The amount of energy released in the fusion of 1H2 to form a 2He4 nucleus will be : [Binding energy per nucleon of 1H2=1.1MeV and 2He4=7MeV]
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Nuclear reaction: 21H2→2He4
Energy released = Binding energy of products − Binding energy of reactants.
Total Binding energy = nucleons x binding energy per nucleon
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Binding energy per nucleon for deuteron 1H2 is 1.1 MeV and that for helium 2He4 is 7 MeV respectively. Energy released in MeV when two deutrons fuse to form a helium nucleus 2He4 is (Write upto two difgits after the decimal point.)
The binding energy of deutron 1H2 is 1.112MeV per nucleon and an alpha particle 2He4 has a binding energy of 7.047MeV per nucleon. Then, in the fusion rection 1H2+1H2→2He4+Q, the energy Q released is (take mass of 1H2=2.01478u; mass of 2He4=4.00338u)
The binding energies per nucleon of deuteron (1H2) and helium atom (2He4) are 1.1MeV and 7MeV. If two deuteron atoms react to form a single helium atom, then the energy released is
Binding energy per nucleon of 1H2 and 2He4 are 1.1MeV and 7.0MeV respectively. Energy released in the process 1H2+1H2→2He4 is
Binding energy per nucleon of 1H2 and 2He4 are 1.1 MeV and 7.0 MeV respectively. Energy released in the process1H2+1H2⟶2He4 is