A radionuclide 'A' decays simultaneously into 'B' and 'C', by α- and β-emission, respectively. The h — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Effective Decay Constant for Parallel Decays When a nuclide $A$ decays simultaneously via two pathways: $$\lambda_\alpha = \frac{\ln 2}{20\ \text{min}}, \quad \lambda_\beta = \frac{\ln 2}{60\ \text{min}}$$ The **effective decay constant** is: $$\lambda_{\text{eff}} = \lambda_\alpha + \lambda_\beta = \frac{\ln 2}{20} + \frac{\ln 2}{60} = \ln 2 \left(\frac{3 + 1}{60}\right) = \frac{4\ln 2}{60}$$ **Effective half-life:** $$t_{1/2,\text{eff}} = \frac{\ln 2}{\lambda_{\text{eff}}} = \frac{60}{4} = 15\ \text{min}$$ Step 2 - Find Time for 87.5% Decay Fraction remaining = $1 - 0.875 = 0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3$ So **3 effective half-lives** have elapsed: $$t = 3 \times 15\ \text{min} = \boxed{45\ \text{min}}$$ Step 3 - Evaluate Options - **(A) 15 min**: Only one effective half-life → 50% decayed. Incorrect. - **(B) 30 min**: Two half-lives → 75% decayed. Incorrect. - **(C) 45 min**: Three half-lives → 87.5% decayed. **Correct.** - **(D) 60 min**: Four half-lives → 93.75% decayed. Incorrect. $$\boxed{\text{Answer: C — 45 minutes}}$$