NEET Physics Nuclei Questions Solved

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Calculate the energy in fusion reaction:

H12+H12H23e+n,where BE of H12=2.3MeV and of H23e=7.73MeV           (2)

Total Binding energy of Initial System 

i.e. H12+H12=(2.23+2.23)MeV           =4.46MeV                              12

Binding energy of Final System i.e. H23e                   12

                            =7.73 MeV

 Hence energy released =7.73MeV-4.46MeV 

                                   =3.27MeV                                 (1)

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