Answer
(a) Balanced equation:
$K_{19}^{40}->Ar_{18}^{40} + \beta_{1}^{0}$
(b) $\lambda = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{1.2*10^{9}} = 5.8*10^{-10} yr^{-1}$
$ln \frac{N_{t}}{N_{0}} = -\lambda t$
Hence, $ln\frac{0.18}{1} = -5.8*10^{-10} t$
Hence, $t = 3*10^{9} years$
Work Step by Step
(a) Balanced equation:
$K_{19}^{40}->Ar_{18}^{40} + \beta_{1}^{0}$
(b) $\lambda = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{1.2*10^{9}} = 5.8*10^{-10} yr^{-1}$
$ln \frac{N_{t}}{N_{0}} = -\lambda t$
Hence, $ln\frac{0.18}{1} = -5.8*10^{-10} t$
Hence, $t = 3*10^{9} years$
