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Firstly, sorry if this question is too basic for this community but seriously I couldn't find any answer. Does it mean the multiplication of the in-plane lattice constants? For example, lets say following geometry file belongs the graphene.

C: Graphite Lattice
1.0
 +2.4410462393  +0.0000000000  +0.0000000000 
 -1.2205231197  +2.1140080551  +0.0000000000 
 +0.0000000000  +0.0000000000 +10.0000000000 
  2
Cartesian
 +0.0000000000  +0.0000000000  +0.0000000000 
 +0.0000000000  +1.4093387034  +0.0000000000 

So its formula unit surface area is: 2.4410462393 x 2.4410462393 = 5.958706742400672. Am I right? If I'm wrong I'd be very appreciated for the correct logic.

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  • $\begingroup$ Can you give an example of where this term is used? $\endgroup$
    – wcw
    Feb 4, 2021 at 19:20
  • 1
    $\begingroup$ Sure. Here it is: pubs.acs.org/doi/abs/10.1021/… $\endgroup$
    – Savir
    Feb 4, 2021 at 19:23

1 Answer 1

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In the paper you linked, the relevant phrase is "$A_{\mathrm{fu}}$ is the per formula unit total surface area of $\mathrm{TiS_2}$."

The "per formula unit surface area" part is equivalent to "surface area per formula unit." So, the full surface area divided by the number of formula units.

For the graphene example, it would be half the in-plane surface area of the cell, $$ \frac{1}{2} \, \mathbf{a}_1 \times \mathbf{a}_2 \approx \frac{1}{2} (2.441)^2 \sin(60\deg) \approx 2.58 , $$ because the cell contains two formula units.

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