A neutron is absorbed by a `._(3)^(6)Li` nucleus with the subsequent emission of an alpha particle.
(i) Write the corresponding nuclear reaction.
(ii) Calculate the energy released, in `MeV`, in this reaction.
[Given: mass `._(3)^(6)Li=6.015126u`, mass (neutron)`=1.0086654 u` mass (alpha particle) `=4.0026044 u` and mass (triton) `=3.010000 u`. Take `i u=931 MeV//c^(2)`]

Explanation

`._(3)Li^(6)+._(0)n^(1)rarr._(2)He^(4)+._(1)H^(3)+Q`
Total initial mass `=6.015126+1.0086654`
`=7.0237914"amu"`
Total initial mass `4.0026044+3.01=7.0126044"amu"`
Mass defect, `Deltam=7.0237914-7.0126044`
`=0.0111870 "amu"`
Energy released, `Q=0.01112870xx931`
`=10.4115 MeV`.
`._(3)Li^(6)+._(0)n^(1)He^(4)+._(1)H^(3)+Q`