Hexagonal Closest Packed Unit Cell Height

[Sculptured]

My attempts to derive the height as a function of radius of the spheres packing the unit cell have failed. The best attempt so far was to create a diagonal line from the top atom to one of the middle ones. With that distance I thought I could use trig to find the component parallel to the height and multiply by two. Of course that didn't work. Any hints or suggestions? Even good 3d visualizations would help.

 

[Gokul43201]

The trick is in realizing that the middle plane of atoms occupy positions directly above the centroids of the triangles in the base plane. This follows directly from a symmetry argument.

#1. Each basal plane has nearest neighbor atoms making equilateral triangles. So, a=2R (where R is the sphere radius).

#2. Each atom at height c/2 above the basal plane is positioned directly above the centroid of the triangles in the base plane. For an equilatreral triangle, the distance from a vertex to the centroid is two-thirds the length of the median.

#3. Each atom in the base plane has a nearest neighbor in this middle plane. So, the distance from the corner atom in the base plane to the nearby atom in the mid-plane is 2R.

#4. Finally, use Pythagoras to write (2R)^2 = (c/2)^2 + (dist. calculated in #2)^2. That should take you to the required ratio.

 

[bsrishu]

I tried to find it with the help of packing efficiency (i.e. 0.74). the number of contributing spheres inside one unit cell is 6. therefore volume of 6 spheres/volume of hexagonal unit cell=0.74
volume of hexagonal unit cell is 6*sqrt(3)*r*r*h
then u will get answer of "h".