... 3, , = a (1) 3 3 a 2 = a (2) The reciprocal lattice vectors are defined by the condition a b = 2πδ ij, which gives

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Domenic Preston
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1 PHZ646: Fall 013 Problem set # 4: Crystal Structure due Monday, 10/14 at the time of the class Instructor: D. L. Maslov maslov@phys.ufl.edu Rm. 114 Office hours: TR 3 pm-4 pm Please help your instructor by doing your work neatly. Every algebraic) final result must be supplemented by a check of units. Without such a check, no more than 75% of the credit will be given even for an otherwise correct solution. P1 0 points Find the primitive vectors of the real a 1 and a ) and reciprocal b 1 and b ) lattices for graphene, as well as positions of the K and K points [see Fig. 3]. Use the carbon-carbon distance a 0.14 nm as the yardstick. Notice that b 1 and b connect the centers of the adjacent hexagons in the reciprocal space. Solution... a 1 = a cos π 3 + a, a sin π ) 3 = a 3 3, ) 3 3 a = a 3, ) 3 1) ) The reciprocal lattice vectors are defined by the condition a b = πδ ij, which gives a 1x b 1x + a 1y b 1y = π a x b 1x + a y b 1y = 0 3) or which gives 3 3 b 1x + b 1y = π a 3 3 b 1x b 1y = 0, 4) b 1 = π a ) 1 3, 1 3 5) By symmetry, b = π a 1 3, 1 ). 6) 3 Notice that b 1 b = 0, 4π/a 3) is along the y-axis. The distance between the K and K points is 1/3 of b 1 b : KK = 4π/a3 3.Therefore, K x = b 1x = π/3a and K y = KK / = π/a3 3, so that K = π/3a)1, 1/ 3). By symmetry, K = π/3a)1, 1/ 3). P 0 points A common building block of high T c superconductors are copper oxide layers.

2 A possible choice of the primitive cell. The cell contains 1 Cu atom and O atoms =O =Cu FIG. 1: Unit cells for a flat D) CuO plane and for a real 3D) CuO sheet. Sketch the Bravais lattice, identify the basis, and define the primitive unit cell for a D CuO plane, as shown in Fig. 4. In a parent compound of high T c superconductors LaCuO 4 ), the oxygen atoms are displaced above and below the plane in an alternating fashion, as shown in Fig. 5. Identify the primitive unit cell for this case. Solution See Figures 1 and. P3 0 points Show that the c/a ratio for an ideal hexagonal close-packed hcp) lattice is 8/ Solution Since the spheres centered at atoms in the basal plane touch each other, a = R, where R is the radius of the sphere. A sphere centered at point A must touch the one centered at point O, therefore, OA = R. If A is the projection of A onto the basal plane, the distance OA is equal to a/)/ cos30 )=a/ 3. From the rectangular triangle OAA, c/ = AA / = OA OA. Therefore, c/a = /3 = 8/3. Sodium transforms from the body-centered cubic bcc) to hcp phase at T = 3 K. Assuming that the number density remains fixed and also that the c/a ratio remains ideal, find the hcp lattice spacing a given that the lattice spacing in the bcc phase is a = 4.3Å.

3 =O above the plane =O below the plane =Cu FIG. : Unit cells for a real 3D) CuO sheet. Solution A bcc lattice can be viewed as a simple cubic lattice with atoms per unit cell. Therefore, the number density in the bcc phase n bcc = /a ) 3. The hcp unit cell of volume 3/)a c also contains atoms, thus n hcp = 4/ 3a c. For an ideal hcp lattice, c = 8a/3 and n hcp = 4/ 8a 3. Equating n bcc and n tcp, we obtain a = a / 1/6 = 3.77Å. P4 0 points Find the volumes of the primitive unit cells of the bcc and fcc cubic lattices. Solution For a bcc lattice, the primitive lattice vectors can be chosen as a 1 = a ŷ + ẑ ˆx) a = a ˆx + ẑ ŷ) a 3 = a ˆx + ŷ ẑ) 7) The volume of the primitive cell Simple vector calculus gives a 1 a = a /)ŷ + ẑ), and V = a 3 /. V = a 1 a a ). 8)

4 fcc lattice: a 1 = a ŷ + ẑ)a = a ˆx + ẑ)a 3 = a ˆx + ŷ) 9) V = a 3 /4. P5 0 points Consider an imperfect crystal, where the nth atom for all n) is displaced from its ideal position, R n, by a random vector S n. We are interested in a diffraction peak around the reciprocal lattice vector G. Assume that the displacements at all sites are small: S n a, where a is the lattice spacing. Find the intensity of the diffraction peak. Hint: Assume that all the three components of S are distributed according to the Gaussian law with zero average P S i ) = 1 πσ exp S i /σ ), where i = x, y, z. Solution The intensity of the scattered wave is proportional to the mod squared of the structure factor averaged over random displacements of atoms I n expig R n ) expig S n ) = n,n expigs n ) expigs n ), 10) where S m is the projection of S m onto G. Assuming that displacements of different atoms are independent and obey Gaussian distributions with dispersions S 0, we obtain expigs n ) = 1 πs0 = = 1 πs0 1 πs0 ds n exp S /S 0 + igs n ) ds n exp 1 ds n exp 1 S 0 S 0 S n is 0GS n ) ) Sn is 0G ) + S 4 0 G )) ) 1 = dz exp z πs0 S0 exp S0G /) = exp S0G /). 11) Therefore, the intensity of the Bragg peak in the direction of G is attenuated by the factor exp G S 0). This factor is known as the Debye-Waller factor.

5 b 1 a 1 K K' a b FIG. 3: Graphene lattice in real left) and reciprocal right) spaces.

6 =O =Cu FIG. 4: Flat D) CuO plane.

7 =O above the plane =O below the plane =Cu FIG. 5: D projection of a real 3D) CuO sheet.

Phys 412 Solid State Physics. Lecturer: Réka Albert

Phys 412 Solid State Physics. Lecturer: Réka Albert Phys 412 Solid State Physics Lecturer: Réka Albert What is a solid? A material that keeps its shape Can be deformed by stress Returns to original shape if it is not strained too much Solid structure