Question

An analysis of a rock shows that relative number of ${}^{87}Sr$ and ${}^{87}Rb$ atoms is $0.052$, i.e., $({}^{87}Sr/{}^{87}Rb=0.052)$. Determine the age of the rock. Given that half life period for $\beta$-decay of $Rb$ to $Sr$ is $4.7\times {10}^{10}$ years.

Answer 1

$\frac{{}^{87}Rb}{{}^{87}Sr}=\frac{x}{y}=\frac{1}{0.052}$
$\frac{x}{x+y}=\frac{1}{1.052}$
$\frac{N}{N_0}=\frac{1}{1.052}$
$\lambda =\frac{2.303}{t}{\log }_{10}\frac{N_0}{N}$
$\frac{0.693}{4.7\times {10}^{10}}=\frac{2.303}{t}{\log }_{10}1.052$
$t=3.43\times {10}^9$ years