Materials Science and Engineering: An Introduction

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

Chapter 2 - Atomic Structure and Interatomic Bonding - Questions and Problems - Page 48: 2.3

Answer

$65.396$ amu

Work Step by Step

We simply take a weighted average: $$\sum_{n=1}^{N} w_{n}m_{n} = M$$ where $w_{n}$ is the atomic average and $m_{n}$ is the atomic weight of a specific zinc isotope, $n$. $M$ is the final average weight that we wish to calculate. $N$ is the total number of isotopes; in this case, it is 5 $.4863(63.929)+.2790(65.926)+.0410(66.927)+.1875(67.925)+.0062(69.925)= 65.396$ The units above are in amu.
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.