Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 18 - Nuclear Chemistry - Questions for Review and Thought - Topical Questions - Page 817b: 29


0.0019 mg

Work Step by Step

The amount of radionuclide at the beginning is $A_{0}=0.56\,mg$. Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{44.5\,day}=0.01557/day$ Time $t=1 \,year=365\,days$ $\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining. $\implies \ln(\frac{0.56\,mg}{A})=0.015573\times365=5.684$ Taking the inverse $\ln$ of both sides, we have $\frac{0.56\,mg}{A}=e^{5.684}=294$ Or $A= \frac{0.56\,mg}{294}=0.0019\,mg $
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