K^40 consists 0.012% of the potassium in nature. The human body comprises 0.35% potassium by weight. — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Mass of Potassium in Human Body $$m_\text{body} = 75\ \text{kg} = 75000\ \text{g}$$ $$m_{\ce{K}} = 75000 \times \frac{0.35}{100} = 262.5\ \text{g}$$ Step 2 - Mass of $\ce{^{40}K}$ $$m_{\ce{^{40}K}} = 262.5 \times \frac{0.012}{100} = 0.0315\ \text{g}$$ Step 3 - Number of $\ce{^{40}K}$ Nuclei $$N = \frac{0.0315\ \text{g}}{40\ \text{g/mol}} \times 6.022 \times 10^{23}\ \text{mol}^{-1}$$ $$N = 7.875 \times 10^{-4} \times 6.022 \times 10^{23} = 4.7423 \times 10^{20}\ \text{nuclei}$$ Step 4 - Convert Half-Life to Seconds $$t_{1/2} = 1.3 \times 10^9\ \text{yr} \times 365.25 \times 24 \times 3600\ \frac{\text{s}}{\text{yr}}$$ $$t_{1/2} = 4.102 \times 10^{16}\ \text{s}$$ Step 5 - Decay Constant $$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69315}{4.102 \times 10^{16}} = 1.6898 \times 10^{-17}\ \text{s}^{-1}$$ Step 6 - Activity $$A = \lambda N = (1.6898 \times 10^{-17}) \times (4.7423 \times 10^{20})$$ $$A \approx \boxed{8017.64\ \text{dps}}$$ Step 7 - Evaluate Options - **(A) 8017.64 dps**: Matches calculation. **Correct.** - **(B) 4008.82 dps**: Exactly half — corresponds to halving the K abundance or K-40 percentage. Incorrect. - **(C) 16035.28 dps**: Exactly double. Incorrect. - **(D) 345.24 dpm**: Wrong unit (dpm) and wrong value. Incorrect. $$\boxed{\text{Answer: A — 8017.64 dps}}$$