Two radioactive elements R and S disintegrate | vedclass

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(C) Let $N_R(t)$ and $N_S(t)$ be the number of atoms of $R$ and $S$ at time $t$.Given initial ratio $\frac{N_{R0}}{N_{S0}} = \frac{2}{1}$.The decay la

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$1:1$

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(C) Let $N_R(t)$ and $N_S(t)$ be the number of atoms of $R$ and $S$ at time $t$.Given initial ratio $\frac{N_{R0}}{N_{S0}} = \frac{2}{1}$.The decay law is $N(t) = N_0 e^{-\lambda t}$.Time $t$ is equal to three half-lives of $R$,so $t = 3 	imes T_{1/2, R} = 3 	imes \frac{\ln 2}{\lambda_R}$.Thus,$\lambda_R t = 3 \ln 2$.For element $S$,the exponent is $\lambda_S t = \lambda_S 	imes \frac{3 \ln 2}{\lambda_R} = 3 \ln 2 	imes \frac{3 	imes 10^{-3}}{4.5 	imes 10^{-3}} = 3 \ln 2 	imes \frac{2}{3} = 2 \ln 2$.Now,the ratio of atoms at time $t$ is:$\frac{N_R(t)}{N_S(t)} = \frac{N_{R0} e^{-\lambda_R t}}{N_{S0} e^{-\lambda_S t}} = \frac{N_{R0}}{N_{S0}} 	imes e^{-(\lambda_R t - \lambda_S t)}$$\frac{N_R(t)}{N_S(t)} = 2 	imes e^{-(3 \ln 2 - 2 \ln 2)} = 2 	imes e^{-\ln 2} = 2 	imes \frac{1}{2} = 1$.Therefore,the ratio is $1:1$.

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