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Two radioacti...

Question

Two radioactive samples $A$ and $B$ have half lives $T_{1} a n d T_{2} (T_{1} > T_{2})$ respectively. At $t = 0$ , the activity of $B$ 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}}{T_{1} + T_{2}}

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Solution

The correct option is A $\frac{T_{1} T_{2}}{T_{1} - T_{2}}$
$\begin{matrix} Let, Rate constants of A and B are K_{1} and K_{2} And, Initial Number of Nuclei are {(N_{0})}_{1} and {(N_{0})}_{2} Given T_{1} and T_{2} are half lifes \begin{matrix} T_{1} = \frac{ln 2}{k_{1}} & and T_{2} = \frac{ln 2}{k_{2}} N_{1} = (N_{0})_{1} e^{- k_{1} t} & N_{1} = K_{1} {(N_{0})}_{1} e^{- k_{1} t} N_{2} = k_{2} {(N_{0})}_{2} e^{- k_{2} t} \end{matrix} \end{matrix}$

$\begin{matrix} At t = 0 Activity of B was twice of A 2 (K_{1} {{N_{0})}_{1} e^{- K_{1} (0)}) = k_{2} {(N_{0})}_{2} e^{- K_{2} (0)} 2 K_{1} {(N_{0})}_{1} = K_{2} {(N_{0})}_{2} - (1) Required time when Activities are equal k_{1} (N_{0}), e^{- K_{1} t} = K_{2} {(N_{8})}_{2} e^{- K_{2} t} \end{matrix}$

$\begin{matrix} From (1) e^{- k_{1} t} = 2 e^{- k_{2} t} e^{(k_{2} - k_{1}) t} = 2 (k_{2} - k_{1}) t = ln 2 (\frac{ln 2}{T_{2}} - \frac{ln 2}{T_{1}}) t = ln 2 t = \frac{T_{1} T_{2}}{T_{1} - T_{2}} \end{matrix}$

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