Question

The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of  \(_2^4 He\) is (Given: helium nucleus mass ≈ 4.0015 u) 

• A. 0.0305 J
• B. 0.0305 erg
• C. 28.4 MeV
• D. 0.061 u

Answer 1

The Correct Option is C

To find the binding energy of the helium nucleus \( (_2^4 \text{He}) \), we first calculate the mass defect and then convert it to energy.

1. Calculate the total mass of individual nucleons (protons and neutrons) in the helium nucleus:
   • The helium nucleus contains 2 protons and 2 neutrons.
   • Mass of 2 protons: \(2 \times 1.0073 \, \text{u} = 2.0146 \, \text{u}\)
   • Mass of 2 neutrons: \(2 \times 1.0087 \, \text{u} = 2.0174 \, \text{u}\)
   • Total mass of nucleons: \(2.0146 \, \text{u} + 2.0174 \, \text{u} = 4.0320 \, \text{u}\)
2. Calculate the mass defect (\( \Delta m \)):
   • Mass defect: \(\Delta m = 4.0320 \, \text{u} - 4.0015 \, \text{u} = 0.0305 \, \text{u}\)
3.  Convert the mass defect to energy using Einstein’s mass-energy equivalence relation, \(E = \Delta m c^2\):
   • Given: \(1 \, \text{u} = 931.5 \, \text{MeV/c}^2\)
   • Binding energy: \(E = 0.0305 \, \text{u} \times 931.5 \, \text{MeV/u} = 28.4 \, \text{MeV}\)

Therefore, the binding energy of the helium nucleus is 28.4 MeV.

Note: The other options (0.0305 J, 0.0305 erg, 0.061 u) are incorrect as they either do not match the units or the calculated energy value of 28.4 MeV.