A radioactive nucleus $\mathrm{n}_2$ 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 $n_1$ ?

<p>Let's work through the problem step by step.</p> <p><p>The decay of a radioactive nucleus is given by the formula: </p> <p>$N(t) = N_0 \, e^{-\lambda t},$ </p> <p>where:</p></p> <p><p>$N(t)$ is the number of nuclei remaining at time $t$,</p></p> <p><p>$N_0$ is the initial number of nuclei,</p></p> <p><p>$\lambda$ is the decay constant.</p></p> <p><p>For nucleus $n_1$ with decay constant $\lambda_1$, its half-life $t_{1/2}$ is: </p> <p>$t_{1/2} = \frac{\ln 2}{\lambda_1}.$</p></p> <p><p>After one half-life of $n_1$, the number of $n_1$ nuclei remaining is: </p> <p>$$N_1 = N_0 \, e^{-\lambda_1 \, t_{1/2}} = N_0 \, e^{-\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-\ln2} = \frac{N_0}{2}.$$</p></p> <p><p>Nucleus $n_2$ has a decay constant that is 3 times that of $n_1$: </p> <p>$\lambda_2 = 3\lambda_1.$</p></p> <p><p>For nucleus $n_2$, the number of nuclei remaining after one half-life of $n_1$ (which is $t = \frac{\ln2}{\lambda_1}$) is: </p> <p>$ N_2 = N_0 \, e^{-\lambda_2 \, t} = N_0\, e^{-3\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-3\ln2} = N_0 \cdot \frac{1}{2^3} = \frac{N_0}{8}. $</p></p> <p><p>Now, we find the ratio of $n_2$ nuclei to $n_1$ nuclei after this time: </p> <p>$ \frac{N_2}{N_1} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1/8}{1/2} = \frac{1}{4}. $</p></p> <p>Therefore, the correct ratio is $\frac{1}{4}$, which corresponds to Option A.</p>