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: 35b


12.7 days

Work Step by Step

If Original activity $N_{0}=100$, then present activity $N=10.0$. Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{3.82\,day}=0.1814/day$ $\ln(\frac{N_{0}}{N})=kt$ where $t$ is the time required. $\implies \ln(\frac{100}{10.0})=2.30=(0.1814/day)\times t$ Or $t= \frac{2.30}{0.1814/day}= 12.7\,days$
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