Author Topic: how to make a Bi2Te3 crystal  (Read 11079 times)

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Offline ramkrishna

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how to make a Bi2Te3 crystal
« on: July 29, 2011, 12:35 »
Hi,
Can anybody tell me how to make a Bi2Te3 crystal or can provide the script. This is not listed in the database of Quantumwise however it is a very important thermo-electric material. Waiting eagerly for your reply.

Thanks
Ramkrishna

Offline zh

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Re: how to make a Bi2Te3 crystal
« Reply #1 on: July 29, 2011, 13:21 »
From the lattice constants, space group and wyckoff positions of Bi2Te3 (they can be found in J. Phys.  Chem.  Solids,  Vol. 24,   pp. 479-485. (1963). please see the attached pdf file), one can easily built its crystal structure using some tools such as VESTA (http://www.geocities.jp/kmo_mma/crystal/en/vesta.html). The fractional coordinates of atoms in the primitive unit cell of Bi2Te3 can be obtained from these tools. Based on these information, the script file of Bi2Te3 crystal can be prepared.

Offline Anders Blom

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Re: how to make a Bi2Te3 crystal
« Reply #2 on: July 29, 2011, 14:25 »
There is an easier way, you can just download a CIF file with the stucture from a database, like http://www.crystallography.net/search.html.

However, a real problem is that there is a bug in ATK which makes it import some trigonal structures incorrectly. I fear this would make it equally impossible to import the structure built in VESTA.

If you give me a few days, I can provide either a CIF file or a ready NanoLanguage structure. In either case, I think it's best to convert it to the corresponding hexagonal setting, which is probably what you need anyway, is that correct? Well, once it's in native NanoLanguage we can rhombohedral too, but since the current Builder in VNL can't cleave anything but cubic systems (11.8 will be able to cleave anything), I think it's best to stick with the hexagonal representation, which you can use directly without cleaving for transport in C (or rotate the axis for other purposes).

The only disadvantage is that the basic crystal in the hex setting is a supercell with 3 rhombohedral units. This will give some zone folding in the band structure, but you may see this in other references too.

For now, for the amusement of all users, I attach a small note I made on the difficulties of correctly importing rhombohedral crystals :)


Offline Anders Blom

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Re: how to make a Bi2Te3 crystal
« Reply #3 on: July 29, 2011, 14:48 »
Ok, VESTA is quite to easy download and install, and you don't actually have to build the crystal from scratch, since it imports a slightly edited CIF file with the structure from the database (attached CIF file). Now, if you export this as a "P1" file from VESTA, you can very easily use the data to construct a NanoLanguage BulkConfiguration in Python. The output is
Quote
Bi2 Te3         4.4169998169         0.0000000000         0.0000000000        -2.2084999084         3.8252340499         0.0000000000         0.0000000000         0.0000000000        29.8400001526      0.000000000         0.000000000         0.187999994  Bi1      0.666666687         0.333333343         0.521333337  Bi1      0.333333343         0.666666687         0.854666650  Bi1      0.000000000         0.000000000         0.597999990  Bi2      0.666666687         0.333333343         0.931333303  Bi2      0.333333343         0.666666687         0.264666677  Bi2      0.000000000         0.000000000         0.388999999  Te1      0.666666687         0.333333343         0.722333312  Te1      0.333333343         0.666666687         0.055666685  Te1      0.000000000         0.000000000         0.800000012  Te2      0.666666687         0.333333343         0.133333325  Te2      0.333333343         0.666666687         0.466666698  Te2      0.000000000         0.000000000         0.000000000  Te3      0.666666687         0.333333343         0.333333343  Te3      0.333333343         0.666666687         0.666666687  Te3
which we by hand can convert to
Code: python
vector_a = [4.417, 0.0, 0.0]*Angstrom
vector_b = [-2.2085, 3.82523, 0.0]*Angstrom
vector_c = [0.0, 0.0, 29.84]*Angstrom
lattice = UnitCell(vector_a, vector_b, vector_c)

elements = [Bismuth,]*6+[Tellurium,]*9

fractional_coordinates = [
     [0.000000000,         0.000000000,         0.187999994], # Bi1
     [0.666666687,         0.333333343,         0.521333337], # Bi1
     [0.333333343,         0.666666687,         0.854666650], # Bi1
     [0.000000000,         0.000000000,         0.597999990], # Bi2
     [0.666666687,         0.333333343,         0.931333303], # Bi2
     [0.333333343,         0.666666687,         0.264666677], # Bi2
     [0.000000000,         0.000000000,         0.388999999], # Te1
     [0.666666687,         0.333333343,         0.722333312], # Te1
     [0.333333343,         0.666666687,         0.055666685], # Te1
     [0.000000000,         0.000000000,         0.800000012], # Te2
     [0.666666687,         0.333333343,         0.133333325], # Te2
     [0.333333343,         0.666666687,         0.466666698], # Te2
     [0.000000000,         0.000000000,         0.000000000], # Te3
     [0.666666687,         0.333333343,         0.333333343], # Te3
     [0.333333343,         0.666666687,         0.666666687]] # Te3

# Set up configuration
bulk_configuration = BulkConfiguration(
    bravais_lattice=lattice,
    elements=elements,
    fractional_coordinates=fractional_coordinates
    )
Have fun! NOTE: This is the hexagonal supercell.
« Last Edit: July 29, 2011, 14:50 by Anders Blom »

Offline Anders Blom

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Re: how to make a Bi2Te3 crystal
« Reply #4 on: July 29, 2011, 15:10 »
Of course, once you've had to import a P1 file like this from VESTA more than 2-3 times by hand, you get tired of it and write a small Custom Builder that does the job :)

Drop the attached script on the "Custom" icon in VNL, then drop the exported P1 file from VESTA on the drop zone, and presto!

So, to summarize:
1) Open CIF file from previous post in VESTA
2) Export as P1 file
3) Import into VNL via the Custom Builder in this post