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Homework # 6 Chapter 6 Kittel Phys 175A Dr. Ray Kwok SJSU Prob. 1 – Kinetic Energy of electron gas Adam Gray Kinetic energy of electron gas. Show that the kinetic energy of a threedimensional gas of N free electrons at 0 K is 3 U 0 = N E f 5 . In Kittel, we’re given the energy of a free electron in this model to be: h2 E k = 2m K 2 To find the mean value of E over the volume of the sphere in K space we use the definition: E 1 ∫ E = Vol ( C ) C For our problem we have h 2 1 ∫∫∫ K 2 • K 2 sinθdkdθdφ E = 3 2m (4 / 3)πK f h 2 1 (4 / 5)πK 5f E = 3 2m (4 / 3)πK f ( ) Which results in the following mean energy per electron of 3 h K E = 2 2 f 5 2 m Comparing to our initial energy equation (12), we can rewrite this as 3 E = E 5 f So for N number of electrons, the total energy would be 3 U 0 = N E f 5 Prob. 2 – Pressure & Bulk modulus Victor 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 (3rd Law 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) ∂U o p = − ∂V 3Nh 3π 2 N Uo = 10m V 2 Solution: −1 2 3 ∂U o 3Nh 2 3π N 3 3π 2 N 2U o p = − =− − 2 = ∂V 10m 3 V V 3V 2 2 (b) Show that the bulk modulus (B) of an electron gas at 0 K is ∂p B = −V ∂V Solution: B= 5 p 10U o = 3 9V 2U o p= 3V 2 ∂U o 2 2U o ∂ 2 2 B = −V + Uo − + Uo − 2 = −V 3V 3V ∂V 3V 3V 3V ∂V 4U o 6U o 10U o B = + = 9V 9V 9V (c) Estimate for potassium (K), using Table 1, the value of the electron gas contribution to B Solution: 10U o 10 3Nε 10 3 1.40 ×10 22 3.15 *1010 dyn 22 eV B= = = = (2.12eV ) = 1.97 *10 3 3 9V 9 5V 9 5 cm cm cm 2 Answer agrees with value from Table 3, chapter 3. Prob. 3 – Chemical potential in 2D Jason Thorsen Show 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: Solution: We know the Fermi-Dirac distribution is given by And we know the identity We know that D(ε) is a constant so take it out of integral: Plug in values and evaluate: QED Prob. 4 – Fermi gas in astrophysics Michael Tuffley (Wavevector is invariant in relativistic limit) (Off by a factor of ~3) Prob. 5 – Liquid H3 Daniel Wolpert 6.5 Liquid He3. The atom He3 has spin ½ and is a fermion. The density of liquid He3 is 0.081g/cm3 near absolute zero. Calculate the Fermi energy εF and the Fermi temperature TF. (.081g/cm3/3.016(g/mole))(6.02x1023atoms/mole) = 1.625x1022 atoms/cm3 2 electrons per molecule = 3.25x1028 electrons/m3 εF = ħ2/2m ( 3π2N/V)2/3 = (1.055x10-34)2/(2(9.11x10-31)) * (29.6*3.25x1028)2/3 = 5.97x10-19 J TF=εF/kb = 5.97x10-19 J/1.38x10-23 J/K TF = 4.3x104 K Prob. 6 – Frequency dependence of σ John Anzaldo Use the equation for the electron drift velocity to show that the conductivity at frequency w is: where Chapter 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 gives and . Chapter 6 #6: Oscillation of Conductivity Solving for gives Multiplying by and gives Recall that J=σ σE=nqv, where q=-e (p. 147148) Chapter 6 #6: Oscillation of Conductivity This gives the relation Solving for gives which is what we set out to prove. Prob. 7 – Dynamic Magnetoconductivity tensor for free electron Michael 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 medium This gives rise to a set of four linear, homogeneous, algebraic equations. For a nontrivial solution to exist, the determinant must vanish (Thornton 497.)1 QED 1Thornton, Stephen T. Classical dynamics of particles and systems. Belmont, CA: Brooks/Cole, 2004. Prob. 8 – Cohesive energy of free electron Fermi gas Gregory Kaminsky Cohesive energy of free electron Fermi gas. We define the dimensionless length rs as r0/aH, where r0 is the radius of a sphere that contains one electron, and aH is 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. Average kinetic energy: Integral of the energy inside the sphere in k space, divided by the volume of the sphere in k space. kF 2 4 π * 4 k dk ∫ h 0 kF 2m ∫ 4π * k 0 h2 2 εF = kF 2m 2 dk 3 = εF 5 Where kF is the wavevector at the Fermi surface and εF is the Fermi energy. Need to show that (3/5) εF = 2.21/rs2 3 3h 2 2 3h 2 3π 2 2 / 3 3 h 2 kF = ( ) = * εF = 2 4 5 5 * 2m 5 * 2m 5 2 mr 0 π * r03 3 9π 4 2/3 2.21 2.21a H2 2.21 * h 4 = 4 2 2 2 = r0 rs e m r0 Converting from Rydbergs. 2.21 * h 4 me 4 h2 4 2 2 * 2 = 2.21 * e m r0 2h 2mr02 Setting them equal to each other and canceling out similar factors. 2/3 3 9π 2.21 = 5 4 That’s true, you can check on the calculator. Q.E.D b)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 r0 is –3e2/r0, or –3/rs in 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. Charge density = δ = 4 3 e e Potential → V = r π *r3 e − e − 3e 2 ro − e 3e 2 2 2 U = ∫∫∫ r sin θdrd θdφ ⋅ ρ ( r ) = ∫ 4 πr dr = 3 ∫ rdr = − ro 0 2r o 4 πr 3 r r 0 o 3 ro c) Show that the coulomb self-energy of the electron distribution in the sphere is 3e2/5r0, or 6/5rs in rydbergs. r1 Energy required to bring a point charge close to a sphere is r2 ρ(d 3 r1 )ρ(d 3 r2 ) r2 r1 Bring a ring close to a charged sphere is r2 ρ(d 3r1 )ρ(4πr22 dr2 ) r2 Stack up all the shell up to r = ro r1 ro ro ρ(d 3 r1 )ρ(4πr22 dr2 ) 2 3 2 2 = 2 πρ d r ( r − r ) 1 o 1 ∫ r2 r2 = r1 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 integral of it: ro 3 2 5 r r r 16 2 2 5 2 2 2 2 2 2 o o o U = ∫ 2πρ (4πr1 dr1 )(ro − r1 ) = 8π ρ [ − ] = π ρ ro 3 5 15 0 With ρ= −e 4 3 πro 3 3e 2 U= 5ro d) The Sum of (b) and (c) gives –1.80/rs for the total coulomb energy per electron. Show that the equilibrium value of rs is 2.45. Will such a material be stable with respect to separated H atoms? Factoring in the Kinetic energy of the electron in a free Fermi gas at 0 K, basically adding in the answer from part (a) to (b) and (c) 2.21 180 . Energy = 2 − rs rs To find the equilibrium value of rs value I took the derivative of Energy with respect to rs d 4.42 180 . (E) = − 3 + 2 drs rs rs 4.42 180 . Setting this equation to zero: d (E) = − 3 + 2 = 0 drs rs rs rs = 2.45 To check whether this is a stable equilibrium, I took the second derivative. d2 13.3 2.6 2 (E) = 4 − drs rs rs3 At E(rs = 2.45), this second d 13.3 2.6 . derivative is positive, so it is dr ( E ) = − 2.454 + 2.453 = 019 s a stable equilibrium. Thus since the second derivative is positive, It is a stable equilibrium, and such a metal will be stable with respect to the separated H atoms at zero Kelvin. The end. Prob. 9 – Static magnetoconductivity Wanshan Li tensor Q: Let a static magnetic A: field B lie along the z a) From (52) p153, the axis, drift velocity is a) find the static current eτ v x = − E x − ωcτv y [1] density j in a static m electric field E. eτ v y = − E y − ωcτv x [2] m b) under strong eτ magnetic field (ωcτ >> 1 vz = − Ez m ), find σ yx , σ xy Problem 9: Current Density (2/3) Plug [2] into [1], and solve for vx ωc eτ 2 eτ − Ex + Ey m vx = m 1 + (ωcτ ) 2 plug [1] into [2], and solve for vy. − ωcτ 2 e eτ Ex − Ey m m vy = 1 + (ωcτ ) 2 From (42) p147, electric current density j x = n(−e)v x Plug in vx, and define electrical conductivity 2 ne τ σo = m Then, σ o ( E x − ωcτE y ) jx = 1 + (ωcτ ) 2 Problem 9: Hall conductivity (3/3) Similarly for jy, jz. In matrix form, jx 1 σo j = ωτ y 2 c ω τ + 1 ( ) c j 0 z − ωcτ 1 0 E x E y 1 + (ωcτ ) 2 E z 0 0 b) Hall conductivity σ yx = σ oωcτ 1 + (ωcτ ) 2 When in high magnetic field, ωcτ >> 1 We have σ yx = σo ωcτ Plug in frequency ωc = eB mc and σ o , then σ yx = nec = −σ xy B Prob. 10 – Maximum surface resistance Nabel Alkhawlani Consider a square sheet of side L, thickness d, and electrical resisitvity ρ. the resistance measured between opposite edges of the sheet is called the surface resistance Rsq = ρ L/Ld = ρ/d, which is independent of the area L2 of the sheet. If we express ρ by 44, then R sq = m n d e 2τ Suppose now that the minimum value of the oscillation time is d determined by the scattering from the surfaces of the sheet, so that, τ ≈ v f where vf is the fermi velocity. Thus the maximum surface resistvity is Rsq ≈ in thickness that mv f nd 2 e 2 R sq ≈ show for a monatomic metal sheet one atom h e2 Sketch of the problem λ~d ρL ρL ρ = = A Ld d ne 2 τ 1 σ= = m ρ d d3 = vol/atom τ= n = 1/d3 = # of atoms/vol vf R sq = R sq = m mv f mv f d = = ne 2 τd ne 2 d 2 e2 With λ ~ d, p = mv = h / λ ~ h / d R sq ≈ h e2 = (6.6 x 10-34)/(1.6 x 10-19)2 = 26 kΩ