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 \[_{3}^{6}Li+\,\,_{0}^{1}n\to _{2}^{4}He+\,\,_{1}^{3}H+Q\]
Mass defect, \[\Delta m=7.0237914-7.0186534\] = 0.005138 amu Energy released, \[Q=0.005138\times 931\,\,MeV\] = 4.78 MeV
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
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