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

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Homework Help Overview

The discussion revolves around determining the nearest neighbor distance in a face-centered cubic (FCC) lattice structure of aluminum, including the number of nearest, next nearest, and third nearest neighbor atoms and their separations. The context includes the lattice constant and atomic weight of aluminum.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to visualize the FCC structure and questions whether the nearest neighbor distance can be calculated using the Pythagorean theorem. Some participants provide references for visual aids and clarify the number of nearest neighbors.

Discussion Status

Participants have shared insights and resources, with some confirming their understanding of the problem. There is an acknowledgment of the difficulty in visualizing the FCC structure, and while some answers have been reached, there is still uncertainty regarding the second nearest neighbors.

Contextual Notes

Participants are navigating assumptions about the structure and the calculations involved, with references to specific pages in a shared file for further clarification. There is a lack of explicit consensus on the second nearest neighbor distance.

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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
 
Last edited:
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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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