Answer
$a=0.32$nm,$c=0.52$nm
Work Step by Step
Atomic weight=24.3g\mol, $\rho=1.74$, and for HCP, r=$\frac{a}{2}$.
$V_{c}=6r^2c\sqrt{3}$
where $\frac{c}{a}=1.624$
$V_{c}=6\times3.266{\sqrt{3}}r^3$
$a=(\frac{V_{c}}{3.7})^{\frac{1}{3}}$=0.32nm
$c=1.624\times0.32=0.52nm$
