A neutron is absorbed by a \( {}^{6}_{3}\mathrm{Li} \) nucleus with subsequent emission of an alpha particle. The energy released in the reaction is
Use mass of \( {}^{6}_{3}\mathrm{Li} = 6.015126\,\mathrm{u} \)
Mass of \( {}^{4}_{2}\mathrm{He} = 4.002604\,\mathrm{u} \)
Mass of \( {}^{1}_{0}\mathrm{n} = 1.008665\,\mathrm{u} \)
Mass of \( {}^{3}_{1}\mathrm{H} = 3.016049\,\mathrm{u} \) and \( 1\,\mathrm{u} = 931\,\mathrm{MeV} \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4.78 MeV
Balancing the mass number (6 + 1 = 4 + 3) and the charge number (3 + 0 = 2 + 1) shows that the second product is tritium:
\( \displaystyle {}^{6}_{3}\mathrm{Li} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{4}_{2}\mathrm{He} + {}^{3}_{1}\mathrm{H} \)
The energy released (the Q-value) equals the mass defect of the reaction times 931 MeV per u.
\( m_{\text{initial}} = 6.015126 + 1.008665 = 7.023791\,\mathrm{u} \)
\( m_{\text{final}} = 4.002604 + 3.016049 = 7.018653\,\mathrm{u} \)
\( \Delta m = 7.023791 - 7.018653 = 0.005138\,\mathrm{u} \)
\( Q = 0.005138 \times 931\,\mathrm{MeV} \approx 4.78\,\mathrm{MeV} \)