The Structure Factor Suggested Reading Pages 303-312 in De Grae f & Mc Henry Pages 59-61 in Engler and Randle 1 Structure Factor (Fhkl) N 2(ihukvlwiji ) Ffehkl i i1 • Describes how atomic arrangement (uvw) influences the intensity of the scattered beam. • It tells us which reflections (i. e., peaks, hkl)to) to expect in a diffraction pattern. 2 Structure Factor (Fhkl) • The amplitude of the resultant wave is given by a ratio of amplitudes: amplitude of the wave scattered by all atoms of a UC F hkl amplitud eof th e wave scatteredb d by one el ectron • The intensity of the diffracted wave is proportional 2 to |Fhkl| . 3 Some Useful Relations ei = e3i = e5i = … = -1 e2i = e4i = e6i = … = +1 eni = (-1)n, where n is anyyg integer eni = e-ni, where n is any integer eix + e-ix =2 cos x Needed for structure factor calculations 4 Fhkl for Simple Cubic N 2(ihukvlwiji ) Ffehkl i • Atom coordinate(s) u,v,w: i1 – 0,0,0 2(00ih k 0) l Ffehkl f No matter what atom coordinates or plane indices you substitute into the structure factor equation for simple cubic crystals, the solution is always non-zero. Thus, all reflections are allowed for simple cubic (primitive) structures. 5 Fhkl for Body Centered Cubic N 2(ihukvlwiji ) Ffehkl i • Atom coordinate(s) u,v,w: i1 – 0,0,0; – ½, ½, ½. hkl 2i 2(0)i 222 Ffefehkl Ffe1 ih k l hkl When h+k+l is even Fhkl = non-zero → reflection. When h+k+l is odd Fhkl = 0 → no reflection. 6 Fhkl for Face Centered Cubic N 2(ihukvlwiji ) Ffehkl i • Atom coordinate(s) u,v,w: i1 – 0,0,0; – ½,½,0; – ½,0,½; – 0,½,½. hk hl k l 222iii 20i 22 22 22 Ffhkl ffe fe f e f e ih k ih l ik l Ffehkl 1 e e 7 Fhkl for Face Centered Cubic ih k ih l ik l Ffehkl 1 e e • Substitute in a few values of hkl and you will find the following: – When h,k,l are unmixed (i.e. all even or all odd), then Fhkl = 4f. [NOTE: zero is considered even] – Fhkl = 0f0 for m ixe did in dices (i.e., a com bina tion o f o dd and even). 8 Fhkl for NaCl Structure N 2(ihukvlwiji ) Ffehkl i • Atom coordinate(s) u,v,w: i1 – Na at 0,0,0 + FC transl.; • 0,0,0; •½,½,0; This means these coordinates • ½0½½,0,½; (u,v,w) • 0,½,½. – Cl at ½,½,½ + FC transl. •½,½,½; ½,½,½ The re-assignment of coordinates is • 11½1,1,½; 00½0,0,½ based upon the equipoint concept in • 1,½,1; 0,½,0 the international tables for • ½,1,1. ½,0,0 crystallography • Substitute these u,v,w values into Fhkl equation. 9 Fhkl for NaCl Structure – cont’d N 2(ihukvlwiji ) Ffehkl i •For Na: i1 2iihkihlikl(()0)()()() ( ) ( ) ( ) feNa e e e ih() k ih () l ik () l feNa 1 e e •For Cl: 2(ih klkh ) 2( ih l ) 2( i k l ) feih() k l e222 e e Cl ih() k l i (22hk lkh ) i ( 2 h 2 l ) i ( 22 kl ) feCl e e e ih() k l i ()()lkh i i() feCl e e e These terms are all positive and even. Whether the exponent is odd or even depends solely on the remaining h, k, and l in each exponent. 10 Fhkl for NaCl Structure – cont’d N 2(ihukvlwiji ) Ffehkl i • Therefore Fhkl: i1 ih() k ihl () ik () l Ffhkl Na 1 e e e ih() k l i ()()()lkh i i feCl e e e which can be simplified to*: ih() k l ih ()()() k ihl ik l Fffehkl Na Cl 1 e e e 11 Fhkl for NaCl Structure When hkl are even Fhkl = 4(fNa + fCl) Primary reflections When hkl are odd Fhkl =4(= 4(fNa - fCl) Superlattice reflections When hkl are mixed Fhkl = 0 NfltiNo reflections 12 (200) 100 90 NaCl CuKα radiation 80 70 (220) %) 60 50 ntensity ( II 40 30 (420) 20 (222) (422) (600) (442) 10 (111) (400) (333) (440) (311) (511) (331) (531) 2θ (°) 0 20 30 40 50 60 70 80 90 100 110 120 13 Fhkl for L12 Crystal Structure N 2(ihukvlwiji ) Ffehkl i • Atom coordinate(s) u,v,w: i1 – 0,0,0; A B – ½,½,0; – ½,0,½; – 0,½,½. hk hl k l 2(0)i 222iii Ffefe22 fe 22 fe 22 hkl AB B B ih() k ih () l ik () l Fffe e e hkl AB 14 Fhkl for L12 Crystal Structure ih() k ih () l ik () l Fffe e e hkl AB (1 0 0) Fhkl = fA + fB(-1-1+1) = fA – fB A B (1 1 0) Fhkl = fA + fB(1-1-1) = fA – fB (1 1 1) Fhkl = fA + fB(1+1+1) = fA +3 fB (2 0 0) Fhkl = fA + fB(1+1+1) = fA +3 fB (2 1 0) Fhkl = fA + fB(-111+1-1) = fA – fB (2 2 0) Fhkl = fA + fB(1+1+1) = fA +3 fB (2 2 1) Fhkl = fA + fB(1-1-1) = fA – fB (3 0 0) Fhkl = fA + fB(-1-1+1) = fA – fB (3 1 0) Fhkl = fA + fB(1-1-1) = fA – fB (3 1 1) Fhkl = fA + fB(1+1+1) = fA +3 fB (2 2 2) Fhkl = fA + fB(1+1+1) = fA +3 fB 15 Example of XRD pattern Intensity (%) from a material with an (111) L1 crystal structure 100 2 A B 90 80 70 60 50 (200) 40 30 (311) (220) 20 (100) (110) 10 (300) (222) (210) (211) (221) (()310) (()320) (321) 2 θ (°) 0 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 16 Fhkl for MoSi2 • Atom positions: c – Mo atoms at 0,0,0; ½,½,½ – Si a toms a t00t 0,0,z; 000,0,z; ½½½½,½,½+z; ½½½½,½,½-z; z=1/3 –MoSi2 is actually body centered tetragonal with a = 3.20 Å and c = 7.86 Å z y x c c c a z x b z z y a x y y x b Viewed down z-axis a b b a Viewed down x-axis Viewed down y-axis 17 Fhkl for MoSi2 N 2 ihu()iji kv lw Ffehkl i i1 Substitute in atom positions: • Mo atoms at 0,0,0; ½,½,½ • Si atoms at 0,0, ; 0,0,z; ½,½,½+z; ½,½,½-z; z=1/3 222iiihk lll hk5 l hkl 2(0)i 2()ii 2() Ffefe 222 fefefe33 226 fe 226 hkl Mo Mo Si Si Si Si lli hk 5lli hk ih k l 2()ii 2() Ff1 e f e33 e e 33 e Mo Si Now we can plug in different values for h k l to determine the structure factor. • For hklh k l = 1001 0 0 5(0) (0) 2(ii(0) ) 2( (0) ) ii10 10 i100 33 33 Ffhkl Mo1 e fe Si e e e 2 Fhkl 0 ffMo(()11)( Si (1111) 0 2 You will soon learn that intensity is proportional to Fhkl ; there is NO REFLECTION! 18 Fhkl for MoSi2 – cont’d Now we can plug in different values for hklh k l to determine the structure factor. •For h k l = 0 0 1 5(1) (1) 2()ii11 2() ii00 00 0 i001 33 33 Ffeehkl Mo fee Si e e ii2 2 fefCOSeMo(1 ) Si (2 (3 ) ) ffMo(1 1) si ( 1 1) 0 2 Fhkl 0 NO REFLECTION! •For h k l = 1 1 0 0iii110 2ii (0) 2 (0) 110 110 Ffeehkl Mo fee Si e e 2(0)(0)22iii fefeeeeMo(1 ) Si ( ) ffMosi(2) (4) 2 Fhkl POSITIVE! YOU WILL SEE A REFLECTION • If you continue for different hklh k l combinations… trends will emerge… this will lead you to the rules for diffraction… h + k + l = even 19 (103) 100 90 MoSi2 CuKα radiation 80 (110) 70 (101) %) 60 50 ntensity ( II 40 (002) (213) 30 (112) (200) 20 (202) (211) (116) (206) 10 (301) (303) (006) (314) (204) (222) (312) 2θ (°) 0 20 30 40 50 60 70 80 90 100 110 120 20 Fhkl for MoSi2 – cont’d 2000 JCPDS -International Centre for Diffraction Data . All rights reserved PCPDFWIN v. 2.1 21 Structure Factor (Fhkl) for HCP N 2ihukvlw()iji Ffhkl ie i1 • Describes how atomic arrangement (uvw) influences the intensity of the scattered beam. i.e., • It tells us which reflections (i.e., peaks, hkl) to expect in a diffraction pattern from a given crystal structure with atoms located at positions u,v,w. 22 • In HCP crystals (like Ru, Zn, Ti, and Mg) the lattice point coordinates are: –0 0 0 121 – 332 • Therefore, the structure factor becomes: hkl2 2i 332 Ffhkl i 1 e •We simppylify this exp ression b y lettin g: hk 21 g 32 which reduces the structure factor to: 2ig Ffehkl i 1 • We can simplify this once more using: from which we find: eeix ix 2cos x 222hkl 2 Ffhkl4cos i 32 Selection rules for HCP 0 when hkn 2 3 and lodd 2 2 fhknli when 2 3 1 and even Fhkl 2 33whenfhknli when 2231 3 1 and odd 2 4fhknli when 2 3 and even For your HW problem, you will need these things to do the structure factor calculation for Ru. HINT: It might save you some time if you already had the I CDD car d f or Ru. List of selection rules for different crystals Crystal Type Bravais Lattice Reflections Present Reflections Absent Simple Primitive, PAny h,k,l None Body-centered Body centered, I h+k+l = even h+k+l = odd Face-centered Face-centered, F h,k,l unmixed h,k,l mixed NaCl FCC h,k,l unmixed h,k,l mixed Zincblende FCC Same as FCC, but if all even h,k,l mixed and if all even and h+k+l4N then absent and h+k+l4N then absent Base-centered Base-centered h,k both even or both odd h,k mixed Hexagonal close-packed Hexagonal h+2k=3N with l even h+2k=3N with l odd h+2k=3N1 with l odd h+2k=3N1 with l even 26 What about solid solution alloys? • If the alloys lack long range order, then you must average the atomic scattering factor.