Answer
$V_{c}=6.64\times10^{-23}\frac{cm^3}{unit cell}$
Work Step by Step
Given that Cobalt(Co) has an HCP unit cell and that the radius of the Co atom is 0.1253 nm:
$\frac{c}{a}=1.623$
$c=3.246R$
$V_{c}=6 R^2 c\sqrt{3}$
Since HCP has a total of 6 atoms:
$V_{c}=6 R^2 c\sqrt{3}= 6\times 3.246\times(0.1253\times10^{-7})^3=6.64\times10^{-23}\frac{cm^3}{unit cell}$
$V_{c}=6.64\times10^{-23}\frac{cm^3}{unit cell}$
