[ase-users] hcp0001 surface of La2O3

[Punit Kumar]

Hello ase-users

I want to make an hcp0001 surface for La and La2O3 from their bulk crystal
structure. I have made bulk crystal structure of La and La2O3 using
'crystal' function from 'spacegroup' module.

>>> la = crystal(['La','La'], [(0., 0., 0.), (1./3.,2./3.,1./4.)],
spacegroup=194, cellpar=[3.77, 3.77, 12.06, 90, 90, 120])

>>> la2o3 = crystal(['La','O'], [(1./3., 2./3., 1./4.), (1./4., 1./4.,
0.)], spacegroup=150, cellpar=[3.93, 3.93, 6.13, 90, 90, 120])

now after making the bulk crystal structure, I tried to make make hcp0001
surface for La and La2O3 using 'surface' function from build module.

>>> la0001_1 = surface('la', (1,1,1), 3, vacuum=10)

But when I compare this surface with the surface

>>>la0001_2 = hcp0001('La', size=(2,2,3), a = 3.77, c = 12.06, vacuum = 10)

these two surfaces are very different from each other. In la0001_2 surface
one can see the layers very clearly while in la0001_1 surface layers are
not visible as in la0001_2 surface. All atoms in the layer of la0001_1 are
showing tag = 0.
The same thing i have done for La2O3. What all I want is to make an hcp0001
surface for La2O3 from its bulk crystal structure.
Can anyone help me in this matter. Any help is highly appreciated.

With Best Regards
Punit
IIT Bombay
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