Assuming that four hydrogen atoms combine to form a helium atom and two positrons, each of mass 0.000549 amu, calculate the energy released.
[Given, `m(""_(1)^(1)H)=1.007825"amu"andm(_(2)^(4)He)=4.002604"amu"`]

Explanation

The reaction involving the combination of four hydrogen atoms is shown below.
Mass defect
`""_(1)^(1)H+"_(1)^(1)H+_(1)^(1)H+_(1)^(1)Hrarr_(2)^(4)He+2_(+i)^(0)e+Q`
`Deltam=4m(""_(1)^(1)H)-m(""_(2)^(4)He)-2m(""_(+1)^(0)e)`
`=4xx1.007825-4.002604-2xx0.000549`
= 0.27598 amu
`therefore` Energy released in the combination
`Q=(Deltam)(931"MeV")" "[therefore"1 amu"-=931"MeV"]`
= (0.027598)(931)MeV = 25.7 MeV