We can also assume the lattice parameter of the cubic unit cell to be a and the that of the extended cell to be a', which means that V' = (a')^3

So is the number of atoms n' in the extended volume V' equal to:

n' = n x (V'/V) ?

The basis of this question arises from the fact that I have the Energy of a system in a terms of eV/atom. And I want to calculate the total electronic energy of the extended volume.

We can assume the unit cell to be conventional for the ease of calculation.


1 Answer 1


Yes, the number of atoms as well as the energy are extensive so they obey the scaling relations $ n \to n'= \frac {V'} {V} n$ and $E \to E'=\frac {V'} {V} E$ where $n$, $E$ and $V$ are the original number of particles, energy and volume, and $n'$, $E'$ and $V'$ are the new number of particles, energy and volume.


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