College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 29 - The Nucleus - Learning Path Questions and Exercises - Exercises - Page 998: 27

Answer

3.13%

Work Step by Step

Half-life of $\,\,^{14}C$=5730 years. Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.2094\times10^{-4}\,y^{-1}$ $t=28650\,y$ Recall that $\ln(\frac{A}{A_{0}})=-\lambda t$ where $A_{0}$ is the amount of carbon-14 at the beginning and $A$ is the amount after 28650 years. $\implies \ln(\frac{A}{A_{0}})=-(1.2094\times10^{-4}\,y^{-1})(28650\,y)=-3.465$ Taking the inverse $\ln$ of both the sides, we have $\frac{A}{A_{0}}=e^{-3.465}=0.0313$ Percentage of the sample remaining= $\frac{A}{A_{0}}\times100\%=0.0313\times100\%=3.13\%$
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