Step 1:

$\overline{){{\mathbf{t}}}_{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}{\mathbf{}}{\mathbf{=}}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{k}}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{k}\mathbf{=}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{{\mathbf{t}}_{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}}\phantom{\rule{0ex}{0ex}}\mathbf{k}\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{693}\mathbf{}}{\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{\times}{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{yr}}$

**k = 5.4579x10 ^{-10} yr^{-1}**

Step 2:

$\overline{){\mathbf{ln}}{\mathbf{}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{t}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{\mathbf{kt}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{ln}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{0}}}}\phantom{\rule{0ex}{0ex}}\mathbf{ln}\mathbf{}\frac{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{t}}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{kt}$

Potassium-40 decays to argon-40 with a half-life of 1.27 x 10^{9} yr.

You may want to reference (Pages 913 - 916)Section 21.4 while completing this problem.

What is the age of a rock in which the mass ratio of ^{40}Ar to ^{40}K is 4.2?

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