Max Packing Fraction of Hexagonal Unit Cell Volume w/ Hard Spheres

[malawi_glenn]

Homework Statement

Calculate the maximum packing fraction of the unit cell volume that can be filled by hard spheres in the Hexagonal structure

Relevant eq: Volume of spheres is number of lattice points multiplied with the maximum volume of one sphere.

The Attempt at a Solution

I know maxium is obtained when c = a i.e when height of the cell is as high as one of the sides in the hexagon. Hence, maximum sphere radius is a/2 (I have shown geometrically that the spheres can touch each other).

Now I am to determine the number of lattice points in this structure, I know that one primitive cell contains totally one lattice point, and a unit cell of a hexagonal structure can be made up by exactly three primitve cells, so the number of lattice points is 3. Is that the correct way to do this?

The rest I can figure out by my self, just are unsure how to determine the number of lattice points.

 

[anathema]

Thank you for your question. You are correct in your approach to determining the maximum packing fraction of hard spheres in a hexagonal structure. The maximum packing fraction can be calculated by dividing the total volume of the spheres by the volume of the unit cell.

In this case, since the spheres are arranged in a hexagonal structure, the volume of the unit cell can be calculated using the formula V = √3a^2c/2, where a is the length of one side of the hexagon and c is the height of the unit cell. As you correctly stated, the maximum packing fraction is achieved when c = a, resulting in a maximum sphere radius of a/2.

To determine the number of lattice points in the unit cell, you are correct in noting that there is one lattice point per primitive cell. In a hexagonal structure, the unit cell is composed of three primitive cells, resulting in a total of three lattice points. Therefore, the maximum packing fraction for a hexagonal structure of hard spheres is 0.74, obtained by dividing the volume of three spheres (3π(a/2)^3) by the volume of the unit cell (√3a^2c/2).

I hope this helps clarify your approach. Please let me know if you have any further questions. Keep up the good work!

 

[m2003]

I would like to first clarify the definition of "packing fraction." The packing fraction of a unit cell is the ratio of the volume occupied by the spherical particles to the total volume of the unit cell. It is a measure of how efficiently the particles are packed within the unit cell.

Based on this definition, the maximum packing fraction of a hexagonal unit cell with hard spheres can be calculated by considering the closest packing arrangement of spheres within the unit cell. This arrangement is known as the hexagonal close-packed (HCP) structure.

In the HCP structure, the spheres are arranged in layers, with each layer offset from the previous one. The spheres in one layer are surrounded by six spheres in the layer below, forming a hexagonal pattern. This arrangement results in a packing fraction of approximately 0.74, which is the maximum achievable packing fraction for hard spheres in a hexagonal unit cell.

To determine the number of lattice points in this structure, we can use the fact that each sphere in the HCP structure is located at the center of a lattice point. Therefore, the number of lattice points is equal to the number of spheres in the unit cell, which is 6. This can also be confirmed by considering the three primitive cells that make up the hexagonal unit cell, with each primitive cell containing one lattice point.

In summary, the maximum packing fraction of a hexagonal unit cell with hard spheres is 0.74, and the number of lattice points in the unit cell is 6.