What is the Lattice Parameter c of an HCP Crystal Structure?

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The discussion focuses on determining the lattice parameter, c, of the hexagonal close-packed (HCP) crystal structure. The user attempts to relate c to the atomic radius, r, using geometric principles involving a triangular pyramid. They initially calculate c as 2*sqrt(3)*r but find it incorrect. Another participant suggests using the Pythagorean theorem to derive c based on the height of the triangular pyramid and the relationship between the sides of the equilateral triangle. The conversation emphasizes the geometric relationships essential for accurately calculating the lattice parameter in HCP structures.
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Homework Statement


I'm trying to figure out the lattice parameter, c, of the HCP crystal structure.

Here are a couple links showing the structure: http://images.google.com/imgres?img...agonal+close+packed&hl=en&safe=off&sa=N&um=1"

Homework Equations


I know that a=2r, where r is the atomic radius of whatever atoms make up the structure. I'm not really sure what else is relevant, it seems to be a problem of geometry and I'm having trouble working it out.

The Attempt at a Solution


I can't really type out my math that I've attempted to do since I am having trouble conceptually. I think that the way to go about doing it is with a triangular pyramid, since with the way that the atoms sit on top of each other you will get an equilateral triangle as your base (sides=a) and then the height is c/2. However I can't seem to get this to work out mathematically. I got c to be 2*sqrt(3)*r, which is incorrect.

Thanks for the help.

Homework Statement


Homework Equations


The Attempt at a Solution

 
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If the height of a triangular piramid is what you seek then a simple method is pytagora with a being the distance between 2 corners.

that is the height of a equilateral triangle is (3^0.5)/2 which means that the pyramid triangle will have a base of 2/3 of the height. That is (3^0.5)/3 so:

Pytagora : a^2(1-3/9)=a^2*c^2 where c=(6/9)^0.5=0.8164
 
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c/a must > 1, and a ~ 2r.
 
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