## Chemistry: The Molecular Science (5th Edition)

$3.6\times10^{9}$ years.
Recall that $\ln(\frac{original\, quantity}{present\, quantity})=\frac{0.693}{t_{1/2}}t$ $\implies \ln(\frac{1}{0.951})=\frac{0.693}{4.9\times10^{10}\,y}\times t$ $\implies 0.05024=(1.414\times10^{-11}\,y^{-1})\times t$ Or $t=\frac{0.05024}{1.414\times10^{-11}\,y^{-1}}=3.6\times10^{9}\,y$