PHY 140A Solid State Physics Solution to Homework 5 Xun Jia1 November 16 2006 1 Email jiaxun physics ucla edu Fall 2006 c Xun Jia November 16 2006 Physics 140A Problem 1 Temperature dependence of the electrical resistivity of a metal The electrical resistivity of a metal is proportional to the probability that an electron is scattered by the vibrating atoms in the lattice and this probability is in turn proportional to the mean square amplitude of vibration of these atoms How does the electrical resistivity of the metal depend on its absolute temperature in the range near room temperature or above where classical statistical mechanics can validly be applied to discuss the vibrations of the atoms in the metal Solution As indicated by the problem the electrical resistivity of a metal is proportional to the probability p that an electron is scattered by the vibrating atoms in the lattice and this probability p is proportional to the mean square amplitude of vibration of these atoms hA2 i Since hA2 i is proportional to the average of potential energy hEp i which is proportional to T due to equipartition theorem Thus the the electrical resistivity is proportional to T Problem 2 Specific heat of anharmonic oscillators Consider a one dimensional oscillator not simple harmonic which is described by a position coordinate x and by a mop2 mentum p and whose energy is given by bx4 where the first term on the 2m right is its kinetic energy and th second term is its potential energy Here m denotes the mass of the oscillator and b is some constant Suppose that this oscillator is in thermal equilibrium with a heat reservoir at a temperature T high enough so that the approximation of classical mechanics is a good one Hint there is no need to evaluate explicitly any integral to answer these questions a What is the mean kinetic energy of this oscillator b What is its mean potential energy c What is its mean total energy d Consider an assembly of weakly interacting particles each vibrating in one dimension so that its energy is given as above What is the specific heart at constant volume per mole of these particles Solution 1 a The mean kinetic energy of this oscillator is hEk i kT due to equipartition 2 theorem 1 Fall 2006 c Xun Jia November 16 2006 Physics 140A b From Boltzmann statistics the mean potential energy is RR hEp i p2 R 4 dpdx bx4 e 2m bx RR p2 4 dpdx e 2m bx dx bx4 e bx R 4 dx e bx4 R 4 dx e bx Z 4 ln dx e bx R dx e bx4 Z Z 1 4 1 4 4 t x ln 1 4 dt e bt ln ln dt e bt 4 1 1 kT 4 4 1 Another way of evaluating this is to use integral by parts to calculating the integral directly R hEp i dx bx4 e bx 4 R dx e bx4 Z 1 1 1 1 4 x e bx4 kT R dx e bx 4 4 4 4 dx e bx4 2 c Thus the mean total energy is 3 hEi hEk i hEp i kT 4 3 d For the weakly interacting particles the internal energy per mole is just the sum of each particles i e 3 3 U NA hEi kNA T RT 4 4 4 where NA is the Avogadro constant Thus the specific heat at constant volume per mole is U 3 cV R 5 T V 4 2 Fall 2006 c Xun Jia November 16 2006 Physics 140A Remark A general equipartition theorem states that in thermal equilibrium the 1 mean value of each term in the energy proportional to 2n is equal to kT where 2n could be any independent variable for example momentum p or coordinate x This is not hard to proof Note here the exponent is 2n which ensures the energy is bounded from below as it should be in any physical context Problem 3 Specific heat of a highly anisotropic solid Consider a solid which has a highly anisotropic crystalline layer structure Each atom in this structure can be regarded as performing simple harmonic oscillations in three dimensions The restoring forces in directions parallel to a layer are very large hence the natural frequencies of oscillations in the x and y direction lying within the plane of a layer are both equal to a value k which is so large that h k 300k the thermal energy kT at room temperature On the other hand the restoring force perpendicular to a layer is quite small hence the frequency of oscillation of an atom in the z direction perpendicular to a layer is so small that h 300k On the basis of this model what is the molar specific heat at constant volume of this solid at 300K Solution Since h k 300k at room temperature T 300K the thermal fluctuation is not high enough to excite the degrees of freedom within the plane i e the degrees of freedom associating with h k is frozen On the other hand 300k h the solid atom can only move along the direction perpendicular to the plane Hence from equipartition theorem the average kinetic and potential energy per 1 atom are both kT and then the average total energy is kT which therefore gives a 2 specific heat per atom of cV k or equivalently cV kNA R is the molar specific heat Problem 4 Kittel 5 1 Singularity in the density of states Ashcroft and Mermin 23 3 essentially the same problem explains that this is called a van Hove singularity a From the dispersion relation derived in Chapter 4 for a monatomic linear lattice of N atoms with nearest neighbor interactions show that the density of modes is D 1 2N 2 m 2 1 2 where m is the maximum frequency b Suppose that an optical phonon branch has the form K 0 AK 2 near K 0 in three dimensions Show that D L 2 3 2 A3 2 0 1 2 for 0 and D 0 for 0 Here the 3 Fall 2006 c Xun Jia November 16 2006 Physics 140A density of modes is discontinuous Solution a A general expression of D in d dimension lattice is a generation of Eqn 35 in Kittel d Z L D d dd K 6 2 shell where the shell refers to a region in K space bounded by two surfaces on which the phonon frequency are and d R In 1 D the shell is just composed by two points and hence shell dd K 2dK Therefore from above equation dK L D 7 d From the dispersion relation of a monatomic linear chain we have dK 1 2 1 d 2 1 a m 2 1 2 d a m cos 2 Ka dK 2 8 thus from above equations note that N L a it follows that D 1 2N 2 m 2 1 2 9 b For the optical branch in 3 D case K 0 AK 2 Now the …