A 0.50 g sample of rock was found to have 2.5 × 10^-6 mol of ₁₉K⁴⁰ (t_1/2 = 1.3 × 10⁹ yr) and 7.5 × 10^-6 mol of ₂₀Ca⁴⁰ How old is the rock?

${ }_{19}{\mathrm{K}}^{40}{⟶}_{20}{\mathrm{Ca}}^{40}{+}_{-1}{\mathrm{e}}^0$

At present 2.5 × 10^-6 7.5 × 10^-6 mole

mole of ₁₉K⁴⁰ consumed = 7.5 × 10^-6 mole initial mole of K⁴⁰ was (7.5 + 2.5) × 10^-6 mole

$\begin{array}{l}\mathrm{t}=\frac{1}{\mathrm{K}}\ln \left(\frac{10\times {10}^{-6}}{2.5\times {10}^{-6}}\right)=\frac{1.3\times {10}^9}{\ln 2}\times \ln (2\\ =1.3\times {10}^9\times 2=2.6\times {10}^9\mathrm{yr}\end{array}$