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(A) Let the decay constant of $n_1$ be $\lambda_1 = \lambda$. Then the decay constant of $n_2$ is $\lambda_2 = 3\lambda$.The number of nuclei remainin

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$1/4$

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(A) Let the decay constant of $n_1$ be $\lambda_1 = \lambda$. Then the decay constant of $n_2$ is $\lambda_2 = 3\lambda$.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For $n_1$: $N_1(t) = N_0 e^{-\lambda t}$.For $n_2$: $N_2(t) = N_0 e^{-3\lambda t}$.The time $t$ is one half-life of $n_1$,so $t = T_{1/2} = \frac{\ln 2}{\lambda}$.Substituting $t$ into the expressions:$N_1(t) = N_0 e^{-\lambda (\frac{\ln 2}{\lambda})} = N_0 e^{-\ln 2} = N_0 / 2$.$N_2(t) = N_0 e^{-3\lambda (\frac{\ln 2}{\lambda})} = N_0 e^{-3 \ln 2} = N_0 (e^{\ln 2})^{-3} = N_0 (2)^{-3} = N_0 / 8$.The ratio $\frac{N_2(t)}{N_1(t)} = \frac{N_0 / 8}{N_0 / 2} = \frac{2}{8} = \frac{1}{4}$.

$A$ radioactive nucleus $n_2$ has $3$ times the decay constant as compared to the decay constant of another radioactive nucleus $n_1$. If initial number of both nuclei are the same,what is the ratio of number of nuclei of $n_2$ to the number of nuclei of $n_1$,after one half-life of $n_1$?

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