Question

Mass spectrometric analysis of potassium and argon atoms in a Moon rock sample shows that the ratio of the number of (stable) ${ }^{40} Ar$ atoms present to the number of (radioactive) ${ }^{40} K$ atoms is $10.3$. Assume that all the argon atoms were produced by the decay of potassium atoms, with a half-life of $1.25 \times 10^{9} yr$. How old is the rock?

• A. $2.95 \times 10^{11}\,yr$
• B. $2.95 \times 10^{9}\,yr$
• C. $4.37 \times 10^{9}\,yr$
• D. $4.37 \times 10^{11}\,yr$

Answer 1

Answer: (C)

$4.37 \times 10^{9}\,yr$

Answer 2

If $N_{0}$ potassium atoms were present at the time the rock was formed by solidification from a molten form, the number of potassium atoms remaining at the time of analysis is, $N_{ K }=N_{0} e^{-\lambda t}$ ...(i) in which $t$ is the age of the rock. For every potassium atom that decays, an argon atom is produced. Thus, the number of argon atoms present at the line of the analysis is $N_{ Ar }=N_{0}-N_{ K }$ ...(ii) We cannot measure $N_{0}$, so let's eliminate it from Eqs. (i) and (ii). We find, after some algebra, that $\lambda t=\ln \left(1+\frac{N_{ Ar }}{N_{ K }}\right),$ in which $N_{ Ar } / N_{ K }$ can be measured. Solving for $t$ $t=\frac{T_{1 / 2} \ln \left(1+N_{ Ar } / N_{ K }\right)}{\ln 2}$ $=\frac{\left(1.25 \times 10^{9} y\right)[\ln (1+10.3)]}{\ln 2}$ $=4.37 \times 10^{9} y$