The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of 42He is (Given helium nucleus mass ≈ 4.0015 u.)
- 0.0305 J
- 0.0305 erg
- 28.4 MeV
- 0.061 u
Answer
(3) 28.4 MeV
Solution
42He nucleus has 2 protons and 2 neutrons.
Mass defect, m = 2mp + 2mn - MHe
= 2(1.0073) + 2(1.0087) - 4.0015
= 4.0320 - 4.0015 = 0.0305 u
Binding energy = m 931.5 MeV
= 0.0305 931.5
28.4 MeV
Answer: (3)