The half-life of ^ 40 K is 1.30 × 10^ 9 y . A 1.00 g sample of pure...

The decay constant $\lambda$ of ^{40}\text{K}$is given by:<br><br>$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{1.30 \times 10^{9} \times 365 \times 24 \times 3600 \text{ s}} = 1.69 \times 10^{-17} \text{ s}^{-1}$<br><br>The activity$A$is$160 \text{ counts s}^{-1}$. The number of ^{40}\text{K}$ atoms in the sample is:

$N_{40} = \dfrac{A}{\lambda} = \dfrac{160}{1.69 \times 10^{-17}} = 9.47 \times 10^{18} \text{ atoms}$

The molar mass of $\text{KCl}$ is approximately $39.1 + 35.5 = 74.6 \text{ g mol}^{-1}$. The total number of potassium atoms in $1.00 \text{ g}$ of $\text{KCl}$ is:

$N_{\text{K}} = \dfrac{1.00}{74.6} \times 6.022 \times 10^{23} = 8.07 \times 10^{21} \text{ atoms}$

The relative abundance of ^{40}\text{K}$in natural potassium is the ratio of the number of ^{40}\text{K}$ atoms to the total number of $\text{K}$ atoms:

$\text{Relative abundance} = \dfrac{N_{40}}{N_{\text{K}}} \times 100\% = \dfrac{9.47 \times 10^{18}}{8.07 \times 10^{21}} \times 100\% = 0.117\%$

Rounding off, we get $0.12\%$.