• # question_answer A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (i) 3.125% (ii) 1% of its original value? OR A neutron is absorbed by $\text{a}\,\,_{3}^{6}$ Li nucleus with subsequent emission of an alpha particle. Write the corresponding nuclear reaction. Calculate the energy released in this reaction. Given,   $m\,(_{3}^{6}Li)=6.015126\,\,amu$             $m\,(_{2}^{4}He)=4.0026044\,\,amu$             $m\,(_{0}^{1}n)=1.0086654\,\,amu$    $m\,(_{0}^{1}H)=3.016049\,\,amu$

(i) As we know, activity of a radioactive sample, $R=\lambda N.$             Thus,    $\frac{R}{{{R}_{0}}}=\frac{N}{{{N}_{0}}}=\frac{3.125}{100}=\frac{1}{32}={{\left( \frac{1}{2} \right)}^{5}}$             Also,     $\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{5}}$             $\therefore$ Number of half-lives, n = 5             $\therefore$ Total time,  $t=n\times T$             Where, T = half-life period             $\Rightarrow$   $t=5\times T=5T$             Therefore, after 5 half-time period activity reduces to 3.125% of initial activity.             (ii) As activity, $\frac{R}{{{R}_{0}}}=\frac{N}{{{N}_{0}}}=\frac{1}{100}$             $\therefore$      $t=\frac{2.303}{\lambda }\log \frac{{{N}_{0}}}{N}=\frac{2.303}{\left( \frac{0.693}{T} \right)}\log \,\,100$             or,        $t=\frac{2.303\times 2\times T}{0.693}=6.65T$             $\therefore$ Total time, t = 6.65T Or $_{3}^{6}Li+\,\,_{0}^{1}n\to _{2}^{4}He+\,\,_{1}^{3}H+Q$
 Initial masses Final masses $m\,\,(_{3}^{6}Li)=6.015126\,\,amu$ $m\,\,(_{2}^{4}He)=4.0026044\,\,amu$ $m\,\,(_{0}^{1}n)=1.0086654\,\,amu$ $m\,\,(_{1}^{3}H)=3.016049\,\,amu$ 7.0237914 amu 7.0186534 amu
Mass defect,       $\Delta m=7.0237914-7.0186534$                         = 0.005138 amu Energy released,  $Q=0.005138\times 931\,\,MeV$                         = 4.78 MeV