What will be the activity after 48 h of the overall excreted material till that time? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Let $N_0$ be initial atoms. The activity of the excreted material accounts for atoms excreted at rate $\lambda_b N(t)$ which subsequently decay radioactively with constant $\lambda_r$. Integrating this gives the activity outside as $A_{out}(T) = A_0 e^{-\lambda_r T}(1 - e^{-\lambda_b T})$. Given $T=48\text{ h}$, radioactive half-lives elapsed = $48/8 = 6$, and biological half-lives elapsed = $48/24 = 2$. $A_{out}(48) = 32 \times (1/2)^6 \times [1 - (1/2)^2] = 32 \times (1/64) \times (3/4) = 0.375 \mu\text{Ci}$. Therefore, correct answer is B.