Author Topic: surface area of unit cell  (Read 180 times)

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Offline sonu

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surface area of unit cell
« on: February 24, 2021, 20:29 »
I want to know the process of calculating the surface area of AC plane of unit cell. Plz help

Offline mlee

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Re: surface area of unit cell
« Reply #1 on: February 26, 2021, 10:50 »
I am not sure about your question.
In the builder, you can use the swap axes if you need.
You can display the information of the active configuration in the builder.

Offline Anders Blom

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Re: surface area of unit cell
« Reply #2 on: March 11, 2021, 01:47 »
So, this is a question of simple geometry, if I understand it right :-)
If you know the A and C vectors in terms of xyz (as shown in the Lattice Parameters plugin) all you need is the area formula area = (ac/2)*sin(beta) where a and c are the lengths of the A and C lattice vectors, and beta the angle between them.

For an example, take the Chromium Bromide (5/3) crystal you can find in the Database. It's a body-centered tetragonal lattice, so when you first open it, it shows lattice constants a=b=5.46 Å, c=10.64 Å and beta = 90, but this doesn't quite help us, as this actually refers to the conventional cell.

Therefore, convert it to a UnitCell, which will show the lattice parameters of the primitive cell: a=b=c=6.57 Å and beta=130.9 degrees. So the area of the AC plane is 6.57^2/2*sin(130.9) or about 16.3 Å^2 (if I got the numbers right; I will leave that as an exercise to verify ;-) ).
« Last Edit: March 11, 2021, 01:50 by Anders Blom »