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(B) Given decay constants are $\lambda_1 = 6\lambda$ and $\lambda_2 = \lambda$.The half-life of $R_2$ is $(t_{1/2})_2 = 1.4 	imes 10^{17} \,s$.The nu

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$4 	imes 10^{16} \,s$

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(B) Given decay constants are $\lambda_1 = 6\lambda$ and $\lambda_2 = \lambda$.The half-life of $R_2$ is $(t_{1/2})_2 = 1.4 	imes 10^{17} \,s$.The number of nuclei remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.For $R_1$: $N_1 = N_0 e^{-\lambda_1 t} = N_0 e^{-6\lambda t}$.For $R_2$: $N_2 = N_0 e^{-\lambda_2 t} = N_0 e^{-\lambda t}$.Given the ratio $\frac{N_2}{N_1} = e$.$\frac{N_0 e^{-\lambda t}}{N_0 e^{-6\lambda t}} = e$.$e^{-\lambda t + 6\lambda t} = e^1$.$e^{5\lambda t} = e^1$.Comparing the exponents: $5\lambda t = 1$, so $t = \frac{1}{5\lambda}$.We know $\lambda = \frac{\ln 2}{(t_{1/2})_2} = \frac{0.7}{1.4 	imes 10^{17}} = 0.5 	imes 10^{-17} \,s^{-1}$.Substituting $\lambda$ into the expression for $t$:$t = \frac{1}{5 	imes 0.5 	imes 10^{-17}} = \frac{1}{2.5 	imes 10^{-17}} = 0.4 	imes 10^{17} \,s = 4 	imes 10^{16} \,s$.

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