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  • Two radioisotopes, A and B are initally present in the ratio of 1: 4, The half lives of A an B are respectively 100 years and 50 years. The time after which the amounts of A and B become equal.

Two radioisotopes, A and B are initally present in the ratio of 1: 4, The half lives of A an B are respectively 100 years and 50 years. The time after which the amounts of A and B become equal.

Option: 1

t=400 \text { years }


Option: 2

t=100 \text { years }


Option: 3

t=225 \text { years }


Option: 4

t=200 \text { years }


Answers (1)

best_answer

\mathrm{we \ have, N= Noe -\lambda t }
\mathrm{\begin{aligned} & N_A=N_0 e^{-\lambda_A t_A} \\ & N_B=N_0 e^{-\lambda_B t_B} \end{aligned} }
let t be the time after which  \mathrm{ N_A=N_B }
\mathrm{ \begin{gathered} t_A=t_B=t \text { (say) }, N_0 e^{-\lambda_A t}=4 N_0 e^{-\lambda_B t} \\ \therefore \quad N_0 e^{-\lambda_A t}=\left(\lambda_B t-\lambda_A t\right) \log _C e \Rightarrow \log _C 4 \\ 2 \log _e \frac{2}{=}\left[\frac{\log _e^2}{T_{B 1 / 2}}-\frac{\log _e^2}{T_{A / 2}}\right] t \quad \because \lambda=\frac{\log _e^2}{T} \\ 2=\left(\frac{1}{50}-\frac{1}{100}\right) t \Rightarrow 2=\left(\frac{2-1}{100}\right) t \end{gathered} }
Now,
\mathrm{ \begin{aligned} & 2=\frac{1}{100} \cdot t \\ & t=200 \text { year } \end{aligned} } Ans.

Posted by

Ritika Harsh

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