Proving Hexagonal Close Packing c/a Ratio is 1.633

[murasame]

Hi all,

I know this has been posted before and I've read the help given but I simply do not understand how to prove Hexagonal Close packing c/a ratio is 1.633.

I referred to this link:
http://www.engr.ku.edu/~rhale/ae510/lecture2/sld013.htm

and have some doubts that require clarificaiton.

1) Why can't I just use one vertical plane of the hexagonal structure, use the diagonal (4 R) and base length (2R) as two sides of a right-angled triangle and subsequently use pythagorean theorem to get the length of c which is the vertical length?

Because if I do, I'll get (4r)^2 = (2r)^2 + c^2

And eventually i get c = (sqrt12)/r and a = 2r and hence the c/a ratio is 1.73 and not 1.633.

2) It can be seen that a unit cell contains 3 layers of atoms, and the centre layer has 3 full atoms , as shown in figure (a).

However, the hard sphere model shows 6 half atoms, instead of 3 full atoms; which simply just doesn't look like figure (a)!

I'm sure I'm missing something, but I'm not sure what it is. Can someone pls englighten me?

Any help would be greatly appreciated, thanks.

 

[Gokul43201]

1) Why can't I just use one vertical plane of the hexagonal structure, use the diagonal (4 R) and base length (2R) as two sides of a right-angled triangle and subsequently use pythagorean theorem to get the length of c which is the vertical length?

Because if I do, I'll get (4r)^2 = (2r)^2 + c^2

And eventually i get c = (sqrt12)/r and a = 2r and hence the c/a ratio is 1.73 and not 1.633.

There does not exist a vertical plane in the hexagonal structure that has 3 neighboring atoms (from which you seem to have got 4R) in a diagonal as well as 2 neighboring atoms in the base (i.e, the same horizontal line).

Edit: The only planes in the hex crystal that have such a combination of atoms would be any of the basal (horizontal) planes. In the basal plane, the distance between parallel lines will be 1.732 times the lattice parameter. In other words, you have only calculated the distance between opposite sides of a regular hexagon, which being twice the height of an equilateral triangle, is naturally a*sqrt(3) !

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