Homework # 6Chapter 6 KittelPhys 175ADr. Ray KwokSJSUKinetic energy of electron gas.Show that the kinetic energy of a three-dimensional gas of N free electrons at 0 K is.fENU=530Prob. 1 – Kinetic Energy of electron gasAdam GrayIn Kittel, we’re given the energy of a free electron in this model to be:To find the mean value of E over the volume of the sphere in K space we use the definition:For our problem we have( )5322232)5/4()3/4(12sin)3/4(12fffKKmEddkdKKKmEππφθθπ=∫∫∫•=hh∫=CECVolE)(1222KmEk=hWhich results in the following mean energy per electron ofComparing to our initial energy equation (12), we can rewrite this asSo for N number of electrons, the total energy would be22253fKmE=hfEE=53fENU=530Prob. 2 – Pressure & Bulk modulusVictor Chikhani(a) Derive a relation connecting the pressure (p) and volume (V) of an electron gas at 0 K.At 0 Kelvin entropy is constant (3rdLaw of Thermo.)When S=0 we can solve for the pressure (p) by taking the partial of the internal energy (Uo) with respect to volume (V)Solution: Uo =3Nh210m3π2NV      23p = −∂Uo∂V      p = −∂Uo∂V      = −3Nh210m233π2NV      −13−3π2NV2      =2Uo3V(b) Show that the bulk modulus (B) of an electron gas at 0 K isSolution:(c) Estimate for potassium (K), using Table 1, the value of the electron gas contribution to BSolution:Answer agrees with value from Table 3, chapter 3.B =5p3=10Uo9VB = −V∂p∂V      p =2Uo3VB = −V23V∂Uo∂V+ Uo∂∂V23V            = −V23V−2Uo3V+ Uo−23V2      B =4Uo9V+6Uo9V      =10Uo9VB =10Uo9V=1093Nε5V=109351.40 ×1022cm3      2.12eV( )=1.97 *1022eVcm3=3.15 *1010dyncm2Show that the chemical potential of a Fermi Gas in two dimensions is given by:for n electrons per unit area. Note: the density of orbitals of a free electron gas in two dimensions is independent of energy: Prob. 3 – Chemical potential in 2DJason ThorsenSolution:We know the Fermi-Dirac distribution is given byAnd we know the identityWe know that D(ε) is a constant so take it out of integral:Plug in values and evaluate:QEDProb. 4 – Fermi gas in astrophysicsMichael Tuffley(Wavevector is invariant in relativistic limit)(Off by a factor of ~3)6.5 Liquid He3. The atom He3has spin ½ and is a fermion. The density of liquid He3is 0.081g/cm3near absolute zero. Calculate the Fermi energy εFand the Fermi temperature TF.TF=εF/kb= 5.97x10-19 J/1.38x10-23J/KTF = 4.3x104K(.081g/cm3/3.016(g/mole))(6.02x1023atoms/mole) = 1.625x1022atoms/cm32 electrons per molecule = 3.25x1028electrons/m3εF= ħ2/2m ( 3π2N/V)2/3= (1.055x10-34)2/(2(9.11x10-31)) * (29.6*3.25x1028)2/3= 5.97x10-19JProb. 5 – Liquid H3Daniel WolpertUse the equation for the electron drift velocity to show that the conductivity at frequency w is: where Prob. 6 – Frequency dependence of σσσσJohn AnzaldoChapter 6 #6: Oscillation of Conductivity Because E oscillates with w as , we find that , where (From Eq. 42 on page 147) Solving gives  Plugging in and gives .Chapter 6 #6: Oscillation of Conductivity Solving for gives Multiplying by and gives Recall that J=σσσσE=nqv, where q=-e (p. 147-148)Chapter 6 #6: Oscillation of Conductivity This gives the relation Solving for giveswhich is what we set out to prove.Prob. 7 – Dynamic Magnetoconductivity tensorfor free electronMichael Tuffley(a) Solve the drift velocity equation (51) to find the components of the magnetoconductivity tensor.QED(b) Show that the dispersion relation for this wave in the medium is:electromagnetic wave equation in a nonmagnetic isotropic mediumThis gives rise to a set of four linear, homogeneous, algebraic equations. For a nontrivial solution to exist, the determinant must vanish (Thornton 497.)1QED1Thornton, Stephen T. Classical dynamics of particles and systems.Belmont, CA: Brooks/Cole, 2004.Cohesive energy of free electron Fermi gas. We define the dimensionless length rsas r0/aH, where r0is the radius of a sphere that contains one electron, and aHis the Bohr radius h2/e2m.a) Show that the average kinetic energy per electron in a free electron Fermi gas at 0 K is 2.21/rs2, where the energy is expressed in rydbergs, with 1 Ry = me4/2h2. Prob. 8 – Cohesive energy of free electronFermi gasGregory KaminskyAverage kinetic energy: Integral of the energy inside the sphere in k space, divided by the volume of the sphere in k space.h2402024435mk dkk dkkk FFFππε**∫∫=εF Fmk=h222Where kFis the wavevector at the Fermi surface and εFis the Fermi energy.Need to show that (3/5) εF= 2.21/rs23535 235 234335294222 2032 32022 3επππF Fmkmrmr= = =h h h* *(*) *//2 212 212 2122044 202... *rare m rsH= =hConverting from Rydbergs. 2 2122 21244 20242202. ** . *hhhe m rmemr=Setting them equal to each other and canceling out similar factors. 2 2135942 3./=πThat’s true, you can check on the calculator.Q.E.Db)Show that the coulomb energy of a point positive charge e interacting with the uniform electron distribution of one electron in the volume of radius r0is –3e2/r0, or –3/rsin rydbergs. Formula for the energy of electron due to point positive charge is just the integral of the potential multiplied by the charge density within the sphere, were interaction is occurring.Potential Ver→ =Charge density = δπ=er433*o2r03o23or022r2e3rdrre3rer34edrr4re)r(ddrdsinrUoo−=−=−ππ=−ρ⋅φθθ=∫∫∫∫∫c) Show that the coulomb self-energy of the electron distribution in the sphere is 3e2/5r0, or 6/5rsin rydbergs. r1r2Energy required to bring a point charge close to a sphere is22313r)rd()rd( ρρr1r2Bring a ring close to a charged sphere is222213r)drr4()rd( πρρr1roStack up all the shell up to r = ro)rr(rd2r)drr4()rd(212o132rrr222213o12−πρ=πρρ∫=This is the potential between a charged sphere and the thick shelf enclosed it.Total energy stored between the 2 charge-density is just the