Síntese de Patterson em Cristalografia

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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 529
In this case there are a total of ten terms to include, h = 0 to 9. Figure 15.1 shows
a plot of Vρ(x) against x; the electron density is at a maximum of 142/V at
x = 0 or x = 1, the boundaries of the unit cell.
0.0 0.2 0.4 0.6 0.8 1.0
−50
0
50
100
150
x
V
ρ(
x)
Figure 15.1
E15B.9(b) �e Patterson synthesis is given by [15B.5–653],
P(r) = 1
V
∑
hk l
∣Fhk l ∣
2 e−2πi(hx+ky+l z)
In this case the structure factors are only given for the x direction so the sum
is just over the index h. Furthermore, because Fh = F−h the summation can be
taken from from h = 0 to h = +∞. Using a similar line of argument to that in
Exercise E15B.7(b), the Patterson synthesis is
VP(x) = F20 + 2
∞
∑
h=1
F2h cos(2πhx)
In this case there are a total of ten terms to include, h = 0 to 9. Figure 15.2 shows
a plot of VP(x) against x. As expected, there strong feature at the origin; this
arises from the separation between each atom and itself.�ere is also a strong
feature at x = 1 which indicate that atoms are separated by 1× a unit along the
x-axis.
E15B.10(b) To constructor the Patterson map, choose the position of one atom to be the
origin.�en add peaks to the map corresponding to vectors joining each pair
of atoms (Fig. 15.3). �e vector between atom A and atom B has the same
magnitude as that between B and A, but points in the opposite direction; the
map therefore includes two symmetry related peaks on either side of the origin.
�e vectors between each atom and itself give a peak at the centre point of the
Patterson map, and the many contributions at this position create an intense
peak.