Half-lives of two radioactive elements \(A\) and \(B\) are 30 minute and 60 minute respectively. Initially the samples have equal number of nuclei. After 120 minute the ratio of decayed numbers of nuclei of \(B\) to that of \(A\) will be

For radioactive decay, amount of sample remaining is given by,
\(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^\lambda\)
\(N_o\) is initial sample and \(\lambda\) is decay constant.
Decay constant, \(\lambda=\left(\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}\right)=\frac{120}{\mathrm{t}_{1 / 2}}\)
... (given)
For element A,
\(\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\frac{120}{30}}=\frac{\mathrm{N}_{\mathrm{o}}}{2^4}\)
...[From(i)]
Amount of sample A decayed,
\(N_A^{\prime}=N_o-N_A=N_o-\frac{N_0}{2^4}=\frac{15}{16} N_0...(i)\)
For element B,
\(\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\frac{120}{60}}=\frac{\mathrm{N}_0}{2^2}..[From(i)]\)
Amount of sample B decayed,
\(\mathrm{N}_{\mathrm{B}}{ }^{\prime}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}-\frac{\mathrm{N}_{\mathrm{o}}}{2^2}=\frac{3}{4} \mathrm{~N}_{\mathrm{O}}...(iii)\)
From (ii) and (iii),