Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

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



Work Step by Step

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 $4.51=\frac{6\times47.87}{(V_{c})\times6.023\times10^{23}}$ $V_{c}=1.06\times{10}^{-28}{m^3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.