1.

Calculate the energy of the following nuclear reaction: `._(1)H^(2)+._(1)H^(3) to ._(2)He^(4) + ._(0)n^(1)+Q` Given: `m(._(1)H^(2))=2.014102u, m(._(1)H^(3))=3.016049u, m(._(2)He^(4))=4.002603u, m(._(0)n^(1))=1.008665u`

Answer» Correct Answer - `17.58MeV`
In the given nuclear reaction,
Mass of reactants = `m(._(1)H^(2))+m(._(1)H^(3))`
`=2.0141202+3.016049`
`=5.030151u`
Mass of products = `m(._(2)H^(4))+m(._(0)n^(1))`
`=4.002603+1.008665`
`=5.011268u`
Mass defect, `Deltam=5.030151u-5.011268u`
`=0.018883u`
Energy released `=0.018883xx931MeV`
`=17.58MeV`


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