At time t = 0,N_1 nuclei of decay constant la | vedclass

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(C) The decay rate of a radioactive sample is given by the law of radioactive decay,which states that the activity $R = \lambda N$. For the first samp

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$N_1 \lambda_1 e^{-\lambda_1 t} + N_2 \lambda_2 e^{-\lambda_2 t}$

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(C) The decay rate of a radioactive sample is given by the law of radioactive decay,which states that the activity $R = \lambda N$. For the first sample,the number of nuclei at time $t$ is $N_1(t) = N_1 e^{-\lambda_1 t}$. Thus,its decay rate is $R_1 = \lambda_1 N_1(t) = \lambda_1 N_1 e^{-\lambda_1 t}$. For the second sample,the number of nuclei at time $t$ is $N_2(t) = N_2 e^{-\lambda_2 t}$. Thus,its decay rate is $R_2 = \lambda_2 N_2(t) = \lambda_2 N_2 e^{-\lambda_2 t}$. Since the samples are mixed,the total decay rate of the mixture is the sum of the individual decay rates: $R_{total} = R_1 + R_2 = N_1 \lambda_1 e^{-\lambda_1 t} + N_2 \lambda_2 e^{-\lambda_2 t}$.

At time $t = 0$,$N_1$ nuclei of decay constant $\lambda_1$ and $N_2$ nuclei of decay constant $\lambda_2$ are mixed. The decay rate of the mixture is:

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