Packing fraction of body-centered cubic lattice - solid state physics

[cameo_demon]

packing fraction of body-centered cubic lattice -- solid state physics

Homework Statement

This is part of a series of short questions (i.e. prove everything in Kittel Ch. 1, Table 2):

Prove that the packing fraction of a BCC (body-centered) cubic lattice is:

1/8 * pi * \sqrt{3}

Homework Equations

packing fraction = volume of a sphere / volume of primitive cell

The Attempt at a Solution

each lattice point (there are two total for BCC) can hold a sphere (or at least part of one) with radius a\sqrt{3} / 2. subbing in:

2 * 4/3 * pi * (a\sqrt{3} / 2) ^{3} / (a^{3}/2)

the \sqrt{3} becomes a 9. how does the \sqrt{3} possibly remain?

my question: how the heck did kittel get 1/8 * pi * \sqrt{3}?

 

[cherreno]

This does not solve your problem at all, but at least clarify how the \sqrt3 survives.
(\sqrt3)^3 = (\sqrt3)^2 * \sqrt3 = 3* \sqrt3.
Nevertheless, you can not get the answer by means of your assupmtion. The packing fraction is the maximum proportion of the available volume in the cell that can be filled with spheres... keep it in mind.

Indeed, the radius of such sphere is half of the nearest-neighbor distance. Try it!

 

[Steve Dutch]

It's basically geometry. Let the unit cell be a cube of side 1. The long diagonal has length sqrt(3). The corner atoms and the central atoms all have their centers on the long diagonal and touch, so their radius is sqrt(3)/4. So what's the volume of an atom, and how many atoms are in a unit cell (remember the corner atoms are shared with neighboring cells)?

 

[lopkiol]

Interesting. What happens if the spheres are allowed to have two different sizes? Would it be possible to exceed the FCC density (0.74)?

 

[paranormal]

The packing fraction of a BCC lattice can be derived by considering the volume occupied by each lattice point and the volume of the primitive cell. As mentioned in the attempt at a solution, each lattice point can hold a sphere with a radius of a√3/2, which is the distance from the center to the corner of the cube. The volume of this sphere is given by V = 4/3 * π * (a√3/2)^3.

Now, the primitive cell of a BCC lattice is a cube with edge length a, which has a volume of a^3. However, only half of each sphere is contained within the primitive cell. This means that the total volume of all the spheres in the primitive cell is 2 * (4/3 * π * (a√3/2)^3).

Therefore, the packing fraction is given by:

(2 * (4/3 * π * (a√3/2)^3)) / (a^3)

= (8/3 * π * a^3 * (√3/2)^3) / (a^3)

= (8/3 * π * (√3/2)^3)

= (8/3 * π * 3√3/8)

= (π * √3/4)

= 1/8 * π * √3

So, in conclusion, the packing fraction of a BCC lattice is indeed 1/8 * π * √3 as stated in Kittel Ch. 1, Table 2.