PHY 140A: Solid State Physics. Solution to Midterm #2

Transcription

1 PHY 140A: Solid State Physics Solution to Midterm #2 TA: Xun Jia 1 December 4, jiaxun@physics.ucla.edu

2 Problem #1 (20pt)(20 points) Use the equation dp dt + p = ee for the electron momentum, τ derived in class in the context of the Drude model, to find the AC conductivity of a metal. Your answer should be of the form σ(ω, n, e, τ). (Please refer Problem #4 in Homework #6.) Problem #2 (20pt) On a per gram basis, which do you expect to have a larger, constant-volume heat capacity C V at room temperature, diamond or sodium? Explain your reasoning. To receive full credit on this problem your argument might include a plot, but do not spend time doing a more detailed calculation than is necessary to answer the question. property carbon (diamond) sodium (bcc) atomic number Z 6 11 atomic weight A number density n, cm mass density r, g/cm Debye temperature, K diamond sodium 3R C V sodium T room T diamond Figure 1: Illustration of the dependence of the molar specific heat of diamond and sodium on temperature. 1

3 The dependence of the molar specific heat of diamond and sodium on temperature is illustrated as in Fig. 1. Since θ sodium < T room, at room temperature, the molar specific heat of sodium is about CV,sodium molar 3R; on the other hand, since θ diamond T room, diamond will have a quite small molar specific heat. Therefore, at room temperature, the molar specific heat (15pt): On a per gram basis, the heat capacity C gram V C molar V,diamond C molar V,sodium (1) = CV molar /A, where A is the atomic weight. The atomic weight of carbon and sodium are of the same order: A = 12 and A = 23, so when converted to per gram basis (5pt): C gram V,sodium = Cmolar V,sodium 23 say, sodium will have larger heat capacity. > Cmolar V,diamond = C gram V,sodium (2) 11 Problem #3 (40pt)You have a monatomic cubic lattice of lattice spacing a = 3Å and sound velocity c = 10 3 m/s. (a). What is the Debye temperature, θ? (Provide a numerical value in degrees.) (b). Say you want to do a lattice heat-capacity experiment in the fully quantummechanical regime (i.e. even the acoustic modes have frozen out ). Your apparatus is capable of reaching liquid-helium temperature (4K). How small does your sample have to be? Give the answer in atoms. Hint: Recall that for a finite sample, the allowed wavevectors are discrete. You are looking for the smallest, non-zero wavevector, as this will give you the lowest energy excitation of the crystal. Relate this energy to the thermal energy to solve the problem. (a). According to the calculation in Kittel, the Debye temperature is θ = hc k B note that V = Na 3, we have: ( ) 1/3 6π 2 N (10pt) (3) V θ = hc ( ) 6π 2 1/3 = 99.3K (5pt) (4) k B a 2

4 dispersion relation = ck ground state first excited state 0 2 /L 4 /L k Figure 2: Dispersion relation. (b). As in Fig. 2, at extremely low temperature where even acoustic modes have frozen out, the discretized nature of energy levels becomes important due to the quantized values of k vector. Since k = 2πn/L, with integer n, the first excited state now corresponds to k = 2π/L. To observe physics in this regime, the thermal energy k B T should be comparable with the energy difference between ground state and first excited state hω = hck = hc2π/l (15pt), i.e., k B T hc 2π L (5) which gives: L a 2π hc k B T a = 40 (6) say, the size of the sample should be about 40 atoms (10pt). Problem #4 (20pt) The figure above left shows the first Brillouin zone in a graphite plane. The figure above right shows some calculations (and data) for the phonon dispersion relation between the points Γ and K in the first Brillouin zone. Consider a graphite sheet a 2D object, but the atoms are allowed to move in 3D. Because there are 2 atoms in the basis (honeycomb structure), we expect six modes in general: AL (acoustic longitudinal), ATo (acoustic transverse out-of plane), ATi (acoustic transverse in-plane), OL(optical longitudinal), OTo (optical transverse out-of-plane), and OTi (optical transverse in-plane). 3

5 (a) Figure 3: (a) The first Brillouin zone in a graphite plane; (b) Some calculations (and data) for the phonon dispersion relation between the points Γ and K in the first Brillouin zone. (b) (a). Draw a graphite lattice plane, and indicate directions corresponding to the three types of modes. Dont worry about the basis for this part. (b). Which acoustic mode will have the lowest speed of sound? (c). Which acoustic mode will have the highest speed of sound? (d). Which optical mode will have the lowest energy near k = 0? (e). What degeneracies do you expect? Label all six modes on the above right figure. Explain how you know all this. (a). The graphite lattice plane is shown in Fig. 4. The wave is propagating along direction k, which is along a bound connecting two neighboring carbon atoms. From the top view, the longitudinal mode L vibrates along direction of k; the transverse in plane mode T ivibrates as in the figure; The transverse out of plane mode vibrates along a direction perpendicular to the plane, as depicted in a 3-D view in Fig. 4(b). (4pt) (b). As the atom vibrating in different modes, the longitudinal motion (L) will have the largest effect coupling constant, since once the atom moves along this direction with a little displacement, the bound connecting neighboring atoms will provide restoring force. 4

6 Ti To k L Ti k L (a) Figure 4: (b) Direction of the three modes for a given wave vector k. (a) top view, (b) 3-D view. The transverse out of plane mode (To) will have lowest coupling constant, since once the atom moves in this direction, all three bounds connecting to this atom will be only stretched by a little amount, and the restoring force will be weak. The transverse in plane mode (Ti) will have an intermediate effect coupling constant. Since the speed of sound will positively correlated with the effective coupling constants, the Acoustic transverse out of plane mode (ATo) will have lowest speed of sound. (3pt) (c). As stated in previous part, the Acoustic longitudinal mode (AL) will have largest Figure 5: Labeling of 6 modes. 5

7 speed of sound. (3pt) (d). The Optical transverse out of plane mode (OTo) will have lowest energy near k = 0 since its effective coupling constant is lowest among all optical modes. (3pt) (e). Optical longitudinal (OL) and Acoustic longitudinal (AL) mode will be degenerate at the Brillioun zone boundary, Optical transverse out of plane (OTo) and Acoustic transverse out of plane (ATo) will be degenerate at the Brillioun zone boundary. (1pt) The 6 modes are labeled as in Fig. 5. (6pt) Remark: For detail of this experiment, please refer to the paper: R. Saito, A. Jorio, A. G. Souza Filho etc., Phys. Rev. Lett., Vol 88, (2002). 6