Register to reply 
Facecentered cubic lattice. 
Share this thread: 
#1
Jan609, 09:46 PM

P: 176

1. The problem statement, all variables and given/known data
An element crystallises in a facecentered 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 secondnearest 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)(r²) = 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. 


#2
Jan709, 11:42 AM

P: 176

^Bump



#3
Jan709, 10:46 PM

P: 176

Any help at all? After reading some articles, I still can't work out this question.



#4
Jan809, 08:15 AM

HW Helper
P: 2,327

Facecentered cubic lattice.
Do you know the meaning of "nearest neighbor" and "second nearest neighbor"?
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. 


#5
Jan809, 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: a³ = (3.55 x 10^10)³ But how could I work out average volume of atom? 


#6
Jan809, 04:57 PM

HW Helper
P: 2,327

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


#7
Jan809, 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)³ / 4 = 11.8 x 10^30 m³ 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 m²) on the (1 1 1) planes: I tried working something out but Im not sure on it, any help for this ? Thanks! 


Register to reply 
Related Discussions  
Packing fraction of bodycentered cubic lattice  solid state physics  Advanced Physics Homework  3  
Shear deformation of diamond cubic lattice  Atomic, Solid State, Comp. Physics  0  
Cubic Lattice Atom Diameter Calculation  Biology, Chemistry & Other Homework  1  
Ratio of Energy in 3D simple cubic lattice  Advanced Physics Homework  0  
Calculate the lattice constant of a bodycentered cubic iron  Advanced Physics Homework  1 