What Is the Nearest Neighbor Distance in an FCC Lattice of Aluminum?

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SUMMARY

The nearest neighbor distance in an FCC (Face-Centered Cubic) lattice of aluminum is calculated to be 2.856 Å, derived from the formula 2r = (sqrt(2) * 4.04 Å) / 2, where 4.04 Å is the lattice constant. The first nearest neighbors total 12 atoms, while the second nearest neighbors are 6 atoms, with a distance equal to the edge length of the unit cell, 4.04 Å. Visual aids and references, such as the provided link to a solid-state introduction, significantly enhance understanding of the FCC structure.

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  • Study the ABCABC stacking pattern in FCC lattices
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Hi all I am new here but i have a major question, iv been told by people that this question is easy but I am just going around in circles with it and i was wondering if you could help.

here is the questionAluminium 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 textbook 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 wrongany 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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hi i got the answer in the end :D
 
Hello Adam!

A Very Good Question.
And, yes it is a bit difficult to visualise this fcc structure. Before understanding how I solved this, you are required to get the following file from which I give reference for further clarification. You will get visualisations here.

http://www.4shared.com/file/62877111/ac7097ff/introduction_to_solid_state_IIT.html

This structure is also known as ccp structure which is formed due to ABCABC_ _ _ pattern of stacking of atoms. (page 4----questions 3 and 4)

The number of first nearest atoms is 12. (page 5----question 7)

The spacing will be just 2r = (sqrt 2 )*4.04/ 2 = 2.856 A (page 7----question 15)
Where 4.04 A is the edge length,a, of the unit cell.

I am sure upto this. But the validity of the following must be checked.

The number of second nearest = 6 (not sure)
The distance between second nearest = edge length = a = 4.04 A (I am sure)
 
hi,
i managed to get the same answers as you :D thank you for the link i will be reading through that file :D thanks for the help
 
Could you please post the answer that you got on the forum?
 

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