The mass percent of carbon in a typical human is \(18 \%\), and the mass percent of \({ }^{14} \mathrm{C}\) in natural carbon is \(1.6 \times 10^{-10} \%\). Assuming a \(180-\mathrm{lb}\) person, how many decay events per second occur in this person due exclusively to the \(\beta\) -particle decay of \({ }^{14} \mathrm{C}\) (for \({ }^{14} \mathrm{C}\), \(t_{1 / 2}=5730\) years)?

We are given that the mass percent of carbon in a typical human is 18%. A 180-pound person would have this mass in kg: \( 180 ~\text{lb} \cdot \frac{1 ~\text{kg}}{2.2046 ~\text{lb}} \approx 81.65 ~\text{kg}\) The mass of carbon in this person is: \( \text{mass of carbon} = 81.65 ~\text{kg} \cdot 18 \% = 14.697 ~ \text{kg} \)

We are given that the mass percent of Carbon-14 in natural carbon is 1.6x10^(-10)%. The mass of Carbon-14 in this person would be: \( \text{mass of Carbon-14} = 14.697 ~\text{kg} \cdot 1.6 \times 10^{-10} \% \approx 2.35152 \times 10^{-8} ~ \text{kg} \) Now we can find the number of Carbon-14 atoms by dividing the mass of Carbon-14 by its atomic mass: \( \text{Number of Carbon-14 atoms} = \frac{\text{mass of Carbon-14}}{\text{atomic mass of Carbon-14}} \) where atomic mass of Carbon-14 is approximately 14 atomic mass unit(AMU): \( 1 ~\text{AMU} \approx 1.66 \times 10^{-27} ~\text{kg} \) Hence, \( \text{Number of Carbon-14 atoms} = \frac{2.35152 \times 10^{-8} ~\text{kg}}{14 \cdot 1.66 \times 10^{-27} ~\text{kg}} \approx 5.09008 \times 10^{23} ~\text{atoms}\)

The decay constant (λ) can be calculated from the half-life (t_(1/2)) using the following formula: \( \lambda = \frac{\ln 2}{t_{1/2}} \) Given that the half-life of Carbon-14 is 5730 years, we first convert it to seconds: \( t_{1/2} = 5730 ~\text{years} \cdot \frac{365.25 ~\text{days}}{1 ~\text{year}} \cdot \frac{24 ~\text{hours}}{1 ~\text{day}} \cdot \frac{60 ~\text{mins}}{1 ~\text{hour}} \cdot \frac{60 ~\text{seconds}}{1 ~\text{min}} \approx 1.8075 \times 10^{11} ~ \text{seconds} \) Now, we can calculate the decay constant: \( \lambda \approx \frac{\ln 2}{1.8075 \times 10^{11} ~\text{seconds}} \approx 3.83465 \times 10^{-12} ~\text{s}^{-1}\)

The number of decay events per second (decays per second) can be calculated by multiplying the number of Carbon-14 atoms by the decay constant: \( \text{Decays per second} = \text{Number of Carbon-14 atoms} \cdot \lambda \) \( \text{Decays per second} \approx 5.09008 \times 10^{23} ~\text{atoms} \cdot 3.83465 \times 10^{-12} ~\text{s}^{-1} \approx 1.95 \times 10^{12} ~\text{decays/s}\) So, there are approximately \(1.95 \times 10^{12}\) decay events per second in a 180-pound person due exclusively to the beta-particle decay of Carbon-14.