Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 22 - Nuclear Chemistry - Exercises - Rates of Decay - Page 885: 55

Answer

12.3 years.

Work Step by Step

If original activity $N_{0}=100$ Then, present activity $N=100-5.5=94.5$ Time $t=1\,year$ $\ln(\frac{N_{0}}{N})=kt$ where $k$ is the rate constant. $\implies \ln(\frac{100}{94.5})=0.05657=k\times1\,y$ Or $k= \frac{0.05657}{1\,y}=0.05657\,y^{-1}$ $t_{1/2}=\frac{0.693}{0.05657\,y^{-1}}=12.3\,y$
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