Lattices Diffraction

$Page 1: Lattices Diffraction$
8/13/2019 Lattices Diffraction.

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Bravais Lattices• Direct Lattice:

– A regular, periodic array of points with aspacing commensurate with the unit cell

dimensions. The environment around eachpoints in a lattice is identical.• The set of direct lattice points can be written as

(vectors): R = t a + ub + vc t ,u ,v integers

• V = volume of unit cell

V = a ! b " c( ) =

b ! c " a( ) = c ! a " b( )

3

Reciprocal Lattices• Reciprocal Lattice: the basis vectors of the reciprocal

lattice are defined as

• The set of Reciprocal Lattice Vectors (RLVs) arewritten as

a* =

b ! c

V ; b* =

c ! a

V ; c* =

a ! b

V ; V = a ! b( )• c

K i = ha*

+ k b* + l c*

h,k , l = integers

e.g., a*= 2!

b " c

V

; ! )

(Note: In physics texts, a factor of 2! is usually included

4



Reciprocal Lattices• Properties of RLVs:

a i a*= b i b*

= c i c*= 1

a*i b = a*

i c = b*i a = b*

i c = c*i b = c*i a = 0

Alternatively, these can be regarded as definitions.

In 2 dimensions, we define the RLVs

a*! b , b

*! a , a i a

*= b i b

*= 1

5

Side by Side (2-D)

• Although they “live in separate universes”, the direct andreciprocal lattices are in rotational “lock-step”.

a

b

Direct Lattice

a!

b!

Reciprocal Lattice

a

b

a!

b!

6



Lattice & Reciprocal LatticeExample: BCC treated as Primitive

a =a

2

x + y ! z( )

b =a

2! x + y + z( )

c =a

2

x ! y + z( )

ax

ay

az

(a/2)(x + y – z)^ ^ ^

(a/2)(x – y + z)^ ^ ^

(a/2)(– x + y + z)^ ^ ^

a

bc

a* =

b ! c

V ; b* =

c ! a

V ; c* =

a ! b

V

V = a ! b( )• c

Use to find Recip. Lattice Basis:

a

7

Lattice & Reciprocal LatticeExample: BCC treated as Primitive

a* =1

a

x + y( )

b* =1

a

y + z( )

c* =1

a

x + z( )

Exercise: Find the reciprocallattice of the reciprocal lattice!

(1/a)y

(1/a)(x + y)^^

(1/a)(x + z)^ ^

(1/a)(y + z)^ ^

c*

a*

b*

(1/a)x

(1/a)z

1/a

If we use the conventional cell (which leaves out half the latticepoints, the reciprocal lattice includes the ‘open’ points too:

a = a x ; b = a y ; c = a z ! a* =

x

a

; b* =

y

a

; c* =

z

a

8



Interrelationships

• The direct lattice can be partitioned such that all

of the lattice points lie on sets of planes.Crystallographers classify these sets of planesusing the intercepts on the unit cell axes cut bythe plane adjacent to the plane through theorigin.

• The intercepts are of the form (1/h, 0, 0), (0,1/k ,0), and (0,0,1/l ).

• These planes are normal to the RLVs

Khkl

= ha* + k b*+ lc*

9

Interrelationships

2a* + 3b*

a*

b*

intercept: (1/2,0)

intercept: (0,1/3)-3a + 2b

note: (-3a + 2b) • (2a* + 3b*) = 0

a

b

10



Interrelationships

Direct Lattice

a

b

Reciprocal Lattice

2a* + b*

intercept: (1/2,0)

intercept: (0,1)

11

Interrelationships

a

b

ˆ ˆd hkl = (1/h)a•Khkl = (1/h) a K cos! a

ˆ

- a unit vector! b

lattice planes

Khkl

Khkl

Khkl

Khkl =

ˆ ˆd hkl = (1/k )b•Khkl = (1/k ) b K cos! b

(0,1/k ,0)

(1/h,0,0)

! a

(hkl ) plane

c

(1/h,0,0)

-(1/h)a + (1/l)c

-(1/h)a + (1/k )ba

b

(0,0,1/l)

(0,1/k , 0)

12



K j •Ri products involve no cross terms

and always yield integers

• For any DLVR

i and any RLVK

j

Ri iK j = t a + ub + vc( ) ! ha*+ k b

*+ lc

*( )= th + uk + vl = m, an integer

! e2" iRi iK j

=1, for all Ri and all K j

13

hkl Planes are normal to K hkl

!1

ha +

1

k b and !

1

ha +

1

lc .

Khkl

! " 1

ha +

1

k b

# $ %

& ' ( = "

h

ha !a* +

k

k b !b*

= "1+1= 0

Khkl

! " 1

ha +

1

lc

# $ %

& ' ( = "

h

ha !a* +

k

k c ! c* = "1+1= 0

!Khkl " to both vectors

!Khkl " to the family of planes

• Proof: Consider the plane that cuts the cell axes at (1/h,0,0), (0,1/k ,0) and (0,0,1/l ).

• Two vectors lie in the plane:

(hkl ) plane

c

(1/h,0,0)

-(1/h)a + (1/l)c

-(1/h)a + (1/k )ba

b

(0,0,1/l)

(0,1/k , 0)

14



• Assume K hkl is the shortest RLV that points in the exact directionit points (i.e., h,k , and l have no common factor.)

• K hkl is normal to the “hkl family of planes”. We can relate thedistance between planes to the magnitude of K hkl :

d hkl

=1

ha ! K

hkl =

a K hkl

hcos"

a =

a

hcos"

a

Where Khkl

is a unit vector parallel to Khkl

.

now, Khkl

=K

hkl

K hkl

=ha

*+ k b

*+ lc

*

K hkl

so d hkl =

1

hK hkl

a ! ha*

+ k b*

+ lc*

( )

#d hkl

=1

K hkl

a

b

ˆ ˆd hkl = (1/h)a•Khkl = (1/h) a K cos! a

ˆ

- a unit vector! b

lattice planes

Khkl

Khkl

Khkl

Khkl =

ˆ ˆd hkl = (1/k )b•Khkl = (1/k ) b K cos! b

(0,1/k ,0)

(1/h

,0,0)

! a

d hkl

=

1

K hkl

=

1

K hkl

To Prove:

15

Diffraction

k =

1

!

• In handling scattering of x-rays (or electrons) by matter, we needto characterize the x-ray beam by specifying its wavelength anddirection of propagation:

• Note: The usual physics definition is

E = E 0 x

H = H 0 y

k = k z

E0 ! H0 =

k

H = H0 exp 2" i(k #r $% t ){ }

ReH = H0 cos 2" (k # r $% t ){ }

E = E0 exp 2" i(k # r $% t ){ }

ReE = E0 cos 2" (k #r $% t ){ }

k =

2!

"

16



Additional Comments• In order to speak of the localization of an electron (or photon),

we must ‘blur’ the specification of the electron’s (photon’s)wavelength, and allow that there is some uncertainty in thewavelength (and frequency and energy):

where we have assumed, for simplicity, that the wave ispropagating along the z -axis and E 0(k ) is a function stronglypeaked near a particular ‘approximate’ wavenumber k.

• Electrons and photons are different in that wave packets forelectrons spread out over time, while photon wave packetsretain their width as they propagate. For more details, see:

E k ( z,t ) = E 0 ( !k )exp 2" i( !k z #$ t ){ }d !k

#%

%

&

http://farside.ph.utexas.edu/teaching/qmech/lectures/node1.html

17

Traveling wave packet• By “adding” up a narrow range of frequencies,a “wave packet” can be constructed and aphoton visualized.

http://farside.ph.utexas.edu/teaching/qmech/lectures/node1.html

18



Elastic Scattering

• When an x-ray is scattered by an atom it may emergein any direction, but unless absorption occurs, it hasthe same wavelength:

• X-ray diffraction for structure determination isconcerned with interference effects that result fromscattering by periodic arrays of atoms.

k = !k =

1

"

19

Interference - “Time Lapse” pictures

This is my own heuristicexercise, not “reality”.

20



Atomic FormFactors: f( 2")

• The “water ripples” concept isnot actually accurate; forwardscattering is actually moreintense.

water ripples

Atomic X-rayscattering

21

Atomic Form Factors: f (2")

Atomic X-rayscattering

• Forward scattering is more intense.• f (2" = 0) = N (number of electrons)

Clegg, p. 24.

22



“Thermal” effects: f #(2")

Scattering at higher angles isattenuated by spreading out

the electron density

• Vibrational motion “spreads out” average electrondensity and causes effective scattering at higherangles to fall off more rapidly; U is the isotropicdisplacement parameter (units of Å2 ).

f j!(2" ) = f

j(2" )•exp

#8$ 2U sin2"

% 2

&

' (

)

* +

0.2 0.4 0.6 0.8

5

10

15

20

Cl–

sin!/" (Å–1)

f

j!(2" )

f

j(2! )

Clegg, p. 24.

23

Interference between scatterersConsider the interference between two scatterers, separated by a

vector d:

k =

k

k

ˆ !k =

!k

!k

k = !k =

1

" or k = " k

k

! k

k

1 2

3 4

d d

d

d

24



Interference between scatterersConsider the interference between two scatterers, separated by a

vector d:

k =

k

k

ˆ !k =

!k

!k

k = !k =

1

" or k = " k

25

Interference between

scatterers• Amplitude of scattered wave

• The phase shift factor depends on thepath length difference =

• For completely constructiveinterference, where n is some integer:

• For completely destructiveinterference

! f 1(2" ) + f

2(2" ) # ( phase shift factor)

k !d + (" ˆ #k ) !d = ( k " ˆ #k ) !d

( k ! ˆ "k ) #d = n$ or (k ! "k )id = n

(k ! "k )id = n +1

2Clegg, Sec. 1-6.

26



• For the general case the amplitude of thescattered wave:

• In general we use a complex form:

A ! f 1(2" ) + f 2 (2" )cos 2# (k $ %k ) • d[ ]

cos 2# (k $ %k ) • d[ ] = 1 when (k $ %k ) • d = n

cos 2# (k $ %k ) • d[ ] = $1 when (k $ %k ) • d = n +1

2

A ! f 1(2" )+ f 2 (2" )exp 2# i(k $ %k ) • d{ }

27

• In a crystal, pairs of translationally equivalentatoms are separated by some direct LatticeVector (R i).

$ If all atoms in the crystal that are related bytranslational symmetry are to scatter X-raysto give constructive interference, we musthave the following:

R i •(k – k #) = n or e2!i(k – k #)•R = 1

where R i is any DLV.

28



The beam is diffracted by a RLV!

This places conditions on the possiblevectors k – k # that are exactly the

same as the conditions used to defineRLVs. We therefore obtain the Lauecondition for diffraction:

k – k # = K j where K j is some RLVk

k

Khkl

k´

29

Bragg View

Clegg, Secs. 1.4-1.5.30



BraggView

31

Optical Transforms; FraunhoferDiffraction

• The diffraction phenomenon candemonstrated with monochromatic visiblelight by use of optical transforms.

• The phenomenon relies on interferenceeffects, like X-ray diffraction, though the

scattering mechanism is different.• The interference phenomenon occurs when

the spacing between scatterers is on thesame order of magnitude as the wavelengthof the light used.

32



33

34



35

36



37

Ewald Construction

• Start with reciprocal latticeand the incident wavevector,k , originating at (h,k ,l ) =(0,0,0).

• Draw a sphere with the tip ofk at the center that containsthe origin, (0,0,0), on the

sphere’s surface.• As the direction of k is

changed, other RLVs (K j’s)

move through the surface –which is the condition that adiffraction peak is observed.

k

(0,0,0)

k

Khkl = k – k!

Khkl

!k

(0,0,0)

k

(0,0,0)

Stout & Jensen, Sec. 2.4.

38



Ewald Construction

c

a

k

!k

k

!k !k

k

(h,0,0)

(0,0,l)

k

!k

kk

!k

k

!k

k

39

Limitations on the Data

• Reflections (K j’s) lie on theEwald sphere to meet the thediffraction condition.

• As the relative orientation ofthe crystal and the incidentX-ray beam are varied, thedirection (but not the length)of k changes, and k can inprincipal point any direction.

• Therefore, the “full sphere” ofpossible reflections hasradius 2|k | = 2k = 2/%. Thespacings between the RLVs(RL points) & 1/a (or 1/b or 1/c).

k1(0,0,0)

k2

40



Getting more Data -shorter %

• Reflections (K j ’s) are limited to

those RLVs no further from theorigin than 2|k | = 2k = 2/%.

• The spacings between the RLVs(& 1/a or 1/b or 1/c) aredetermined by the unit celldimensions.

• For example, if the X-raywavelength is shortened, morereflections fall “in thesphere” (e.g., %Cu = 1.54056

(K'1), % Mo = 0.71073 (K'1).

k Cu

(0,0,0)

k Mo

Stout & Jensen, Sec. 2.4.

41

k ! "k = K j # ! "k = K j ! k

# "k i "k = (K j ! k) i (K j ! k)

k2= K j

2! 2kiK j + "k

2

$kiK j =1

2K j

2

k K j cos(% 2 !& )

sin(& )

! "## $##

= K j

2 # 2k sin& = K j #

2

' sin& = K j

# 2

K j

sin& = ' # 2n Khkl

sin& = ' # 2d hkl sin& = n' Bragg's Law

In the last line we've used the fact that K j is a multiple of Khkl , which is the

shortest RLV for which h, k , and l have no common factors.

i.e., K j = nKhkl , where n = integer.

Use to getLaue equations

42



Laue Equations

n! = 2d hkl

sin" ;

K hkl

2= (4 !

2 )sin2" where K

hkl = ha

*+ k b

*+ l c

* (***)

or [1 d 2] = (4 ! 2 )sin2

" , since K hkl

2= [1 d 2]

The “Laue Equations” for each crystal class can be derived from the form (***).

Cubic

sin2! = A(h2

+ k 2+ l

2 ); A = " 2 4a2 or [1 d

2 ] = (h2+ k

2+ l

2 ) a2

Tetragonal

sin2! = A(h

2+ k

2) +Cl

2; A = "

24a

2; C = "

24c

2

or [1 d 2] = (h2

+ k 2 ) a

2+ l

2c

2

Hexagonal

sin2! = A(h2

+ hk + k 2) + Cl

2; A = " 2 3a2 ; C = "

2 4c2

or [1 d 2] = 4(h2

+ k 2 ) 3a2

+ l 2c

2

Orthorhombicsin

2! = Ah

2+ Bk

2+Cl

2; A = "

24a

2; B = "

24b

2; C = "

24c

2

or [1 d 2] = h

2a

2+ k

2b

2+ l

2c

2

43

Laue Equations,cont.

K hkl

2= (4 !

2 )sin2"

where K hkl

= ha*+ k b

*+ l c

* (***)

Monoclinic

sin2! = Ah2+ Bk

2+Cl

2" Dhl

A = # 2 (4a2 sin2 $ ); B = # 2 4b2 ;

C = # 2 (4c2

sin2 $ ); D = (# 2 cos$ ) (2acsin

2 $ )

or [1 d 2] = h

2 (a2 sin2 $ ) + k 2b

2+ l

2 (c2 sin2 $ ) " 2hl cos$ (acsin2 $ )

Triclinic

sin2! = (# 2 4)[1 d 2]

1

d 2 =

h

a

h

acos% cos$

k

b1 cos&

l

ccos& 1

+k

b

1h

acos$

cos% k

bcos&

cos$ l

c1

+k

b

1 1h

a

cos& 1k

b

cos$ cos& l

c

1 cos& cos$

cos& 1 cos&

cos$ cos& 1

44



Results

• The distances between lattice planes are onthe order of lengths of unit cells (~10 Å)

• 50 Å ! dhkl ! 0.5 Å for most “small moleculestructures”

• Two most common sources:

%Cu = 1.540562 Å (K'1)

%Mo = 0.71073 Å (K'1)

45

Neutron Diffraction• In neutron diffraction, so-called thermal neutrons areusually used.

• Thermal neutrons have deBroglie wavelengths that arecomparable to typical X-ray wavelengths:%Cu = 1.5418 Å,

%Mo = 0.71069 Å,

%n = h/ pn = h[3mnk BT ]-1/2 = 1.45 Å

• A major advantage of neutron diffraction is in the verydifferent form factors, which have magnitudes that donot scale with atomic number (good for many lightatoms, especially deuterium).

• Neutron spin (magnetic moment) can be exploited to probe magnetic ordering.

46



Electron Diffraction• In high-energy electron

diffraction, electrons with 100keV kinetic energies arecommonly used

• e – ’s have short deBrogliewavelengths:%Cu = 1.5418 Å,

%Mo = 0.71069 Å,

%e,100keV = h/ p = h[2me E ]-1/2 = 0.0039 Å

• Ewald sphere is very large incomparison with reciprocallattice spacings. Therefore, alarge section of any plane tangent

to the Ewald sphere is effectively“on the sphere” at the same time.

47

Summary• The geometrical aspects of a crystal (celldimensions, translational symmetry, etc.)determine the direct lattice.

• The direct lattice, in turn, determines thereciprocal lattice.

• There is a one-to-one correspondencebetween the RLVs and the vectors through

which the incident radiation is diffracted.• The geometry of the diffraction pattern isdetermined by the cell dimensions andsymmetry - no specific structural detailsbeyond symmetry and dimensions of the unitcell affect the “ positions” of diffraction peaks.

48



Multiple Atom Structures

d

A

B

d

A

B

d

A

B

d

A

B

d

A

B

d

A

B

A

d

A

B

d

A

B

A A A

B B B

• Within translationally related setsof atoms, the conditions forconstructive interference will besatisfied simultaneously. However,different sets of atoms will

generally diffract the X-ray withdifferent phases.

• Example: For the “crystal” shownhere, when the set of A atoms allscatter the X-ray constructivelywith respect to each other, the setof B atoms will also scatter the X-ray constructively. But the the sets{Ai } and {Bi } will generally scatterwith different phases.

• The resultant diffraction intensitieswill be determined by the sum of

the diffraction amplitudes due to{Ai } and {Bi } - including the effect ofdifferent phases!

49

How do data determine “Structures” ?• The positions of diffraction peaks tell us only

about the lattice parameters.

• The intensities of the peaks tell us about thenature and positions of the atoms within theunit cell.

• The central problem of crystallography is in

working backwards from the peak intensitiesto locations and identities of atoms in the unitcell.

Easier: How do “Structures” determine data?

50



Structure Factors• Atoms distributed throughout the unit cell, with

fixed positions with respect to each other, scatter

X-rays that produce interference patterns. Amplitude of scattered wave =

A ! S K

= f j "(K)

!"#

exp 2# iKid j { }geometric factor

! "$$ #$$ j =1

n

$

atomic form factor

(determined by chargedistribution of the jth atom)

51

The Structure Factor expression& Systematic Absences

• Symmetry operations place restrictions onhow atoms in the cell must be distributed.(Physically, perhaps it is the other wayaround!)

• This means that there are symmetryrelationships between the list of f j’s and d j’s.

A ! S K

= f j " (2# )

form factor !"#

$ exp 2% iKid j { }geometric factor

! "$$ #$$ j =1

n

&

52



Recall: BCC treated as Primitive

a =a

2

x + y ! z( )

b =a

2! x + y + z( )

c =a

2

x ! y + z( )

ax

ay

az

(a/2)(x + y – z)^ ^ ^

(a/2)(x – y + z)^ ^ ^

(a/2)(– x + y + z)^ ^ ^

a

bc

a* =

b ! c

V ; b* =

c ! a

V ; c* =

a ! b

V

V = a ! b( )• c

Use to find Recip. Lattice Basis:

a

53

BCC treated with conventional cell (2 atoms/cell)

a

a = a x

b = a y

c = a z

!

"#

$# % a* =

x

a

; b* =

y

a

; c* =

z

a

Atomic Positions:

d1 = 0

d2 =

1

2 (a + b + c

)Form Factors:

f 1!(2" ) = f 2

! (2" ) = ! f Mo

ax

ay

az

(a/2)(x + y – z)^ ^ ^

(a/2)(x – y + z)^ ^ ^

(a/2)(– x + y + z)^ ^ ^

a

bc

(a/2)(x + y + z)^ ^ ^

54



• All recip. Lattice vectors are of the form

Khkl

= ha!+ k b

!+ lc

!

S hkl = f Mo ! exp 2" i(hi0 + k i0 + li0{ } + exp 2" i h

2

+k

2

+l

2

# $ %

& ' (

)*

+

,-

.

/

0

12

3

4

S hkl = f Mo ! 1+ exp " i h + k + l( ){ }/0 23

exp i! (h + k + l){ } = 1 " S hkl = 2 f Mo

exp i! (h + k + l){ } = "1 # S hkl

= 0

If h + k + l =

even no. :

odd no. :

! if h + k + l = odd, reflection will not be observed.

55

Recall: BCC treated as Primitive

a* =1

a

x + y( )

b* =1

a

y + z( )

c* =1

a

x + z( )(1/a)y

(1/a)(x + y)^^

(1/a)(x + z)^ ^

(1/a)(y + z)^ ^

c*

a*

b*

(1/a)x

(1/a)z

1/a

If we use the conventional cell (which leaves out half the latticepoints, the reciprocal lattice includes the ‘open’ points too:

a = a x ; b = a y ; c = a z ! a* =

x

a

; b* =

y

a

; c* =

z

a

56



Example: NaCl

Na:

dNa1 = 0

dNa2 = 1

2 (a + b) 1

2, 1

2,0( )

dNa3

= 1

2

(a + c) 1

2

,0, 1

2( )dNa4 = 1

2 (b + c) 0, 1

2, 1

2( )

!

"

##

$##

Cl:

Cl1 1

2,0,0( )

Cl2 0, 1

2,0( )

Cl3 0,0, 1

2( )Cl4 1

2, 1

2, 1

2( )

!

"

##

$

#

#

57

S hkl = f Na 1+ exp 2! i h

2+

k

2

" # $

% & '

()*

+,-+ exp 2! i

h

2+

l

2

" # $

% & '

()*

+,-+ exp 2! i

k

2+

l

2

" # $

% & '

()*

+,-

.

/0

1

23

Khkl

• 0,0,0( ) Khkl

• 1

2, 1

2,0( ) K

hkl • 1

2,0, 1

2( ) Khkl

• 0, 1

2, 1

2( )

+ f Cl

exp 2! i h

2

" # $ % & '

()*

+,-+ exp 2! i

k

2

" # $ % & '

()*

+,-+ exp 2! i

l

2

" # $ % & '

()*

+,-

+ exp 2! i h

2+

k

2+

l

2

" # $

% & '

()*

+,-

.

/

0000

1

2

3333Cl1 Cl2 Cl3

Cl4

S hkl = f Na 1+ exp ! i h + k ( ){ }+ exp ! i h + l( ){ }+ exp ! i k + l( ){ }"# $%

+ f Cl exp ! i h( ){ }+ exp ! i k ( ){ }+ exp ! i l( ){ }+ exp ! i h + k + l( ){ }"# $%

= f Na 1+ (&1)h+k + (&1)h+l

+ (&1)k +l"# $% + f Cl (&1)h+ (&1)k

+ (&1)l+ (&1)h+k +l"# $%

1

2

,0,0

( ) 0, 1

2

,0

( ) 0,0, 1

2( ) 1

2 , 1

2 . 1

2( )

58



S hkl = f Na

i 4 + f Cl

i 4 = 4 f Na

+ f Cl( )

S hkl = f Nai

4 + f Cli

!4( ) = 4 f Na

! f Cl( )

S hkl = f

Nai 0 + f

Cli 0 = 0

S hkl

= 0

S hkl

= 0

S hkl

= 0

S hkl

= 0

S hkl

= 0

h k l

even even even

odd odd oddodd even even

even odd even

even even odd

odd odd even

odd even odd

even odd odd

Systematic extinctionsdue to face centering

c*

a*

(0,0,2)

b*

(2,0,0)

(0,2,0)

(0,2,2)

(2,2,2)

(1,1,1)

(2,2,0)

59

Other systematic absences

• Example: Systematic absences expected fora crystal with a 41 screw axis.

x, y, z41

! " ! y, x, z + 14

41 ! " ! x, y, z + 1

2

41 ! " ! y, x, z + 3

4

41 screw axis along c

S hkl ! f p p

"

exp 2# i hx + ky + lz( ){ }+ exp 2# i $hy + kx + lz( ){ }ie

# i 2( )l

+ exp 2# i $hx $ ky + lz( ){ }ie # i( )l

+ exp 2# i hy $ kx + lz( ){ }ie 3# i 2( )l

%

&

''

(

''

)

*

''

+

''

(Every atom belongs to set of 4 symmetry-related atoms.)

( p refers to summation over the sets of symmetry-related atoms.)

60



… and l = odd

S hkl ! f p p

"#

$ %

&

' ( exp 2) ilz( ){ } i 1+ e) i 2

+ e) i+ e

3) i 2*+ ,- = 0

1st

term2nd

term3rd

term4th

term

… and l = 2, 6, 10 …

S hkl ! f p p

"#

$ %

&

' ( exp 2) ilz( ){ } i 1+ e

) i+1+ e

) i*+ ,- = 0

(-1) (-1)

… and l = 4, 8, 12 …

S hkl ! ! 1+1+1+1[ ] (non zero)

Consider h = k = 0

For h = k = 0, only reflections with l = 4, 8, 12 … will be observed.

61

Other systematic absences

• Example: crystal with an a-glide ( to a 41 axis.

(41 screw axis along c, $ a glide || to the ab-plane)

This will be zero when the term in brackets is zero:

If l = 0,then absences occur where h = odd.

For b-glide ( to c-axis, find absences where k = odd (l = 0).

x, y, z a! glide " # """ x +

12 , y, z

a! glide " # """ x + 1, y, z

S hkl $ f p p

% exp 2& i hx + ky + lz( ){ }

+ exp 2& i h x + 1 2( ) + ky ! lz( ){ }'()

*)

+,)

-)

exp 2! i hx + ky + lz( ){ } = "

exp 2! i h x + 1 2( )+ ky"

lz( ){ }exp 2! i lz( ){ } = " exp 2! i h 2 " lz( ){ } = "e! ih exp 2! i "lz( ){ }

exp 2! ilz{ } = "("1)h exp "2! ilz{ }

62



Symmetry of the Diffraction Pattern;Equivalent Reflections - Laue Symmetry

The pattern of single crystal diffraction “spots” is oftenreferred to as the intensity-weighted reciprocal space.Including the intensities, what symmetries should theintensity-weighted reciprocal space exhibit? – In the absence of absorption and/or “anomalous scattering”,

reflections related to each other by inversion in reciprocal space,(h,k,l ) and (-h,-k,-l ), should be of equal intensity. This is calledFriedel’s Law and it applies even for noncentrosymmetric spacegroups (no inversion center).

The Patterson symmetry adds an inversion center if thespace group is acentric. Therefore, in centrosymmetriccases, the Patterson symmetry is the same as the

symmetry of the diffraction pattern.

63

Symmetry of the Diffraction Pattern;

Equivalent Reflections - Laue Symmetry

The translational parts of symmetryoperations can be ignored in findingreflections that are expected to beequivalent. – Screw operations have same symmetry

effect as simple rotational point groupoperations

– Glide operations have the same symmetryeffect as simple mirror planes

See: Table 11.4, p. 382 in Cotton.

64



Tl(1+ x )BaSrCa(1– x )Cu2O(7-)) Superconductor

• Within the disordermodel ( x = 0.22), the

space group issymmorphic: P4/mmm.

• The reciprocal latticeintensities reflect eachsymmetry operation

• No systematicabsences

a = 3.8234 Å; c = 12.384 Å

65

Tl(1+ x )BaSrCa(1- x )Cu2O(7-)) Superconductor

• Simulated electrondiffraction

• 001 plane.• The reciprocal lattice

intensities reflecteach symmetryoperation


66



Tl(1+ x )BaSrCa(1- x )Cu2O(7-)) Superconductor

• Simulated electrondiffraction pattern

• 100 plane.• The reciprocal

lattice intensitiesreflect eachsymmetryoperation


67

Tl(1+ x )BaSrCa(1- x )Cu2O(7-)) Superconductor • Simulated electron

diffraction pattern• 110 plane.• The reciprocal

lattice intensitiesreflect each

symmetryoperation• No systematic

absences

68



Cs-Leucite, low symmetry (298K)

• At high temp (473K),

Cs-leucite(CsAlSi2O6) adopts acubic structure(space group: Ia3d),a = 13.7062Å.

• At room temperature,the symmetry islower

• What is it?

69

Leucite (298K), l = 0 plane

• Simulated electrondiffraction pattern

• What symmetriesare possible?

70



Leucite (298K), h = 0 plane

• What symmetries aresuggested?

71


• Adjacent spots havea separation of0.153 Å –1.

• 1/(0.153 Å –1) = 6.536 Å.

What does thisnumber mean?

72




• Intense peaks areindexed. (How wasthis done?)

73

Powder Diffraction• In principle, powder diffraction is nothing

more than a ‘spatially averaged’ version ofsingle-crystal diffraction.

• Much of the rich spatial information onegathers in single-crystal data is lost in

powder diffraction, only the 2" informationremains.• However, once (if) the cell can be inferred ,

powder diffraction data is still quitepowerful.

74



NaCl: h = 0 and h = k planes

• The circles correspond to reflections on the

same “cone”, i.e., with the same 2" values.

75

Laue Equations• These are relations connecting observed 2"

values and indices: Recall:

earlier we found

or

a!i a = b

!i b = c

!i c = 1

a!i b = a

!i c = b

!i a = b

!i c = c

!i a = c

!i b = 0

Khkl =

2

! sin"

KhkliKhkl

=

4

! 2 sin2"

sin2! =

" 2

4K

hkl

2=

"

2d hkl

#

$ %&

' (

2

76



• Any RLV is written as Khkl

= ha!+ k b

!+ lc

!

Khkl

i Khkl

= h2a!i a

!+ k

2b!i b

!+ l

2c!i c

!

+ 2hk a!i b

!+ 2hla

!i c

!+ 2klb

!ic

!=

4

"

2sin

2#

a!=

1

a x ; b

!=

1

b y ; c

!=

1

c z

K 2=

h2

a2 +

k 2

b2 +

l2

c2 " sin

2# =

$

2

% & ' ( ) *

2h

2

a2 +

k 2

b2 +

l2

c2

%

& '(

) *

• Special cases: Orthorhombic

77

• Cubic and Tetragonal - special cases oforthorhombic:

• Hexagonal

! = " = 90° # = 120

°

out of plane of paper c

a = a( 3

2 x ! 1

2 y)

b = a y

c = c z

V = a i (b ! c) =

3

2a

2c

sin2! =

"

2a

# $ %

& ' (

2

(h2+ k

2+ l

2) cubic

sin2! = "

2 h2+ k

2

4a2

+l

2

4c2

#

$ %&

' ( tetragonal

78



• Reciprocal lattice

• Laue equation

a!=b " cV

=ac x

3 2( )a2c

=2

3a x

b! = 1a

x3+ y# $ % & ' (

; c! = 1c

z

sin2! =

" 2

4KiK =

" 2

4h

2a

# 2

+ k 2b

# 2

+ l2c# 2

+ 2hk a#i b

#$%&

'()

sin2! = "

2

3a2 h

2+ k

2+ hk ( )+ "

2

4c2 l2

79

Single Crystal Data Collection

and Structure Solution1) Mount Crystal and Collect Data (SMART)

2) Obtain Unit Cell Parameters (SMART, Cell Now)

4) Data Reduction and Corrections (SAINT, SADABS, Xprep)

5) Solve Structure (SHELX)

6) Complete Structure (SHELX)

7) Refine structure model (SHELX)

8) Interoperate results (You)

3) Measure Intensities (SAINT)

80



Data Collection

• Mount crystal

• Center Crystal

81

Collecting a Hemisphere of Data Using

SMART

• Set up for 4 different * scans

• Center of detector held at -28º 2"

– This will collect data from 0º to 56º 2"

• + is held at a different angle for each

scan

• , is fixed at 54.9º

• Each scan will rotate the crystal 180ºin *

– Data collected every 0.3º for 20seconds

82



Collecting a Matrix Using SMART• Collect “snap shots” of

portions of the Ewald sphere

• The peaks can then be

indexed to the give cellparameters• The decision to collect a full

hemisphere is based on thedata obtained from the matrix – Stats on how well the peaks fit

the indexed cell – Number of indexed peaks

– Shape and size of peaks – If the cell parameters are

reasonable for your targetcompound Snap shot of Ewald sphere

(reciprocal lattice)

83

Index Data using SMART orCell Now

• SMART will “harvest” reflections – Coordinates of every peak (above a minimum

intensity) will be written to a file

• Index the reflections – SMART

• Elegant vector searching – Fast, but does not handle twinning

– Cell Now• Brute force vector searching

– Computationally demanding, but handles twinning

• hkl ’s will be assigned to every indexed peak

84



Data Reduction Using Saint

• Input into Saint – Laue Class

– Cell parameters – All frame data

• Coordinates of the peaks indexed from SMART or Cell Now• Intensity and coordinates of every pixel of data collected

• Integrates all peaks and makes some corrections

• Key Output from Saint – Intensity of all reflections

85

86



Absorption Correction withSADABS

• Input – coordinates and integrated intensities for all

indexed peaks – Laue Class – Indexed faces if determined

• Compares intensities of all equivalentreflections to construct an absorption surfaceand corrects intensities accordingly

• Output

– hkl ’s with corrected intensities

87

Re-index And Intensity

Normalization With Xprep• Input

– Coordinates and corrected intensities for allreflections

– Speculated contents of unit cell

• Looks for systematic absences and relative

intensities of equivalent reflections• Output

– Space group with normalized intensities orequivalent reflections

88



Structure Solution with ShelxUsing Direct Methods

• Input – Space group – Coordinates and intensities of all reflections

– Speculated contents of unit cell

• Uses Direct or Patterson Methods

• Output – Unit cell coordinates of atoms in real space

89

Structure Refinement Using Least

Squares Method in Shelx Modify initial output from directmethods using chemical sense

Refine UsingLeast Squares

Is all the electron densityaccounted for?(Low R values)

Yes NO

Documents

Lattices Diffraction