固体物理习题解答：Chapter 6 problems

1.Kinetic energy of electron gasThe kinetic energy of one electron ismkk222Then the kinetic energy of the electron gas at 0K isFFkFkkkkkkkkVkmVdkkkVmmkUFF35202222258541 4/222VVkVNkF332)2(and ,3 FFFNkVU535 322.Pressure and bulk modulus of an electron gas(a)The relation connecting the pressure and volume of an electron gas at 0KTVUpThe pressure p isAt T=0K,3/23/2220 325353AVVNmNNUFThen VUAVVUpT323203/5orconst.320UpVwhere 3/222)3(103NNmA(b)The bulk modulus of an electron gas at 0KThe bulk modulus is defined as:VpVBAt T=0K,VUp320VUAVAVVVVpVB910910 3203/53/5(c)Estimate the bulk modulus B for potassiumFrom Table 1,the electron concentration of potassium322cm 104.1/VNnThe Fermi energy erg 106.112.2eV 12.212F31012220erg/cm 102.3 106.112.2104.132 32/53910910FFnVNVUBCompare the value to Table 3.3(page 59)B=0.0321012 dyn/cm23.Chemical potential in two dimensionsFirstly we calculate the density of state per unit area for a 2d electron Fermi gas.Consider a two dimension lattice with the area LL.There is one allowed value of k per area of(2/L)2 in k space.The total number of modes with the wavevector less than k is222222222222mAkAkLLkNThe electron concentration 2/mANnThe density of state per unit area is2)(mddnDThe chemical potential is given by0)()(fDdnWe have1/)exp(1)(TkfBTkeTmkedxTkmTkmdnBBTkxBBB1ln 1 1/)exp(1 2202Therefore1expln2TmknTkBB4.Fermi gases in astrophysics(a)Estimate the number of electrons in the SunThe main element in the Sun is hydrogen.The mass of a hydrogen atom is 1.71024 g.The number of electrons in the Sun is57243310107.1102HSunmmNIf all these electrons are contained in a white dwarf star with radius 2109 cm,the Fermi energy iseV103eV 106.1/105 erg105)102(41014.391092)10(3/432412883/23957282273/2322RNmF(b)In the limit mc2,kcpcThe Fermi energyckFF3/13/1233 VNVNkF3/13VNcF(a)Estimate the Fermi energy of the electron gas in a pulsarIf all these electrons are contained in a pulsar with radius 10km=106 cm,the Fermi energy iseV106erg109 3/)10(410103103 3843/1365710273/1VNcFThis energy is much bigger than the energy release in the reaction n p+e,with is only 0.8106 eV.5.Liquid He3The atom He3 has spin and is a fermion.The density of liquid He3 d=0.081g/cm3,the mass of the He3 mHe3 is mHe3 3mp 31.71024 5.1 1024g.Then the concentration of the liquid He3 atoms n is32224cm106.1101.5081.03HemdnThe Fermi energy iseV104erg106 106.114.33101.52)10(324163/2222242273/2223nmHeFThe Fermi temperature TF=F/kB 5 K.6.Frequency dependence of the electron conductivityConsider the electrons driven in an electron E(t)=E0exp(it).The equation of motion m(dv/dt+v/)=eE gives:)exp()/(0tieEvdtdvmThe solution of this equation is v(t)=v0exp(it).Thus we haveeEimv)(12)(11 )1(/i.e.imEeimeEvThe electric current density j isEimnenevj22)(11Therefore the conductivity is222)(11)0()(11)(iimneEjwhere mne2)0(7.Dynamic magnetoconductivity tensor for free electrons(a)The component of the dynamic magnetoconductivity tensor Consider the electrons with concentration n in the static magnetic field and the electric field zB).exp(0tiEEzzxyyyxxeEvdtdmvcBEevdtdmvcBEevdtdm111The drift velocity equations areThe velocity v=(vx,vy,vz)is)exp()exp()exp(000tivvtivvtivvzzyyxxSubstituted into the drift velocity equations,then zzxcyyycxxEmeivvEmeivvEmeiv111111where .mceBcIn the limit 1/,ii11Then we havezzxcyyycxxEmievvEmeivvEmeivThe solutions are:zzxcycyycxcxEmievEmeEmievEmeEmiev122122)()(Assume c,then we obtainzzxcyyycxxEmievEmeEmievEmeEmiev22The current density vnejzpzzypxpcxcyyypcxpycxxEiEmniejEiEEmneEmniejEEiEmneEmniej4444422222222222222i.e.where we definemnep224The Ohm lawEjTherefore the magnetoconductivity tensor is4 0 0 0 4 40 4 4 2222222pppcpcpzzzyzxyzyyyxxzxyxxiiii.e.2224 and4 pcyxxypzzyyxxi(b)From the Maxwell equations,an electromagnetic wave in a dielectric material22221dtEdcEwhere the dielectric function tensor is related to:)/4(1 iConsider the electromagnetic wave with wavevector.zkkThen we haveyExEtkziEEyx(expThe Maxwell equation becomesEicEcEk)41(122222041 4 4 4 41 4 4 4 4102222yxzzxzzxyzyyyxxzxyxxyxEEiiiiiiiiicEkEki.e.Substitute the components of inside,we obtain0)(0)(2222222222ypxcpycpxpEkcEiEiEkcThe determinant should be zero0)()(2222222222pcpcppkciikcTherefore the dispersion relation is cppkc22222As a given frequency there are two modes of propagation with different wavevectors and different velocities.The two modes correspond to circularly polarized waves.Because a linearly polarized wave can be decomposed into two circularly polarized waves,it follows that the plane of polarization of a linearly polarized wave will be rotated by the magnetic field.This is the magneto-optic effect.8.Cohesive energy of free electron Fermi gas(a)The kinetic energy U of the free electron Fermi gas at 0K isFNU53Then the average kinetic energy per electron is3/222310353/nmNUFThe electron concentration n is313010)(433/4HsarrVnThen3/22223/23223/22249103 )(4331033103HsHsamrarmnm222422Ry 1 HmameRy 21.2Ry 4953 223/2ssrr(b)The charge density of the uniform electron charge distribute is130103/4reeVThe coulomb energy of a point positive charge e interacting with this uniform electron distribution in the volume of radius r0 is:0202302123414300redrrrredVrerrrHaeme22Ry 1 224Ry 3 1sr(c)The charge Q in the sphere with radius r is:330334rrerQThe charge dQ in the shell between r and r+dr is:drrredQ2303The coulomb energy of the interacting between Q and dQ isdrrrerQdQd46023The coulomb self-energy of the electron distribution in the sphere with radius r0 isRy 56 5330204602200srrrrredrrred(d)The total coulomb energy per electron C is Ry 8.1Ry 56321sssCrrrThe total energy per electron is the sum of the kinetic energy and the coulomb energy.Ry 8.121.22ssCrrAt equilibrium state,08.121.2223sssrrdrdThen we obtain the equilibrium value of rs=22.21/1.8=2.46The equilibrium energyRy 37.0 Ry 46.28.146.221.2Ry 8.121.222ssrrThis energy is higher than the ground state of an electron around a hydrogen atom(1 Ry).Therefore the separated H atoms are more stable.9.Static magnetoconductivity tensorzzxyyyxxeEvdtdmvcBEevdtdmvcBEevdtdm111Consider a static magnetic field along z axis.The drift velocity equations areFor the static current 0dtvdThen zzxcyxyyycxyxxEmevvEmevcBEmevvEmevcBEmevmceBc whereThe solutions of these equations arezzyxccyycxcxEmevEEmevEEmev22)(11)(11The current density venjzzzyxccyxccyycxcycxcxEEmnejEEEEmnejEEEEmnej0220222022)(1)(11)(1)(11mne20 whereThe Ohm lawEjzyxcccczyxEEE)(jjj2201 0 00 1 0 1)(1In the high magnetic field limit,c 1Bneccccxyyx020)(110.Maximum surface resistanceFor a monatomic metal sheet one atom in thickness,the electron concentration n is ,where a is the diameter of the atom.33/1/1danThe surface resistivity is222/edmvendmvRFFsqThe electrons are limited in the thickness of d,therefore the de Broglie wavelength is approximately d.then we havedmvF/32101.4/eRsq