Two radioactive nuclei A and B both convert i | vedclass

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(B) The rate of disintegration is given by $R = \lambda N$,where $\lambda = \frac{\ln 2}{T_{1/2}}$.For nucleus $A$: $T_{1/2, A} = 1 \, min$,so $\lambd

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$6$

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(B) The rate of disintegration is given by $R = \lambda N$,where $\lambda = \frac{\ln 2}{T_{1/2}}$.For nucleus $A$: $T_{1/2, A} = 1 \, min$,so $\lambda_A = \frac{\ln 2}{1} = \ln 2 \, min^{-1}$.For nucleus $B$: $T_{1/2, B} = 2 \, min$,so $\lambda_B = \frac{\ln 2}{2} \, min^{-1}$.The number of nuclei at time $t$ is $N_A(t) = 4N_0 e^{-\lambda_A t}$ and $N_B(t) = N_0 e^{-\lambda_B t}$.The rates of disintegration are $R_A = \lambda_A N_A = \lambda_A (4N_0 e^{-\lambda_A t})$ and $R_B = \lambda_B N_B = \lambda_B (N_0 e^{-\lambda_B t})$.Setting $R_A = R_B$:$4 \lambda_A N_0 e^{-\lambda_A t} = \lambda_B N_0 e^{-\lambda_B t}$$4 (\ln 2) e^{-(\ln 2) t} = \frac{\ln 2}{2} e^{-(\frac{\ln 2}{2}) t}$$8 = \frac{e^{-(\frac{\ln 2}{2}) t}}{e^{-(\ln 2) t}} = e^{(\ln 2 - \frac{\ln 2}{2}) t} = e^{(\frac{\ln 2}{2}) t}$Taking the natural logarithm on both sides:$\ln 8 = \frac{\ln 2}{2} t$$3 \ln 2 = \frac{\ln 2}{2} t$$t = 6 \, min$.

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