$X_{40} X_{2}^{240} K$ has a total half-life of $1.25 \times 1 0^{9}$ years, and only $10.7%$ of its decays produce $X_{40} X_{2}^{240} Ar$ (the rest give $X_{40} X_{2}^{240} Ca$ ). A mineral had no argon at formation and now has the mole ratio $X_{40} X_{2}^{240} Ar : X_{40} X_{2}^{240} K = 0.0310$ . Use $ln 2 = 0.693$ . The age of the mineral, in units of $1 0^{9}$ years, to two decimal places, is

Numerical answer type, answer in x 10^9 years, tolerance 0.02.

Answer: 0.46 x 10^9 years

Let

K_{0}

be the original K-40. Atoms decayed

= K_{0} - K

, and

Ar = 0.107 (K_{0} - K)

. So

Ar / K = 0.107 (K_{0} / K - 1)

, giving

K_{0} / K = 1 + 0.0310/0.107 = 1.2897

λ = 0.693/1.25 = 0.5544 (1 0^{9} y)^{- 1}

, so

t = ln (1.2897) / λ = 0.46 \times 1 0^{9}

years.