In 1935, a British scientist named Arthur Patterson developed the Patterson method to address the phase problem by using the value of IF_{hkl} I^{2} directly from the X-ray diffraction experiments. He modified eq48 by replacing F_{hkl} with IF_{hkl} I^{2} to give a density map in the Patterson space of uvw:
The density map is related to IF_{hkl} I^{2}, which is dependent on the interatomic vectors of atom pairs (see question below). In other words, the peaks on the Patterson map correspond to (the end points of) interatomic vectors.
Question
Show that IF_{hkl} I^{2 }is dependent on the interatomic vectors of atom pairs.
Answer
Substituting eq29 in eq36 and extending the equation in three dimensions with a = r and letting h = s – s_{0}, we have
Since
where r_{j }and r_{k }are the position vectors of atoms j and k respectively, which makes r_{j} – r_{k} the interatomic vector of the j–k atom pair.
Whereas the structure factor F(h) depends on the position vector r of the atom, which is reference to some origin, IF(h)I^{2 }depends on the difference between the position vectors of the j and k atoms in the unit cell (interatomic vector), which is independent of the origin. For each r_{j} – r_{k }vector, there is also a r_{k} – r_{j }vector in eq52, i.e. one vector pointing from atom j to atom k and another from atom k to atom j, with both vectors appearing on the Patterson map.
Furthermore, a peak on the Patterson map has a maximum height that is proportional to the product of the scattering factors of the pair of atoms associated with the interatomic vector. Since the maximum value of the scattering factor of an atom is the number of electrons in the atom, the height of a peak on the Patterson map provides important information of the nature of the pair of atoms contributing to the peak. This information is useful for reconstructing the density map of the unit cell.
Question
Show that a peak on the Patterson map has a maximum height that is proportional to the product of the scattering factors of the pair of atoms associated with the interatomic vector.
Answer
Comparing eq48 with eq29 and eq51, the vector form of eq48 is
The vector form of eq50 is therefore
where u is the three-dimensional lattice vector in Patterson space.
Substitute eq52 in eq53
The maximum value of P(u) is when u = (r_{j }– r_{k}),
The diagram below shows three different crystal lattices and their corresponding Patterson maps. Figure Ia has two similar atoms in the unit cell. The peaks in the corresponding Patterson map are at the ends of the interatomic vector (r_{2} – r_{1}), which is denoted by 1-2, and (r_{1} – r_{2}), which is denoted by 2-1. There is also a slightly more intense peak at the origin of the Patterson map that is due to the interatomic vector of atom 1 onto itself (r_{1} – r_{1}) as well as that of atom 2 onto itself (r_{2} – r_{2}).
By the same logic, a unit cell of three atoms (figure IIa) has nine peaks (N^{2}= 3^{2}), three of which are superimposed at the origin, resulting in a relatively intense peak there. The unit cell with five atoms (figure IIIa) has twenty-five peaks (N^{2}= 5^{2}) in the Patterson map (figure IIIb) at the following positions:
Position | Interatomic vectors | Superimposed peaks |
1 | c → a | 1 |
2 | d → a,
c → b |
2 |
3 | d → b | 1 |
4 |
a → b, d → c |
2 |
5 | a → c | 1 |
6 |
b → c, a → d |
2 |
7 | b → d | 1 |
8 | b → a,
c → d |
2 |
9 | e → a | 1 |
10 | c → e | 1 |
11 | e → b | 1 |
12 | d → e | 1 |
13 | a → e | 1 |
14 | e → c | 1 |
15 | b → e | 1 |
16 | e → d | 1 |
O | a → a,
b → b, c → c, d → d, e → e |
5 |
Total | 25 |
Even though some Patterson maps have semblance to the structures of compounds being investigated, they can be very complicated and have to be deconstructed fully to deduce the structures. This used to be done manually, which is a painfully arduous process. In modern days, a computer does the work by considering the symmetry of the Patterson function, analyzing the peak intensities and noting the presence of heavy atoms.