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I have read that height of HCP arrangement is given by $\mathrm{4r\sqrt{\frac23}}$ as distance between consecutive layers is $\mathrm{2r\sqrt{\frac23}}$(in both FCC and HCP). If that is the case, then what is the height of FCC arrangement? Can it be $\mathrm{3r\sqrt{\frac23}}$? I highly doubt that FCC is an "abc" arrangement and not "ab" arrangement. (r=radius of sphere)

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My first instinct was to say that FCC has no height, because that's a cubic cell and it only has one parameter.

Then again, you didn't ask about the cell. The question was about the FCC arrangement, and as such, it makes perfect sense (if maybe in imperfect wording, as "height" is not quite the word to be used here). What is the "vertical" distance from a layer to the next layer with identical placement, really?

Well, it is as simple as that: like you said, HCP is a two-layer packing, hence its height is twice the distance between the layers. On the other hand, FCC (when thought of in terms of hexagonal layers) is a three-layer packing, hence its "height" is three times the distance between the layers. You also said that the distance between the adjacent layers is $2\sqrt{2\over3}r$, which is correct.

See where this is going?

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