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: 2.4

Answer

Fraction of $m_{In^{113}}=.0430$ Fraction of $m_{In^{115}}=.9570$

Work Step by Step

The fraction of naturally occurring $In^{113}$ will be denoted by $f$. The fraction of naturally occurring $In^{115}$ will be denoted by $t$. Since there are only two naturally occurring isotopes, $$f+t=1$$ Hence, $t=1-f$. We calculate the weighted average via $$fm_{In^{113}}+(1-f)m_{In^{115}}=114.818$$ See problem 2.3 for the weighted average formula. Rearranging, we have $$f=\frac{114.818-m_{In^{115}}}{m_{In^{113}}-m_{In^{115}}}$$ Plugging in $m_{In^{113}}=112.904$ and $m_{In^{115}}=114.904$, we find $f=.0430$ and, hence, $t=1-f=.9570$
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.