MSCI406 HOMEWORK 04 6-1: Kinetic energy of electron gas. Show that the kinetic energy of a three-dimensional gas of N free electrons at 0K is U0 = 53 N F 2 The energy eigenvalues are εk = 2m k 2 . The mean value over the volume of a sphere in k space is 2 2 k 2 dk·k 2 hεi = 2m k2 dk = 35 · 2m kF2 = 53 εF . The total energy of N electrons is U0 = N · 53 εF . 6-2: Pressure and bulk modulus of an electron gas. (a) Derive a relation connecting the pressure and volume of an electron gas at 0K. Hint: Use the result of Problem 6.1 and the relation between F and electron concentration. The result may be written as p = 23 (U0 /V ). (b) Show that the bulk modulus B = −V (∂p/∂V ) of an electron gas at 0K is B = 5p/3 = 10U0 /9V . (c) Estimate for potassium, using Table 6.1, the value of the electron gas contribution to B. a. In general p = −∂U/∂V at constant entropy. At absolute zero all processes are at constant 2 2/3 2 entropy (the Third law), so that p = −dU0 /dV , where U0 = 53 N εF = 35 N 2m 3πV N , whence p = 23 · UV0 . 2 U0 2 dp 0 b. Bulk modulus B = −V dV = V − 23 VU02 + 3V2 dU = 3 · V + 32 dV c. For K, B ≈ 3GP a, very close to 3.2GP a experimental data. U0 V = 10 U0 . 9 V 6-6: Frequency dependence of the electrical conductivity. Use the equation m (dv/dt + v/τ ) = −eE for the electron drift velocity v to show that the conductivity at free frequency ω 1+iωτ is σ(ω) = σ(0) 1+(ωτ )2 , where σ(0) = ne2 τ /m. eE/m Let E, v vary as e−iωt . Then, v = − −iω+(1/τ = − eτmE · ) j = n(−e)v = ne2 τ m · 1+iωτ , 1+(ωτ )2 and the electric current density is, 1+iωτ E. 1+(ωτ )2 7-1: Square lattice, free electron energies. (a) Show for a simple square lattice (two dimensions) that the kinetic energy of a free electron at a corner of the first zone is higher than that of an electron at midpoint of a side face of the zone by a factor of 2. (b) What is the corresponding factor for a simple cubic lattice (three dimensions)? (c) What bearing might the result of (b) have on the conductivity of divalent metals? a. The wavevector at the corner is longer than the wavevector at the midpoint of a side by the √ √ 2 factor 2. As ε ∝ k 2 for a free electron, the energy is higher by 2 = 2. √ 2 b. In three dimensions the energy at a corner is higher by 3 than at the midpoint of a frace. c. Unless the band gap at the midpoint of a face is larger than the kinetic energy difference between this point and a corner, the electrons will spill over into the second zone in preference to filling up the corner states in the first zone. Divalent elements under these conditions will be metals and not insulators. 2 7-3: Kronig-Penney model. (a) For the delta-function potential and with P 1, find at k = 0 the energy of the lowest energy band. (b) For the same problem find the band gap at k = π/a. a. At k = 0 the determinantal equation is (P/Ka) sin Ka + cos Ka = 1. In the limit of small positive P this equation will have a solution only when Ka 1. Expand the sine and cosine to obtain in lowest order P ' 21 (Ka)2 . The energy is ε = ~2 K 2 /2m ' ~2 P/ma2 . b. At K = π/a, the determinantal equation is (P/Ka) sin Ka+cos Ka = −1. In the same limit this equation has solutions Ka = π + δ, where δ 1. We expand to obtain (P/π)(−δ) + (−1 + 12 δ 2 ) = −1, which has the solution δ = 0 and δ = 2P/π. The energy gap is Eg = (~2 /2ma2 )(2πδ) ' (~2 /2ma2 )(4P ). F-5: A triangular wave is represented by f (x) = x at 0 < x < π, and f (x) = −x at −π < x < 0, Find Fourier series. Do the coefficients Ah , Bh (and Ch ) decrease faster or slower, with the number h, than in the Problems F1 or F4? Follow (F2,3) to obtain Bh = 0, Ah = 2 πh2 (−1)h − 1 , that is f (x) = π 2 − 4 π P h=1,3,5,7,... cos hx , h2 so the coefficients Ah ∝ h−2 decrease faster than for the problem F1 or F4, because this function is smoother. F-6: If the coefficients Ah , Bh are known for a function f (x), find the coefficients in the Fourier series for df /dx and for d2 f /dx2 . Try the higher derivatives, and exercise with the exponential representation Eq.(F4), that is the Ch . f fh = − 2πh Ah , A fh = 2πh Bh , for f 0 ; B fh = − 2πh 2 Bh , Differentiate F1 or F4 directly to obtain B x a a a f f fh = − 2πh 2 Ah , for fxx ” and, C fh = i2πh Ch , and C fh (n) = i2πh n Ch . fh = − 2πh 2 Ch , . . . , C A a a a a