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: 60

Answer

0.00236

Work Step by Step

$\ln (\frac{A}{A_{0}})=-kt=-\frac{0.693}{t_{1/2}}\times t$ $=-\frac{0.693}{5730\,y}\times50000\, y=-6.04712$ $\implies \frac{A}{A_{0}}=e^{-6.04712}=0.00236$
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