# Chapter 3 - The Structure of Crystalline Solids - Questions and Problems - Page 98: 3.12b

$a=0.296$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 $c=1.58\times0.296=0.468nm$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.