Hi all im new here but i have a major question, iv been told by people that this question is easy but im just going around in circles with it and i was wondering if you could help.(adsbygoogle = window.adsbygoogle || []).push({});

here is the question

Aluminium cystallises in the fcc structure with a single atom basis. the lattice constant is 4.04 A and the atomic weight of aluminium is 27 a.m.u

a) deduce the number of nearest, next nearest and third nearest neighbour atoms and their separations.

now i understand that the lattice constant corresponds to the length of each side of the cube. My text book says that the fcc lattice has one atom at each corner and one in the centre of each face of the cube making for a total of 4 atoms in the cube. Now can the spacing from one corner atom to a face centred atom be found using simple Pythagoras meaning that the separation to the nearest atom would be sqrt(2.02^2 + 2.02^2)?? or am i visualising all this wrong

any help on what this structure actually looks like would be really helpfull as i find it hard to visualise.

thanks

adam

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# Homework Help: Help needed with Crystal structure

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