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Half Life and Average Life

Two radioacti...

Question

Two radioactive sample A and B have half $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$ respectively. At $T = 0$ , the activity of a B was twice the activity of A. Their activity will become equal after a time :

A

\frac{T_{1} T_{2}}{T_{1} - T_{2}}

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B

\frac{T_{1} - T_{2}}{2}

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C

\frac{T_{1} + T_{2}}{2}

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D

\frac{T_{1} T_{2}}{12}

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Solution

The correct option is A $\frac{T_{1} T_{2}}{T_{1} - T_{2}}$
Consider initial activity of nuclei A and B be $A_{0}$ and $2 A_{0}$ respectively at a time $^{'} t^{'}$ their activity is given by:-
$A_{1} A_{0} e - λ t$ [for A ] $A_{2} = 2 A_{0} e - λ t$ [for B]
where $λ = \frac{l n 2}{T}$ [T=half -life]
When $A_{1} = A_{2}$ at time 't'
$= A_{0} e^{-} \frac{l n 2}{T_{1}} t = 2 A_{0} e^{-} \frac{l n 2}{T_{2}} t$ [ $λ_{1} = \frac{m_{2}}{T_{1}}$ and $λ_{2} = \frac{l n 2}{T_{2}}]$
Take log of both sides ,
$\frac{- m 2}{T_{1}} t = \frac{- m 2}{T_{2}} + l n 2$
$= \frac{- t}{T_{1}} = \frac{- t}{T_{2}} + 1$
$= t = \frac{T_{1} T_{2}}{T_{1} - T_{2}}$

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