# Let us say that height of HCP is h then what is the distance of tetrahedral voids [closed]

In a hexaoal cose-packedstructue et $h$ be the spacing between equivalent basal planes. Then what are the "heights" of the tetrahedral voids lying between any two adjacent, equivalent basal planes, as distances from either plane?

The book gives the answer $h/8$ and $(7h)/8$. How is this derived? My (the original asker's) expectation, based on how I understand the hcp structure, was that the heights would be $h/4$ and $(3h)/4$.

• @andselisk it might mean the base of a single unit cell Jan 12 '18 at 14:22
• What is the base of a single unit cell then? This is not a crystallographic term by any means and carries no information. I suppose it's an attempt to denote $xy$ plane. Jan 12 '18 at 14:26
• Colleagues, please refrain from close-voting. This question deserves an answer which should be broader than the question itself. In particular, it should explain why the given answer of h/8 and 7h/8 is wrong, and in what sense it may be considered right (sort of). Jan 12 '18 at 15:10
• This is a great reason why we should be allowed to proactively vote to prevent closure. I got my answer in but intend to vote to reopen if the q gets closed. Jan 12 '18 at 17:26
• I down-voted this. Even though there might be people who understand your cryptic wording, it is hard to tell that we are actually talking about a crystal structure, or whatever. As such I think this question is utterly useless except for yourself.In a general sense I think it is unclear. An if you start a question with 'the answer is' - then why are you even asking? If you are asking whether a book is wrong include the source. Jan 13 '18 at 6:19

Actually $h/8$ and $(7h)/8$ are correct ... but not the whole truth. TL, DR the correct set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$.
Here is how to get $h/8$ and $(7h)/8$:
Think of the unit cell as consisting of a "base" having parts of seven atoms, then three middle atoms entirely inside the cell, then the top layer having parts of seven more atoms. Define $h$ as the height from the middle of the base atoms to the middle of the top atoms. Then each of the three atoms at height $h/2$ above the base is the common vertex of two tetrahedea. One has its base at the base of the larger cell, and since the center of the tetrahedron is one fourth of the way from the base to the apex it's height is $(h/2)/4=h/8$. The second of the two tetrahedea is upside down, with its base at the top of the cell, and the height of its center is $h-(h/8)=(7h)/8$.
Now for the curveball: these cannot be all solutions. We forgot to include the possibility of voids involving atoms from surrounding cells! If we include these surrounding atoms we discover that there is a layer of tetrahedral voids at $h/8$ above the mid-level, matching those at $h/8$ above the base, and another layer at $h/8$ below the midlevel matching those just below the top. These additional voids are hard to "see" because they involve atoms in surrounding cells. Nevertheless they are there, the complete solution set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$, and an equal number of tetrahedral voids (three per cell) lie at each of these heights.