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Calculate the energy released in the fol...

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

Text Solution

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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 +3.016049 =7.018653 amu
Mass difference =(7.023791-7.018653)=0.005138 amu
Energy released `=0.005138xx931 MeV` =4.783 MeV
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