For the parallel radioactive decay: <br> A B + 40 MeV <br> A C + 80 MeV <br> the average energy rele — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

The fraction decaying via path 1 is $\frac{\lambda_1}{\lambda_1 + \lambda_2} = \frac{0.05}{0.20} = 0.25$. Path 2 fraction is $\frac{0.15}{0.20} = 0.75$. Average energy = $(0.25 \times 40 \text{ MeV}) + (0.75 \times 80 \text{ MeV}) = 10 + 60 = 70 \text{ MeV}$. Therefore, correct answer is B.