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 i1

• 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

ei = e3i = e5i = … = -1 e2i = e4i = e6i = … = +1 eni = (-1)n, where n is anyyg integer eni = e-ni, 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: i1 – 0,0,0

2(00ih  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: i1 – 0,0,0; – ½, ½, ½.

hkl 2i  2(0)i 222 Ffefehkl   Ffe1 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: i1 – 0,0,0; – ½,½,0; – ½,0,½; – 0,½,½.

hk  hl  k l 222iii   20i  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 mi xe 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: i1 – 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: i1 2iihkihlikl(()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 ) ih() k l 222 feCl   e e e  

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: i1

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 ( 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: i1 – 0,0,0; A B – ½,½,0; – ½,0,½; – 0,½,½.

hk hl k l 2(0)i 222iii       Ffefe22  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 at t00 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 i1

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

222iiihk  lll hk5 l hkl  2(0)i   2()ii 2()      Ffefe222  fefefe33 226  fe 226 hkl Mo Mo Si Si Si Si     

ll 5ll ih k l 2()ii 2() ihk  ihk  Ff1 e  f e33  e  e33  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  i100 33 33  Ffhkl Mo1 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 i001 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

0iii110 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 ( 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 2ihukvlw()iji  Ffhkl  ie i1 • 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 2i 332 Ffhkl i 1 e   •We simppylify this ex pression b y lettin g: hk 21 g   32 which reduces the structure factor to:

2ig Ffehkl i 1 

• We can simplify this once more using:

from which we find: eeix  ix  2cos x

222hkl 2 Ffhkl4cos 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 th e 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+l4N then absent and h+k+l4N 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=3N1 with l odd h+2k=3N1 with l even

26 What about solid solution alloys?

• If the alloys lack long range order, then you must average the atomic scattering factor.

falloy =xAfA + xBfB

where xn is an atomic fraction for the atomic constituent

27 Exercises

•For CaF2 calculate the structure factor and determine the selection rules for allowed reflections.

28