A radioactive sample $S_{1}$ having an activity of $5 μ C i$ has twice the number of nuclei as another sample $S_{2}$ which has an activity of $10 μ C i$ . The half lives of $S_{1}$ and $S_{2}$ can be

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A radioactive sampleS1 having activity A1 has twice the number of nuclei as another sample S2 of activity A2. If A2=2A1. Then ratio of half life of S1 to the half life of S2 is:

The correct option is D 4A1=ln2T12N0A2=ln2T2N0Dividing the both Equations, we get14=T2T14=T1T2The Ratio of half lives is 4

A radioactive sample $S_{1}$ having activity $A 1$ has twice the number of nuclei as another sample $S_{2}$ of activity $A_{2}$ . If $A_{2} = 2 A_{1}$ . Then ratio of half life of $S_{1}$ to the half life of $S_{2}$ is:

The correct option is D $4$
$A_{1} = \frac{l n 2}{T_{1}} 2 N_{0}$

$A_{2} = \frac{l n 2}{T_{2}} N_{0}$

Dividing the both Equations, we get

$\frac{1}{4} = \frac{T_{2}}{T_{1}}$

$4 = \frac{T_{1}}{T_{2}}$

The Ratio of half lives is 4