Suggested Reading. Pages in Engler and Randle

Transcription

1 The Structure Factor Suggested Reading Pages in DeGraef & McHenry Pages in Engler and Randle 1

2 Structure Factor (F ) N i1 1 2 i( hu kv lw ) F fe i i j i Describes how atomic arrangement (uvw) influences the intensity of the scattered beam. It tells us which reflections (i.e., peaks, ) to expect in a diffraction pattern. 2

3 Structure Factor (F ) The amplitude of the resultant wave is given by a ratio of amplitudes: F amplitude of the wave scattered by all atoms of a UC amplitude of the wave scattered by one electron The intensity of the diffracted wave is proportional to F 2. 3

4 Some Useful Relations e i = e 3i = e 5i = = -1 e 2i = e 4i = e 6i = = +1 e ni = (-1) n, where n is any integer e ni = e -ni, where n is any integer e ix + e -ix =2 cos x Needed for structure factor calculations 4

5 F for Simple Cubic Atom coordinate(s) u,v,w: 0,0,0 N i1 2 ihu ( kv lw) F fie i j i 2 i (0 h 0 k 0 l F ) fe 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

6 F for Body Centered Cubic Atom coordinate(s) u,v,w: 0,0,0; ½, ½, ½. N i1 2 ihu ( kv lw) F fie i j i h k l 2 i 2 i(0) F fe fe F f 1e i When h+k+l is even F = non-zero reflection. When h+k+l is odd F = 0 no reflection. 6

7 F for Face Centered Cubic Atom coordinate(s) u,v,w: 0,0,0; ½,½,0; ½,0,½; 0,½,½. N i1 2 ihu ( kv lw) F fie i j i 2i h k 2i h l 2i k l 2 i F fe fe fe fe F f 1e e e ih k ih l ik l 7

8 F for Face Centered Cubic F f 1e e e ih k ih l ik l Substitute in a few values of and you will find the following: When h,k,l are unmixed (i.e. all even or all odd), then F = 4f. [NOTE: zero is considered even] F = 0f for mixed dindices (i.e., a combination of odd and even). 8

9 F for NaCl Structure Atom coordinate(s) u,v,w: Na at 0,0,0 + FC transl.; 0,0,0; ½,½,0; This means these coordinates ½0½ ½,0,½; (u,v,w) 0,½,½. N i1 2 ihu ( kv lw) F fie i j i Cl at ½,½,½ + FC transl. ½,½,½; ½,½,½ 11½ 1,1,½; 00½ 0,0,½ 1,½,1; 0,½,0 ½,1,1. ½,0,0 The re-assignment of coordinates is based upon the equipoint concept in the international tables for crystallography Substitute these u,v,w values into F equation. 9

10 F for NaCl Structure cont d For Na: 2 i(0) i( hk) i( hl) i( kl) f e e e e Na f 1e e e Na i( hk) i( hl) i( kl) N i1 2 ihu ( kv lw) F fie i j i For Cl: Cl 2 ( l ) 2 ( k ( ) ) 2 ( h i i hk i h l i kl) f e e e e Cl Cl i( ) i( 2h2kl) i( 2hk2l) i( h2k2l) f e e e e f e e e e i( ) i( l) i( k) i( h) 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

11 F for NaCl Structure cont d Therefore F : N i1 2 ihu ( kv lw) F fie i j i F f e e e i( hk) i( hl) i( kl) Na 1 Cl f e e e e i( ) i( l) i( k) i( h) which can be simplified to * : i( ) i( h k) i( h l) i( k l) 1 F f f e e e e Na Cl 11

12 F for NaCl Structure When are even F = 4(f Na + f Cl ) Primary reflections When are odd F =4(f Na - f Cl ) Superlattice reflections When are mixed F = 0 No reflections 12

13 100 (200) NaCl CuKα radiation 70 %) Intensity ( (220) 30 (420) 20 (222) (422) (600) (442) 10 (111) (400) (333) (440) (311) (331) (511) (531) θ ( ) 13

14 F for L1 2 Crystal Structure Atom coordinate(s) u,v,w: 0,0,0; A B ½,½,0; ½,0,½; 0,½,½. N i1 2 ihu ( kv lw) F fie i j i h k h l k l 2 i (0) 2i 2i 2i F f Ae fbe fbe fbe 2 2 ih ( k) ih ( l) ik ( l) F fa fb e e e 14

15 F for L1 2 Crystal Structure ih ( k) ih ( l) ik ( l) F fa fb e e e (1 0 0) F = f A + f B (-1-1+1) = f A f B (1 1 0) F = f A + f B (1-1-1) = f A f B A B (1 1 1) F = f A + f B (1+1+1) = f A +3 f B (2 0 0) F = f A + f B (1+1+1) = f A +3 f B (2 1 0) F = f A + f B (-1+1-1) 1 1) = f A f B (2 2 0) F = f A + f B (1+1+1) = f A +3 f B (2 2 1) F = f A + f B (1-1-1) = f A f B (3 0 0) F = f A + f B (-1-1+1) = f A f B (3 1 0) F = f A + f B (1-1-1) = f A f B (3 1 1) F = f A + f B (1+1+1) = f A +3 f B (2 2 2) F = f A + f B (1+1+1) = f A +3 f B 15

16 Example of XRD pattern from a material with an 100 (111) L1 2 crystal structure A B Intensity (%) (200) (220) (311) (100) 10 (110) (300) (222) (210) (211) (221) (310) (320) (321) 2 θ ( )

17 F for MoSi 2 Atom positions: Mo atoms at 0,0,0; ½,½,½ c Si atoms at t00 0,0,z; 00 0,0,z; ½½½ ½,½,½+z; ½½½ ½,½,½-z; z=1/3 MoSi 2 is actually body centered tetragonal with a = 3.20 Å and c = 7.86 Å z x y c c a c z x b z x y z y x b y a Viewed down z-axis a b b a Viewed down x-axis Viewed down y-axis 17

18 F for MoSi 2 i1 2 i( hu kv lw ) i j i 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 F N fe i i h k l 2 (0) 2 ( ) 2 ( ) l l i h k l i h k l i i i F f e f e f e f e f e f e Mo Mo Si Si Si Si 2 ( l ) 2 ( ) l l l i h k l i i i h k i h k 3 F f 1e f e e e e Mo Si Now we can plug in different values for h k l to determine the structure factor. For = (0) (0) 5(0) (0) i( ) 2 i( ) i 1 0 i i 3 3 F fmo 1 e fsi e e e e f (1 1) f (1111) 0 Mo Si F 2 0 You will soon learn that intensity is proportional to 2 F hk l ; there is NO REFLECTION! 18

19 F for MoSi 2 cont d Now we can plug in different values for to determine the structure factor. For h k l = (1) (1) i001 2 i( ) 2 i( ) i i 3 3 F fmo e e fsi e e e e 2 F f e f COS e i 2 2i Mo(1 ) Si (2 ( 3 ) ) f (1 1) f ( 1 1) 0 Mo 0 NO REFLECTION! For h k l = si 0 i i (0) 2 i (0) i i F f e e f e e e e Mo Si 2 F 2 i (0) (0) 2i 2i Mo(1 ) Si ( ) f e f e e e e f Mo (2) f (4) si POSITIVE! YOU WILL SEE A REFLECTION If you continue for different combinations combinations trends will emerge this will lead you to the rules for diffraction h + k + l = even 19

20 100 (103) MoSi 2 CuKα radiation 70 (101) (110) %) Intensity ( (002) (213) 20 (112) (200) (202) (211) (116) (206) 10 (301) (303) (006) (204) (222) (312) (314) θ ( ) 20

21 F for MoSi 2 cont d 2000 JCPDS-International Centre for Diffraction Data. All rights reserved PCPDFWIN v

22 Structure Factor (F ) for HCP F N i1 fe i 2 i ( hu kv lw ) i j i Describes how atomic arrangement (uvw) influences the intensity of the scattered beam. i.e., It tells us which reflections (i.e., peaks, ) to expect in a diffraction pattern from a given crystal structure with atoms located at positions u,v,w. 22

23 In HCP crystals (like Ru, Zn, Ti, and Mg) the lattice point coordinates are: Therefore, the structure factor becomes: h 2k l 2 i F fi 1 e

24 We simplify this expression by letting: g h 2k which reduces the structure factor to: 2 ig F 1 fi e We can simplify this once more using: from which we find: ix ix e e 2cos x cos i F f h k l 3 2

25 Selection rules for HCP F 2 0 when h 2k 3 n and l odd 2 fi when h 2k 3n 1 and l even 2 3 fi when h 2 k 3 n 1 and l odd 2 4 fi when h2k 3 n and l 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 ICDD card for Ru.

26 List of selection rules for different crystals Crystal Type Bravais Lattice Reflections Present Reflections Absent Simple Primitive, P Any 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 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=3n1 with l odd h+2k=3n1 with l even h,k,l mixed and if all even and h+k+l4n then absent h+2k=3n with l odd 26

27 What about solid solution alloys? If the alloys lack long range order, then you must average the atomic scattering factor. f alloy =x A f A + x B f B where x n is an atomic fraction for the atomic constituent 27

28 Exercises For CaF 2 calculate the structure factor and determine the selection rules for allowed reflections. 28