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(D) The activity $A$ is given by $A = \lambda N = \frac{0.693}{T_{1/2}} N$.Since the initial number of atoms $N_0$ for $A$ and $B$ are equal,the initi

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$150 	ext{ dis/min}$

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(D) The activity $A$ is given by $A = \lambda N = \frac{0.693}{T_{1/2}} N$.Since the initial number of atoms $N_0$ for $A$ and $B$ are equal,the initial activity $A_0$ is inversely proportional to the half-life $T_{1/2}$.Therefore,$\frac{A_0(A)}{A_0(B)} = \frac{T_{1/2}(B)}{T_{1/2}(A)} = \frac{2 	ext{ days}}{1 	ext{ day}} = 2$.Given $A_0(A) + A_0(B) = 1200 	ext{ dis/min}$.Substituting $A_0(A) = 2 A_0(B)$,we get $2 A_0(B) + A_0(B) = 1200$,which implies $3 A_0(B) = 1200$,so $A_0(B) = 400 	ext{ dis/min}$ and $A_0(A) = 800 	ext{ dis/min}$.After $t = 4 	ext{ days}$,the activity of $A$ is $A(A) = \frac{A_0(A)}{2^{t/T_{1/2}(A)}} = \frac{800}{2^{4/1}} = \frac{800}{16} = 50 	ext{ dis/min}$.The activity of $B$ is $A(B) = \frac{A_0(B)}{2^{t/T_{1/2}(B)}} = \frac{400}{2^{4/2}} = \frac{400}{4} = 100 	ext{ dis/min}$.The total activity after $4 	ext{ days}$ is $50 + 100 = 150 	ext{ dis/min}$.

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