Explanation

(c) 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,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N_{K} = \ N_{0}e^{- \lambda t}$ $\ \ \ \ \ \ \ \ldots.(i)$
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
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N_{Ar}\ = \ N_{0} - N_{K}\ \ \ \ \ \ldots(ii)$
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$\ t$
$t\ = \frac{T_{1/2}ln(1 + N_{Ar}/N_{K})}{In\ 2}$
$= \frac{\left( 1.25\ \times 10^{9}y \right))\lbrack\ln(1 + 10.3)\rbrack}{\ln 2} = 4.37\ \times 10^{9}\ yr.$