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(D) The initial activity is $A_0 = 2.4 	imes 10^5 	ext{ Bq}$.The half-life $T_{1/2} = 8 	ext{ days} = 8 	imes 24 = 192 	ext{ hours}$.The decay co

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(D) The initial activity is $A_0 = 2.4 	imes 10^5 	ext{ Bq}$.The half-life $T_{1/2} = 8 	ext{ days} = 8 	imes 24 = 192 	ext{ hours}$.The decay constant $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.7}{192} 	ext{ h}^{-1}$.After time $t = 11.5 	ext{ hours}$,the activity of the total blood volume $V$ is $A(t) = A_0 e^{-\lambda t}$.Using the approximation $e^{-x} \approx 1 - x$ for small $x$:$A(t) \approx A_0 (1 - \lambda t) = 2.4 	imes 10^5 	imes (1 - \frac{0.7 	imes 11.5}{192}) \approx 2.4 	imes 10^5 	imes (1 - 0.0419) \approx 2.4 	imes 10^5 	imes 0.9581 = 2.299 	imes 10^5 	ext{ Bq}$.The activity of $2.5 	ext{ ml}$ of blood is $115 	ext{ Bq}$.Therefore,the total activity $A(t)$ is related to the sample activity $A_s$ by $A(t) = A_s 	imes \frac{V}{V_s}$,where $V_s = 2.5 	ext{ ml}$.$V = \frac{A(t) 	imes V_s}{A_s} = \frac{2.299 	imes 10^5 	imes 2.5 	ext{ ml}}{115} \approx 4997.8 	ext{ ml} \approx 5 	ext{ liters}$.

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