At a given instant,say t = 0,two radioactive | vedclass

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(C) The activity of a radioactive substance at time $t$ is given by $R = R_0 e^{-\lambda t}$.Given that at $t = 0$,$R_{A,0} = R_{B,0} = R_0$.The ratio

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$\frac{\ln 2}{4}$

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(C) The activity of a radioactive substance at time $t$ is given by $R = R_0 e^{-\lambda t}$.Given that at $t = 0$,$R_{A,0} = R_{B,0} = R_0$.The ratio of activities at time $t$ is $\frac{R_B}{R_A} = \frac{R_0 e^{-\lambda_B t}}{R_0 e^{-\lambda_A t}} = e^{-(\lambda_B - \lambda_A)t}$.According to the problem,this ratio is $e^{-3t}$.Comparing the exponents,we get $\lambda_B - \lambda_A = 3$.The decay constant $\lambda$ is related to half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}}$.Given $T_{A,1/2} = \ln 2$,so $\lambda_A = \frac{\ln 2}{\ln 2} = 1$.Substituting $\lambda_A$ into the equation: $\lambda_B - 1 = 3 \Rightarrow \lambda_B = 4$.Since $\lambda_B = \frac{\ln 2}{T_{B,1/2}}$,we have $4 = \frac{\ln 2}{T_{B,1/2}}$.Therefore,$T_{B,1/2} = \frac{\ln 2}{4}$.

At a given instant,say $t = 0$,two radioactive substances $A$ and $B$ have equal activities. The ratio $\frac{R_B}{R_A}$ of their activities after time $t$ decays with time $t$ as $e^{-3t}$. If the half-life of $A$ is $\ln 2$,the half-life of $B$ is:

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