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) !

More later...

 

[charng]

it is important to understand the concepts and principles behind the phenomena we are studying and to be able to prove them through rigorous experimentation and analysis. In the case of the hexagonal close packing c/a ratio, there are several steps and considerations that need to be taken into account in order to prove that the ratio is indeed 1.633.

Firstly, it is important to understand the concept of hexagonal close packing and how it relates to the arrangement of atoms in a crystal lattice. This arrangement involves stacking layers of atoms in a hexagonal pattern, with each layer being offset from the previous one. This creates a structure with a c/a ratio, which is the ratio of the height (c) to the length of the sides (a) of the unit cell.

To prove the c/a ratio of 1.633, one cannot simply use a right-angled triangle and the pythagorean theorem as suggested in the link provided. This approach does not take into account the arrangement of atoms in the hexagonal lattice and will not yield an accurate result.

Instead, a more rigorous approach would involve calculating the number of atoms in a unit cell, taking into account the arrangement of atoms in each layer. This can be done using the hard sphere model, which takes into account the size and arrangement of atoms within a crystal lattice.

Furthermore, it is important to note that the unit cell contains 3 layers of atoms, with each layer having a different arrangement of atoms. The centre layer, as shown in figure (a), has 3 full atoms, while the top and bottom layers have 6 half atoms each. This accounts for the 18 atoms in a unit cell, as shown in the hard sphere model.

In conclusion, proving the hexagonal close packing c/a ratio of 1.633 requires a thorough understanding of the concept and a rigorous approach to calculating the number of atoms in a unit cell. It is important to consider the arrangement of atoms in the lattice and to use appropriate models and equations to arrive at an accurate result. I hope this helps clarify any doubts or confusion.