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Q.

The ratio of activities of two radionuclides X and Y in a mixture at time t = 0 was found to be 8 : 1. After two hour's the ratio of activities become 1 : 1. If the $t_{1 / 2}$ of radionuclide X is 20 min., find the $t_{1 / 2}$ [in minutes] of radionuclide Y.

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answer is 40.

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Detailed Solution

$\begin{array}{l} \frac{N_{x} λ_{1}}{N_{y} λ_{2}} = \frac{8}{1} \\ N_{x} λ_{1} e^{- λ_{1} t} = N_{x} λ_{2} e^{- λ_{2} t} \\ \ln 8 = (λ_{1} - λ_{2}) \cdot t \\ \frac{1}{120} \ln 8 = [\frac{\ln 2}{20} - \frac{\ln 2}{t_{1 / 2}}] \\ \Rightarrow \frac{\ln 2}{40} - \frac{\ln 2}{20} = - \frac{\ln 2}{t_{1 / 2}} \\ \Rightarrow t_{1 / 2} = 40 \end{array}$

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