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Assuming that four hydrogen atom combine to form a helium atom and two positrons, each of mass 0.00549u, calculate the energy released. Given `m(._(1)H^(1))=1.007825u, m(._(2)He^(4))=4.002604u`. |
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Answer» Correct Answer - `25.7MeV` The nuclear reaction is `4._(1)H^(1) to ._(2)He^(4)+2 ._(+1)e^(0)+Q` Total initial mass `=4xx1.007825=4.031300u` Total final mass `=4.002604+2xx0.000549` `=4.003702u` Mass defect, `Deltam=4.031300-4.003702` `0.027598u` Total energy released `=0.027598xx931MeV` `=25.7MeV` |
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