## Materials Science and Engineering: An Introduction

$V_{c}=1.06\times{10}^{-28}{m^3}$
Given Atomic weight=47.87g\mol $\rho=4.51$ and $\rho=\frac{nA}{{V_{c}}{N_{A}}}$ For HCP, n=6. $N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avogadro's number $4.51=\frac{6\times47.87}{(V_{c})\times6.023\times10^{23}}$ $V_{c}=1.06\times{10}^{-28}{m^3}$