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(D) The rate of disintegration is given by $R = \lambda N$,where $\lambda = \frac{\ln 2}{T_{1/2}}$ and $N = N_0 e^{-\lambda t}$.For source $A$: $T_{1/

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$1 : 1$

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(D) The rate of disintegration is given by $R = \lambda N$,where $\lambda = \frac{\ln 2}{T_{1/2}}$ and $N = N_0 e^{-\lambda t}$.For source $A$: $T_{1/2, A} = 2 \ hr$,so $\lambda_A = \frac{\ln 2}{2}$.For source $B$: $T_{1/2, B} = 4 \ hr$,so $\lambda_B = \frac{\ln 2}{4}$.At $t = 8 \ hr$,the number of atoms remaining are:$N_A = N_0 e^{-\lambda_A t} = N_0 e^{-(\frac{\ln 2}{2}) 	imes 8} = N_0 e^{-4 \ln 2} = N_0 (2)^{-4} = \frac{N_0}{16}$.$N_B = N_0 e^{-\lambda_B t} = N_0 e^{-(\frac{\ln 2}{4}) 	imes 8} = N_0 e^{-2 \ln 2} = N_0 (2)^{-2} = \frac{N_0}{4}$.The rates of disintegration are:$R_A = \lambda_A N_A = (\frac{\ln 2}{2}) 	imes \frac{N_0}{16} = \frac{N_0 \ln 2}{32}$.$R_B = \lambda_B N_B = (\frac{\ln 2}{4}) 	imes \frac{N_0}{4} = \frac{N_0 \ln 2}{16}$.The ratio $R_A : R_B = \frac{N_0 \ln 2}{32} : \frac{N_0 \ln 2}{16} = \frac{1}{32} : \frac{1}{16} = 1 : 2$.

The radioactive sources $A$ and $B$ have half-lives of $2 \ hr$ and $4 \ hr$ respectively,and initially contain the same number of radioactive atoms. At the end of $8 \ hr$,what is the ratio of their rates of disintegration?

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