QuantumATK Forum

QuantumATK => General Questions and Answers => Topic started by: sonu on February 24, 2021, 20:29

Title: surface area of unit cell
Post by: sonu 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

Title: Re: surface area of unit cell
Post by: mlee 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.

Title: Re: surface area of unit cell
Post by: Anders Blom 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 ;-) ).