Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 19 - Nuclear Chemistry - Worked Example - Page 819: 9

Answer

44.5 days

Work Step by Step

Recall that $\ln(\frac{R_{0}}{R})=kt=\frac{0.693}{t_{1/2}}t$ $\implies \ln(\frac{16800}{10860})=0.43629=\frac{0.693}{t_{1/2}}\times28.0\,d$ $\implies t_{1/2}=\frac{0.693\times28.0\,d}{0.43629}=44.5\,d$
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