- Home
- Standard 12
- Physics
At a given instant, say $t = 0,$ two radioactive substances $A$ and $B$ have equal activates. The ratio $\frac{{{R_B}}}{{{R_A}}}$ of their activities. The ratio $\frac{{{R_B}}}{{{R_A}}}$ of their activates after time $t$ itself decays with time $t$ as $e^{-3t}.$ If the half-life of $A$ is $ln2,$ the half-life of $B$ is
$4\,ln2$
$\frac {ln2}{2}$
$\frac {ln2}{4}$
$2\,ln2$
Solution
${\text{R}} = {{\text{R}}_{\text{o}}}{{\text{e}}^{ – \lambda {\text{t}}}}$
$\therefore$ $\frac{{{\text{R}}_{\text{B}}}}{{{\text{R}}_{\text{A}}}}=\frac{{{\text{R}}_{\text{o}}}{{\text{e}}^{-{{\lambda }_{\text{B}}}\text{t}}}}{{{\text{R}}_{\text{o}}}{{\text{e}}^{-{{\lambda }_{\text{B}}}t}}}$ $={{\text{e}}^{-\left( {{\lambda }_{B}}-{{\lambda }_{\text{A}}} \right)\text{t}}}={{\text{e}}^{-3\text{t}}}$
${\Rightarrow} {\lambda_{B}-\lambda_{A}=3} $
$\Rightarrow $ $\frac{\ell {{n}^{2}}}{{{T}_{B}}}-\frac{\ell n2}{\ell n2}=3$
${\Rightarrow \quad T_{B}=\frac{\ell n 2}{4}}$
