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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`.

Text Solution

Verified by Experts

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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Calculate the energy released during the combination of four hydrogen atoms and forming a helium atom along with two positrons. Use the atomic masses given as follows: m(""_(1)H^(1)) = 1.007824 u, m(""_(2)He^(4)) = 4.002604 u and mass of each positrons = 0.000548 u.

Why two hydrogen atoms combine to form H_(2) but two helium atoms do not combine to form He_(2) ?

Knowledge Check

  • The masses of neutron and proton are 1.0087 a.m.u. and 1.0073 a.m.u. respectively. If the neutrons and protons combine to form a helium nucleus (alpha particle) of mass 4.0015 a.m.u. The binding energy of the helium nucleus will be (1 a.m.u. = 931 MeV) .

    A
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    B
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    C
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    D
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