10/14/2013. Structure Factors and Fourier Synthesis. Structure Factors. Superposition of Waves. Vector Notation of a Wave. Superposition of Waves

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1 /4/2 Structure actors and Structure actors ( kl) kl : function of reflection indices and atom positions in unit cell: all atoms scatter waves in kl direction to develop equations tat represent ow waves add: wave represented as vector moving wit constant angular velocity: amplitude is magnitude of vector pase is angle (δ) vector makes wit orizontal at t = Vector Notation of a Wave C A f O B 2 OB = f cos OC = f sin f 2 = OB 2 + OC 2 cycle = 2 radians x 2 f f waves of same frequency, but different pase (δ), add to give a new wave, wit a different amplitude and pase (from eiter) but te same frequency. f 4 4 x = f cos 4 x 2 = f 2 cos ( + ) f f 2 x = x + x 2 f 2 x 2 x f x = f cos x 2 = f 2 cos ( + ) x x r = f cos + f 2 cos ( + ) cos cos sin sin = ( f + f 2 cos ) cos f 2 sin sin x = o cos ( + ) = f cos + f 2 cos ( + ) cos = f + f 2 cos trigometric identity y = o sin = f 2 sin

2 f f /4/2 f 2 2 f f 2 f f 2 cos f 2 f 2 sin cos = x = f + f 2 cos y = f 2 sin = sin f f f 2 f 2 f 2 f 2 x = f cos + f 2 cos + f cos = f j cos j j = y = f sin + f 2 sin + f sin = f j sin j j = A B = (x 2 + y 2 ) /2 = j f j cos j 2 + j f j sin j 2 /2 f 2 f f 2 = tan j f j sin j j f j cos j = tan B A Structure actors Atom Pases consider ( ) Miller planes obvious (from symbols) tat structure factor, kl, results from N waves scattered in ( k l) direction by N atoms in unit cell eac wave as an amplitude to f j (scattering factor) and a pase j wit respect to te origin of unit cell to calculate s, need to know s in terms of atomic position and indices of reflection needed (,, ) = 2x = 2(.) atom at (.,, ) is exactly out of pase () wrt atom at (,, ) (½ cycle) atom at (,, ) as a pase difference of 2 wrt atom at (,, ) ( cycle) 2

3 /4/2 (,, ) Atom Pases consider (2 ) Miller planes = 2x = 22(.2) atom at (.2,, ) is exactly out of pase () wrt atom at (,, ) (½ cycle) atom at (,, ) as a pase difference of 4 wrt atom at (,, ) (2 cycles) Atom Pases or, for te general case: an atom at x, y, z in te unit cell as a pase (wrt a ypotetical atom at (,, )) for a given reflection ( kl) of f j cos j j = f j sin j j = = 2 (x + ky + lz) = A kl = f j cos 2(x j + ky j + lz j ) = B kl = f j sin 2(x j + ky j + lz j ) kl = (A kl2 + B kl2 ) /2 kl = tan B kl A kl f f 2 f 2 Structure actor as a Complex Number y a a + ib f = a + ib = f (cos + i sin ) = f e i kl = j f j e 2i (xj + kyj + lzj) f b x a units along x b units along iy exponential form of structure factor two components (magnitude and pase) in one equation: note: kl not kl Electron Density instead of kl coming from N atoms scattering in (kl) direction, can also consider te scattering from individual e s: wavelets scattered from e ((x, y, z)dv), were = # e /V atom scattering: e scattering: fe 2i (xj + kyj + lzj) (x, y, z) e 2i (x + ky + lz) dv unit cell scattering; all atoms: kl = j f j e 2i (xj + kyj + lzj) unit cell scattering; all e s: kl = (x, y, z) e 2i (x + ky + lz) dv ourier Transform ourier Transform in NMR kl = (x, y, z) e 2i (x + ky + lz) dv expression to calculate structure factors from known e density to solve crystal structure, also need to perform inverse operation: e density from a set of crystal data, kl since e density is periodic, can be represented by ourier Series time tis wave is made of Jean Baptiste Josep ourier renc (68-8) tese waves add to frequency

4 /4/2 ourier Transform in Crystallograpy T structure factors kl electron density (x, y, z) ourier Analysis box-car function general form of -D ourier series f(x) = C = C e 2ix (cos 2x + i sin 2x) were is an index of te term # only cos terms in box-car function centrosymmetric:.. y = /4 f(x) = f( x)cos( x) = cos(x); sin( x) = sin(x) f(x) = y = /4 y = cos (2) x y = / cos (2) x y = / cos (2) x C cos 2x y = / cos (2) x y = / cos (2) x y = / cos (2) x y = / cos (2) x.... 4

5 /4/2... y = cos (2) x. y = / cos (2) x y = / cos (2) x. y = / cos (2) x y = / cos (2) x. y = / cos (2) x

6 /4/2... y = / cos (2) x

7 /4/ D periodic e density in a crystal could be represented by:... f(x) = C e 2ix (x, y, z) =C k l e 2i ( x + k y + l z) k l kl = (x, y, z) e 2i (xj + kyj + lzj) dv kl = C k l e 2i ( x + k y + l z) e 2i (x + ky + lz) dv k l... kl = k l C k l e 2i ((+ )x + (k+k )y + (l+l )z) dv integration of tis cyclic function is, except for case wen: = k = k l = l kl = C k l dv = VC k l ; C k l = /V kl (x, y, z) = V kl e 2i (x + ky + lz) k l kl = kl e 2i pase in cycles kl = ourier Transform (x, y, z) e 2i (x + ky + lz) dv kl = j f j e 2i (xj + kyj + lzj) (x, y, z) = (x, y, z) = V V kl e 2i e 2i (x + ky + lz) k l kl e 2i (x + ky + lz ) k l finally: expanding exponential into cos and sin terms and assuming riedel s Law (sin cancels for riedel pairs): (x, y, z) = V kl cos2 (x + ky + lz kl ) k l (x, y, z) = V kl cos2 (x + ky + lz kl ) k l

8 /4/2 Structure actors: -D Example 2 atoms a = Å CuK Scattering actor: f C sin/ = /2d = /2a x C =.8.8 = x C2 x C (+ ) = x C2 centrosymmetric kl = j f j e 2i (xj + kyj + lzj) 2 = f j (cos 2x j + i sin 2x j ) j = = f j e 2i xj = f C [cos 2x C + i sin 2x C + cos 2 ( x C ) + i sin 2 ( x C )] cos( x) = cos(x) sin( x) = sin(x) = tan = 2 f C cos 2x C pase due to position (radians) j f C sin x j = = or sign of + or on j f C cos x j 2 j = scattering power (in direction of ) = ± = ±2 = ± d = Å d =. Å sin/ = sin/ = d =. Å sin/ =. d = a/ Structure actors: -D Example 2 atoms sin ( = / 2d) sin/ f C 6. ± (4.4 o ). 4. ±2.4 (8.8 o ).2 6. = 2 f C cos 2x C x C =.8 f o sin/ sin ( = / 2d) sin/ f C 6. (4.4 o )..4 (8.8 o ).2.2 (. o ). 4.2 ± ±2 ± ±4 ± ±6 ± ±8 ± ± ± ±2 Structure actors: -D Example 2 atoms (. o ).2 (22.6 o ).2 (2. o ). (2.6 o ). (8. o ).4 (4. o ).4 (. o ). (. o ). (6. o ) Structure actors: -D Example 2 atoms Electron Density: -D Example 2 atoms = 4 = = 2 = = = = 2 = = 4 (I) / a ourier syntesis on sould give te structure: (x) = (x) = L L L e 2ix since = (cos 2x i sin 2x) riedel s Law (x) = (cos 2x i sin 2x) + + (cos 2x i sin 2x) 8

9 /4/2 Electron Density: -D Example 2 atoms L (x) = (cos 2x i sin 2x) + + (x) = (cos 2x i sin 2x) since: cos( x) = cos(x) sin( x) = sin(x) + 2 cos 2x L = total e in unit cell cos 2x cos 2x cos 2x cos 2x cos 2x

10 /4/2 6 cos 2x cos 2x cos 2x cos 2x cos 2x cos 2x

11 /4/2 2 cos 2x (2 cos 2x) / for = (x) (2 cos 2x) / for = (2 cos 2x) / for = 4 (x) (x) (2 cos 2x) / for = (2 cos 2x) / for = 6 (x) (x)

12 /4/2 (2 cos 2x) / for = (2 cos 2x) / for = 8 (x) (x) (2 cos 2x) / for = (2 cos 2x) / for = (x) (x) (2 cos 2x) / for = (2 cos 2x) / for = 2 (x) (x)

13 /4/2 ( + 2 cos 2x) / for = 2 ( + 2 cos 2x) /.8.8 2x data (x) (x) ½ data D Structure te -D calculation of (x, y, z) is called: -D Electron Density ourier syntesis ourier map e density map note ripples around te atom (Au) because of finite data Systematic Absences G kl = m e2i (xm + kym + lzm) example: consider 2 atoms related by 2-fold screw axis along b: x, y, z x, y + ½, z G kl kl = e 2i (x + ky + lz) + e 2i (x ky k/2 + lz) G kl = e 2i (x + lz) e 2i ky + e 2i (x + lz) e 2iky e 2ik/2 G kl = e 2i ky (e 2i (x + lz) + e 2i (x + lz) e 2i k/2 ) looking at k reflections G k = e 2i ky ( + e ik ) G k G k = j f j e 2i (xj + kyj + lzj) kl = n f n m e 2i (xm,n + kym,n + lzm,n) m = # asym units n = atoms in asym unit = 2e 2i ky for k = 2n (even) = for k = 2n + (odd)