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Assuming that four hydrogen atom combine...

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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The correct Answer is:
`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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