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Estimate the amount of energy released in the nuclear fusion reaction: `_(1)H^(2)+._(1)H^(2)rarr._(2)He^(2)+._(0)n^(1)` Given that `M(._(1)H^(2))=2.0141u, M(._(2)He^(3))=3.0160u` `m_(n)=1.0087u` , where `1u=1.661xx10^(-27)kg` . Express your answer in units of MeV. |
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Answer» Here, mass of two deutrons `=2xx2.0141=4.0282u` Mass of `_(2)He^(3)` and neutron `=3.0160+1.0087=4.0247u` Loss of mass `= Deltam=4.0282-4.0247` `=0.0035u=0.0035xx1.661xx10^(-27)kg` Energy released `E=(Deltam)c^(2)` `=0.0035xx1.661xx10^(-27)(3xx10^(8))^(2)"joule"` `=52.2xx10^(-14)"joule"` As `1MeV =1.602xx10^(-13)"joule:` ` :. ` Energy released `=(52.2xx10^(-14))/(1.602xx10^(-13))=3.26MeV` |
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