Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning
ISBN 10: 0534420125
ISBN 13: 978-0-53442-012-3

Chapter 21 - Nuclear Chemistry - Questions and Exercises - Exercises - Page 934: 21.42


0.757 or 75.7%

Work Step by Step

Half-life $t_{1/2}$=5730 years Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.2094\times10^{-4}\,y^{-1}$ $t=2300\,y$ Recall that $\ln(\frac{A}{A_{0}})=-k t$ where $A_{0}$ is the amount of sample at the beginning and $A$ is the amount of sample after time $t$. $\implies \ln(\frac{A}{A_{0}})=-(1.2094\times10^{-4}\,y^{-1})(2300\,y)=-0.278$ Taking the inverse $\ln$ of both the sides, we have $\frac{A}{A_{0}}=e^{-0.278}=0.757$
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