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Face-centered cubic lattice.

by hhhmortal
Tags: cubic, facecentered, lattice
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hhhmortal
#1
Jan6-09, 09:46 PM
P: 176
1. The problem statement, all variables and given/known data

An element crystallises in a face-centered cubic lattice with a basis group of two atoms at [000] and [1/4 1/4 1/4]. The lattice constant is 3.55 x 10^-10 m

(Q1) How many nearest and second-nearest neighbours does each atom have?

(Q2) Calculate the average volume per atom

2. Relevant equations



3. The attempt at a solution

Ok, well I tried doing (Q2) I cant make sense of (Q1) though.

For (Q2) I used equation :

(4/3)(pi)(r) = volume of atom

and r = {√3}/4 x lattice constant


But I dont think this is the case since I want the average volume per atom?

Thanks.
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hhhmortal
#2
Jan7-09, 11:42 AM
P: 176
^Bump
hhhmortal
#3
Jan7-09, 10:46 PM
P: 176
Any help at all? After reading some articles, I still can't work out this question.

turin
#4
Jan8-09, 08:15 AM
HW Helper
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P: 2,327
Face-centered cubic lattice.

Do you know the meaning of "nearest neighbor" and "second nearest neighbor"?

Quote Quote by hhhmortal View Post
and r = {√3}/4 x lattice constant
Why do you think this is true?

Try to do it the other way around. What is the volume of a cell? How many atoms are there per cell? Answering the second question may be a bit tricky.
hhhmortal
#5
Jan8-09, 12:30 PM
P: 176
I dont understand what "nearest neighbours and second nearest neighbours" is, especially for this fcc.

And I think that comes from the fact that the length of the diagonal is √3. and a line from one corner to the center corner passes through 4r. so

a = 4r / √3

Are there 4 atoms in this FCC? 3 from the face centered atoms and 1 from the corners.

I calculated the volume of cell to be: a = (3.55 x 10^-10)


But how could I work out average volume of atom?
turin
#6
Jan8-09, 04:57 PM
HW Helper
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P: 2,327
Quote Quote by hhhmortal View Post
I dont understand what "nearest neighbours and second nearest neighbours" is, especially for this fcc.
For a given point, what is the closest possible other point? That other point is a nearest neighbor. There may be more than one nearest neighbor. For example, for a simple cubic, every atom has six nearest neighbors. Can you see why? For the second nearest neighbors, you simply form a sphere, whose radius is the distance to the centers of the nearest neighbors, centered on the atom of consideration, and then you start looking outside that sphere for the centers of atoms that are closest to it. If I'm figuring correctly, each atom of a simple cubic has eight second nearest neighbors.

I suggest to draw a picture, but be careful when considering the consequence of 2-D to 3-D.

Unfortunately, I don't remember if you must count atoms or lattice sites. I will try to find the answer to this and get back to you.



Quote Quote by hhhmortal View Post
... a line from one corner to the center corner passes through 4r.
It's been a while for me, but isn't that for a BCC?



Quote Quote by hhhmortal View Post
Are there 4 atoms in this FCC? 3 from the face centered atoms and 1 from the corners.
Maybe you know how to do this better than me. I just weight each corner as 1/8 of a lattic site (b/c corners are shared with 8 cells) and each face-centered site as 1/2 (b/c faces are shared with 2 cells). Then, there are 8 corners and 6 faces. So, I get

1/8 x 8 + 1/2 x 6 = 4

Of course, then you also need to consider how many atoms there are per lattice site.



Quote Quote by hhhmortal View Post
I calculated the volume of cell to be: a = (3.55 x 10^-10)
I'll take your word for it.



Quote Quote by hhhmortal View Post
But how could I work out average volume of atom?
If, on average, N atoms occupy a volume of V, then 1 atom, on average, occupies a volume of ...
hhhmortal
#7
Jan8-09, 10:58 PM
P: 176
Ok well I got it!

From your help I was able to assume that nearest neighbours would be 4 and then second nearest 12 I believe right?

And then you were right in saying that its not 4r, in actual fact its 2√2r

So then I was able to calculate the average volume of an atom to be:

(3.55 x 10^-10) / 4 = 11.8 x 10^-30 m

I'm pretty sure this is the correct value.

I was also able to calculate the packing factor to be 74%.


But then I have a last question which is:

What is the number density of atoms (number per m) on the (1 1 1) planes:

I tried working something out but Im not sure on it, any help for this ?

Thanks!


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