For the parallel radioactive decay: A --(lambda_1 = 0.05 min^-1)--> B + 40 MeV, A --(lambda_2 = 0.15 — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Parallel (Branched) Radioactive Decay Setup Nucleus A decays simultaneously via two independent channels: $$\text{A} \xrightarrow{\lambda_1 = 0.05\ \text{min}^{-1}} \text{B} + 40\ \text{MeV}$$ $$\text{A} \xrightarrow{\lambda_2 = 0.15\ \text{min}^{-1}} \text{C} + 80\ \text{MeV}$$ Step 2 - Weighted Average Energy Formula The fraction of atoms decaying via each channel is proportional to the decay constant for that channel: $$f_1 = \frac{\lambda_1}{\lambda_1 + \lambda_2}, \qquad f_2 = \frac{\lambda_2}{\lambda_1 + \lambda_2}$$ The average energy released per decay of A: $$E_\text{avg} = f_1 E_1 + f_2 E_2 = \frac{\lambda_1 E_1 + \lambda_2 E_2}{\lambda_1 + \lambda_2}$$ Step 3 - Substitute Values $$E_\text{avg} = \frac{(0.05)(40) + (0.15)(80)}{0.05 + 0.15}$$ Numerator: $$= \frac{2.0 + 12.0}{0.20} = \frac{14.0}{0.20}$$ $$E_\text{avg} = \boxed{70\ \text{MeV}}$$ Step 4 - Why Not the Other Options - **(A) 60 MeV**: Simple arithmetic mean $\frac{40+80}{2} = 60$. Incorrect — ignores that the two channels have different rates ($\lambda_1 \neq \lambda_2$). - **(B) 70 MeV**: Correct weighted average, giving more weight to the faster channel ($\lambda_2 = 0.15$) and its higher energy (80 MeV). **Correct.** - **(C) 66.67 MeV**: Incorrect algebraic manipulation. - **(D) 75 MeV**: Incorrect. $$\boxed{\text{Answer: B — }70\ \text{MeV}}$$