Answer
$V_{c}=4.3\times10^{-23}\frac{cm^3}{unit cell}$
$\rho=2.07\frac{g}{cm^3}$
Work Step by Step
Given Beryllium (Be) has an HCP unit cell and the radius of the Be atom is 0.1143 nm.
$\frac{c}{a}=1.568$
$c=3.13R$
$V_{c}=6 R^2 c\sqrt{2}$
Since HCP has a total of 6 atoms, we see that A=9.01.
So, the total volume of sphere is:
$V_{c}=6 R^2 c\sqrt{2}= 6\times 3.13\times(0.1143\times10^{-7})^3=4.3\times10^{-23}\frac{cm^3}{unit cell}$
$V_{c}=4.3\times10^{-23}\frac{cm^3}{unit cell}$
The density of the cell is:
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
$\rho=\frac{6\times9.01}{{4.3\times10^{-23}}\times{6.023\times10^{23}}}=2.07\frac{g}{cm^3}$
