Chap 3 Scattering and structures

Transcription

1 Chap 3 Scattering and structures Dept of Phys M.C. Chang

2 Von Laue was struck in 1912 by the intuition that X-ray might scatter off crystals in the way that ordinary light scatters off a diffraction grating. He discussed For example, For NaCl, the thermal fluctuation is expected to be cm ~ the wavelength of X-ray 10-9 cm (Marder, p.43) Now we know that thermal fluctuation would only broaden the diffraction peaks, but not distroy them. Laue did not actually do the experiment himself. Rather, he persuaded a couple of graduate students to do the experiment for him. Laue then set an example that has inspired PIs ever since he was given all the credit!

3 Scattering from an array of atoms (Von Laue, 1912) The same analysis applies to EM wave, electron wave, neutron wave etc. Scattering off an atom at the origin: 1914 ikr e scattered wave ψ( r ) fa ( θ) for large r r atomic form factor: Fourier transform of charge distribution n(ρ) ( ) i k Δ ρ f Δ k = dve n( ρ), Δk k ' k a

4 The atom form factor iq fa ( q) dve ρ = n( ρ) 10 electrons tighter ~ Kr 36 ~ Ar 18

5 Scattering off an atom not at the origin ik r R e scattered wave ψ ( r) fa ( Δk) e r R r R r rˆ R O( r ) r R r ikr e ψ ( r) fa ( Δk) e r Δ i k R ik R R Two-atom scattering ikr e ψ ( r) fa e + e r ( Δ i k R Δ i k R) 1 2

6 N-atom scattering: one dimensional case iδk a + e iδk 2 a + e iδk 3 a + e iδk 4a iδk 0 ψ e + e 1 exp( inδk a) ψ f = 1 exp( iδk a) N Δ i k na a e n= 1 For large N, ψ is nonzero only when Δk a = 2 π h ( h is an integer) 2π ψ 2 aδk

7 N-atom scattering (3D case, neglect multiple scatterings) For a Bravais lattice i k R ψ ( r) f e, R = na + n a + n a. a Δ R The lattice-sum can be separated R Δ i k R e e e e Δ i k n1a1 i k n2a2 i k n3a3 = Δ Δ n1 n 2 n 3 0 only when Δk a1 = 2 π h, Δk a2 = 2 π k, Δk a = 2 πl. 3 Δ i k R e = Nδ δ δ = Nδ Δka 1,2 πh Δka 2,2 πk Δka 3,2 πl ΔkG, R Number of atoms in the crystal Laue s diffraction condition hkl See 2 pages later

8 Scattering from a crystal with basis d j : location of the j-th atom in a unit cell Eg., a d = 0, d = ax 1 2 atomic form factor for the j-th atom p Δ i k ( R+ d j ) ψ ( r ) faje R j= 1 = p Δ i kr Δ i k dj e faje R j= 1 = Nδ S( Δk), ΔkG hkl structure factor p i k dj S( k) f e Δ Δ = j= 1 aj

9 Laue s diffraction condition Δk a1 = 2 π h, Δk a2 = 2 π k, Δk a = 2 πl. 3 What are the k s that satisfy this condition? Simplest case: h=1, k, l=0 a2 a3 Δk b1 = 2π a ( a a ) Similarly, for k=1 (the others 0), we have a3 a1 Δk b2 = 2π a ( a a ) Δk b = for l=1 (the others 0), we have a1 a2 2π a ( a a ) i In general, when hkl,, are nonzero, Δk a1 = 2 π h, Δk a2 = 2 π k, Δk a3 = 2 πl. The soultion is just Δ k = hb + kb + lb G hkl That is, the set of solutions form a lattice with primitive vectors b 1, b 2, and b 3 (reciprocal lattice)

10 Reciprocal lattice ( 倒晶格 ) (direct) lattice reciprocal lattice primitive vectors a 1,a 2,a 3 primitive vectors b 1,b 2,b 3 Def. 1 b1 a1 = 2π, b1 a2 = b1 a3 = 0, b2 a2 = 2π, b2 a3 = b2 a1 = 0, b a = 2π, b a = b a = b a a because of orthogonality, then use b a = π to determine the constant. Def. 2 a2 a3 b1 = 2π a1 ( a2 a3), a3 a1 b2 = 2π a1 ( a2 a3), a1 a2 b3 = 2π a ( a a ) The reciprocal of a reciprocal lattice is the direct lattice (obvious from Def.1)

11 Ex: Simple cubic lattice z z a y 2π/a y x a1 = ax, a2 = ay, a3 = az. a ( a a ) = a x a2 a3 2π b a a a a x 1 = 2π =, 1 ( 2 3) a3 a1 2π b a a a a y 2 = 2π =, 1 ( 2 3) a1 a2 2π b a a a a z 3 = 2π =. 1 ( 2 3) F 3 2 b1 b2 b3 = H G πi ( ) a K J When the direct lattice rotates, its reciprocal lattice rotates the same amount as well.

12 FCC lattice BCC lattice z z a x y 4π/a x y b a a1 = x + y, 2 ab a2 = y + z g, 2 ab a3 = z + x g. 2 3 a ( a a ) = a / g a2 a3 4π 1 b1 = 2π = b x+ y z g, a1 ( a2 a3) a 2 a3 a1 4π 1 b2 = 2π = b x+ y+ z g, a1 ( a2 a3) a 2 a1 a2 4π 1 b3 = 2π = b x y+ z g. a ( a a ) a 2 b ( b b ) π a = F H G I K J 3

13 Miller indices An Indexed PbSO 4 Crystal The Miller indices (h,k,l) for a crystal plane rules: no need to be primitive vectors 1. 取截距 ( 以 a 1, a 2, a 3 為單位 ) 得 (x, y, z) 2. 取倒數 (1/x,1/y,1/z) x=2 3. 通分成互質整數 (h,k,l) a 3 z=1 a 2 a 1 (h,k,l)=(3,2,6) y=3

14 Cubic crystals (including bcc, fcc etc) [1,1,1] Square bracket [h,k,l] refers to the direction ha 1 +ka 2 +la 3, instead of a crystal plane. For cubic crystals, [h,k,l] direction (h,k,l) planes

15 Diamond structure (eg. C, Si or Ge) Termination of 3 low-index surfaces: {h,k,l} = (h,k,l)-plane + those equivalent to it by crystal symmetry <h,k,l>= [h,k,l]-direction + those equivalent to it by crystal symmetry

16 Miller Indices for hexagonal lattice (i, j, k, l) corresponding to the I, J, K, L axes below L advantage? J I K [Courtesy of M.F.Yang at Tunhai Univ.] 1. (1 0 0) plane <-> ( ) [2 1 0] vector <-> [ ] [2 1 0] vector (1 0 0) plane, or we can say [ ] vector ( ) plane. (1,0,1,1) 2. For two side faces, they can be (1 0 0) (-1 1 0), or ( ) ( ), which belong to the same { } k = - (i+j)

17 Geometrical relation between G hkl vector and (hkl) planes ( hkl,, ) planes Ghkl hb1+ kb2 + lb3 Pf: m m v = a a h l m m v = a a k l G v = hkl 1 Ghkl v2 = 0 0 m/h m/l v 2 a v 3 1 a 1 a 2 m/k G hkl ( h, k, l)-plane

18 Inter-plane distance (hkl) lattice planes G hkl R d hkl Ghkl R = 2 π n ( n Z) Gˆ hkl R = 2 π n/ Ghkl inter-plane distance d = 2 π / G hkl hkl For a cubic lattice G = hb + kb + lb hkl = 2 π a d = hkl b hx+ ky+ lz a h + k + l g In general, planes with higher index have smaller inter-plane distance

19 Laue s diffraction condition k = k + G hkl Given an incident k, want to find a k that satisfies this condition (under the constraint k = k ) One problem: there are infinitely many G hkl s. It s convenient to solve it graphically using the Ewald construction (Ewald 構圖法 ) k k G More than one (or none) solutions may be found.

20 Laue s condition = Braggs condition From the Laue condition, we have k G hkl = G 2 hkl k G hkl Given k and G hkl, we can find the diffracted wave vector k G hkl k k θ θ (k vector points to the plane bi-secting the G hkl vector) a(hkl)-lattice plane It s easy to see that θ=θ because k = k. By using 2ksinθ = G = hkl 2π n d 2π and k =, λ 2d sin θ = nλ. Bragg s diffraction condition hkl hkl

21 Braggs view of the diffraction (1912, father and son) You can view the same phenomena from 2 (or more) different angles, and each can get you a Nobel prize! 25 Treat the lattice as a stack of lattice planes 1915 mirror-like reflection from crystal planes when 2dsinθ = nλ Difference from the usual mirror reflection: λ > 2d, no reflection λ < 2d, reflection only at certain angles Measure λ, θ get distance between crystal planes d

22 Rotating crystal method 2dsinθ = nλ

23 Powder method

24 The structure factor for ighkl d j Shkl (,, ) fe aj a crystal with basis: j= 1 f a is the atomic form factor Example: fcc lattice = cubic lattice with a 4-point basis d1 = 0, a1 a2 a2 a3 a3 a1 d2 = +, d3 = +, d4 = Ghkl = hb1+ kb2 + lb3 Shkl (,, ) = f a 1+ e + e + e iπ( h+ k) iπ( k+ l) iπ( l+ h) = 4f a when h,k,l are all odd or all even = 0 otherwise Eliminates all the points in the reciprocal cubic lattice with S=0. The result is a bcc lattice, as it should be! p

25 Homework: Find out the structure factor of the honeycomb structure, then draw its reciprocal structure. Different points in the reciprocal structure may have different structure factors. Draw a larger dots if the associated S 2 is larger.

26 Watson and Crick, 1953 (See Double helix, highly recommended) Franklin and Gosling, 1953 simulation

27 a ribosome hemoglobin Kendrew Perutz Hodgkin Braggs, 1914 Shen et al, Phys Today Mar, 2006

28 Marder, p.65