Question

A neutron is absorbed by a `._3Li^6` nucleus with subsequent emission of an alpha particle. Write the corresponding nuclear reaction. Calculate the energy released in this reaction.
`m(._3Li^6)=6.015126u, m(._2He^4)=4.0026044u`
`m(._0n^1)=1.0086654u, m(._1He^3)=3.016049u`
Take `1u=931MeV`.

Answer 1

The nuclear reaction is given by
`._3Li^6+ ._0n^1 to ._2He^4+ ._1He^3`
Mass defect,
`Deltam=m(Li^6)+m(._0n^1)-m(He^4)-m(H^3)`
`=6.015126+1.0086654-4.00026044-3.016049`
`=7.0237914-7.0186534`
`Delta m=0.005138u`.
Energy realeased =`Delta m xx931MeV`
`=0.005138xx931`
`=4.783MeV`