Chemistry: An Introduction to General, Organic, and Biological Chemistry (12th Edition)

Published by Prentice Hall
ISBN 10: 0321908449
ISBN 13: 978-0-32190-844-5

Chapter 5 - Section 5.6 - Nuclear Fission and Fusion - Understanding the Concepts - Page 162: 5.65c

Answer

9.0 days

Work Step by Step

Original amount $A_{0}=4.8\,mg$ Present amount $A=1.2\,mg$ Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\,d}=0.154\,d^{-1}$ $\ln(\frac{A_{0}}{A})=kt$ $\implies \ln(\frac{4.8}{1.2})=1.386=0.154\,d^{-1}(t)$ Then, $t=\frac{1.386}{0.154\,d^{-1}}=9.0\,d$
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