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Consider the reaction `_1^2H+_1^2H=_2^4He+Q`. Mass of the deuterium atom`=2.0141u`. Mass of helium atom `=4.0024u`. This is a nuclear………reaction in which the energy Q released is…………MeV.

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Correct Answer - B::D
Q=(`Deltam` in atomic mass unit)`xx931.4 MeV`
`=(2xxmass of `_1H^2-mass of `_2He^4)xx931.4 MeV`
`=(2xx2.014-4.0024)xx93.14 MeV`
`Q~~24MeV`

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