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 935: 21.53



Work Step by Step

Half-life $t_{1/2}$=30.17 years Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{30.17\,y}=0.02297\,y^{-1}$ $t=100\,y$ 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 the sample after time $t$. $\implies \ln(\frac{A}{A_{0}})=-(0.02297\,y^{-1})(100\,y)=-2.297$ Taking the inverse $\ln$ of both the sides, we have $\frac{A}{A_{0}}=e^{-2.297}=0.10$
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