Dear Sanjay,

I'll try to help you. First, lets define the primitive vectors, that I'll denote as b1, b2 and b3:

b1 = a( sqrt(3)/2, 1/2, 0)

b2 = a(-sqrt(3)/2, 1/2, 0)

b3 = c(0, 0, 1)

In the above expressions, a is the "in plane" lattice parameters and c is the lattice parameter along the z direction. Now we have to define the positions of the basis atoms: different from the graphene, we have now FOUR atoms in the basis, two for the graphene and two for the BN. Try:

B    -0.7075   0.0000   0.0000

N     0.7075   0.0000   0.0000

C    -0.7075   0.0000   d

C     0.7075   0.0000   d

where d is the distance betwenn the BN and graphene plans (d < c). You should research the d value in the literature. Try the article:

http://arxiv.org/pdf/1204.2030v1.pdf

In fact, it is intteresting to put c = 10 angstrons in order to avoid the interaction between the planes and their periodic images. The value 0.7075 is the half of the C-C distance in graphene, in order that we are constraining the B-N distance be the same.

I hope be help. If does not work, you can contact me directly in my e-mail: [email protected]

Best regards and sorry for the english level, but I lose my glasses yesterday!

Adriano

Dear Sanjay,

if you want different cell dimensions for both layers, you have to modify the positions of the B and N atoms. Notice that for the C atoms, the value 0.7075 was taken as half of C-C distance. In this way, you have to put the half of the B-N distance for the B and N atoms. I don't know this distance, but if it is 1.20 Angstrons, the positions would be:

B   -0.600   0.000   d

N    0.600   0.000   d

I hope this can help you. Regards,

Adriano

i was searching for the answer related to it.

Thanks to everyone who has answered it .