The t_1/2 of Pb^212 is 8.0 h. It undergoes decay to its daughter (unstable) element Bi^212 of half-l — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Setup the Consecutive Decay (Bateman Equation) The decay chain is: $\ce{^{212}_{82}Pb} \xrightarrow{\lambda_A} \ce{^{212}_{83}Bi} \xrightarrow{\lambda_B} \text{stable}$ The daughter Bi-212 reaches maximum activity when its rate of formation equals its rate of decay, i.e., when $\frac{dN_B}{dt} = 0$. At maximum, the Bateman condition gives: $$t_{\max} = \frac{\ln(\lambda_B / \lambda_A)}{\lambda_B - \lambda_A}$$ Step 2 - Convert Half-Lives to Decay Constants (in minutes) $$t_{1/2}(\text{Pb}^{212}) = 8.0\ \text{h} = 480\ \text{min}$$ $$\lambda_A = \frac{\ln 2}{480} = \frac{0.6931}{480} = 1.444 \times 10^{-3}\ \text{min}^{-1}$$ $$t_{1/2}(\text{Bi}^{212}) = 60.0\ \text{min}$$ $$\lambda_B = \frac{\ln 2}{60.0} = \frac{0.6931}{60.0} = 1.155 \times 10^{-2}\ \text{min}^{-1}$$ Step 3 - Calculate $t_{\max}$ $$t_{\max} = \frac{\ln\!\left(\dfrac{\lambda_B}{\lambda_A}\right)}{\lambda_B - \lambda_A} = \frac{\ln\!\left(\dfrac{1.155 \times 10^{-2}}{1.444 \times 10^{-3}}\right)}{(1.155 - 0.1444) \times 10^{-2}}$$ $$= \frac{\ln(8.0)}{1.011 \times 10^{-2}} = \frac{2.079}{1.011 \times 10^{-2}} \approx \boxed{205.7\ \text{min}}$$ Step 4 - Evaluate Options - **(A) 205.7 min**: Matches our calculation. **Correct.** - **(B) 3.429 min**: Incorrect; this would arise from inverting the ratio. - **(C) 60.0 min**: This is just the half-life of Bi-212, not the time of maximum activity. - **(D) 67.5 min**: Incorrect numerical result. $$\boxed{\text{Answer: A — 205.7 min}}$$