Chemistry (7th Edition)

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

Chapter 19 - Nuclear Chemistry - Section Problems - Page 837: 57

Answer

63 years.

Work Step by Step

Number of moles $n=\frac{\text{mass in grams}}{\text{molar mass}}=\frac{1.0\times10^{-9}\,g}{44\,g/mol}=2.2727\times10^{-11}\,mol$ Number of particles $N=n\times\text{Avogadro number}$ $=2.2727\times10^{-11}\times6.022\times10^{23}=1.3686\times10^{13}$ Decay rate= $4.8\times10^{3}\,dps=kN=k\times1.3686\times10^{13}$ $\implies$ Rate constant $ k=\frac{4.8\times10^{3}/s}{1.3686\times10^{13}}=3.507\times10^{-10}\,s^{-1}$ $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{3.507\times10^{-10}\,s^{-1}}=1.976\times10^{9}\,s$ $=1.976\times10^{9}\,s\times\frac{1\,h}{3600\,s}\times\frac{1\,d}{24\,h}\times\frac{1\,y}{365\,d}=63\,y$
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