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`.
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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`
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