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(C) Given the half-lives $T_1 = 1400 \, years$ and $T_2 = 700 \, years$.The decay constants are $\lambda_1 = \frac{\ln 2}{1400} \, year^{-1}$ and $\la

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$740$

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(C) Given the half-lives $T_1 = 1400 \, years$ and $T_2 = 700 \, years$.The decay constants are $\lambda_1 = \frac{\ln 2}{1400} \, year^{-1}$ and $\lambda_2 = \frac{\ln 2}{700} \, year^{-1}$.The net decay constant is $\lambda_{net} = \lambda_1 + \lambda_2 = \ln 2 \left( \frac{1}{1400} + \frac{1}{700} ight) = \ln 2 \left( \frac{1+2}{1400} ight) = \frac{3 \ln 2}{1400} \, year^{-1}$.Let the initial number of nuclei be $N_0$. We want to find the time $t$ when $N(t) = \frac{N_0}{3}$.Using the radioactive decay law $N(t) = N_0 e^{-\lambda_{net} t}$, we have $\frac{N_0}{3} = N_0 e^{-\lambda_{net} t}$.$\frac{1}{3} = e^{-\lambda_{net} t} \implies \ln(3) = \lambda_{net} t$.Substituting the values: $1.1 = \left( \frac{3 	imes 0.693}{1400} ight) t$.$t = \frac{1.1 	imes 1400}{3 	imes 0.693} \approx \frac{1540}{2.079} \approx 740.7 \, years$.Thus, the time is approximately $740 \, years$.

$A$ radioactive material decays by simultaneous emissions of two particles with half-lives of $1400 \, years$ and $700 \, years$ respectively. What will be the time after which one-third of the material remains? (Take $\ln 3 = 1.1$) (In $years$)

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