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,   \[m\,(_{3}^{6}Li)=6.015126\,\,amu\]

\[m\,(_{2}^{4}He)=4.0026044\,\,amu\]

\[m\,(_{0}^{1}n)=1.0086654\,\,amu\]

\[m\,(_{0}^{1}H)=3.016049\,\,amu\]

 

Answer:

            (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