Half-lives of two radioactive elements $A$ and $B$ are 30 minute and 60 minute respectively. Initially the…
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
$1: 15$
$1: 4$
$4: 5$
$5: 4$
Solution
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),