اگر یادتون باشه استاد سر کلاس کریستال  فاکتور اتمی چیده شدن شبکه ی هگزاگونال رو به عنوان تمرین داد و فکر کنم هیشکی نتونست حلش کنه.

For the hexagonal close-packed (HCP) structure the derivation is similar. The side length of the hexagon will be denoted as a while the height of the hexagon will be denoted as c. Then:

a = 2 \times r


c = (\sqrt{\frac{2}{3}})(4r).

It is then possible to calculate the APF as follows:

\mathrm{APF} = \frac{N_\mathrm{atoms} V_\mathrm{atom}}{V_\mathrm{crystal}}


= \frac{6 (4/3)\pi r^3}{[(3\sqrt{3})/2](a^2)(c)}


= \frac{6 (4/3)\pi r^3}{[(3\sqrt{3})/2](2r)^2(\sqrt{\frac{2}{3}})(4r)}


= \frac{6 (4/3)\pi r^3}{[(3\sqrt{3})/2](\sqrt{\frac{2}{3}})(16r^3)}
= \frac{\pi}{\sqrt{18}}


\approx 0.74.\,\!