Question

Calculate the energy released in the following reaction
`._3Li^6 + ._0n^1 to ._2He^4 + ._1H^3`

Answer 1

Mass of `._3Li^6`= 6.015126 amu
Mass of `._1H^3`= 3.016049 amu
Mass of `._2He^4`= 4.002604 amu
Mass of `._0n^1`= 1.008665 amu
Total mass of reactants =6.015126+1.008665 =7.023791 amu
Total mass of products = 4.002604+3.16049=7.018653 amu
Mass difference =(7.023791-7.018653) =0.005138 amu
Energy released = 0.005138 x 931 MeV
=4.783 MeV