A radioactive nucleus \(\mathrm{n}_2\) has 3 times the decay constant as compared to the decay constant of another radio…

The number of remaining nuclei after time t can be given by:

\(N(t)={N}_{0}{e}^{−\lambda t}\)

For half-life \({t}_{1/2}\) of \({n}_{1}\), we know that:

\({t}_{1/2}=\frac{\ln 2}{{\lambda }_{1}}\)

At the end of one half-life, the number of \({n}_{1}\) nuclei left will be:

\({N}_{1}=\frac{{N}_{0}}{2}\)

For nucleus \({n}_{2}\), which has a decay constant 3 times that of \({n}_{1}\) the remaining number of nuclei will be:

\({N}_{2}={N}_{0}{e}^{-3}{\lambda }_{1}t\)

Since \(t={t}_{1/2}=\frac{\ln 2}{{\lambda }_{1}}\), we substitute it into the equation for \({N}_{2}\) :

\({N}_{2}={N}_{0}{e}^{-3{\lambda }_{1}\times \frac{\ln 2}{{\lambda }_{1}}}={N}_{0}{e}^{-3\ln 2}={N}_{0}{\left(\frac{1}{2}\right)}^{3}=\frac{{N}_{0}}{8}\)

Now, the ratio of the number of nuclei of \({n}_{2}\) to \({n}_{1}\) after one half-life of \({n}_{1}\) is:

\(\frac{{N}_{2}}{{N}_{1}}=\frac{\frac{{N}_{0}}{8}}{\frac{{N}_{0}}{2}}=\frac{1}{4}\)