I am making struct file for Mg (magnesium) using wien2k_13.
I have following information of Mg from mincryst
(http://database.iem.ac.ru/mincryst/s_carta.php?MAGNESIUM+2671)
structure : Hexagonal
Spacegroup: P6(3)/mmc with spacegroup No. 194.
lattice parameter a=b= 3.2095, alpha=beta= 90, gama = 12
No.
P6/mmc is NOT a rhombohedral space group, but a hexagonal one.
Thus NO conversion of positions .
So just give a,a,c,alpha,beta,gamma (120) and
1/3,2/3,0.25 as position of an atom.
The second atom at 2/3,1/3,.75 will be created automatically if you use
the spacegroup symbol.
PS:
thanks Peter for your kind help. I use the atomic positions as you suggested
like 1/3 2/3 0.25. it works properly and generated second atom as expected.
Thanks and regards
Saurabh Singh
IIT Mandi
On Monday, January 27, 2014 2:49 PM, Peter Blaha
wrote:
No.
P6/mmc is NOT a rhombohe
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