## Materials Science and Engineering: An Introduction

$V_{c}=6.64\times10^{-23}\frac{cm^3}{unit cell}$
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: $V_{c}=6 R^2 c\sqrt{3}= 6\times 3.246\times(0.1253\times10^{-7})^3=6.64\times10^{-23}\frac{cm^3}{unit cell}$ $V_{c}=6.64\times10^{-23}\frac{cm^3}{unit cell}$