Recent content by Mathew Murdock

If the question is "the percentage of", I won't ask you guys :))))

Ah I get it now, thank you :D

Why do I have to divide it by n^3, could you explain it? is that the relative's meaning? I thought that it was done. Thank you

I still cannot figure it out by now, please have a look at my new solution and I am looking forward to hearing from you

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...

So do I get your point of view now?

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

I mean my final answer is 3n^2-9n+7

I got it wrong, the n surface must be (6 (n-2)^2)/2. Thank you for helping me

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...

I am a physics student from Vietnam