The phase problem

A problem with the procedure outlined above is that the observed intensity I_hkl is proportional to the square modulus |Fhkl|₂, so we cannot say whether we should use +|Fhkl |or−|F_hkl |in the sum in eqn 20.8. In fact, the difficulty is more severe for non-centrosymmetric unit cells because, if we write Fh_kl as the complex number |F_hkl |e^iα, whereαis the phase of F_hkl and |Fhkl |is its magnitude, then the intensity lets us deter mine |F_hkl |but tells us nothing of its phase, which may lie anywhere from 0 to 2π. This ambiguity is called the phase problem; its consequences are illustrated by comparing the two plots in Fig. 20.24. Some way must be found to assign phases to the structure factors, for otherwise the sum for ρ cannot be evaluated and the method would be useless. The phase problem can be overcome to some extent by a variety of methods. One procedure that is widely used for inorganic materials with a reasonably small number of atoms in a unit cell and for organic molecules with a small number of heavy atoms is the Patterson synthesis. Instead of the structure factors Fhkl, the values of |Fhkl|₂, which can be obtained without ambiguity from the intensities, are used in an expression that resembles eqn 20.8:

P(r) = ∑_hkl | Fhkl |²e^{−2πi(hx+ky+lz)}

The outcome of a Patterson synthesis is a map of the vector separations of the atoms (the distances and directions between atoms) in the unit cell. Thus, if atom A is at the coordinates (xA,yA,zA) and atom B is at (xB,yB,zB), then there will be a peak at (xA − xB, yA −yB,zA−zB) in the Patterson map. There will also be a peak at the negative of these coordinates, because there is a vector from B to A as well as a vector from A to B. The height of the peak in the map is proportional to the product of the atomic numbers of the two atoms, ZAZB. For example, if the unit cell has the structure shown in Fig. 20.25a, the Patterson synthesis would be the map shown in Fig. 20.25b, where the location of each spot relative to the origin gives the separation and relative orientation of each pair of atoms in the original structure. Heavy atoms dominate the scattering because their scattering factors are large, of the order of their atomic numbers, and their locations may be deduced quite readily. The sign of Fhkl can now be calculated from the locations of the heavy atoms in the unit cell, and to a high probability the phase calculated for them will be the same as the phase for the entire unit cell. To see why this is so, we have to note that a structure fac tor of a centrosymmetric cell has the form

F= (±) f_heavy + (±) f_light + (±) f_light + · · ·

where fheavy is the scattering factor of the heavy atom and flight the scattering factors of the light atoms. The flight are all much smaller than fheavy, and their phases are more or less random if the atoms are distributed throughout the unit cell. Therefore, the net effect of the flight is to change F only slightly from fheavy, and we can be reasonably confident that F will have the same sign as that calculated from the location of the heavy atom. This phase can then be combined with the observed |F| (from the reflection intensity) to perform a Fourier synthesis of the full electron density in the unit cell, and hence to locate the light atoms as well as the heavy atoms. Modern structural analyses make extensive use of direct methods. Direct methods are based on the possibility of treating the atoms in a unit cell as being virtually randomly distributed (from the radiation’s point of view), and then using statistical techniques to compute the probabilities that the phases have a particular value. It is possible to deduce relations between some structure factors and sums (and sums of squares) of others, which have the effect of constraining the phases to particular values (with high probability, so long as the structure factors are large). For example, the Sayre probability relation has the form sign of F _{h+h′,k+k′,l+l}′ is probably equal to (sign of Fhkl) × (sign of F _h′k′l′)

For example, if F₁₂₂ and F₂₃₂ are both large and negative, then it is highly likely that F₃₅₄, provided it is large, will be positive.

Fig. 20.25 The Patterson synthesis corresponding to the pattern in (a) is the pattern in (b). The distance and orientation of each spot from the origin gives the orientation and separation of one atom–atom separation in (a). Some of the typical distances and their contribution to (b) are shown as R1, etc.

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مواضيع ذات صلة

Energetics

Structure

Close packing

Structure refinement

Bragg’s law

X-ray diffraction

Lattices and unit cells

Surface pressure

Membrane formation

Micelle formation

The electrical double layer

Classification and preparation