Solid State Physics 460- Lecture 5 Diffraction and the Reciprocal Lattice Continued (Kittel Ch. 2)

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1 Solid State Physics 460- Lecture 5 Diffraction and the Reciprocal Lattice Continued (Kittel Ch. 2) Ewald Construction 2θ k out k in G Physics 460 F 2006 Lect 5 1

2 Recall from previous lectures Definition of a crystal Lattice + Basis Reciprocal lattice Lattice in Fourier space (reciprocal space) Diffraction from crystals Bragg Condition 2d sin θ = n λ Diffraction and the reciprocal lattice Today: Diffraction and the reciprocal lattice continued Ewald construction and the Brillouin Zone (BZ) Physics 460 F 2006 Lect 5 2

3 Recall from Lecture 3 Summary: Real and Reciprocal lattices Crystal lattice of translations: T(n 1,n 2, ) = n 1 a 1 +n 2 a 2 + n 3 a 3 Reciprocal lattice: G(m 1,m 2, ) = m 1 b 1 +m 2 b 2 + m 3 b 3, where b i. aj = 2π δ ij, where δ ij = 1, δ ij = 0, i j Any periodic function can be written f(r) = Σ G f G exp( i G. r) Information about the basis for the actual crystal is in the values of the Fourier coefficients f G = (1/V cell ) cell d 3 r f(r) exp( - i G. r) Kittel Ch. 2 Primitive vectors Physics 460 F 2006 Lect 5 3

4 Recall from Lecture 3 Bragg Scattering Law λ λ θ θ d 2 d sin θ Condition for constructive interference (Diffraction): 2d sin θ = n λ Maximum λ = 2d Only waves with λ smaller than 2d can satisfy the Bragg scattering law for diffraction For a typical crystal the maximum d ~ nm, so that λ < ~ nm Physics 460 F 2006 Lect 5 4

5 Recall from Lecture 3 Scattering and Fourier Analysis λ k in k out λ k = 2π/λ Note that k is a vector in reciprocal space with k = 2π/λ The in and out waves have the form: exp( i k in. r -i ωt) and exp( i k out. r -i ωt) If the incoming wave drives the electron density, which then radiates waves, the amplitude of the outgoing wave is proportional to: space dr n(r) exp(i (k in - k out ). r) d Physics 460 F 2006 Lect 5 5

6 Recall from Lecture 3 Scattering and Fourier Analysis λ k= G k in k out λ k = 2π/λ d Define k = k in -k out Then we know from Fourier analysis that (1/V cell ) cell dr n(r) exp(- i k. r) = ng only if k = G, where G is a reciprocal lattice vector Otherwise the integral vanishes Note: These statements are for a perfect crystal of size infinity. See prob. Kittel 2.4 for a finite crystal where the scattering is peaked at k = G with a finite width. Physics 460 F 2006 Lect 5 6

7 Recall from Lecture 3 λ Elastic Scattering k= G λ k = 2π/λ k in k out d For elastic scattering (energy the same for in and out waves) k in = k out, or k in2 = k out2 = k in + G 2 Then one arrives at the condition for diffraction: (using - G in expression above) G Is any one of the 2 k in. G = G 2 recip. lattice vectors Equivalent to the Bragg condition see next lecture Physics 460 F 2006 Lect 5 7

8 Ewald Construction G Is any one of the recip. lattice vectors Condition for diffraction: k out = k in + G 2θ k out k in G and 2 k in. G = G 2 = 2 k in G sin θ G = 2 k in sin θ (note sine function, not cosine) Why? Discussed in class Physics 460 F 2006 Lect 5 8

9 Equivalent to Bragg Condition From last slide, 2θ k out k in G G = 2 k in sin θ But k in = 2π/λ, and G = n (2π/d), where d = spacing between planes (see homework, Kittel prob. 2-1) Bragg condition 2d sin θ = n λ Physics 460 F 2006 Lect 5 9

10 Geometric Construction of Diffraction Conditions Recall k in k out = G and k in = k out Consequence of condition 2 k in. G = G 2 b 2 -k out G b 1 k in The vector k in (and k out ) lies along the perpendicular bisecting plane of a Gvector One example is shown Physics 460 F 2006 Lect 5 10

11 Diffraction and the Brillouin Zone Brillouin Zone - (BZ) - the Wigner-Seitz cell of the reciprocal lattice Formed by perpendicular bisectors of G vectors b 2 -k out G b 1 k in Special Role of Brillouin Zone Diffraction occurs only for k on surface of Brillouin Zone No diffraction occurs for any k inside the first Brillouin Zone Important later in course Brillouin Zone Physics 460 F 2006 Lect 5 11

12 Comparison of diffraction from different lattices The Bragg condition can also be written G = 2 k in sin θ sin θ = (λ /4π) G Thus the ratios of the sines of the angles for diffraction are given by: sin θ 1 / sin θ 2 = G 1 / G 2 Each type of lattice has characteristic ratios the positions of diffraction peaks as a function of sin θ Simple scaling with λ Physics 460 F 2006 Lect 5 12

13 Experimental Powder Pattern Diffraction peaks at angles satisfying the Bragg condition Experimental example Differences for imperfect powder averages Reciprocal Lattice units Physics 460 F 2006 Lect 5 13

14 Comparison of diffraction from different lattices Ratios sin θ i / sin θ 0 = G i / G 0, where θ 0 is the lowest angle peak (smallest G) Easiest to give ratios of squares G i 2 / G 0 2 Simple Cubic lattice (G in units of 2π/a) G i G i 2 ratio 1,0, ,1, ,1, ,0, ,1,0 5 5 Physics 460 F 2006 Lect 5 14

15 Comparison of diffraction from different lattices - continued FCC real space lattice (G in units of 2π/a) G i G i 2 ratio 1,1, ,0,0 4 4/3 2,2,0 8 8/3 3,1, /3 2,2, ,0,0 16 1/3 BCC real space lattice (G in units of 2π/a) G i G i 2 ratio 1,1, ,0, ,1, ,2, ,1, ,2, Same ratios as Simple cubic! Physics 460 F 2006 Lect 5 15

16 See Kittel Fig. 17 Example of KCl, KBr KCl and KBr have fcc structure expect fcc powder patterns Ι KBr (111) (200) (220) 2 θ (311) (222) But KCl has a special feature K+ and Cl- have the same number of electrons, they scatter x-rays almost the same ---- thus KCl has a pattern like simple cubic Ι KCl (111) (200) (220) (311) (222) Why does this happen? 2 θ Physics 460 F 2006 Lect 5 16

17 Comparison of diffraction from different lattices - continued Lower symmetry lattices Example - Orthorhombic G = (n 1 2π/a 1, n 2 2π/a 2, n 3 2π/a 3 ) Lengths of G s are in general not any special numbers since the a s can be in any ratios Many lines in diffraction pattern because of many different values of G Hexagonal - length along c axis not related to lengths perpendicular to c axis Physics 460 F 2006 Lect 5 17

18 Fourier Analysis of the basis The intensity of the diffraction at each G is proportional to the square of the amplitude of the Fourier component n G = (1/V cell ) cell dr n(r) exp(- i G. r) It is also possible to regard the crystal density n(r) as a sum of atomic-like densities n atom (r - R i ), centered at point R i n(r) = all i n atom i (r - R i ) Then also n G = i in cell space dr n atom i (r - R i ) exp(- i G. r) Cell Physics 460 F 2006 Lect 5 18

19 One atom per cell and Form Factor Then one can set R i = 0 and n G is the Fourier transform of one atom density n G = space dr n atom (r) exp(- i G. r) n atom (r) r Called Form Factor Example in Kittel n G Values of G for a particular crystal G Physics 460 F 2006 Lect 5 19

20 More than one atom per cell n G = i in cell space dr n atom i ( r - R i ) exp(- i G. r) = i in cell exp(- i G. Ri ) space dr n atom i ( r - R i ) exp(- i G. (r - Ri ) ) = i in cell exp(- i G. Ri ) space dr n atom i ( r) exp(- i G. r) = i in cell exp(- i G. Ri ) n G atom i Interpretation: Structure Factor = Form factor n atom i G x phase factor exp(- ig. Ri ) for each atom in unit cell Physics 460 F 2006 Lect 5 20

21 Structure factor Often the basis contains more than one atom that is same element, e.g., diamond structure Then n G atom i = n G atom is the same for each i and n G = i in cell exp(- i G. Ri ) n G atom i = n G atom i in cell exp(- i G. Ri ) Define pure structure factor S 0 G = (1/N cell ) i in cell exp(- i G. Ri ) where N cell = number of atoms in cell Then n G = N 0 S 0 G n G atom NOTE - Kittel defines n G to be the structure factor Physics 460 F 2006 Lect 5 21

22 Body Centered Cubic viewed as Simple Cubic with 2 points per cell S 0 G = (1/2) i =1,2 exp(- i G. Ri ) = (1/2) ( 1 + exp(- i G. R2 ) = (1/2) exp(- i G. R2 /2) [exp( i G. R2 /2) + exp(- i G. R2 /2) ] = exp(- i G. R2 /2) cos ( G. R2 /2) Result: If G = (v 1 v 2 v 3 ) 2π/a S 0 G = 1 if sum of integers is even S 0 G = 0 if sum is odd a 3 a a 1 a 2 Points at R 1 = (0,0,0) ; R 2 = (1,1,1) a/2 Same as we found before! FCC reciprocal lattice Physics 460 F 2006 Lect 5 22

23 Face Centered Cubic viewed as Simple Cubic with 4 points per cell S 0 G = (1/4) i =1,4 exp(- i G. Ri ) Result: If G = (v 1 v 2 v 3 ) 2π/a then a 3 S 0 G = 1 if all integers are odd or all are even S 0 G = 0 otherwise Same as we found before! BCC reciprocal lattice a a 1 a 2 Points at (0,0,0) ; (1,1,0) a/2 ; (1,0,1) a/2 ; (0,1,1) a/2 Physics 460 F 2006 Lect 5 23

24 Structure factor for diamond Example: diamond structure S 0 G = (1/2) i =1,2 exp(- i G. Ri ) R 1 = + (1/8, 1/8, 1/8)a R 2 = - (1/8, 1/8, 1/8)a Homework problem Similar approach would apply to a graphite plane Physics 460 F 2006 Lect 5 24

25 Summary - Diffraction & Recip. Lattice Bragg Condition for diffraction Fourier Analysis and the Reciprocal Lattice G(m 1,m 2, ) = m 1 b 1 + m 2 b 2 + m 3 b 3, where the b s are primitive vectors defined by b i. aj = 2π δ ij, where δ ij = 1, δ ij = 0, i j Examples of Reciprocal lattice: fcc, bcc,... Ewald Construction Diffraction for k in, k out in planes - perp. bisectors of G s Defines Brillouin Zone - no diffraction in first BZ Information about the actual crystal is in the values of the Fourier coefficients f G f G = (1/V cell ) cell dr f(r) exp(- i G. r) Form factor, Pure Structure factor Physics 460 F 2006 Lect 5 25

26 Quasicrystals Not periodic in sense described before Example a crystal with periodicity a with a density wave that is a different period a with a /a not a rational number n(x) = n 1 cos(2πx/a) + n 2 cos(2πx/a ) never repeats! Examples in higher dimensions Orientation order without translational order Penrose Tiles Five fold symmetry in x-ray patterns. Physics 460 F 2006 Lect 5 26

27 Penrose Tiles Many examples Nice WWW sites See this site for a Java program for Penrose tiles Physics 460 F 2006 Lect 5 27

28 Next Time Crystal Binding (Chapter 3) Physics 460 F 2006 Lect 5 28