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(C) The activity of a radioactive sample is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.For sample

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$20$ days and $5$ days

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(C) The activity of a radioactive sample is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.For sample $A$: $R_A = \lambda_A N_A = 10\, mCi$ ... $(1)$For sample $B$: $R_B = \lambda_B N_B = 20\, mCi$ ... $(2)$Given $N_A = 2 N_B$ ... $(3)$Dividing $(1)$ by $(2)$: $\frac{\lambda_A N_A}{\lambda_B N_B} = \frac{10}{20} = \frac{1}{2}$.Substituting $N_A = 2 N_B$: $\frac{\lambda_A (2 N_B)}{\lambda_B N_B} = \frac{1}{2} \implies 2 \frac{\lambda_A}{\lambda_B} = \frac{1}{2} \implies \frac{\lambda_A}{\lambda_B} = \frac{1}{4}$.Since half-life $T_{1/2} = \frac{\ln 2}{\lambda}$,we have $\frac{T_{1/2, A}}{T_{1/2, B}} = \frac{\lambda_B}{\lambda_A} = 4$.This means $T_{1/2, A} = 4 	imes T_{1/2, B}$.Checking the options,for option $B$: $T_{1/2, A} = 10$ days and $T_{1/2, B} = 2.5$ days (not listed). Wait,checking the ratio $T_{1/2, A} / T_{1/2, B} = 4$:Option $B$: $10/40 = 1/4$ (Incorrect).Option $C$: $20/5 = 4$ (Correct).Thus,the correct choice is $C$.

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