References for this Topic ✿Clegg, “Crystal Structure Determination”. Lattices, Reciprocal Lattices ✿ Stout & Jensen, “X-Ray Structure Determination”, 2nd Edition. Instrumentation discussion is completely and Diffraction outdated, but still a good text on the subject. ✿ A more authoritative general reference: Giacovazzo, et Chem 634 al., “Fundamentals of Crystallography”, IUCr Texts on T. Hughbanks Crystallography. ✿ MIT has a good site (MIT Open Courseware): http:// ocw.mit.edu/OcwWeb/Chemistry/ 5-841Fall-2006/LectureNotes/index.htm 1 2 Bravais Lattices • Direct Lattice: Reciprocal Lattices – A regular, periodic array of points with a • Reciprocal Lattice: the basis vectors of the reciprocal spacing commensurate with the unit cell lattice are defined as dimensions. The environment around each b × c c × a a × b points in a lattice is identical. a* = ; b* = ; c* = ; V = (a × b)• c • The set of direct lattice points can be written as V V V (vectors): R = ta + ub + vc t,u,v integers (Note: In physics texts, a factor of 2π is usually included b × c • V = volume of unit cell e.g., a* = 2π ; ! ) V V = a ⋅ b × c = • The set of Reciprocal Lattice Vectors (RLVs) are ( ) written as * * * K i = ha + kb + lc b⋅ c × a = c⋅ a × b ( ) ( ) h,k,l = integers 3 4 Side by Side (2-D) b b∗ Reciprocal Lattices a a∗ Reciprocal Lattice • Properties of RLVs: DirectLattice a i a* = b i b* = c i c* = 1 a* i b = a* i c = b* i a = b* i c = c* i b = c* i a = 0 b∗ Alternatively, these can be regarded as definitions. a∗ b In 2 dimensions, we define the RLVs a a* ⊥ b , b* ⊥ a , a i a* = b i b* = 1 • Although they “live in separate universes”, the direct and reciprocal lattices are in rotational “lock-step”. 5 6 Lattice & Reciprocal Lattice Lattice & Reciprocal Lattice Example: BCC treated as Primitive Example: BCC treated as Primitive ^ ^ (a/2)(–x^ + y^ + ^z) (1/a)(y + z) a ^ 1 a = (xˆ + yˆ − zˆ) az a* = (xˆ + yˆ ) 2 a (1/a)z^ a 1 b = −xˆ + yˆ + zˆ b* = yˆ + zˆ 2 ( ) b a a ( ) c ^ ^ b* a 1 (1/a)(x + z) c* 1/a c = (xˆ − yˆ + zˆ) c* = (xˆ + zˆ) ^ 2 (a/2)(x^ – y^ + z^) a (1/a)y ay^ (1/a)x^ a* Use to find Recip. Lattice Basis: ax^ Exercise: Find the reciprocal lattice of the reciprocal lattice! (1/a)(x^ + y^) (a/2)(x^ + y^ – z^) b × c c × a a × b a* = ; b* = ; c* = a If we use the conventional cell (which leaves out half the lattice V V V points, the reciprocal lattice includes the ‘open’ points too: V = a × b • c ( ) xˆ yˆ zˆ a = axˆ ; b = ayˆ ; c = azˆ ⇒ a* = ; b* = ; c* = a a a 7 8 Interrelationships Interrelationships • The direct lattice can be partitioned such that all of the lattice points lie on sets of planes. Crystallographers classify these sets of planes using the intercepts on the unit cell axes cut by the plane adjacent to the plane through the origin. • The intercepts are of the form (1/h, 0, 0), (0,1/k, b b* 0), and (0,0,1/l). a 2a* + 3b* • These planes are normal to the RLVs a* intercept: (1/2,0) * * * -3a+ 2b intercept: (0,1/3) K hkl = ha + kb + lc note: (-3a + 2b)•(2a* + 3b*) = 0 9 10 Interrelationships Interrelationships Reciprocal Lattice lattice planes DirectLattice b K hkl ˆ K hkl K hkl = K hkl θb - aunitvector (0,1/k,0) θa a c b (1/h,0,0) ˆ ˆ dhkl =(1/h)a•K hkl = (1/h)a K cosθa 2a* + b* ˆ ˆ a (hkl) plane dhkl =(1/k)b•K hkl = (1/k)b K cosθb (0,0,1/l) intercept: (1/2,0) (0,1/k, 0) intercept: (0,1) -(1/h)a + (1/l)c b (1/h,0,0) a -(1/h)a + (1/k)b 11 12 hkl Planes are normal to Khkl Kj•Ri products involve no cross terms and always yield integers c (hkl) plane • For any DLV Ri and any RLV Kj (0,0,1/l) (0,1/k, 0) R K ta ub vc ha* kb* lc* -(1/h)a + (1/l)c i i j = ( + + )⋅( + + ) b (1/h,0,0) = th + uk + vl = m, an integer a -(1/h)a + (1/k)b • Proof: Consider the plane that cuts the cell axes at (1/h,0,0), e2πiRiiK j 1, for all R and all K (0,1/k,0) and (0,0,1/l). ⇒ = i j 1 1 1 1 • Two vectors lie in the plane: − a + b and − a + c . h k h l ⎛ 1 1 ⎞ h k K a b a a* b b* 1 1 0 hkl ⋅⎜ − + ⎟ = − ⋅ + ⋅ = − + = ⎝ h k ⎠ h k ∴K hkl ⊥ to both vectors ⎛ 1 1 ⎞ h k ∴K ⊥ to the family of planes K ⋅ − a + c = − a ⋅a* + c⋅c* = −1+1= 0 hkl hkl ⎝⎜ h l ⎠⎟ h k 13 14 1 1 To Prove: dhkl = = K Khkl Diffraction hkl • Assume K is the shortest RLV that points in the exact direction hkl • In handling scattering of x-rays (or electrons) by matter, we need it points (i.e., h,k, and l have no common factor.) to characterize the x-ray beam by specifying its wavelength and • Khkl is normal to the “hkl family of planes”. We can relate the direction of propagation: E E xˆ distance between planes to the magnitude of Khkl: = 0 H H yˆ lattice planes = 0 k = kzˆ ˆ Eˆ Hˆ kˆ 1 a K a b 0 × 0 = d a Kˆ hkl cos cos hkl = ⋅ hkl = θa = θa K H H exp 2 i(k r t) h h h hkl = 0 { π ⋅ −ν } ReH H cos 2 (k r t) Where Kˆ is a unit vector parallel to K . = 0 { π ⋅ −ν } hkl hkl ˆ K hkl 1 K hkl = k = * * * E = E exp 2πi(k ⋅r −νt) K ha + kb + lc K hkl 0 { } λ ˆ hkl θb now, K hkl = = a - aunitvector ReE E cos 2 (k r t) (0,1/k,0) θa = 0 { π ⋅ −ν } Khkl Khkl • Note: The usual physics definition is 1 * * * (1/h,0,0) ˆ ˆ so d = a ⋅ ha + kb + lc d =(1/h)a•K = (1/h)a K cosθa 2π hkl hK ( ) hkl hkl k hkl ˆ ˆ = dhkl =(1/k)b•K hkl = (1/k)b K cosθb λ 1 ∴dhkl = Khkl 15 16 Additional Comments Traveling wave packet • By “adding” up a narrow range of frequencies, • In order to speak of the localization of an electron (or photon), we must ‘blur’ the specification of the electron’s (photon’s) a “wave packet” can be constructed and a wavelength, and allow that there is some uncertainty in the photon visualized. wavelength (and frequency and energy): ∞ E (z,t) = E (k′)exp 2πi(k′z −νt) dk′ k ∫−∞ 0 { } where we have assumed, for simplicity, that the wave is propagating along the z-axis and E0(k) is a function strongly peaked near a particular ‘approximate’ wavenumber k. • Electrons and photons are different in that wave packets for electrons spread out over time, while photon wave packets retain their width as they propagate. For more details, see: http://farside.ph.utexas.edu/teaching/qmech/lectures/node1.html http://farside.ph.utexas.edu/teaching/qmech/lectures/node1.html 17 18 Interference - “Time Lapse” pictures Elastic Scattering • When an x-ray is scattered by an atom it may emerge in any direction, but unless absorption occurs, it has the same wavelength: 1 k = k′ = λ • X-ray diffraction for structure determination is concerned with interference effects that result from scattering by periodic arrays of atoms. This is my own heuristic exercise, not “reality”. 19 20 Atomic Form Atomic X-ray Atomic Form Factors: f (2θ) scattering Factors: • Forward scattering is more intense in “real f(2θ) atoms” with diffuse electron clouds. If all the electron density were concentrated very close to the nucleus, 2θ dependence would disappear. • f (2θ = 0) = N (number of electrons) water ripples • The “water ripples” concept is not actually accurate; forward Atomic X-ray scattering is actually more scattering intense. Clegg, p. 24. 21 22 “Thermal” effects: f ′(2θ) Interference between scatterers • Vibrational motion “spreads out” average electron Consider the interference between two scatterers, separated by a vector d: density and causes effective scattering at higher 1 2 angles to fall off more rapidly; U is the isotropic k displacement parameter (units of Å2 ). 20 2 2 ⎛ ⎞ d d −8π U sin θ ˆ k – f ′(2 ) f (2 ) •exp k = Cl j θ = j θ ⎜ 2 ⎟ λ k 15 ⎝ ⎠ ˆ k′ k′ = 3 4 k′ k f j (2θ) Again, scattering at higher angles 10 1 is attenuated by spreading out the k = k′ = or kˆ = λk λ f ′(2 ) electron density ′ 5 j θ d k d 0.2 0.4 0.6 0.8 sinθ/λ (Å–1) Clegg, p. 24. 23 24 Interference between Interference between scatterers scatterers Consider the interference between two scatterers, separated by a • Amplitude of scattered wave vector d: ∝ f1(2θ) + f2 (2θ)⋅( phase shift factor) k kˆ = k • The phase shift factor depends on the path length difference = k′ kˆ ′ = k′ kˆ ⋅d + (−kˆ ′)⋅d = (kˆ − kˆ ′)⋅d 1 k = k′ = or kˆ = λk λ • For completely constructive interference, where n is some integer: (kˆ − kˆ ′)⋅d = nλ or (k − k′) i d = n • For completely destructive interference 1 (k − k′) i d = n + Clegg, Sec.