The atom $40 \mathrm {~K}$ decays radioactively to the gas $40 \mathrm { Ar }$. An ancient rock is found to contain enough $40 \mathrm { Ar }$ gas to indicate that $77.0 \%$ of the $40 \mathrm {~K}$ in the rock has decayed to $40 \mathrm { Ar }$ ince the rock solidified. (Any previous argon would have boiled out of liquid rock.) How long ago did the rock solidify? The half-life of $40 \mathrm {~K}$ is $1.25$ billion years.

Question

Quizplus · Accepted Answer

The decay of $40 \mathrm {~K}$ to $40 \mathrm { Ar }$ follows first-order kinetics, and the time elapsed can be calculated using the formula $t = \frac{\ln(\frac{N_0}{N})}{k}$, where $N_0$ is the initial amount of $40 \mathrm {~K}$, $N$ is the remaining amount of $40 \mathrm {~K}$, and $k$ is the decay constant. Given that $77.0\%$ of $40 \mathrm {~K}$ has decayed, $23.0\%$ remains. The half-life ($t_{1/2}$) of $40 \mathrm {~K}$ is $1.25$ billion years, and the decay constant $k$ can be calculated using $k = \frac{\ln(2)}{t_{1/2}}$. Substituting the given values and solving for $t$ yields an age of approximately $2.65$ billion years for the rock.

The Atom 40 K40 \mathrm {~K}40 K Decays Radioactively to the Gas 40Ar40 \mathrm { Ar }40Ar

The Atom $40 \mathrm {~K}$ Decays Radioactively to the Gas $40 \mathrm { Ar }$