Answer
$a=0.296$nm,$c=0.468$nm
Work Step by Step
Given the atomic weight=47.87g\mol, $\rho=4.51$, and r=$\frac{a}{2}$ for HCP:
$V_{c}=6r^2c\sqrt{3}$
where $\frac{c}{a}=1.58$
$V_{c}=6\times3.266{\sqrt{3}}r^3$
$a=(\frac{V_{c}}{4.105})^{\frac{1}{3}}$=0.296nm
$c=1.58\times0.296=0.468nm$
