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(C) The rate of disintegration $R$ is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms remain

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$\sqrt{2} : 1$

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(C) The rate of disintegration $R$ is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms remaining at time $t$.Given $T_{1/2, A} = 2 \ hr$ and $T_{1/2, B} = 4 \ hr$. The decay constants are $\lambda_A = \frac{\ln 2}{2}$ and $\lambda_B = \frac{\ln 2}{4}$.Initially,$N_A(0) = N_B(0) = N_0$.At $t = 2 \ hr$,the number of atoms remaining is $N(t) = N_0 e^{-\lambda t} = N_0 (1/2)^{t/T_{1/2}}$.For source $A$: $N_A(2) = N_0 (1/2)^{2/2} = N_0/2$.For source $B$: $N_B(2) = N_0 (1/2)^{2/4} = N_0/\sqrt{2}$.The rates of disintegration are $R_A = \lambda_A N_A$ and $R_B = \lambda_B N_B$.Ratio $\frac{R_A}{R_B} = \frac{\lambda_A N_A}{\lambda_B N_B} = \frac{(\ln 2 / 2) 	imes (N_0 / 2)}{(\ln 2 / 4) 	imes (N_0 / \sqrt{2})} = \frac{1/4}{1/(4\sqrt{2})} = \sqrt{2}$.Thus,the ratio is $\sqrt{2} : 1$.

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 $2 \ hr$,their rates of disintegration are in the ratio:

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