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: 34


34.8 days

Work Step by Step

If Original activity $N_{0}=100$, then present activity $N=5.0$. Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{8.04\,day}=0.086194/day$ $\ln(\frac{N_{0}}{N})=kt$ where $t$ is the time required. $\implies \ln(\frac{100}{5.0})=2.9957=(0.086194/day)\times t$ Or $t= \frac{2.9957}{0.086194/day}= 34.8\,days$
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