Ok let me start again. The total atoms is n^3, the atoms lie inside the cube is (n-2)^3, then the atoms on the outer layer is n^3 - (n-2)^3 = 6n^2-12n+8, for which yield 8 for n=2, 26 for n=3 etc... Whats wrong with my solution?
And mjc123 says about relative number, can you help me figure it...
I mean the low-coordinated atoms made me confused and therefor I thought that I had to calculate the atoms that are entirely inside the unit cells, so there were terms 1/8, 1/4, 1/2 in my answer. You guys then explained that we just basically find the number of atoms on the outer layer, and...
ok I got what you mean by now, just find the whole atoms, minus the non low coordinated atoms and we will get the answer.
So as you said, I have made difficulties myself? :)
I just thought that if it is just about calculating the "area" of this, the teacher does not need to give me this homework...
Thank you for helping me. I just thought that if it is as you have pointed out for me, then the problem seems to be easier than I thought, and that makes me a little bit nervous
Do I do it right guys?
The relative amount of low-coordinated atoms is
The number of atoms at 8 corners: n_corner=8 1/8=1.
The number of atoms at 12 edges: n_edge=12 (n-2)/4=3n-6.
The number of atoms at 6 surfaces: n_surface=6 (n^2-4n)/2=3n^2-12n.
The relative amount of low-coordinated atoms is...