The masses of a neutron and a proton are 1.0087 u and 1.0073 u respectively. If two neutrons and two protons combine to form a helium nucleus (alpha particle) of mass 4.0015 u, the binding energy of the helium nucleus will be (take \( 1\,\mathrm{u}=931\,\mathrm{MeV}/c^{2} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 28.4 MeV
The binding energy of a nucleus is the energy equivalent of its mass defect \( \Delta M \). The mass defect is the amount by which the mass of the nucleus is less than the total mass of its separate nucleons. So \( E_b=\Delta M\,c^{2} \).
The total mass of two neutrons and two protons is:
\( \begin{aligned} 2m_n+2m_p &= 2\times1.0087\,\mathrm{u}+2\times1.0073\,\mathrm{u} \\ &= 2.0174\,\mathrm{u}+2.0146\,\mathrm{u} \\ &= 4.0320\,\mathrm{u} \end{aligned} \)
The mass defect is:
\( \begin{aligned} \Delta M &= (2m_n+2m_p)-M_{\mathrm{He}} \\ &= 4.0320\,\mathrm{u}-4.0015\,\mathrm{u} \\ &= 0.0305\,\mathrm{u} \end{aligned} \)
With \( 1\,\mathrm{u}=931\,\mathrm{MeV}/c^{2} \), the binding energy is:
\( \begin{aligned} E_b &= \Delta M\,c^{2} \\ &= 0.0305\times931\,\mathrm{MeV} \\ &\approx 28.4\,\mathrm{MeV} \end{aligned} \)