The mass of proton, neutron and helium nucleus are respectively 1.0073 u, 1.0087 u and 4.0015 u. The binding energy of helium nucleus is:

We need to find the binding energy of a helium-4 nucleus given the masses of proton, neutron, and helium nucleus. The mass of a proton is $$m_p = 1.0073$$ u, the mass of a neutron is $$m_n = 1.0087$$ u, and the mass of the helium nucleus is $$m_{He} = 4.0015$$ u.

Helium-4 contains two protons and two neutrons, so the total mass of the individual nucleons is $$ M_{nucleons} = 2m_p + 2m_n = 2(1.0073) + 2(1.0087) = 2.0146 + 2.0174 = 4.0320 \text{ u} $$

The mass defect, which is the difference between the total mass of the individual nucleons and the actual nuclear mass, is $$ \Delta m = M_{nucleons} - m_{He} = 4.0320 - 4.0015 = 0.0305 \text{ u} $$

Using Einstein’s mass-energy equivalence, where 1 u = 931.5 MeV/c$$^2$$, the binding energy is $$ BE = \Delta m \times 931.5 = 0.0305 \times 931.5 \approx 28.41 \text{ MeV} $$

The binding energy of the helium nucleus is approximately 28.4 MeV. The correct answer is Option B: 28.4 MeV.