A radioactive sample S_1 having an activity 5 | vedclass

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(A) The activity $A$ of a radioactive sample is given by $A = \lambda N = \frac{\ln 2}{T_{1/2}} N$,where $N$ is the number of nuclei and $T_{1/2}$ is

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$20$ years and $5$ years,respectively

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(A) The activity $A$ of a radioactive sample is given by $A = \lambda N = \frac{\ln 2}{T_{1/2}} N$,where $N$ is the number of nuclei and $T_{1/2}$ is the half-life.For sample $S_1$: $A_1 = 5 \mu Ci$ and $N_1 = 2N_0$.Thus,$5 = \frac{\ln 2}{T_1} (2N_0) \implies \frac{\ln 2}{T_1} = \frac{2.5}{N_0}$.For sample $S_2$: $A_2 = 10 \mu Ci$ and $N_2 = N_0$.Thus,$10 = \frac{\ln 2}{T_2} (N_0) \implies \frac{\ln 2}{T_2} = \frac{10}{N_0}$.Dividing the two expressions: $\frac{T_2}{T_1} = \frac{2.5}{10} = \frac{1}{4}$.Therefore,$T_1 = 4T_2$.Checking the options,if $T_2 = 5$ years,then $T_1 = 20$ years. This matches option $A$.

$A$ radioactive sample $S_1$ having an activity $5 \mu Ci$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10 \mu Ci$. The half-lives of $S_1$ and $S_2$ can be

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