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A radioactive material of half-life T was kept in a nuclear reactor at two different instants. The quantity kept second time was twice of that kept first time. If now their present activities are $A_{1}$ and $A_{2}$ respectively then their age difference equals

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$\frac{T}{\ln 2} \ln \frac{2 A_{1}}{A_{2}}$

$T \ln \frac{A_{1}}{A_{2}}$

$\frac{T}{\ln 2} \ln \frac{A_{2}}{2 A_{1}}$

$T \ln \frac{A_{2}}{2 A_{1}}$

answer is C.

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Detailed Solution

Activity $A \propto$ Number of atoms

$A_{1} = {A_{o^{e}}}^{- λ t_{1}}$

$∴ t_{1} = \frac{1}{λ} \ln (\frac{A_{0}}{A_{1}}) = \frac{T}{\ln 2} \ln \frac{A_{0}}{A_{1}}$

$A_{2} = 2 A_{0} e^{- λ t_{2}}$

$t_{2} = \frac{T}{\ln 2} \ln (\frac{2 A_{0}}{A_{2}})$

$t_{1} - t_{2} = \frac{T}{\ln 2} (\frac{A_{0}}{A_{1}} \times \frac{A_{2}}{2 A_{0}}) = \frac{T}{\ln 2} \ln (\frac{A_{2}}{2 A_{1}})$

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