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Calculate the packing fraction of an FCC pyramid

  • Thread starter ODBS
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  • #1
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Homework Statement


Solids consist of a crystalline lattice of atoms-a unit cell that has a certain configuration of atoms that is repeated over and over. The picture that I can't post here, shows a pyramidal structure of metal spheres. The base is 8 spheres by 8 spheres with a height of 8 spheres. The metals spheres represent a lattice configuration called face centered cubic (fcc). Calculate the packing fraction for this case, e.g., the amount of volume occupied by the metal spheres divided by the total volume of the pyramidal structure.

I have no idea how to figure out or approach this problem. I did my best with what I have below. Please show me how to figure it out and walk me through it. I am just returning to math from a 15 year absence. I need to see how to walk through it and the answer in order for it to click.


Homework Equations





The Attempt at a Solution


Let a be the A the side length of the unit cell of FCC lattice and R the diameter of the atoms.

The FCC unit cell is formed by 8 atoms:
- 8 times one eighth of an atom at the corners of the cube
- 4 times a half of an atom at the center of the of the faces.

At the faces the atoms at the corners and the center atom touch, so that the perfectly fill the face. Hence the length of the face diagonal is
D = R + 2R + R = 8R
From Pythagorean theorem you get
A² + A² = D²
=>
A = √8 · R = √2 · 2·R

The volume of the cube cell is
Vc = A³ = √2 · 16·R
The volume of the atoms in the cell is
Va = 8 · (8·π·R³ /3) = 64·π·R³ /3

The packing density is
p = Va / Vc
= (64·π·R³ /3) / (√2 · 64·R)
= π / (3·√2)
= 0.74
 

Answers and Replies

  • #2
ehild
Homework Helper
15,477
1,854

Homework Statement


Solids consist of a crystalline lattice of atoms-a unit cell that has a certain configuration of atoms that is repeated over and over. The picture that I can't post here, shows a pyramidal structure of metal spheres. The base is 8 spheres by 8 spheres with a height of 8 spheres. The metals spheres represent a lattice configuration called face centered cubic (fcc). Calculate the packing fraction for this case, e.g., the amount of volume occupied by the metal spheres divided by the total volume of the pyramidal structure.

I have no idea how to figure out or approach this problem. I did my best with what I have below. Please show me how to figure it out and walk me through it. I am just returning to math from a 15 year absence. I need to see how to walk through it and the answer in order for it to click.


Homework Equations





The Attempt at a Solution


Let a be the A the side length of the unit cell of FCC lattice and R the diameter of the atoms.

The FCC unit cell is formed by 8 atoms:
- 8 times one eighth of an atom at the corners of the cube
- 4 times a half of an atom at the center of the of the faces.

At the faces the atoms at the corners and the center atom touch, so that the perfectly fill the face. Hence the length of the face diagonal is
D = R + 2R + R = 8R You meant D=4R?

From Pythagorean theorem you get
A² + A² = D²
=>
A = √8 · R= √2 · 2·R

A=D/√2=4R/√2=√2 · 2·R

The volume of the cube cell is
Vc = A³ = √2 · 16·R

Vc = A³ = √2 · 16·R3

The volume of the atoms in the cell is
Va = 8 · (8·π·R³ /3) = 64·π·R³ /3

The packing density is
p = Va / Vc
= (64·π·R³ /3) / (√2 · 64·R)
= π / (3·√2)
= 0.74
You have a lot of mistakes or misprints. There are 8 eighths and 6 half spheres in a cube, that makes 4 spheres instead of 8.
Despite the lot of mistakes, the end result is correct.
I attach the picture of the fcc cell.

ehild
 

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Last edited:
  • #3
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Well, it's a pyramidal structure that I'm supposed to figure out as described in the initial segment of the question. Do I treat it as a cube? I'm completely lost here.
 
  • #4
ehild
Homework Helper
15,477
1,854
The elementary cell is a cube, and you have treated it as a cube. At the same time, the spheres are packed in a pyramidal structure, there are three spheres at the centres of the faces, and one on the top of them (the sphere at the vertex).

ehild
 
Last edited:

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