Calculate the packing fraction of an FCC pyramid

[ODBS]

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

 

[ehild]

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

 

[ODBS]

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.

 

[ehild]

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

 

[Nickie]

Therefore, the packing fraction of an FCC pyramid is approximately 0.74 or 74%. This means that 74% of the total volume of the pyramidal structure is occupied by the metal spheres.

To calculate the packing fraction, we first need to determine the volume of the unit cell and the volume of the atoms within the cell. From the given information, we know that the unit cell has a side length of √2 · 2·R and contains 8 atoms.

The volume of the unit cell can be calculated using the formula V = A³, where A is the side length. So, Vc = (√2 · 2·R)³ = √2 · 8·R³.

Next, we need to calculate the volume of the atoms within the unit cell. Since the atoms are arranged in a face-centered cubic lattice, there are 8 atoms at the corners of the cube and 4 atoms at the center of each face. The atoms at the corners only contribute 1/8 of their volume to the unit cell, while the atoms at the center of the faces contribute 1/2 of their volume. So, the total volume of the atoms in the unit cell is Va = 8 · (8·π·R³ /3) = 64·π·R³ /3.

Finally, we can calculate the packing fraction by dividing the volume of the atoms by the volume of the unit cell. This gives us p = Va / Vc = (64·π·R³ /3) / (√2 · 8·R³) = π / (3·√2) = 0.74.

In conclusion, the packing fraction of an FCC pyramid is approximately 0.74 or 74%. This means that 74% of the total volume of the pyramidal structure is occupied by the metal spheres, while the remaining 26% is empty space.