omega d. h cut omega by e to the power h cut omega KT-1 d omega. So this is your form and now we put x equals to h cut omega d by KT and just we just put this here we will get u will be L by pi 1 by Vs K square T square by h cut 0 to omega d h cut omega by KT by e to the power h cut omega KT-1 into h cut by KT into d omega. We just put the h cut by d omega. h cut KT, KT h cut I am just introducing them and I got hx equals to h cut omega d by KT and omega d equals to theta d by T. So you will get L by pi this omega d. Now if we write them as x K square T square by h cut square this is K square 0 to x suppose this will be x by sorry x by e to the power x-1 dx. This is x equals to omega d KT just here. If we replace x omega d by x and then this omega d will be x and x I can write also as this is theta d by T because x equals to your theta d by T here I can get. So for T tends to 0, x will tends to infinity. So we will have u equals to L by pi 1 by Vs K square T square by h into 0 to infinity x e to the power x-1 dx and this value we know this is pi square by 6. So if we put the value here this is this value is pi square by 6. So if we put the pi square by 6 we get u equals to sorry pi square by 3 L K square by h mu s into T square and again we got theta d is h mu s by 2 K. So u will be given as pi square by 3 into L K square into 2 K by h mu s I am just forming it like that into a 2 K into T square. By substituting everything we will get pi square by 3 L K 2a T square 1 by theta d this is T square. So C V will tell you del T this will be just pi square by 3 the T square 2 will come this 2 2 will cancel out this 2 and this 2 will cancel out. So it will be 1 by theta d L K by 2a into 2 T and this 2 2 has already cancelled out. So C V is your pi square K T by 3a theta d this is your problem and you can take L equals to 1 because a is very small and n is a large number. So we can always consider this L equals to 1 if we consider that it is pi square K T by 3a theta d. This is your option C. .