If the masses of a proton, neutron and nucleus of alpha-particle are \( 1.00728 \) u, \( 1.00867 \) u and \( 4.00151 \) u, respectively, the binding energy of the alpha-particle will be: (Take \( 1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^{2} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 28.3 \) MeV
The binding energy is the energy equivalent of the mass defect: the mass of the separate nucleons minus the mass of the nucleus. An alpha particle has 2 protons and 2 neutrons.
\( 2m_p = 2\times1.00728 = 2.01456\,\mathrm{u} \)
\( 2m_n = 2\times1.00867 = 2.01734\,\mathrm{u} \)
\( \Delta m = 2m_p + 2m_n - m_\alpha = 2.01456 + 2.01734 - 4.00151 = 0.03039\,\mathrm{u} \)
\( E_b = \Delta m \times 931.5\,\mathrm{MeV\,u^{-1}} = 0.03039\times931.5 \approx 28.3\,\mathrm{MeV} \)