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) 4$^{\circ}$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

The Correct Option is C

If$N_{o}$ 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_0e^{-\lambda t}\hspace15mm ...(i)$ in which tis 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 time of the analysis is $\hspace15mm N_{Ar}=N_0-N_K\hspace15mm ...(ii)$ We cannot measure NO, so Iet's eliminate it from Eqs. (i) and (ii). We find, after some algebra, that $ \lambda t=In\Bigg(1+\frac{N_{Ar}}{N_K}\Bigg)$ In which $N_{Ar}/ N_K$ can be measured. Solving for t $=\frac{\left(1.25\times10^{9}\right)\left[ln\left(1+10.3\right)\right]}{ln 2} =4.37\times10^{9}yr$