PHY 140A: Solid State Physics. Solution to Midterm #1

Transcription

1 PHY 140A: Solid Stte Physics Solution to Midterm #1 TA: Xun Ji 1 October 24, Emil: jixun@physics.ucl.edu

2 Problem #1 (20pt)Clculte the pcking frction of the body-centered cubic lttice. Solution: (Plese refer Problem #3 in Homework #1.) For bcc lttice, consider the body digonl, we hve 3 = 4r, thus = 4r/ (4pt) (1) Moreover, there re 8 spheres t the corners, nd 1/8 of ech is contined in the cube; there re 1 spheres t the center, which is totlly contined in the cube; thus the number of spheres in the cube is n = = (4pt) (2) 8 The volume of the cube nd the volume of ech sphere re: V c = (4pt) (3) V s = 4π 3 r (4pt) (4) Therefore, the pcking frction is: nv s 3π = = (4pt) (5) V c 8 Problem #2 (40pt)Experimentlists hve creted room-temperture superconductor, nd you ( theorist) think its structure is simple cubic with 6.0Å lttice constnt. The experimentlists hve powder Debye-Scherrer cmer with 1.5Å X-ry source. Clculte the ngulr positions φ of the first 3 diffrction rings for shre of the Nobel prize. (). Drw picture tht illustrtes the diffrction condition relting G ( reciprocl lttice vector), k (the wve vector of the incident X-ry), nd k (the wve vector of the scttered X-ry). (b). Use the picture in prt () to derive reltionship between G, k, nd the scttering ngle φ. (c). Write symbolic expression for G. (d). Use the results of prts ()-(c) to clculte the ngulr positions φ of the first 3 diffrction rings. Your nswer should be numeric, but plese leve ny nontrivil trigonometric functions, squre roots, etc. unevluted. 1

3 Solution: (). As in the Fig. :(8pt) k' G k Figure 1: Illustrtion of the diffrction condition. (b). In the cse of elstic scttering, k = k. From bove figure, the reltionship mong them re: 2 k sin φ 2 = G (8pt) (6) (c). The reciprocl lttice for sc is lso n sc. The expression of G is: G = 2π v v v (8pt) (7) (d). Since k = 2π, from bove, we hve the expression for φ: λ φ = 2 rcsin λ v1 2 + v2 2 + v (4pt) (8) 2 For the simple cubic in this problem, the reciprocl lttice is lso simple cubic. The first three rings of diffrction pek correspond to the cse (v 1, v 2, v 3 ) = (100), (110), nd (111). Substitute into Eqn. (8) with the vlue of wve length λ, nd lttice constnt, we hve: φ 1 = 2 rcsin( 1 8 ) (4pt) 2 φ 2 = 2 rcsin( 8 ) (4pt) 3 φ 3 = 2 rcsin( 8 ) (4pt) (9) Problem #3 Copper oxide lyers: The common building blocks for most high temperture superconductors re copper oxide lyers, s depicted below. Assume the distnce between 2

4 the copper toms (filled circles) is. Sor simplicity, let us lso ssume tht in the third dimension these CuO 2 lyers re simply stcked with spcing c, nd there re no other toms in the crystl. In the first pproximtion the lyers hve four-fold symmetry; the crystl is tetrgonl. (). For the crystl in Fig. 2: sketch the Brvis lttice. indicte possible set of primitive vectors. nd describe the bsis. On the digrm, indicte primitive cell. (b). In some compounds closely relted to the high-temperture superconductors one finds, t closer inspection, tht the CuO 2 lttice is ctully not flt, but tht oxygen toms re moved smll mount out of the plne ( up or down ) in n lternting fshion (see Fig. 3, + mens up nd - mens down.) On the digrm, indicte: the primitive cell. the lttice spcing. nd the bsis for the distorted crystl. (c). Wht is the reciprocl lttice the the new (distorted) Brvis lttice? Solution: Clculte G(v 1, v 2, v 3 ). Drw the reciprocl lttice plne with v 3 = 0. Indicte the primitive reciprocl lttice vector b 1 nd b 2 of the undistorted lttice on the sme drwing. Describe (qulittively) wht hppens in the reciprocl lttice, nd the X-ry diffrction pttern, s the distortion is decresed grdully to zero. (). For the crystl: Sketch the Brvis lttice. As shown in the dsh line in Fig. 2. The Brvis lttice in this 2d plne is squre structure.(2pt) Construct the coordinte s in the figure, then the primitive vectors re: 1 = x, 2 = y, 3 = cz (3pt) (10) The bsis cn be chosen s the copper tom t (0,0) nd the oxygen toms t (0, /2) nd (/2, 0), s mrked out by solid curve in the figure. 3

5 y x 2 1 Figure 2: CuO 2 plne of the superconductor. One primitive cell is the squre with four copper toms t its corners. There re mny other eqully correct choices, for exmple, the squre with copper tom t the center. Both re illustrted in the figure. (b). For the distorted crystl: The primitive cell is shown s the shded re in Fig. 3:(3pt) Now the primitive vector re: 1 = (x + y), 2 = (x y), 3 = cz (11) then the lttice spcing is: = 1 = 2 = (3pt) (12) The bsis for the distorted lttice cn be chosen s mrked out by the solid curve in the figure, tht is, the copper toms t (0, 0) nd (, 0), nd the oxygen toms t (/2, 0), (3/2, 0), (, /2), nd (, /2).(4pt) (c). For the new reciprocl lttice: 4

6 y x ' ' Figure 3: The distorted CuO 2 plne of the superconductor. From the expression of i, we hve: b 1 = (2π) ( 2 3 ) = π (x + y) b 2 = (2π) ( 2 3 ) = π (x y) (13) therefore, the reciprocl vector: b 3 = (2π) ( 2 3 ) = 2π z π G(v 1, v 2, v 3 ) = v 1 (x + y) + v π 2 (x y) + v 2π 3 z (5pt) (14) At v 3 = 0, the reciprocl lttice plne is shown s in the Fig. 4:(5pt) For the undistorted lttice, from Eqn. (10) it is esy to get: nd they re drwn in the Fig. 4.(5pt) b 1 = 2π x, b 2 = 2π y (15) In Fig. 4, the lttice points in the center of ech squre belong to the reciprocl lttice of the distorted lttice, but do not belong to the undistorted 5

7 y x b' 1 b' 2 / b 2 b 1 Figure 4: The reciprocl lttice in v 3 = 0 plne. lttice. Since ech reciprocl lttice point corresponds to diffrction pek in the X-ry diffrction pttern, s the distortion is decresed grdully to zero, the lttice points in the center of ech squre dispper, nd the diffrction pek corresponding to those points will get weker nd weker, nd finlly dispper. In other words, the structure fctor of the reciprocl lttice points in the centers of the squres goes to zero s the distortion decreses. For the peks corresponding to the rest of reciprocl lttice, which belong to both distorted nd undistorted lttice, the pek will get stronger. (5pt) 6