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# Pack Man

The most densely-packed planes for FCC are along the diagonals of the faces of the cube. Thus if the side length is $a$, $\sqrt{2}a = 4\mbox{ radii} \therefore a = \frac{4}{\sqrt{2}}r$. So the fill is $4\mbox{ atoms}/a^3 = \frac{\sqrt{2}\pi}{6} = 74\%\;.$

For BCC, the most densely-packed plane is the inside diagonal, $\therefore \sqrt{3}a = 4\mbox{ radii} \therefore a = \frac{4}{\sqrt{2}} r$. So the fill is $2\mbox{ atoms}/a^3 = \frac{\sqrt{3}\pi}{8} = 68\%\;.$

If you find the FCC side length in terms of the atomic radius of copper, you will find that the answer is very close to the experimentally-found density of copper, $8920\textrm{ kg/m}^3.$

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The most densely-packed planes for FCC are along the diagonals of the faces of the cube. Thus if the side length is $a$, $\sqrt{2}a = 4\mbox{ radii} \therefore a = \frac{4}{\sqrt{2}}r$. So the fill is $4\mbox{ atoms}/a^3 = \frac{\sqrt{2}\pi}{6} = 74\%\;.$

For BCC, the most densely-packed plane is the inside diagonal, $\therefore \sqrt{3}a = 4\mbox{ radii} \therefore a = \frac{4}{\sqrt{2}} r$. So the fill is $2\mbox{ atoms}/a^3 = \frac{\sqrt{3}\pi}{8} = 68\%\;.$

If you find the FCC side length in terms of the atomic radius of copper, you will find that the answer is very close to the experimentally-found density of copper, $8920\textrm{ kg/m}^3.$