Lattices, Reciprocal Lattices and Diffraction Chem 634 T. Hughbanks 1 References for this Topic ! Clegg, “Crystal Structure Determination”. ! Stout & Jensen, “X-Ray Structure Determination ”, 2nd Edition. Instrumentation discussion is completely outdated, but still a good text on the subject. ! A more authoritative general reference: Giacovazzo, et al., “Fundamentals of Crystallography”, IUCr Texts on Crystallography. ! MIT has a good site (MIT Open Courseware): http:// ocw.mit.edu/OcwWeb/Chemistry/ 5-841Fall-2006/LectureNotes/index.htm 2
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• 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 ).
• 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:
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
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
• 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:
• 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).
• 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.
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.
• 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.
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