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青州's Question
Posted 2 months ago

In a certain Patterson map for an orthorhombic crystal, space group
P222, with coordinates of equivalent positions at x,y,z;x,y,z,x,y,1/2+z,x,y,1/2zx, y, z ; x, y,-z,-x,-y, 1 / 2+z,-x, y, 1 / 2-z, there
were high peaks at u,v,wu, v, w of 0,0,0.74;0.6,0,0;0,0.36,0.250,0,0.74 ; 0.6,0,0 ; 0,0.36,0.25, assuming one heavy atom per
asymmetric unit, find the coordinates of this heavy atom.
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Answer from Sia
Posted 2 months ago
Key Concept
Patterson Map Analysis
In a Patterson map, peaks correspond to vectors between atom pairs. The peak at u,v,w=0,0,0.74u,v,w = 0,0,0.74 suggests a vector along the cc-axis, which is half the unit cell length in an orthorhombic P222 space group, indicating the heavy atom is at z=0.37z = 0.37. The peaks at 0.6,0,00.6,0,0 and 0,0.36,0.250,0.36,0.25 correspond to vectors from the origin to the heavy atom's xx and yy coordinates, respectively. Thus, the heavy atom's coordinates are approximately (0.3,0.18,0.37)(0.3, 0.18, 0.37).

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