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: 57

Answer

5200 years.

Work Step by Step

$N_{0}=16.1/min \cdot g\,C$ $N= 8.6/min \cdot g\,C$ Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.2094\times10^{-4}/y$ $\ln(\frac{N_{0}}{N})=kt$ $\implies \ln(\frac{16.1}{8.6})=0.627=$$(1.2094\times10^{-4}/y)\times t$ $\implies t=\frac{0.627}{1.2094\times10^{-4}\,y^{-1}}=5200\,y$
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