Question

The half-life of ⁴⁰K is 1.30 × 10⁹ y. A sample of 1.00 g of pure KCI gives 160 counts s⁻¹. Calculate the relative abundance of ⁴⁰K (fraction of ⁴⁰K present) in natural potassium.

Answer 1

Given:
Half-life period of ⁴⁰K, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = 1.30 × 10⁹ years
Count given by 1 g of pure KCI, A = 160 counts/s

Disintegration constant, $\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$
Now, activity, A = λN

$$\begin{gathered} \Rightarrow 160=\frac{0.693}{t_{1/2}}\times N \\ \Rightarrow 160=\left(\frac{0.693}{1.30\times {10}^9\times 365\times 86400}\right)\times N \\ \Rightarrow N=\frac{160\times 1.30\times 365\times 86400\times {10}^9}{0.693} \\ \Rightarrow N=9.5\times {10}^{18} \end{gathered}$$
6.023 × 10²³ atoms are present in 40 gm.

$$\begin{gathered} \mathrm{Thus,}\mathrm{9.5\; \times \; 1018}\mathrm{atoms}\mathrm{will}\mathrm{be}\mathrm{present}in\frac{40\times 9.5\times {10}^{18}}{6.023\times {10}^{23}}\mathrm{gm}. \\ =\frac{4\times 9.5\times {10}^{-4}}{6.023}\mathrm{gm} \\ =6.309\times {10}^{-4} \\ =0.00063\mathrm{gm} \end{gathered}$$
Relative abundance of ⁴⁰K in natural potassium = (2 × 0.00063 × 100)% = 0.12%