Answer
$R=0.160nm$
Work Step by Step
Given that Magnesium(Mg) has an HCP unit cell and that the Mg atomic density is $\rho= 1.74 \frac{g}{cm^3}$, it follows:
$\frac{c}{a}=1.624$
$c=3.246R$
$V_{c}=6 R^2 c\sqrt{3}$
As we know HCP has a total of 6 atoms:
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
$1.74=\frac{6\times24.31}{{12\sqrt{3}\times{(1.624\times10^{-7})R^3}}\times{6.023\times10^{23}}}$
$R=1.60\times 10^{-8}{cm}=0.160nm$
