# Radioactive decay of ""^40K atoms in an igneous rock has resulted in a ratio of 25 percent ""^40K atoms to 75 percent ""^40Ar and ""^40Ca atoms. How many years old is this rock?

May 31, 2017

2.5 x ${10}^{9}$ yrs old

#### Explanation:

Given K-40 remaining = 25% of original K-40 @ 100%
${t}_{\frac{1}{2}} = 1.251 x {10}^{9} y r s$ => $k = \frac{0.693}{t} _ \left(\frac{1}{2}\right)$
$k = \frac{0.693}{1.251 x {10}^{9}} y r {s}^{-} 1 = 5.54 x {10}^{-} 10 y r {s}^{-} 1$

From C_("final") = C_("initial")e^-"k.t"; t = time of decay

Solving for time of decay (t) = ln(C_("final")/C_"initial")/-k

$t = \left(\ln \frac{\frac{25}{100}}{-} \left(5.54 x {10}^{-} 10\right)\right)$ years old = 2.5 x ${10}^{9}$ yrs old