> Like Rp, Rwp, Chi2, RB are respectively from 3.96, 5.33, 2.68, 5.80 > to 3.35, 4.42, 1.85, 3.07.
This should pass a test of statistical significance. That does not mean it really is preferred orientation, but if it is not preferred orientation, then you probably want to find out what it is. > By the way, I haven't known how to calulate the site radius so far. I > have read the other two papers > following the suggestion of Dr Alan Hewat, but these two papers > didn't tell the calculation course in > details. Is there some programs which can calculate it? > .... > ----------------------------------------------------------------- > D atoms Environment Site radius Occupancies > (angstrom) (angstrom) (%) > ------------------------------------------------------------------ > D1 2b 1 Ni1 1.589 0.387 16.0% > 3 Ni3 1.640 > D2 2b 1 Ni2 1.606 0.377 15.5% > 3 Ni3 1.623 > D3 6c 1 La 2.595 0.548 23.0% > 1 La 2.461 > 2 Ni3 1.654 > ... D1 has one neighbour Ni which is 1.589 angstrom away and 3 neighbours Ni3 which are 1.640 Angstrom away. The distance from the centre of the D1 atom to the surface of the neighbouring atoms is just the distance minus the radius of the atoms, ie: 1x0.347 and 3x0.398, as you noted. If you assume the site is tetrahedral, what is the largest sphere which can fit in the space which the three neighbours enclose? A quick and dirty approximation is (0.347+3*0.398)/4=0.385. Presumably the extra 0.02 comes from working it out properly, so let's try to do that... General equation of a sphere centred at [a,b,c] in cartesian co-ords is: (x-a)^2+(y-b)^2+(z-c)^2=r^2. Substitute in points for the x,y,z co-ordinates of the closest bit of the Ni surfaces. In general you are going to need four points to define a cavity in this way. This is easiest if you take a unit cube and put the points on alternate corners of the cube. Thus the points are at [u,u,u],[-1,-1,1],[-1,1,-1],[1,-1,-1], where u=(0.347/0.398) and one unit in this space is 0.398/sqrt(3). (This is a shortcut, I've assumed the tetrahedral angles are equal and the closest Ni site is displaced along the body diagonal of the cube, convince yourself about u by drawing it out). u^2-2ua+a^2 + u^2-2ub+b^2 + u^2-2uc+c^2 = r^2 ; [ u, u, u] ; eq 1 1 +2a +a^2 + 1 +2b +b^2 + 1 -2c +c^2 = r^2 ; [-1,-1, 1] ; eq 2 etc. By symmetry a=b=c. You could solve it without noticing that, but life is too short. eq 1 = r^2 = eq 2 = r^2 3(u^2-2ua+a^2) = r^2 = 3 + 2a + 3a^2 = r^2 a = 3(u^2-1)/(6u+2) = -0.0995 So "a" is a small negative number - the centre of the sphere is displaced away from the closest neighbour - as we would expect. Finally r^2 = 3 + 2a + 3a^2 = 2.8307 So that the "site radius" is sqrt(2.8307)*0.398/sqrt(3)=0.3866 Which rounds to the 0.387 given in the paper you cite. I leave the others to your own pencil and paper. The point is that the site radius is a function of only the neighbouring atoms - it has nothing to do what is placed on the site, or where you put it within the cavity. A computer program which told you this number would not seem to help any more than a paper which tells you this number, although I can see that it is tedious to calculate in the general case. Hope this helps, Jon
