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Crystal Structure Factor Calculations

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Crystal Structure Factor Calculations Crystal Structure factor calculations (A) X-Ray Scattering by an Atom ❑ The conventional UC has lattice points as the vertices ❑ There may or may not be atoms located at the lattice points ❑ The shape of the UC is a parallelepiped in 3D ❑ There may be additional atoms in the UC due to two reasons: ➢ The chosen UC is non-primitive ➢ The additional atoms may be part of the motif Scattering by the Unit cell (uc) ▪ Coherent Scattering ▪ Unit Cell (UC) is representative of the crystal structure ▪ Scattered waves from various atoms in the UC interfere to create the diffraction pattern The wave scattered from the middle plane is out of phase with the ones scattered from top and bottom planes a AC = d = h00 h MCN :: AC :: RBS :: AB :: x AB x x = = AC a h = MCN = 2d Sin( ) = R1R2 h00 2 AB x = = = RBS = = 2 R1R3 AC a h 2 x x = = 2 h x R1R3 → fractional coordinate → x R R = 2 hx a a 1 3 h a Extending to 3D =2 (h x + k y + l z ) Independent of the shape of UC Note: R1 is from corner atoms and R3 is from atoms in additional positions in UC In complex notation =2 (h x + k y + l z ) E== Aei fe i[2 ( h x++ k y l z )] ▪ If atom B is different from atom A → the amplitudes must be weighed by the respective atomic scattering factors (f) ▪ The resultant amplitude of all the waves scattered by all the atoms in the UC gives the scattering factor for the unit cell ▪ The unit cell scattering factor is called the Structure Factor (F) Scattering by an unit cell = f(position of the atoms, atomic scattering factors) Amplitude of wave scattered by all atoms in uc F = Structure Factor = 2 Amplitude of wave scattered by an electron I F nn hkl i i[2 ( h x++ k y l z )] j j j j For n atoms in the UC Fn== f j e f j e jj==11 Structure factor is independent of the shape and size of the unit cell If the UC distorts so do the planes in it!! Structure factor calculations eni = (−1)n Simple Cubic (odd n) i A Atom at (0,0,0) and equivalent positions e = −1 e(even n) i = +1 eni = e−ni ei + e−i = Cos( ) 2 ij i[2 ( h x j++ k y j l z j )] F== fjj e f e F= f ei[2 ( h 0 + k 0 + l 0)] = f e 0 = f 2 2 F = f F is independent of the scattering plane (h k l) B Atom at (0,0,0) & (½, ½, 0) and equivalent positions C- centred Orthorhombic ij i[2 ( h x j++ k y j l z j )] F== fjj e f e 11 i[2 ( h + k + l 0)] F=+ f ei[2 ( h 0 + k 0 + l 0)] f e 22 hk+ i[ 2 ( )] =f e0 + f e2 = f[1 + ei ( h+ k ) ] Real F = 2 f F 2 = 4 f 2 F = f [1+ ei (h+k ) ] e.g. (001), (110), (112); (021), (022), (023) F = 0 F 2 = 0 e.g. (100), (101), (102); (031), (032), (033) F is independent of the ‘l’ index ▪ If the blue planes are scattering in phase then on C- centering the red planes will scatter out of phase (with the blue planes- as they bisect them) and hence the (210) reflection will become extinct ▪ This analysis is consistent with the extinction rules: (h + k) odd is absent ▪ In case of the (310) planes no new translationally equivalent planes are added on lattice centering this reflection cannot go missing. ▪ This analysis is consistent with the extinction rules: (h + k) even is present Body centred C Atom at (0,0,0) & (½, ½, ½) and equivalent positions Orthorhombic ij i[2 ( h x j++ k y j l z j )] F== fjj e f e 1 1 1 i[2 ( h + k + l )] F=+ f ei[2 ( h 0 + k 0 + l 0)] f e 2 2 2 h++ k l i[ 2 ( )] =f e0 + f e2 = f[1 + ei ( h++ k l ) ] Real F = 2 f F 2 = 4 f 2 F = f [1+ ei (h+k +l) ] e.g. (110), (200), (211); (220), (022), (310) F = 0 F 2 = 0 e.g. (100), (001), (111); (210), (032), (133) D Atom at (0,0,0) & (½, ½, 0) and equivalent positions Face Centred Cubic (½, ½, 0), (½, 0, ½), (0, ½, ½) ij i[2 ( h x j++ k y j l z j )] F== fjj e f e h+k k +l l+h i[2 ( )] i[2 ( )] i[2 ( )] i[2 (0)] F = f e + e 2 + e 2 + e 2 i (h+k ) i (k +l) i (l+h) = f [1+ e + e + e ] Real F = f [1+ ei (h+k ) + ei (k +l) + ei (l+h) ] (h, k, l) unmixed F = 4 f F 2 =16 f 2 e.g. (111), (200), (220), (333), (420) (h, k, l) mixed F = 0 F 2 = 0 e.g. (100), (211); (210), (032), (033) Two odd and one even (e.g. 112); two even and one odd (e.g. 122) Mixed indices Two odd and one even (e.g. 112); two even and one odd (e.g. 122) Mixed indices CASE h k l A o o e B o e e CASEA:[1+ei()()() e + e i o + e i o ][1111]0 = + − − = CASEB:[1+ei()()() o + e i e + e i o ][1111]0 = − + − = (h, k, l) mixed e.g. (100), (211); (210), (032), (033) Unmixed indices All odd (e.g. 111); all even (e.g. 222) Unmixed indices CASE h k l A o o o B e e e CASEA:[1+eeei()()() e + i e + i e ][1111]4 = + + + = F = 0 F 2 = 0 CASEB:[1+eeei()()() e + i e + i e ][1111]4 = + + + = (h, k, l) unmixed F = 4 f F 2 =16 f 2 e.g. (111), (200), (220), (333), (420) + → (½, ½, 0), (½, 0, ½), (0, ½, ½) E Na at (0,0,0) + Face Centering Translations − Cl at (½, 0, 0) + FCT → (0, ½, 0), (0, 0, ½), (½, ½, ½) NaCl: h+k k +l l+h Face Centred Cubic i[2 ( )] i[2 ( )] i[2 ( )] i[2 (0)] 2 2 2 F = f + e + e + e + e + Na h k l h+k +l i[2 ( )] i[2 ( )] i[2 ( )] i[2 ( )] 2 2 2 2 f − e + e + e + e Cl i (h+k ) i (k +l) i (l+h) F = f + [1+ e + e + e ]+ Na i (h) i (k ) i (l) i (h+k +l) f − [e + e + e + e ] Cl i (h+k ) i (k +l) i (l+h) F = f + [1+ e + e + e ]+ Na i (h+k +l) i (−k −l) i (−l−h) i (−h−k ) f − e [e + e + e +1] Cl i (h+k +l) i (h+k ) i (k +l) i (l+h) F = [ f + + f − e ][1+ e + e + e ] Na Cl i (h+k+l) i (h+k) i (k+l) i (l+h) F =[ f + + f − e ][1+e +e +e ] Na Cl F = [Term −1][Term − 2] Zero for mixed indices Mixed indices CASE h k l Mixed indices A o o e B o e e CASE A : Term − 2 = [1+ ei (e) + ei (o) + ei (o) ] = [1+1−1−1] = 0 CASE B : Term − 2 = [1+ ei (o) + ei (e) + ei (o) ] = [1−1+1−1] = 0 (h, k, l) mixed e.g. (100), (211); (210), (032), (033) F = 0 F 2 = 0 Unmixed indices CASE h k l Unmixed indices A o o o B e e e CASE A : Term − 2 = [1+ ei (e) + ei (e) + ei (e) ] = [1+1+1+1] = 4 CASE B : Term − 2 = [1+ ei (e) + ei (e) + ei (e) ] = [1+1+1+1] = 4 i (h+k +l) F = 4[ f + + f − e ] (h, k, l) unmixed Na Cl e.g. (111), (222); (133), (244) 2 2 F = 4[ f + + f − ] If (h + k + l) is even F =16[ f + + f − ] Na Cl Na Cl e.g. (222),(244) 2 2 F = 4[ f + − f − ] If (h + k + l) is odd F =16[ f + − f − ] Na Cl Na Cl e.g. (111), (133) Presence of additional atoms/ions/molecules in the UC can alter the intensities of some of the reflections NiAl: Simple Cubic (B2- ordered structure) F Al at (0, 0, 0) Ni at (½, ½, ½) Click here to know more about ordered structures 1 1 1 i[2 ( h + k + l )] i[2 ( h 0 + k 0 + l 0)] 2 2 2 F=+ fAl e f Ni e h++ k l i[ 2 ( )] 02 i [ h++ k l ] =fAl e + f Ni e = f Al + f Ni e Real SC 22 F=+ fAl f Ni F=+() fAl f Ni i [] h++ k l e.g. (110), (200), (211), (220), (310) F=+ fAl f Ni e 22 F=− fAl f Ni F=−() fAl f Ni e.g. (100), (111), (210), (032), (133) ▪ When the central atom is identical to the corner ones− we have the BCC case. This term is zero for BCC ▪ This implies that (h+k+l) even reflections are only present in BCC. Reciprocal lattice/crystal of NiAl Click here to know more about e.g. (110), (200), (211); (220), (310) 2 I+() fAl f Ni 2 I−() fAl f Ni e.g. (100), (111), (210), (032), (133) Click here to know more about ordered structures Al Atom at (0,0,0) G Simple Cubic (L12 ordered structure) Ni atom at (½, ½, 0) and equivalent positions (½, ½, 0), (½, 0, ½), (0, ½, ½) h+ k k + l l + h i[2( )] i [2( )] i [2( )] i[2 ( 0)] 2 2 2 F= fAl e + f Ni e + e + e =f + f[ ei()()() h+ k + e i k + l + e i l + h ] Al Ni Real Ni Al i()()() h+ k i k + l i l + h F= fAl + f Ni[] e + e + e F=+ f3 f 22 (h, k, l) unmixed Al Ni F=+( fAl 3 f Ni ) h,k,l → all even or all odd e.g.
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