Question

The binding energy of deuteron \({}_1^2H\) is 1 MeV per nucleon and an α – particle \({}_2^4He\) has a binding energy of 7 MeV per nucleon. Then in the fusion reaction\({}_1^2H + {}_1^2H \to {}_2^4He + Q\), the energy Q released is
1. 1 MeV
2. 11.9 MeV
3. 24 MeV
4. 931 MeV

Answer 1

Correct Answer - Option 3 : 24 MeV

Mass of ₁H² = 2.01478 a.m.u.      

Mass of ₂He⁴ = 4.00388 a.m.u.                                                                                   

Mass of two deuterium = 2 × 2.01478 = 4.02956                

Energy equivalent to 2₁H²= 4.02956 × 1 MeV = 4.03 MeV                                              

Energy equivalent for \({}_2^4He\) = 4.00388 × 7 MeV = 28.03 MeV                                         

Energy released = 28.03 – 4.03 = 24 MeV