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
Question
For the parallel radioactive decay: <br> A $\xrightarrow{\lambda_1 = 0.05 \text{ min}^{-1}}$ B + 40 MeV <br> A $\xrightarrow{\lambda_2 = 0.15 \text{ min}^{-1}}$ C + 80 MeV <br> the average energy released per atom decay of 'A' is
Answer: B
💡 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.
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